Ace Legona

\(\left(x^2+x\right)\left(x^2+x+1\right)=6\)

Đặt \(t=x^2+x\) thì ta có:

\(t\left(t+1\right)=6\Leftrightarrow t^2+t-6=0\)

\(\Leftrightarrow\left(t-2\right)\left(t+3\right)=0\)\(\Rightarrow\left[{}\begin{matrix}t-2=0\\t+3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}t=2\\t=-3\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x^2+x=2\\x^2+x=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2+x-2=0\\x^2+x+3=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x+2\right)=0\\x^2+x+\dfrac{1}{4}+\dfrac{11}{4}=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}>0\end{matrix}\right.\)

Sửa lại đề \(\dfrac{a}{a'}=\dfrac{b}{b'}=\dfrac{c}{c'}\) (cái này có trong CHTT rồi nhé nhưng giờ bỗng dưng rảnh làm lại luôn đỡ mất công tìm)

Áp dụng BĐT Cauchy-Schwarz ta có:

\(VP^2=\left(a+b+c\right)\left(a'+b'+c'\right)\)

\(\ge\left(\sqrt{a\cdot a'}+\sqrt{b\cdot b'}+\sqrt{c\cdot c'}\right)=VT^2\)

Tức là \(VP\ge VT\)

Xảy ra khi \(\dfrac{a}{a'}=\dfrac{b}{b'}=\dfrac{c}{c'}\)

\(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}=2x+2\)

\(pt\Leftrightarrow\sqrt{2\left(x+1\right)\left(x+3\right)}+\sqrt{\left(x-1\right)\left(x+1\right)}-2\left(x+1\right)=0\)

\(\Leftrightarrow\sqrt{x+1}\left(\sqrt{2\left(x+3\right)}+\sqrt{x-1}-2\sqrt{x+1}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x+1}=0\\\sqrt{2\left(x+3\right)}+\sqrt{x-1}-2\sqrt{x+1}=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\\sqrt{2\left(x+3\right)}+\sqrt{x-1}=2\sqrt{x+1}\end{matrix}\right.\)

Xét \(\sqrt{2\left(x+3\right)}+\sqrt{x-1}=2\sqrt{x+1}\)

\(\Leftrightarrow2\left(x+3\right)+x-1+2\sqrt{2\left(x+3\right)\left(x-1\right)}=4\left(x+1\right)\)

\(\Leftrightarrow2\sqrt{2\left(x+3\right)\left(x-1\right)}=x-1\)

\(\Leftrightarrow8\left(x+3\right)\left(x-1\right)-\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(7x+25\right)=0\Rightarrow x=1\) (Thỏa)

Vậy nghiệm của pt là \(x=±1\)

sao mình ko thấy hại não nhỉ chắc não mịn quá rồi :v

Bài 1:

\(x^3-x^2-x+1=\sqrt{4x+3}+\sqrt{3x^2+10x+6}\)

\(pt\Leftrightarrow x^3-x^2-4x-2=\sqrt{4x+3}-\left(x+1\right)+\sqrt{3x^2+10x+6}-\left(2x+2\right)\)

\(\Leftrightarrow x^3-x^2-4x-2=\dfrac{4x+3-\left(x+1\right)^2}{\sqrt{4x+3}+x+1}+\dfrac{3x^2+10x+6-\left(2x+2\right)^2}{\sqrt{3x^2+10x+6}+2x+2}\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-2x-2\right)=\dfrac{-\left(x^2-2x-2\right)}{\sqrt{4x+3}+x+1}+\dfrac{-\left(x^2-2x-2\right)}{\sqrt{3x^2+10x+6}+2x+2}\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-2x-2\right)+\dfrac{x^2-2x-2}{\sqrt{4x+3}+x+1}+\dfrac{x^2-2x-2}{\sqrt{3x^2+10x+6}+2x+2}=0\)

\(\Leftrightarrow\left(x^2-2x-2\right)\left(\left(x+1\right)+\dfrac{1}{\sqrt{4x+3}+x+1}+\dfrac{1}{\sqrt{3x^2+10x+6}+2x+2}\right)=0\)

Dễ thấy: \(\left(x+1\right)+\dfrac{1}{\sqrt{4x+3}+x+1}+\dfrac{1}{\sqrt{3x^2+10x+6}+2x+2}>0\) (ơn trời dễ thấy thật :v)

\(\Rightarrow x^2-2x-2=0\Rightarrow x=\dfrac{2\pm\sqrt{12}}{2}\)

Bài 3:

Từ \(a+b+c=1\Rightarrow2a+2b+2c=2\)

\(\Rightarrow\left(a+b\right)+\left(b+c\right)+\left(c+a\right)=2\)

Ta có: \(\dfrac{ab+c}{a+b}=\dfrac{a\left(a+b+c\right)+bc}{b+c}=\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}\)

Nên viết lại BĐT cần chứng minh là:

\(\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{b+c}+\dfrac{\left(a+c\right)\left(a+b\right)}{c+a}\ge2\)

Đặt \(\left\{{}\begin{matrix}a+b=x\\b+c=y\\c+a=z\end{matrix}\right.\left(x,y,z>0\right)\)thì có:

\(\dfrac{xy}{z}+\dfrac{xz}{y}+\dfrac{yz}{x}\ge2\forall\left\{{}\begin{matrix}x,y,z>0\\x+y+z=2\end{matrix}\right.\)

BĐT cuối đúng theo AM-GM

\(\sqrt{x-2}+\sqrt{4-x}+\sqrt{2x-5}=2x^2-5x\)

ĐK:\(\dfrac{5}{2}\le x\le4\)

\(pt\Leftrightarrow\sqrt{x-2}-1+\sqrt{4-x}-1+\sqrt{2x-5}-1=2x^2-5x-3\)

\(\Leftrightarrow\dfrac{x-2-1}{\sqrt{x-2}+1}+\dfrac{4-x-1}{\sqrt{4-x}+1}+\dfrac{2x-5-1}{\sqrt{2x-5}+1}=\left(x-3\right)\left(2x+1\right)\)

\(\Leftrightarrow\dfrac{x-3}{\sqrt{x-2}+1}-\dfrac{x-3}{\sqrt{4-x}+1}+\dfrac{2\left(x-3\right)}{\sqrt{2x-5}+1}-\left(x-3\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(\dfrac{1}{\sqrt{x-2}+1}-\dfrac{1}{\sqrt{4-x}+1}+\dfrac{2}{\sqrt{2x-5}+1}-2x-1\right)=0\)

Dễ thấy: \(\dfrac{1}{\sqrt{x-2}+1}-\dfrac{1}{\sqrt{4-x}+1}+\dfrac{2}{\sqrt{2x-5}+1}-2x-1< 0\forall\dfrac{5}{2}\le x\le4\)

\(\Rightarrow x-3=0\Rightarrow x=3\)

\(P=\dfrac{bc\sqrt{a-2}+ac\sqrt[3]{b-3}+ab\sqrt[4]{c-6}}{abc}\)

\(=\dfrac{\sqrt{a-2}}{a}+\dfrac{\sqrt[3]{b-3}}{b}+\dfrac{\sqrt[4]{c-6}}{c}\)

Áp dụng BĐT AM-GM ta có:

\(=\dfrac{\sqrt{2\left(a-2\right)}}{\sqrt{2}a}+\dfrac{\sqrt[3]{2\left(b-3\right)}}{\sqrt[3]{2}b}+\dfrac{\sqrt[4]{2\left(c-6\right)}}{\sqrt[4]{2}c}\)

\(\le\dfrac{\dfrac{2+a-2}{2}}{\sqrt{2}a}+\dfrac{\dfrac{2+b-3+1}{3}}{\sqrt[3]{2}b}+\dfrac{\dfrac{2+c-6+1+1+1+1}{4}}{\sqrt[4]{2}c}\)

\(=\dfrac{\dfrac{a}{2}}{\sqrt{2}a}+\dfrac{\dfrac{b}{3}}{\sqrt[3]{2}b}+\dfrac{\dfrac{c}{4}}{\sqrt[4]{2}c}=\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt[3]{2}}+\dfrac{1}{4\sqrt[4]{2}}\)

Từ \(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}\) và \(\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{y}{12}=\dfrac{z}{15}\)

\(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8+12-15}=\dfrac{10}{5}=2\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=2\Rightarrow x=2\cdot8=16\\\dfrac{y}{12}=2\Rightarrow y=2\cdot12=24\\\dfrac{z}{15}=2\Rightarrow z=2\cdot15=30\end{matrix}\right.\)

Ta có BĐT \(3\le ab+bc+ca\le a^2+b^2+c^2\)

Và BĐT: \(3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)

\(\Rightarrow a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}\)

\(\le\sqrt{9}=3\le a^2+b^2+c^2\)

Áp dụng BĐT Cauchy-Schwarz ta có:

\(VT^2=\left(\sqrt{a+3}+\sqrt{b+3}+\sqrt{c+3}\right)^2\)

\(\le\left(1+1+1\right)\left(a+b+c+9\right)\)

\(\le\left(a^2+b^2+c^2\right)\left[a^2+b^2+c^2+3\left(a^2+b^2+c^2\right)\right]\)

\(=4\left(a^2+b^2+c^2\right)=VP^2\)

Xảy ra khi \(a=b=c=1\)

Đặt \(T=\left(a+b\right)\left(b+c\right)\left(c+a\right)>0\)

\(BDT\Leftrightarrow\dfrac{a^2+bc}{b+c}+\dfrac{b^2+ca}{c+a}+\dfrac{c^2+ab}{a+b}\ge a+b+c\)

\(\Leftrightarrow\dfrac{a^2+bc}{b+c}-a+\dfrac{b^2+ca}{c+a}-b+\dfrac{c^2+ab}{a+b}-c\ge0\)

\(\Leftrightarrow\dfrac{a^2+bc-ab-ac}{b+c}+\dfrac{b^2+ac-ab-bc}{a+c}+\dfrac{c^2+ab-ac-bc}{a+b}\ge0\)

\(\Leftrightarrow\dfrac{\left(a-b\right)\left(a-c\right)}{b+c}+\dfrac{\left(b-a\right)\left(b-c\right)}{a+c}+\dfrac{\left(c-a\right)\left(c-b\right)}{a+b}\ge0\)

\(\Leftrightarrow\dfrac{\left(a^2-b^2\right)\left(a^2-c^2\right)+\left(b^2-a^2\right)\left(b^2-c^2\right)+\left(c^2-a^2\right)\left(c^2-b^2\right)}{T}\ge0\)

\(\Leftrightarrow\dfrac{a^4+b^4+c^4-b^2c^2-c^2a^2-a^2b^2}{T}\ge0\)

\(\Leftrightarrow\dfrac{\left(a^2-b^2\right)^2+\left(b^2-c^2\right)^2+\left(c^2-a^2\right)^2}{2T}\ge0\)

Xảy ra khi \(a=b=c\)

Bữa trước ko để ý a,b,c ko âm với ngược dấu sai thê thảm =))

Dự đoán \(a=b=1\) và \(c=0\) thì tính được \(2+\frac{1}{\sqrt2}\)

Ta sẽ chứng minh nó là GTNN.Thật vậy cần chứng minh

\(\sqrt{\dfrac{ab+bc+ca}{a^2+b^2}}+\sqrt{\dfrac{ab+bc+ca}{b^2+c^2}}+\sqrt{\dfrac{ab+bc+ca}{a^2+c^2}}\ge2+\dfrac{1}{\sqrt{2}}\)

Khôn mất tính tổng quá giả sử \(c=\min\{a,b,c\}\). Khi đó:

\(\dfrac{ab+ac+bc}{a^2+b^2}-\dfrac{(a+c)(b+c)}{(a+c)^2+(b+c)^2}=\dfrac{c(a+b+2c)(2ab+ac+bc)}{a^2+b^2)((a+c)^2+(b+c)^2}\ge0\)

Tương tự cũng có:

\(\dfrac{ab+ac+bc}{a^2+c^2}-\dfrac{b+c}{a+c}=\dfrac{c(2ab+ac-c^2)}{(a+c)(a^2+c^2)}\ge0\)

Và \(\dfrac{ab+ac+bc}{b^2+c^2}-\dfrac{a+c}{b+c}=\dfrac{c(2ab+bc-c^2)}{(b+c)(b^2+c^2)}\ge0\)

Đặt \(\dfrac{a+c}{b+c}=x^2;\dfrac{b+c}{a+c}=y^2\left(x,y>0\right)\)\(\Rightarrow xy=1\) và ta có:

\(x+y+\dfrac{1}{\sqrt{x^2+y^2}}\ge2+\dfrac{1}{\sqrt{2}}\)

\(\Leftrightarrow x+y-2\sqrt{xy}\ge\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{x^2+y^2}}\)

\(\Leftrightarrow(\sqrt{x}-\sqrt{y})^2\ge\dfrac{(x-y)^2}{\sqrt{2(x^2+y^2)}(\sqrt{x^2+y^2}+\sqrt{2})}\)

\(\Leftrightarrow\sqrt{2(x^2+y^2)}(\sqrt{x^2+y^2}+\sqrt{2})\ge(\sqrt{x}+\sqrt{y})^2\)

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\sqrt{2(x^2+y^2)}=\sqrt{(1^2+1^2)(x^2+y^2)}\ge x+y\)

\(=\dfrac{1}{2}(1^2+1^2)((\sqrt{x})^2+(\sqrt{y})^2)\ge\dfrac{1}{2}(\sqrt{x}+\sqrt{y})^2\)

Vậy cần chứng minh \(\sqrt{x^2+y^2}+\sqrt{2}\ge2\)

Đúng theo AM-GM:\(\sqrt{x^2+y^2}+\sqrt{2}\ge\sqrt{2xy}+\sqrt{2}=2\sqrt{2}>2\)

\(x\left(2x-1\right)+\dfrac{1}{3}-\dfrac{2}{3}x=0\)

\(\Leftrightarrow x\left(2x-1\right)+\dfrac{1}{3}\left(1-2x\right)=0\)

\(\Leftrightarrow x\left(2x-1\right)-\dfrac{1}{3}\left(2x-1\right)=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{3}\right)\left(2x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=0\\2x-1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Ta có BĐT phụ \(\dfrac{1+\sqrt{a}}{1-a}\ge4a+1\)

\(\Leftrightarrow-\dfrac{\sqrt{a}\left(2\sqrt{a}-1\right)^2}{\sqrt{a}-1}\ge0\forall\dfrac{1}{4}< a< 0\)

Tương tự cho 3 BĐT còn lại ta cũng có:

\(\dfrac{1+\sqrt{b}}{1-b}\ge4b+1;\dfrac{1+\sqrt{c}}{1-c}\ge4c+1;\dfrac{1+\sqrt{d}}{1-d}\ge4d+1\)

Cộng theo vế 4 BĐT trên ta có:

\(VT\ge4\left(a+b+c+d\right)+4=8=VP\)

Xảy ra khi \(a=b=c=d=\dfrac{1}{4}\)

Ta thấy: Xét \(22\) có: \(2*2=\text{_}4\) và \(2+2=4\) suy ra \(\text{_}=4\) tức \(44\)

Xét \(44\) có: \(4*4=1\text{_}6\) và \(4+4=8\) suy ra \(\text{_}=8\) tức \(168\)

Xét \(186\) có: \(1*8*6=\text{_}4\text{_}8\) và \(1+8+6=15\) suy ra \(\text{_}=1\) và \(\text{_}=5\) tức \(1458\)

Xét \(1458\) có: \(1*4*5*8=1\text{_}6\text{_}0\) và \(1+4+5+8=18\) suy ra \(\text{_}=1\) và \(\text{_}=8\) tức \(11680\)

Xét \(11680\) có: \(1*1*6*8*0=\text{_} .\text{_}0\) (dấu . để phân biệt giữa 2 số cần tìm là " _ " ) và \(1+1+6+8=16\) suy ra \(\text{_}=1\) và \(\text{_}=6\) tức \(160\)

P/s:cực dị ứng với dạng này, hiểu thì dễ mà nói lại cho người khác hiểu thì hơi khó :V

\(M=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)

Đặt \(t=x^2+5x+4\) thì ta có:

\(=t\left(t+2\right)+1=t^2+2t+1\)

\(=\left(t+1\right)^2=\left(x^2+5x+5\right)^2\)

Mà \(x\in Z\) suy ra \(\left(x^2+5x+5\right)^2\) là số chính phương của 1 số nguyên

a)\(\sqrt{x^2+48}=4x-3+\sqrt{x^2+35}\)

\(\Leftrightarrow\sqrt{x^2+48}-7=4x-4+\sqrt{x^2+35}-6\)

\(\Leftrightarrow\dfrac{x^2+48-49}{\sqrt{x^2+48}+7}=4\left(x-1\right)+\dfrac{x^2+35-36}{\sqrt{x^2+35}+6}\)

\(\Leftrightarrow\dfrac{x^2-1}{\sqrt{x^2+48}+7}-4\left(x-1\right)-\dfrac{x^2-1}{\sqrt{x^2+35}+6}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\dfrac{x+1}{\sqrt{x^2+48}+7}-4-\dfrac{x+1}{\sqrt{x^2+35}+6}\right)=0\)

\(\Rightarrow x-1=0\Rightarrow x=1\)

b)\(\left(\sqrt{x-1}+1\right)^3+2\sqrt{x-1}=2-x\)

\(pt\Leftrightarrow\left(\sqrt{x-1}+1\right)^3-1+2\sqrt{x-1}=1-x\)

\(\Leftrightarrow\left(\sqrt{x-1}+1-1\right)\left(\left(\sqrt{x-1}+1\right)^6+\left(\sqrt{x-1}+1\right)^3+1\right)+2\sqrt{x-1}-\left(1-x\right)=0\)

\(\Leftrightarrow\sqrt{x-1}\left(\left(\sqrt{x-1}+1\right)^6+\left(\sqrt{x-1}+1\right)^3+1\right)+2\sqrt{x-1}+x-1=0\)

\(\Leftrightarrow\sqrt{x-1}\left(\left(\sqrt{x-1}+1\right)^6+\left(\sqrt{x-1}+1\right)^3+3+\sqrt{x-1}\right)=0\)

Dễ thấy: \(\left(\sqrt{x-1}+1\right)^6+\left(\sqrt{x-1}+1\right)^3+3+\sqrt{x-1}>0\)

\(\Rightarrow\sqrt{x-1}=0\Rightarrow x-1=0\Rightarrow x=1\)

Ta thấy: \(\left\{{}\begin{matrix}\left(3x+1\right)^6\ge0\\\left|1-y\right|\ge0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}2\left(3x+1\right)^6\ge0\\3\left|1-y\right|^3\ge0\end{matrix}\right.\)

\(\Rightarrow2\left(3x+1\right)^6+3\left|1-y\right|^3\ge0\)

\(\Rightarrow2\left(3x+1\right)^6+3\left|1-y\right|^3+2\ge2\)

\(\Rightarrow\dfrac{1}{2\left(3x+1\right)^6+3\left|1-y\right|^3+2}\le\dfrac{1}{2}\)

\(\Rightarrow\dfrac{3}{2\left(3x+1\right)^6+3\left|1-y\right|^3+2}\le\dfrac{3}{2}\)

Xảy ra khi \(\left\{{}\begin{matrix}2\left(3x+1\right)^6=0\\3\left|1-y\right|^3=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{3}\\y=1\end{matrix}\right.\)

\(A=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)

\(=\dfrac{5\sqrt{2}+7-5\sqrt{2}-7}{\sqrt[3]{\left(5\sqrt{2}+7\right)^2}+\sqrt[3]{\left(5\sqrt{2}-7\right)^2}+\sqrt[3]{5\sqrt{2}+7}\cdot\sqrt[3]{5\sqrt{2}-7}}\)

\(=\dfrac{0}{\sqrt[3]{\left(5\sqrt{2}+7\right)^2}+\sqrt[3]{\left(5\sqrt{2}-7\right)^2}+\sqrt[3]{5\sqrt{2}+7}\cdot\sqrt[3]{5\sqrt{2}-7}}=0\)

P/s: Vẫn viết được mà lần sau ko bấm được thì vào latex gõ \sqrt[3]{ biểu thức }

Áp dụng BĐT AM-GM ta có:

\(\left\{{}\begin{matrix}1+x\ge2\sqrt{x}\\x+y\ge2\sqrt{xy}\\1+y\ge2\sqrt{y}\end{matrix}\right.\)

Cộng theo vế 3 BĐT trên ta có:

\(2\left(1+x+y\right)\ge2\left(\sqrt{x}+\sqrt{y}+\sqrt{xy}\right)\)

\(\Leftrightarrow VT=1+x+y\ge\sqrt{x}+\sqrt{y}+\sqrt{xy}=VP\)

Xảy ra khi \(\left\{{}\begin{matrix}1+x=2\sqrt{x}\\x+y=2\sqrt{xy}\\1+y=2\sqrt{y}\end{matrix}\right.\)\(\Rightarrow x=y=1\)

Khi đó \(P=x^2+y^2=1^2+1^2=2\)

Và \(Q=x^{2009}+y^{2009}=1^{2009}+1^{2009}=2\)

-BĐT hoán vị TA là cyclic viết tắt là cyc \(\sum_{cyc}a=a+b+c\)

-BĐT đối xứng TA là symmetric viết tắt là sym \(\sum_{sym}f(a,b,c)=f(a,b,c)+f(b,c,a)+f(c,a,b)\)

Còn 1 thể loại nữa là \(\sum_{perm} f(a,b,c) = f(a,b,c) + f(a,c,b) + f(b,c,a) + f(b,a,c) + f(c,a,b) + f(c,b,a).\) cái này t cx ko rõ lắm =))

\(A=1\cdot2+2\cdot3+...+68\cdot69+69\cdot70\)

\(3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+...+69\cdot70\cdot\left(71-68\right)\)

\(3A=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+...+69\cdot70\cdot71-68\cdot69\cdot70\)

\(3A=69\cdot70\cdot71\Rightarrow A=\dfrac{69\cdot70\cdot71}{3}=114310\)

\(\left(50^2+48^2+...+2^2\right)-\left(49^2+47^2+...+1^2\right)\)

\(=50^2+48^2+...+2^2-49^2-47^2-...-1^2\)

\(=\left(50^2-49^2\right)+\left(48^2-47^2\right)+...+\left(2^2-1^2\right)\)

\(=\left(50-49\right)\left(50+49\right)+\left(48-47\right)\left(48+47\right)+...+\left(2-1\right)\left(2+1\right)\)

\(=50+49+48+47+...+2+1\)

\(=\dfrac{50\left(50+1\right)}{2}=\dfrac{50\cdot51}{2}=1275\)

\(x^3-2x^2-4xy^2+x\)

\(=x\left(x^2-2x-4y^2+1\right)\)

\(=x\left(-4y^2+x^2-2x+1\right)\)

\(=x\left[-\left(2y\right)^2-\left(x-1\right)^2\right]\)

\(=-x\left(2y-x+1\right)\left(2y+x-1\right)\)

\(K=-3x^2-y^2+8x-2xy+2\)

\(=\left(-2x^2+8x-8\right)+\left(-x^2-2xy-y^2\right)+10\)

\(=-2\left(x^2-4x+4\right)-\left(x^2+2xy+y^2\right)+10\)

\(=-2\left(x-2\right)^2-\left(x+y\right)^2+10\ge10\)

Xảy ra khi \(\left\{{}\begin{matrix}-2\left(x-2\right)^2=0\\-\left(x+y\right)^2=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x-2=0\\x=-y\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

\(P=\dfrac{1}{x}+\dfrac{4}{y}+\dfrac{9}{z}=\dfrac{1^2}{x}+\dfrac{2^2}{y}+\dfrac{3^2}{z}\)

\(\ge\dfrac{\left(1+2+3\right)^2}{x+y+z}=\dfrac{36}{4}=9\)

Xảy ra khi \(x=\dfrac{2}{3};y=\dfrac{4}{3};z=2\)

Áp dụng BĐT Cauchy-Schwarz và BĐT AM-GM ta có:

\(S=\dfrac{1}{xy}+\dfrac{1}{xz}\ge\dfrac{4}{xy+xz}=\dfrac{4}{x\left(y+z\right)}\)

\(\ge\dfrac{4}{\dfrac{\left(x+y+z\right)^2}{4}}=\dfrac{16}{\left(x+y+z\right)^2}=1\)

Xảy ra khi \(x=2;y=z=1\)

\(5+x+2\sqrt{\left(4-x\right)\left(2x-2\right)}=4\left(\sqrt{4-x}+\sqrt{2x-2}\right)\)

ĐK:\(1\le x\le 4\)

Đặt \(\left\{{}\begin{matrix}\sqrt{4-x}=a\\\sqrt{2x-2}=b\end{matrix}\right.\)\(\left(a,b\ge0\right)\) thì ta có:

\(a^2+b^2+3+2ab=4\left(a+b\right)\)

\(\Leftrightarrow\left(a+b\right)^2-4\left(a+b\right)=-3\)

\(\Leftrightarrow\left(a+b\right)\left(a+b-4\right)=-3\)

\(\Rightarrow\left[{}\begin{matrix}b=1-a\\b=3-a\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\sqrt{2x-2}=1-\sqrt{4-x}\\\sqrt{2x-2}=3-\sqrt{4-x}\end{matrix}\right.\)\(\Rightarrow x=3\)

\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)+2008\)

\(=\left(x^2-5x+4\right)\left(x^2-5x+6\right)+2008\)

Đặt \(t=x^2-5x+4\) thì ta có:

\(t\left(t+2\right)+2008=t^2+2t+2008\)

\(=t^2+2t+1+2007\)

\(=\left(t+1\right)^2+2007\ge2007\)

Xảy ra khi \(t=-1\Rightarrow x^2-5x+4=-1\Rightarrow\)\(x=\dfrac{5\pm\sqrt{5}}{2}\)

Từ \(2x=3y=4z\Rightarrow\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{4}}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{4}}=\dfrac{x+y-z}{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}=\dfrac{28}{\dfrac{7}{12}}=48\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{\dfrac{1}{2}}=48\Rightarrow x=48\cdot\dfrac{1}{2}=24\\\dfrac{y}{\dfrac{1}{3}}=48\Rightarrow y=48\cdot\dfrac{1}{3}=16\\\dfrac{z}{\dfrac{1}{4}}=48\Rightarrow z=48\cdot\dfrac{1}{4}=12\end{matrix}\right.\)

Bài 1:

\(x^2+3y^2+2z^2-2x+12y+4z+15=0\)

\(\Leftrightarrow x^2-2x+1+3y^2+12y+12+2z^2+4z+2=0\)

\(\Leftrightarrow\left(x-1\right)^2+3\left(y^2+4y+4\right)+2\left(z^2+2z+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2+3\left(y+2\right)^2+2\left(z+1\right)^2=0\)

Dễ thấy: \(\left(x-1\right)^2+3\left(y+2\right)^2+2\left(z+1\right)^2\ge0\)

Xảy ra khi \(\left\{{}\begin{matrix}\left(x-1\right)^2=0\\3\left(y+2\right)^2=0\\2\left(z+1\right)^2=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\\z=-1\end{matrix}\right.\)

Bài 2:

a)\(A=x^2-4xy+5y^2+10x-22y+28\)

\(=x^2-4xy+10x+4y^2-20y+25+y^2-2y+1+2\)

\(=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)

Xảy ra khi \(\left\{{}\begin{matrix}\left(x-2y+5\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)

b)\(B=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)+15\)

\(=\left(x^2-5x+4\right)\left(x^2-5x+6\right)+15\)

Đặt \(t=x^2-5x+4\) thì ta có:

\(t\left(t+2\right)+15=t^2+2t+1+14\)

\(=\left(t+1\right)^2+14\ge14\)

Xảy ra khi \(t=-1 \)\(\Rightarrow x^2-5x+4=-1\Rightarrow x=\dfrac{5\pm\sqrt{5}}{2}\)

\(3-x+\sqrt{x^2-3x+2}=0\)

\(pt\Leftrightarrow\dfrac{7}{3}-x+\sqrt{x^2-3x+2}-\dfrac{2}{3}=0\)

\(\Leftrightarrow\dfrac{7}{3}-x+\dfrac{x^2-3x+2-\dfrac{4}{9}}{\sqrt{x^2-3x+2}+\dfrac{2}{3}}=0\)

\(\Leftrightarrow-\left(x-\dfrac{7}{3}\right)+\dfrac{\left(x-\dfrac{7}{3}\right)\left(x-\dfrac{2}{3}\right)}{\sqrt{x^2-3x+2}+\dfrac{2}{3}}=0\)

\(\Leftrightarrow\left(x-\dfrac{7}{3}\right)\left(-1+\dfrac{x-\dfrac{2}{3}}{\sqrt{x^2-3x+2}+\dfrac{2}{3}}\right)=0\)

Dễ thấy: \(-1+\dfrac{x-\dfrac{2}{3}}{\sqrt{x^2-3x+2}+\dfrac{2}{3}}>0\forall\left[{}\begin{matrix}x\ge1\\x\ge2\end{matrix}\right.\)

\(\Rightarrow x-\dfrac{7}{3}=0\Rightarrow x=\dfrac{7}{3}\)

\(\sqrt{7-x}+\sqrt{x+1}=x^2-6x+13\)

Áp dụng BĐT Cauchy-Schwarz ta có:

\(VT^2=\left(\sqrt{7-x}+\sqrt{x+1}\right)^2\)

\(\le\left(1+1\right)\left(7-x+x+1\right)=16\)

\(\Rightarrow VT^2\le16\Rightarrow VT\le4\)

Lại có: \(VP=x^2-6x+13\)

\(=x^2-6x+9+4=\left(x-3\right)^2+4\ge4\)

Suy ra \(VT\le VP=4\) xảy ra khi \(VT=VP=4\)

\(\Rightarrow\left(x-3\right)^2+4=4\Rightarrow x-3=0\Rightarrow x=3\)

Áp dụng BĐT Cauchy-Schwarz ta có;

\(\left(1^2+1^2+1^2\right)\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)

\(\Rightarrow3\left(a^2+b^2+c^2\right)\ge2^2=4\)

\(\Rightarrow K=a^2+b^2+c^2\ge\dfrac{4}{3}\)

Xảy ra khi \(a=b=c=\dfrac{2}{3}\)

Từ \(2\left(a^2+b^2\right)=\left(a-b\right)^2\)

\(\Leftrightarrow2a^2+2b^2=a^2-2ab+b^2\)

\(\Leftrightarrow2a^2+2b^2-a^2+2ab-b^2=0\)

\(\Leftrightarrow a^2+2ab+b^2=0\Leftrightarrow\left(a+b\right)^2=0\)

\(\Leftrightarrow a+b=0\Leftrightarrow a=-b\)

Tức là \(a;b\) là 2 số đối nhau