Ace Legona

\(x^2-2xy-4z^2+y^2\)

\(=x^2-2xy+y^2-4z^2\)

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

\(=\left(x-y\right)^2-\left(2z\right)^2\)

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

Bài 1:

\(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)

\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\)

\(=-\sqrt{1}+\sqrt{100}=100-1=99\)

Bài 2:

\(\dfrac{\sqrt{3x-1}}{\sqrt{5x+2}}=7\)

\(pt\Leftrightarrow\sqrt{\dfrac{3x-1}{5x+2}}=7\)\(\Leftrightarrow\dfrac{3x-1}{5x+2}=49\)

\(\Leftrightarrow3x-1=49\left(5x+2\right)\)

\(\Leftrightarrow3x-1-245x-98=0\)

\(\Leftrightarrow-242x-99=0\Rightarrow x=-\dfrac{9}{22}\)

a)\(\left|x-1\right|+\left|\left(x-1\right)\left(x-2\right)\right|=0\)

\(pt\Leftrightarrow\left|x-1\right|\left(\left|x-2\right|+1\right)=0\)

Dễ thấy: \(\left|x-2\right|+1\ge1>0\) (loại)

\(\Rightarrow\left|x-1\right|=0\Rightarrow x-1=0\Rightarrow x=1\)

b)\(\left|x\right|+\left|y-1\right|+\left|z-3\right|=0\)

Dễ thấy: \(\left\{{}\begin{matrix}\left|x\right|\ge0\\\left|y-1\right|\ge0\\\left|z-3\right|\ge0\end{matrix}\right.\)

\(\Rightarrow\left|x\right|+\left|y-1\right|+\left|z-3\right|\ge0\)

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

c)\(\left|x-1\right|=x-1\)

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

\(\Rightarrow\left[{}\begin{matrix}x\in R\\x=1\end{matrix}\right.\)

\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\)

\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)

\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)

\(2A-A=\left(1+\dfrac{1}{2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\right)\)

\(A=1-\dfrac{1}{2^{10}}\)

\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2\)

\(2A=2\left(2^2-2^3+2^4-...-2^{97}+2^{98}-2^{99}+2^{100}\right)\)

\(2A=2^3-2^4+...+2^{99}-2^{100}+2^{101}\)

\(2A+A=\left(2^3-2^4+...+2^{101}\right)+\left(2^2-2^3+...+2^{100}\right)\)

\(3A=2^{101}+4\Rightarrow A=\dfrac{2^{101}+4}{3}\)

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

\(\Leftrightarrow\sqrt{x}-1+2\sqrt{x+3}-4+\sqrt{x^2+3}-2=0\)

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

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

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

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

Dễ thấy: \(\dfrac{x-1}{\sqrt{x}+1}+\dfrac{4}{2\sqrt{x+3}+4}+\dfrac{x+1}{\sqrt{x^2+3}+2}>0\)

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

bài này chỉ ở dạng trung trung thôi, có 2 cái link 1 tổng quát 2 hiệu quát ko biết giúp j dc ko

a)Tại \(x=87;y=13\) thì

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

\(=\left(87-13\right)\left(87+13\right)=74\cdot100=7400\)

b)Tại \(x=\dfrac{1}{3}\) thì

\(B=9x^2-6x+1=\left(x-\dfrac{1}{3}\right)^2\)

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

c)Tại \(x=1;y=2\) thì

\(C=4x^2-12xy+9y^2=\left(2x-3y\right)^2\)

\(=\left(2\cdot1-3\cdot2\right)^2=\left(-4\right)^2=16\)

Ta có:

\(\sqrt{1+\dfrac{1}{a^2}+\dfrac{1}{\left(a+1\right)^2}}=\sqrt{1+\dfrac{1}{a^2}+\dfrac{1}{\left(a+1\right)^2}+\dfrac{2\left(a+1-a-1\right)}{a\left(a+1\right)}}\)

\(=\sqrt{1+\dfrac{1}{a^2}+\dfrac{1}{\left(a+1\right)^2}+2\cdot1\cdot\dfrac{1}{a}-2\cdot1\cdot\dfrac{1}{a+1}-2\cdot\dfrac{1}{a}\cdot\dfrac{1}{a+1}}\)

\(=\sqrt{\left(1+\dfrac{1}{a}-\dfrac{1}{a-1}\right)^2}=1+\dfrac{1}{a}-\dfrac{1}{a-1}\)

a)\(3\left(\sqrt{2x^2+1}-1\right)=x\left(1+3x+8\sqrt{2x^2+1}\right)\)

\(\Leftrightarrow3\left(\dfrac{2x^2+1-1}{\sqrt{2x^2+1}+1}\right)-x\left(1+3x+8\sqrt{2x^2+1}\right)=0\)

\(\Leftrightarrow\dfrac{6x^2}{\sqrt{2x^2+1}+1}-x\left(1+3x+8\sqrt{2x^2+1}\right)=0\)

\(\Leftrightarrow x\left(\dfrac{6x}{\sqrt{2x^2+1}+1}-\left(1+3x+8\sqrt{2x^2+1}\right)\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\\dfrac{6x}{\sqrt{2x^2+1}+1}=1+3x+8\sqrt{2x^2+1}\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}a=\sqrt{2x^2+1}\\b=3x\end{matrix}\right.\left(a>0\right)\) thì

\(pt\left(2\right)\Leftrightarrow\)\(\dfrac{2b}{a+1}=1+b+8a\)

\(\Rightarrow\left\{{}\begin{matrix}a=-17\\b=120\end{matrix}\right.;\left\{{}\begin{matrix}a=-8\\b=49\end{matrix}\right.;\left\{{}\begin{matrix}a=-5\\b=26\end{matrix}\right.;\left\{{}\begin{matrix}a=-2\\b=5\end{matrix}\right.;\left\{{}\begin{matrix}a=-0\\b=1\end{matrix}\right.\) (loại vì \(a>0\))

Hay pt vô nghiệm

ko liên hợp được nữa vì bản chất liên hợp là xuất hiện nhân tử chung :v, mà đề đã có \(\sqrt{x}\) là nhân tử chung rồi mà :v

Ta có: \(x^2-21x-100\)

\(=\left(x-25\right)\left(x+4\right)\)

Lại có: \(x^2-21x-100=\left(x+a\right)\left(x+b\right)\)

\(\Rightarrow\left(x-25\right)\left(x+4\right)=\left(x+a\right)\left(x+b\right)\)

Đồng nhất 2 vế ta có:

\(\left\{{}\begin{matrix}x-25=x+a\\x+4=x+b\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=-25\\b=4\end{matrix}\right.\)

\(\Rightarrow\left|a-b\right|=\left|-25-4\right|=\left|-29\right|=29\)

P.s:thực ra nếu xếp \(x^2-21x-100=\left(x+4\right)\left(x-25\right)\) thì đồng nhất à \(\left\{{}\begin{matrix}x+4=x+a\\x-25=x+b\end{matrix}\right.\) kq |a-b| vẫn ko thay đổi nên mk xét 1 trg hợp thôi

Từ \(\dfrac{2bz-3cy}{a}=\dfrac{3cx-az}{2b}=\dfrac{ay-2bx}{3c}\)

\(\Rightarrow\dfrac{2abz-3acy}{a^2}=\dfrac{6bcx-2abz}{4b^2}=\dfrac{3acy-6bcx}{9c^2}\)

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

\(\dfrac{2bz-3cy}{a}=\dfrac{3cx-az}{2b}=\dfrac{ay-2bx}{3c}\)

\(=\dfrac{2abz-3acy}{a^2}=\dfrac{6bcx-2abz}{4b^2}=\dfrac{3acy-6bcx}{9c^2}\)

\(=\dfrac{2abz-3acy+6bcx-2abz+3acy-6bcx}{a^2+4b^2+9c^2}=0\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{2bz-3cy}{a}=0\\\dfrac{3cx-az}{2b}=0\\\dfrac{ay-2bx}{3c}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}2bz=3cy\\3cx=az\\ay=2bx\end{matrix}\right.\)\(\Rightarrow\dfrac{x}{a}=\dfrac{y}{2b}=\dfrac{z}{3c}\)

Another way

BĐT trên thuần nhất nên ta chuẩn hóa \(x^2+y^2+z^2=1\)

\(BDT\Leftrightarrow\dfrac{xyz\left(x+y+z+1\right)}{xy+yz+xz}\le\dfrac{3+\sqrt{3}}{9}\)

Áp dụng BĐT AM-GM ta có:

\(\left\{{}\begin{matrix}xy+yz+xz\ge3\sqrt{x^2y^2z^2}\\\sqrt[3]{xyz}\le\sqrt{\dfrac{x^2+y^2+z^2}{3}}\Rightarrow xyz\le\sqrt{\dfrac{\left(x^2+y^2+z^2\right)^3}{27}}=\dfrac{1}{\sqrt{27}}\\x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}=\sqrt{3}\end{matrix}\right.\)

\(\Rightarrow A\le\dfrac{\sqrt[3]{xyz}\left(x+y+z+\sqrt{x^2+y^2+z^2}\right)}{3\left(x^2+y^2+z^2\right)}\)

\(=\dfrac{\sqrt[3]{xyz}\left(x+y+z+1\right)}{3}\le\dfrac{3+\sqrt{3}}{9}\)

\(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{x\left(x+2\right)}=\dfrac{2011}{2013}\)

\(\Leftrightarrow2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{x\left(x+2\right)}\right)=\dfrac{2011}{2013}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x-2}=\dfrac{2011}{4026}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x-2}=\dfrac{2011}{4026}\)\(\Leftrightarrow\dfrac{1}{x-2}=\dfrac{1}{2013}\)

\(\Rightarrow x-2=2013\Rightarrow x=2015\)

\(VT=\sqrt{4-a}+\sqrt{4-b}+\sqrt{4-c}\)

Ta có BĐT phụ \(\sqrt{4-a}>-\dfrac{1}{2}a+2\)

\(\Leftrightarrow-\dfrac{1}{4}a\left(a-4\right)>0\forall0< a< 4\) (đúng)

Tương tự cho 2 BĐT còn lại cũng có:

\(\sqrt{4-b}>-\dfrac{1}{2}b+2;\sqrt{4-c}>-\dfrac{1}{2}c+2\)

Cộng theo vế 3 BĐT trên ta có:

\(VT>-\dfrac{1}{2}\left(a+b+c\right)+6=4=VP\)

Ta có: \(VP=\left(x^2+x-2\right)\left(x^2+cx+d\right)\)

\(=x^4+\left(c+1\right)x^3+\left(d+c-2\right)x^2+\left(d-2c\right)x-2d\)

Và \(VT=x^4+x^3-x^2+ax+b\)

Đồng nhất 2 đa thức trên ta có:

\(\left\{{}\begin{matrix}\left(c+1\right)x^3=x^3\\\left(d+c-2\right)x^2=-x^2\\\left(d-2c\right)x=ax\\-2d=b\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}c+1=1\\d+c-2=-1\\d-2c=a\\-2d=b\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}c=0\\d-2=-1\\d=a\\b=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}c=0\\d=1\\d=a\\b=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=d=1\\b=c=0\end{matrix}\right.\)

\(x+y=12\)

\(\sqrt{5x^2+4x}-\sqrt{x^2-3x-18}=5\sqrt{x}\)

ĐK:\(x\ge 6\)

\(pt\Leftrightarrow\left(\sqrt{5x^2+4x}-21\right)-\left(\sqrt{x^2-3x-18}-6\right)=5\sqrt{x}-15\)

\(\Leftrightarrow\dfrac{5x^2+4x-441}{\sqrt{5x^2+4x}+21}-\dfrac{x^2-3x-18-36}{\sqrt{x^2-3x-18}+6}=\dfrac{25x-225}{5\sqrt{x}+15}\)

\(\Leftrightarrow\dfrac{\left(x-9\right)\left(5x+49\right)}{\sqrt{5x^2+4x}+21}-\dfrac{\left(x-9\right)\left(x+6\right)}{\sqrt{x^2-3x-18}+6}-\dfrac{25\left(x-9\right)}{5\sqrt{x}+15}=0\)

\(\Leftrightarrow\left(x-9\right)\left(\dfrac{5x+49}{\sqrt{5x^2+4x}+21}-\dfrac{x+6}{\sqrt{x^2-3x-18}+6}-\dfrac{x-9}{5\sqrt{x}+15}\right)=0\)

Dễ thấy: \(\dfrac{5x+49}{\sqrt{5x^2+4x}+21}-\dfrac{x+6}{\sqrt{x^2-3x-18}+6}-\dfrac{x-9}{5\sqrt{x}+15}>0\)

\(\Rightarrow x-9=0\Rightarrow x=9\)

\(a^3+2c=3ab\)

\(\Leftrightarrow\left(x+y\right)^3+2\left(x^3+y^3\right)=3\left(x+y\right)\left(x^2+y^2\right)\)

Ta có: \(VT=\left(x+y\right)^3+2\left(x^3+y^3\right)\)

\(=\left(x+y\right)^3+2\left(x+y\right)\left(x^2-xy+y^2\right)\)

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

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

\(=\left(x+y\right)\left(3x^2+3y^2\right)=3\left(x+y\right)\left(x^2+y^2\right)=VP\)

Hay ta có ĐPCM

Sửa đề \(3x^2\left(ax^2-2bx-3c\right)=3x^4-12x^3+27x^2\)

\(\Leftrightarrow3x^2\left(ax^2-2bx-3c\right)=3x^2\left(x^2-4x+9\right)\)

Đồng nhất 2 đa thức ta có:

\(\left\{{}\begin{matrix}ax^2=x^2\\-2bx=-4x\\-3c=9\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=1\\b=-2\\c=-3\end{matrix}\right.\)

Ta thấy: \(\sqrt{-x^2-6x-8}=\sqrt{-x^2-6x-9+1}\)

\(=\sqrt{-\left(x^2+6x+9\right)+1}=\sqrt{-\left(x+3\right)^2+1}\le1\)

Và \(\sqrt{2-x^2+2x}=\sqrt{-x^2+2x+2}=\sqrt{-x^2+2x-1+3}\)

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

Cộng theo vế 2 BĐT trên ta có:

\(VT\le VP\) xảy ra khi \(\left\{{}\begin{matrix}\sqrt{-\left(x-1\right)^2+3}=\sqrt{3}\\\sqrt{-\left(x+3\right)^2+1}=1\end{matrix}\right.\)

Lại có x không thể tồn tại 2 giá trị cùng lúc

Suy ra vô nghiệm