Ace Legona

gì mà khác 0 thay x=3 là rõ

Áp dụng BĐT AM-GM ta có:

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

Tương tự cho 2 BĐT còn lại ta cũng có:

\(\dfrac{y+1}{1+z^2}\ge y+1-\dfrac{z\left(y+1\right)}{2};\dfrac{z+1}{1+x^2}\ge z+1-\dfrac{x\left(z+1\right)}{2}\)

Cộng theo vế 3 BĐT trên ta có:

\(Q\ge\left(x+y+z+3\right)-\dfrac{x\left(z+1\right)+y\left(x+1\right)+z\left(y+1\right)}{2}\)

\(=6-\dfrac{xy+yz+xz+x+y+z}{2}\)

\(\ge6-\dfrac{\dfrac{\left(x+y+z\right)^2}{3}+3}{2}=6-3=3\)

Đẳng thức xảy ra khi \(x=y=z=1\)

\(x^2+2y^2-2xy+4y+4=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2+4y+4\right)=0\)

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

Dễ thấy: \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\\\left(y+2\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow\left(x-y\right)^2+\left(y+2\right)^2\ge0\)

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

câu hỏi tt ko thiếu mà mấy dạng này thì thêm bớt nt từ \(x^2;x^3;x^4;...\) đến số mà có nt cao nhất r` đặt ra là xong dễ ẹc

a)\(x-x^2=-x^2+x-\dfrac{1}{4}+\dfrac{1}{4}\)

\(=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)

Đẳng thức xảy ra khi \(x=\dfrac{1}{2}\)

b)\(4x-x^2+3=-x^2+4x-4+7\)

\(=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)

Đẳng thức xảy ra khi \(x=2\)

\(x^3-4x^2-12x+27\)

\(=x^3-7x^2+9x+3x^2-21x+27\)

\(=x\left(x^2-7x+9\right)+3\left(x^2-7x+9\right)\)

\(=\left(x+3\right)\left(x^2-7x+9\right)\)

giúp lần cuối lần sau tự nhìn vào bài này mà nghĩ

Bài 1:

\(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=63\)

\(\Leftrightarrow x^2-1=63\)

\(\Leftrightarrow x^2=64\Rightarrow x=\pm8\)

Bào 2:

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

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

\(\Leftrightarrow x^2-4x+4=x^2+2x-3\)

\(\Leftrightarrow7-6x=0\Leftrightarrow6x=7\Rightarrow x=\dfrac{7}{6}\)

\(\left(x^3-4\right)^3=\left(\sqrt[3]{\left(x^2+4\right)^2}+4\right)^2\)

\(pt\Leftrightarrow x^9-12x^6+48x^3-64=\left(\sqrt[3]{\left(x^2+4\right)^2}\right)^2+8\sqrt[3]{\left(x^2+4\right)^2}+16\)

\(\Leftrightarrow x^9-12x^6+48x^3-128=\left(\sqrt[3]{\left(x^2+4\right)^2}\right)^2-16+8\sqrt[3]{\left(x^2+4\right)^2}-32\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)\left(x^6-4x^3+16\right)=\dfrac{\left(x^2+4\right)^4-4096}{\left(\sqrt[3]{\left(x^2+4\right)^2}\right)^2+16}+\dfrac{512\left(x^2+4\right)^2-32768}{8\sqrt[3]{\left(x^2+4\right)^2}+32}\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)\left(x^6-4x^3+16\right)=\dfrac{\left(x-2\right)\left(x+2\right)\left(x^2+12\right)\left(x^4+8x^2+80\right)}{\left(\sqrt[3]{\left(x^2+4\right)^2}\right)^2+16}+\dfrac{512\left(x-2\right)\left(x+2\right)\left(x^2+12\right)}{8\sqrt[3]{\left(x^2+4\right)^2}+32}\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)\left(x^6-4x^3+16\right)-\dfrac{\left(x-2\right)\left(x+2\right)\left(x^2+12\right)\left(x^4+8x^2+80\right)}{\left(\sqrt[3]{\left(x^2+4\right)^2}\right)^2+16}-\dfrac{512\left(x-2\right)\left(x+2\right)\left(x^2+12\right)}{8\sqrt[3]{\left(x^2+4\right)^2}+32}=0\)

\(\Leftrightarrow\left(x-2\right)\left[\left(x^2+2x+4\right)\left(x^6-4x^3+16\right)-\dfrac{\left(x+2\right)\left(x^2+12\right)\left(x^4+8x^2+80\right)}{\left(\sqrt[3]{\left(x^2+4\right)^2}\right)^2+16}-\dfrac{512\left(x+2\right)\left(x^2+12\right)}{8\sqrt[3]{\left(x^2+4\right)^2}+32}\right]=0\)

Dễ thấy: \(\left(x^2+2x+4\right)\left(x^6-4x^3+16\right)-\dfrac{\left(x+2\right)\left(x^2+12\right)\left(x^4+8x^2+80\right)}{\left(\sqrt[3]{\left(x^2+4\right)^2}\right)^2+16}-\dfrac{512\left(x+2\right)\left(x^2+12\right)}{8\sqrt[3]{\left(x^2+4\right)^2}+32}=0\) vô nghiệm

\(\Rightarrow x-2=0\Rightarrow x=2\)

P.s: dễ thấy thật :v

\(A=\frac{1}{3}(3(x+y)-1)^2-\frac{301}{3}\)

\(x^2+2y^2+2xy-2y+1=0\)

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

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

Dễ thấy: \(\left\{{}\begin{matrix}\left(x+y\right)^2\ge0\\\left(y-1\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow\left(x+y\right)^2+\left(y-1\right)^2\ge0\)

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

ý tưởng : nhóm với ad |a|+|b|>=|a+b| sao cho VT>= VP

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

\(xy-2x-3y=5\)

\(\Leftrightarrow x\left(y-2\right)-\left(3y-6\right)=11\)

\(\Leftrightarrow x\left(y-2\right)-3\left(y-2\right)=11\)

\(\Leftrightarrow\left(x-3\right)\left(y-2\right)=11\)

\(\Rightarrow\left\{{}\begin{matrix}x-3=1\\y-2=11\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x-3=11\\y-2=1\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x-3=-11\\y-2=-1\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x-3=-1\\y-2=-11\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=4\\y=13\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=14\\y=3\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=-8\\y=1\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=2\\y=-9\end{matrix}\right.\)

tự làm đi thớt bài này quá dễ rồi

\(A=\dfrac{3}{5\cdot10}+\dfrac{3}{10\cdot15}+...+\dfrac{3}{95\cdot100}\)

\(=\dfrac{3}{5}\left(\dfrac{5}{5\cdot10}+\dfrac{5}{10\cdot15}+...+\dfrac{5}{95\cdot100}\right)\)

\(=\dfrac{3}{5}\left(\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{15}+...+\dfrac{1}{95}-\dfrac{1}{100}\right)\)

\(=\dfrac{3}{5}\left(\dfrac{1}{5}-\dfrac{1}{100}\right)\)\(=\dfrac{3}{5}\cdot\dfrac{19}{100}=\dfrac{57}{500}\)

nhóm vào rồi áp dụng BĐT $$|a|+|b|\ge|a+b|$$

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

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

\(=\left(2-x\right)\left(z-y\right)\) với \(x=2;y=1,007;z=-0,006\) thì

\(A=\left(2-2\right)\left(-0,006-1,007\right)=0\)

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

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

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

\(=\left(z-y\right)\left(m-x\right)\) với \(x=18,3;y=24,6;z=10,6;m=-31,7\) thì

\(B=\left(10,6-24,6\right)\left(-31,7-18,3\right)=700\)

làm tương tự bài cũ đi làm lại mệt thấy 3`

Ta có: \(\dfrac{b}{\sqrt{a+b}-\sqrt{a-b}}=\dfrac{b}{\dfrac{\left(a+b\right)-\left(a-b\right)}{\sqrt{a+b}+\sqrt{a-b}}}\)

\(=\dfrac{b}{\dfrac{a+b-a+b}{\sqrt{a+b}+\sqrt{a-b}}}=\dfrac{\sqrt{a+b}+\sqrt{a-b}}{b}\)

Và \(\dfrac{c}{\sqrt{a+c}-\sqrt{a-c}}=\dfrac{c}{\dfrac{\left(a+c\right)-\left(a-c\right)}{\sqrt{a+c}+\sqrt{a-c}}}\)

\(=\dfrac{c}{\dfrac{a+c-a+c}{\sqrt{a+c}+\sqrt{a-c}}}=\dfrac{\sqrt{a+c}+\sqrt{a-c}}{c}\)

Từ \(a>b>c>0\) thì \(\left\{{}\begin{matrix}a+b>a+c\\a-b>a-c\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\sqrt{a+b}>\sqrt{a+c}\\\sqrt{a-b}>\sqrt{a-c}\end{matrix}\right.\)

\(\Rightarrow\sqrt{a+b}+\sqrt{a-b}>\sqrt{a+c}+\sqrt{a-c}\)

\(\Rightarrow\dfrac{\sqrt{a+b}+\sqrt{a-b}}{b}< \dfrac{\sqrt{a+c}+\sqrt{a-c}}{c}\left(b>c>0\right)\)

Hay ta có ĐPCM

\(\left\{{}\begin{matrix}x\left(x-y\right)=\dfrac{3}{10}\\y\left(x-y\right)=-\dfrac{3}{10}\end{matrix}\right.\)

Cộng theo vế 2 pt trên ta có:

\(x\left(x-y\right)+y\left(x-y\right)=\dfrac{-3}{10}+\dfrac{3}{10}=0\)

\(\Leftrightarrow\left(x+y\right)\left(x-y\right)=0\)\(\Rightarrow\left[{}\begin{matrix}x=-y\\x=y\end{matrix}\right.\)

*)Xét \(x=-y\) thì \(x\left(x-y\right)=\dfrac{3}{10}\Leftrightarrow-y\left(-y-y\right)=\dfrac{3}{10}\)

\(\Leftrightarrow2y^2=\dfrac{3}{10}\Leftrightarrow y^2=\dfrac{3}{20}\Rightarrow y=\pm\sqrt{\dfrac{3}{20}}\)\(\Rightarrow x=\mp\sqrt{\dfrac{3}{20}}\)

*)Xét \(x=y\) thì \(x\left(x-y\right)=\dfrac{3}{10}\Leftrightarrow y\left(y-y\right)=\dfrac{3}{10}\)

\(\Leftrightarrow0=\dfrac{3}{10}\) (loại)

Vậy \(x=\mp\sqrt{\dfrac{3}{20}};y=\pm\sqrt{\dfrac{3}{20}}\)

\(\dfrac{b}{\sqrt{a+b}-\sqrt{a-b}}=\dfrac{b}{\dfrac{a+b-a-b}{\sqrt{a+b}+\sqrt{a-b}}}=\dfrac{b}{\dfrac{0}{\sqrt{a+b}+\sqrt{a-b}}}\rightarrow\varnothing\)

Ta có:

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

Từ \(a^3+b^3+c^3=0\Rightarrow a^3+2c^3=c^3-b^3\), thì:

\(b^3(c^3-b^3)+3a^3c^3=-b^6+c^3(b^3+3a^3)\)

Và từ \(a^3+b^3+c^3=0\Rightarrow b^3+3a^3=2a^3-c^3\)

Suy ra \(-b^6+c^3(2a^3-c^3)=-(b^3-c^3)^2\le 0\)

Gọi 3 số tự nhiên liên tiếp đó lần lượt là \(a;a+1;a+2\left(a;a+1;a+2\in N\right)\)

Theo bài ra ta có:

\(a\left(a+1\right)+\left(a+1\right)\left(a+2\right)+a\left(a+2\right)=26\)

\(\Leftrightarrow a^2+a+a^2+3a+2+a^2+2a=26\)

\(\Leftrightarrow3a^2+6a+2=26\)

\(\Leftrightarrow3a^2+6a-24=0\)

\(\Leftrightarrow a^2+2a-8=0\)

\(\Leftrightarrow\left(a+4\right)\left(a-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a+4=0\\a-2=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}a=-4\left(loai\right)\\a=2\end{matrix}\right.\)

Vậy 3 số đó lần lượt là \(2;3;4\)

\(\left(x^{2n}+x^ny^n+y^{2n}\right)\left(x^n-y^n\right)\left(x^{3n}+y^{3n}\right)\)

\(=\left(x^{3n}-y^{3n}\right)\left(x^{3n}+y^{3n}\right)\)

\(=x^{6n}-y^{6n}\)

sai đề

Ta có: \(P\left(x\right)=x^5+ax^4+bx^3+cx^2+dx+e\)

Suy ra \(P\left(1\right)=1^5+a\cdot1^4+b\cdot1^3+c\cdot1^2+d\cdot1+e=1\)

\(\Rightarrow a+b+c+d+e=0\)

\(P\left(2\right)=2^5+a\cdot2^4+b\cdot2^3+c\cdot2^2+d\cdot2+e=4\)

\(\Rightarrow16a+8b+4c+2d+e+28=0\)

\(P\left(3\right)=3^5+a\cdot3^4+b\cdot3^3+c\cdot3^2+d\cdot3+e=9\)

\(\Rightarrow81a+27b+9c+3d+e+234=0\)

\(P\left(4\right)=4^5+a\cdot4^4+b\cdot4^3+c\cdot4^2+d\cdot4+e=16\)

\(\Rightarrow256a+64b+16c+4d+e+1008=0\)

\(P\left(5\right)=5^5+a\cdot5^4+b\cdot5^3+c\cdot5^2+d\cdot5+e=25\)

\(\Rightarrow625a+125b+25c+5d+e+999=0\)

Thay lẫn lộn vào nhau đi nhé