\(4x^2+8x+2=0\)\(\Leftrightarrow4x^2+8x+4-2=0\)\(\Leftrightarrow4\left(x^2+2x+1\right)-2=0\)\(\Leftrightarrow4\left(x+1\right)^2-\left(\sqrt{2}\right)^2=0\)\(\Leftrightarrow\left[2\left(x+1\right)\right]^2-\left(\sqrt{2}\right)^2=0\)\(\Leftrightarrow\left[2\left(x+1\right)+\sqrt{2}\right]\left[2\left(x+1\right)-\sqrt{2}\right]=0\)\(\Leftrightarrow\left(2x+2+\sqrt{2}\right)\left(2x+2-\sqrt{2}\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}2x+2+\sqrt{2}=0\\2x+2-\sqrt{2}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-2-\sqrt{2}}{2}\\x=\frac{-2+\sqrt{2}}{2}\end{cases}}\)

Vậy tập nghiệm của pt đã cho là \(S=\left\{\frac{-2\pm\sqrt{2}}{2}\right\}\)

ơ ơ ko lm đc 

thay 2 => 4 may ra còn lm đc

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

Điều kiện: \(x\ne2;x\ne0\)

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

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

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

\(\Leftrightarrow2x^2+3x-4=2x^2-4x\)

\(\Leftrightarrow x=4\)

\(\dfrac{x-1}{3}-\dfrac{2x+3}{6}\le x\)

\(\Leftrightarrow6\left(x-1\right)-3\left(2x+3\right)\le6.3x\)

\(\Leftrightarrow6x-6-6x-9\le18x\)

\(\Leftrightarrow18x\ge-15\)

\(\Leftrightarrow x\ge-\dfrac{5}{6}\)

ko thây đề bn ơi

\(\left(a^3+b\right)\left(\dfrac{1}{a}+b\right)\ge\left(a+b\right)^2\Rightarrow\dfrac{1}{a^3+b}\le\dfrac{\dfrac{1}{a}+b}{\left(a+b\right)^2}=\dfrac{ab+1}{a\left(a+b\right)^2}\)

Tương tự: \(\dfrac{1}{b^3+a}\le\dfrac{ab+1}{b\left(a+b\right)^2}\)

\(\Rightarrow P\le\left(a+b\right)\left(\dfrac{ab+1}{a\left(a+b\right)^2}+\dfrac{ab+1}{b\left(a+b\right)^2}\right)-\dfrac{1}{ab}\)

\(P\le\dfrac{\left(ab+1\right)}{a+b}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)-\dfrac{1}{ab}=\dfrac{ab+1}{ab}-\dfrac{1}{ab}=1\)

\(P_{max}=1\) khi \(a=b=1\)

Với mọi x;y dương ta có:

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

Áp dụng:

\(VT\le\dfrac{1}{ab\left(a+b\right)+1}+\dfrac{1}{bc\left(b+c\right)+1}+\dfrac{1}{ca\left(c+a\right)+1}\)

\(VT\le\dfrac{abc}{ab\left(a+b\right)+abc}+\dfrac{abc}{bc\left(b+c\right)+abc}+\dfrac{abc}{ca\left(c+a\right)+abc}\)

\(VT\le\dfrac{c}{a+b+c}+\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}=1\)

Dấu "=" xảy ra khi \(a=b=c=1\)

HF x ??

Xét \(\Delta\text{A}BC\)có :

 \(ED//\text{A}C\left(gt\right)\)

\(\Rightarrow\frac{BE}{\text{A}B}=\frac{DE}{\text{A}C}\)

\(\Rightarrow\frac{BE}{ED}=\frac{\text{A}B}{\text{A}C}(1)\)

Có : AD là phân giác góc \(B\text{A}C\)

=> góc \(B\text{A}D\)=  góc \(C\text{A}D\) 

Có : \(ED//\text{A}C\left(gt\right)\)

=> góc \(\text{A}DE\)=  góc \(C\text{A}D\) 

mà  góc \(B\text{A}D\)=  góc \(C\text{A}D\) ( cmt)

=> góc \(\text{A}DE\)=  góc \(B\text{A}D\)

=> \(\Delta ED\text{A}\) cân tại E

=> \(ED=E\text{A}\)

Cộng mỗi vế của (1) với 1, ta có : 

\(1+\frac{\text{A}B}{\text{A}C}=\frac{BE}{ED}+1\)

=>\(\frac{\text{A}B}{\text{A}B}+\frac{\text{A}B}{\text{A}C}=\frac{BE}{ED}+\frac{ED}{ED}\)

mà \(ED=E\text{A}\left(cmt\right)\)

=>\(\frac{\text{A}B}{\text{A}B}+\frac{\text{A}B}{\text{A}C}=\frac{BE}{ED}+\frac{E\text{A}}{ED}\)

=>\(\frac{\text{A}B}{\text{A}B}+\frac{\text{A}B}{\text{A}C}=\frac{\text{A}B}{ED}\)

=>\(\frac{1}{\text{A}B}+\frac{1}{\text{A}C}=\frac{1}{ED}\)

mà  ​​\(ED=E\text{A}\left(cmt\right)\)

=> \(\frac{1}{\text{A}B}+\frac{1}{\text{A}C}=\frac{1}{E\text{A}}\left(đpcm\right)\)

Không mất tính tổng quát, giả sử \(x^2\ge y^2\)

suy ra \(1=x^2+y^2-xy\ge y^2\)

\(\Rightarrow y\in\left\{-1,0,1\right\}\)(vì \(y\)nguyên) 

Với \(y=-1\): \(x^2+1+x=1\Leftrightarrow x^2+x=0\Leftrightarrow\orbr{\begin{cases}x=-1\\x=0\end{cases}}\)(thỏa mãn) 

Với \(y=0\): \(x^2=1\Leftrightarrow x=\pm1\)

Với \(y=1\): \(x^2+1-x=1\Leftrightarrow x^2-x=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}\)(thỏa mãn)