\(A=x^2-xy+y^2\)

\(\Rightarrow A=x^2-xy+\dfrac{1}{4}y^2-\dfrac{1}{4}y^2+y^2\)

\(\Rightarrow A=\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2\)

mà \(\left(x-\dfrac{1}{2}y\right)^2\ge0;\dfrac{3}{4}y^2\ge0\)

\(\Rightarrow A=\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2\ge0\)

\(\Rightarrow\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2>0\) với mọi x,y không đồng thời bằng 0

\(\dfrac{2x}{x-1}+\dfrac{3\left(x+1\right)}{x}=5\left(\text{đ}k\text{x}\text{đ}:x\ne1\right)\\ \Leftrightarrow\dfrac{2x^2}{x\left(x-1\right)}+\dfrac{3\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)}=\dfrac{5x\left(x-1\right)}{x\left(x-1\right)}\\ \Rightarrow2x^2+\left(3x+3\right)\left(x-1\right)=5x^2-5x\\ \Leftrightarrow2x^2+3x^2-3x+3x-3=5x^2-5x\\ \Leftrightarrow5x^2-3-5x^2+5x=0\\ \Leftrightarrow5x-3=0\\ \Leftrightarrow5x=3\\ \Leftrightarrow x=\dfrac{3}{5}\)

\(b,\left|1-2x\right|=2x-1\) `(1)`

Nếu `1-2x ≥0<=> 2x≥1<=>x≥`\(\dfrac{1}{2}\)  thì biểu thức `(1)` trở thành

`1-2x=2x-1`

`<=> 1+1=2x+2x`

`<=> 2=4x`

`<=> -4x=-2`

`<=>x=` \(\dfrac{-2}{-4}=\dfrac{1}{2}\) ( thoả mãn đk )

Nếu `1-2x <0<=> 2x<1<=>x<`\(\dfrac{1}{2}\) thì biểu thức `(1)` trở thành

`-(1-2x)=2x-1`

`<=>-1+2x=2x-1`

`<=> 2x-2x=-1+1`

`<=>0=0` ( luôn đúng )

`c,`

\(\dfrac{2\left(x+1\right)}{3}-2\ge\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{4\left(x+1\right)}{6}-\dfrac{2}{6}\ge\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow4x+4-2\ge3x-6\\ \Leftrightarrow4x+2\ge3x-6\\ \Leftrightarrow4x-3x\ge-6-2\\ \Leftrightarrow x\ge-8\)

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

\(\dfrac{x\cdot2x}{x\left(x-1\right)}+\dfrac{3\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)}=5\)

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

\(\dfrac{2x^2}{x^2-x}+\dfrac{3x^2-3}{x^2-x}=5\)

\(\dfrac{2x^2+3x^2-3}{x^2-x}=\dfrac{5x^2-3}{x^2-x}=5\)

\(\Rightarrow5x^2-3=5\left(x^2-x\right)=5x^2-5x\)

\(\Rightarrow3=5x\)

\(x=\dfrac{3}{5}\)

b) \(\left|1-2x\right|=2x-1\)

TH1: \(1>2x\)

\(\Rightarrow\left[{}\begin{matrix}1-2x>0\\2x-1< 0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left|1-2x\right|>0\\2x-1< 0\end{matrix}\right.\) => Vô lí

TH2: \(1\le2x\)

\(\Rightarrow\left[{}\begin{matrix}1-2x\le0\Rightarrow\left|1-2x\right|\ge0\\2x-1\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left|1-2x\right|=2x-1\ge0\)

\(\Leftrightarrow2x-1\ge0\Rightarrow2x-1+1=2x\ge0+1=1\)

\(\Leftrightarrow\dfrac{2x}{2}=x\ge\dfrac{1}{2}\)

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

\(\Rightarrow2a^2+2b^2=a^2-2ab+b^2\)

\(\Rightarrow a^2+b^2+2ab=0\)

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

\(\Rightarrow a+b=0\Rightarrow a=-b\Rightarrow dpcm\)

\(A=x^2+x+1\)

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

\(A=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

mà \(\left(x+\dfrac{1}{2}\right)^2\ge0\)

\(\Rightarrow A=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>\dfrac{3}{4}>0\) với mọi x

\(\Rightarrow Dpcm\)

Bạn xem lại đề

\(\left(x^2-25\right)^2-\left(x+5\right)^2=0\)

\(\Leftrightarrow\left[x^2-5^2\right]^2-\left(x+5\right)^2=0\)

\(\Leftrightarrow\left[\left(x+5\right)\left(x-5\right)\right]^2-\left(x+5\right)^2=0\)

\(\Leftrightarrow\left(x+5\right)^2\left(x-5\right)^2-\left(x+5\right)^2=0\)

\(\Leftrightarrow\left(x+5\right)^2\left[\left(x-5\right)^2-1\right]=0\)

\(\Leftrightarrow\left(x+5\right)^2\left[\left(x-5\right)+1\right]\left[\left(x-5\right)-1\right]=0\)

\(\Leftrightarrow\left(x+5\right)^2\left(x-4\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x+5\right)^2=0\\x-4=0\\x-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x=4\\x=6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=4\\x=6\end{matrix}\right.\)

Vậy: \(S=\left\{-5;6;4\right\}\)

Ta có ( x² - 25 )² - ( x + 5 )² = 0

Vì ( x² - 25 )² ≥ 0 ; ( x + 5 )² ≥ 0 

⇒ ( x² - 25 )² - ( x + 5 )² ≥ 0

Dấu " = " xảy ra khi 

\(\left[{}\begin{matrix}\left(x^2-25\right)^2=0\\\left(x+5\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\pm5\\x=-5\end{matrix}\right.\Rightarrow x=-5\)

Vậy x = 5 

\(x^3-4x^2-9x+36=0\)

\(x^2\left(x-4\right)-9\left(x-4\right)=0\)

\(\left(x-4\right)\left(x^2-9\right)=0\)\(\)

\(\Rightarrow x-4=0\) hay \(x^2-9=0\)

\(\Rightarrow x=4\) hay \(x^2=9=3^2\)

\(\Rightarrow x=4\) hay \(x=\pm3\)

⇔x²(x-4) -9(x-4) = 0

⇔(x-4).(x-3).(x+3) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\\x=-3\end{matrix}\right.\)

Ta có

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

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

\(\Rightarrow K=\left(x+y\right)^3-3xy\left(x+y\right)+3xy\) Với x+y=1

\(\Rightarrow K=1^3-3xy+3xy=1\)