Chả biết đúng hay sai :v làm thử 

\(a)\) Với \(\hept{\begin{cases}x+1\ge0\\x\ge0\\x-1\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge-1\\x\ge0\\x\ge1\end{cases}\Leftrightarrow}x\ge1}\) ta có : 

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

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

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

\(\Leftrightarrow\)\(3x^2=0\)

\(\Leftrightarrow\)\(x^2=0\)

\(\Leftrightarrow\)\(x=0\) ( không thỏa mãn ) 

Với \(\hept{\begin{cases}x+1< 0\\x< 0\\x-1< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< -1\\x< 0\\x< 1\end{cases}\Leftrightarrow}x< -1}\) ta có : 

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

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

Đến đây giải giống như trên nha bạn 

\(\Leftrightarrow\)\(x=0\) ( không thỏa mãn ) 

Vậy không có giá trị x thỏa mãn đề bài 

Chúc bạn học tốt ~ 

\(\left(x-1\right)^2+2\left(x-1\right)\left(3x+1\right)+\left(3x+1\right)^2-16x^2\)

\(=\left(x-1+3x+1\right)^2-16x^2\)

\(=\left(4x\right)^2-16x^2\)

\(=16x^2-16x^2\)

\(=0\)

\(5x\left(3x^2-4x+2\right)\)

\(=5x.3x^2-5x.4x+5x.2\)

\(=15x^3-20x^2+10x\)

hk tốt

^^

Đặt \(\frac{a-b}{c}=x;\frac{b-c}{a}=y;\frac{c-a}{b}=z\Rightarrow\frac{c}{a-b}=\frac{1}{x};\frac{a}{b-c}=\frac{1}{y};\frac{b}{c-a}=\frac{1}{z}\)

Vì a+b+c=0 => a=-b-c ; b=-c-a ; c=-a-b 

                         a³+b³+c³=3abc

Ta có: \(\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3+\frac{x+z}{y}+\frac{x+y}{z}+\frac{y+z}{x}\)

Lại có: \(\frac{x+z}{y}=\left(x+z\right)\cdot\frac{1}{y}=\left(\frac{a-b}{c}+\frac{c-a}{b}\right)\cdot\frac{a}{b-c}=\frac{ab-b^2+c^2-ac}{bc}\cdot\frac{a}{b-c}\)

\(=\frac{a\left(b-c\right)-\left(b-c\right)\left(b+c\right)}{bc}\cdot\frac{a}{b-c}=\frac{\left(a-b-c\right)\left(b-c\right)}{bc}\cdot\frac{a}{b-c}=\frac{a\left(a+a\right)}{bc}=\frac{2a^2}{bc}=\frac{2a^3}{abc}\)

Tượng tự \(\frac{x+y}{z}=\frac{2b^3}{abc};\frac{y+z}{x}=\frac{2c^3}{abc}\)

Do đó \(\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)=3+\frac{2a^3+2b^3+2c^3}{abc}=3+\frac{2\left(a^3+b^3+c^3\right)}{abc}=3+\frac{2.3abc}{abc}=9\)

=>đpcm

Sao phải phức tạp thế?

bạn dùng chủ yếu là hdt thôi còn tam thức bậc 2 thì có thẻ dùng máy tính được nha

\(a,4x^2-4x+1=0\)

\(\Leftrightarrow\left(2x\right)^2+2.2x.1+1^2=0\)

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

\(\Leftrightarrow2x+1=0\)

\(\Leftrightarrow2x=-1\)

\(\Leftrightarrow x=\frac{-1}{2}\)

\(b,4x^2-4x-8=0\)

\(\Leftrightarrow4\left(x^2-x-2\right)=0\)

\(\Leftrightarrow x^2-x-2=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)

\(c,\left(3x-4\right)^2-14\left(3x-4\right)\left(6+3x\right)+49\left(3x+6\right)=16\)

\(\Leftrightarrow\left(3x-4\right)^2-14\left(3x-4\right)\left(3x+6\right)\left(3x+6\right)+49=16\)

\(\Leftrightarrow\left(3x-4\right)^2-14\left(3x-4\right)\left(3x+6\right)+49\left(3x+6\right)=16\)

\(\Leftrightarrow9x^2-24x+16-126x^2-252x+168x+336+147x+294=16\)

\(\Leftrightarrow-117x^2+39x+646=16\)

\(\Leftrightarrow117x^2-39x-646+16=0\)

\(\Leftrightarrow117x^2-39x+630=0\)

\(\Leftrightarrow...\)

a) \(\left(2x\right)^2-2.2.x+1=\left(2x-1\right)^2\)

b) \(\left(2x\right)^2-2.2.x+1-9=\left(2x-1\right)^2-3^2\)

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

c)