X^3  .  x  - 3 . x^3 = -3 +x 

x^3 . ( x - 3 ) = ( x-3 ) .1

=> x^3 = 1 

=> x = 1 

k cho mình nhé

sao lại sai bạn

Áp dụng BĐT Cosi dạng engel ta có:

\(\frac{1}{x^2+2yz}+\frac{1}{y^2+2zx}+\frac{1}{z^2+2xy}\ge\frac{\left(1+1+1\right)^2}{x^2+2xy+y^2+2zx+z^2+2xy}=\frac{9}{\left(x+y+z\right)^2}=9\) (vì x+y+z=1)

Dấu "=" xảy ra <=> \(x=y=z=\frac{1}{3}\)

\(\frac{1}{x^2+2yz}+\frac{1}{y^2+2zx}+\frac{1}{z^2+xy}\ge\frac{\left(1+1+1\right)^2}{x^2+y^2+z^2+2xy+2yz+2zx}\)

                                                                  \(=\frac{9}{\left(x+y+z^2\right)}=\frac{9}{1}=9\)

Dấu "=" xảy ra khi x=y=z=1/3

      \(2x^2+y^2+z^2-2x-2xy+2z+2=0\)

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

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

Ta có: \(\hept{\begin{cases}\left(x-y\right)^2\ge0\forall x;y\\\left(x-1\right)^2\ge0\forall x\\\left(z+1\right)^2\ge0\forall z\end{cases}\Rightarrow\left(x-y\right)^2+\left(x-1\right)^2+\left(z+1\right)^2\ge0\forall x;y;z}\)

Do đó: \(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(x-1\right)^2=0\\\left(z+1\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x-y=0\\x-1=0\\z+1=0\end{cases}\Rightarrow}\hept{\begin{cases}y=1\\x=1\\z=-1\end{cases}}}\)

Vậy \(x+y+z=1+1+\left(-1\right)=2\)

Chúc bạn học tốt.

a) \(3x\left(x-2\right)-4x+8=0\)

\(\Leftrightarrow3x\left(x-2\right)-4\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(3x-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\3x-4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{4}{3}\end{cases}}\)

b) \(3\left(2x-1\right)^2+2-4x=0\)

\(\Leftrightarrow3\left(2x-1\right)^2-2\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(6x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\6x-5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{5}{6}\end{cases}}\)

DỄ QUÁ!!!!!!!!!!!!!!!!!!!

TUi HK BIẾT LÀM

TÍCH CHO TUI ĐI

THANKS

3[2x-1]-5[x-3]+[3x-4]=24-6

\(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)+2\left(x+2\right)\left(x-3\right)+10\)

\(=-5x^2-2x+16+4\left(x-2\right)+2\left(x+2\right)\left(x-2\right)+10\)

\(=-5x^2-2x+16+4x-8+2\left(x+2\right)\left(x-3\right)+10\)

\(=-5x^2-2x+16+4x-8+2x^2-2x-12+10\)

\(=-3x^2+6\)

Nguyễn Văn Tường lần sau viết rõ đề hộ mk nhé.

\(\left|8-x\right|=x^2-x\)

<=> \(\orbr{\begin{cases}8-x=x^2-x\\8-x=x-x^2\end{cases}}\)

<=> \(\orbr{\begin{cases}8=x^2\\8=2x-x^2\end{cases}}\)

<=> \(\orbr{\begin{cases}x=\pm2\sqrt{2}\\x\left(2-x\right)=8\end{cases}}\)

Tới đây bạn tự giải nhé,.

ta có: |8-x|=x²-x

=> \(\orbr{\begin{cases}8-x=x^2-x\\8-x=x-x^2\end{cases}}\) 

(+) 8-x=x²-x 

<=> x²=8 <=> x=\(\sqrt{8}\)

(+) 8-x=x-x²

<=> x²-2x+8=0

<=> x²-2x+1+7 =0

<=> (x-1)²+7=0

mà (x-1)²\(\ge\) 0 \(\forall\)x nên (x-1)²+7>0

=> ptvn

vậy phương trình đã cho có 1 nghiệm là x=\(\sqrt{8}\)