a, \(x^3+3x^2-\left(x+3\right)=0\Leftrightarrow x^2\left(x+3\right)-\left(x+3\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+3\right)=0\Leftrightarrow x=1;x=-1;x=-3\)

b, \(15x-5+6x^2-2x=0\Leftrightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)

\(\Leftrightarrow\left(2x+5\right)\left(3x-1\right)=0\Leftrightarrow x=-\frac{5}{2};x=\frac{1}{3}\)

c, \(5x-2-25x^2+10x=0\)

\(\Leftrightarrow\left(5x-2\right)-5x\left(5x-2\right)=0\Leftrightarrow\left(1-5x\right)\left(5x-2\right)=0\Leftrightarrow x=\frac{2}{5};x=\frac{1}{5}\)

= 1/5 nha

Trả lời:

\(3x\left(x-7\right)+2x-14=0\)

\(\Leftrightarrow3x\left(x-7\right)+2\left(x-7\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(3x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\3x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-\frac{2}{3}\end{cases}}}\)

Vậy x = 7; x = - 2/3 là nghiệm của pt.

sửa đề :

 \(3x\left(x-7\right)+2x-14=0\Leftrightarrow3x\left(x-7\right)+2\left(x-7\right)=0\)

\(\Leftrightarrow\left(3x+2\right)\left(x-7\right)=0\Leftrightarrow x=-\frac{2}{3};x=7\)

\(x^2y+y^2x+x^2z+z^2x+y^2z+z^2y+2xyz\)..

\(=\left(x^2y+z^2y+2xyz\right)+\left(y^2x+y^2z\right)+\left(z^2x+x^2z\right)\).

\(=y\left(x+z\right)^2+y^2\left(x+z\right)+xz\left(x+z\right)\)

\(=\left(xy+yz\right)\left(x+z\right)+\left(x+z\right)\left(y^2+xz\right)\).

\(=\left(x+z\right)\left(xy+yz+y^2+xz\right)\).

\(=\left(x+z\right)\left[x\left(y+z\right)+y\left(y+z\right)\right]\).

\(=\left(x+z\right)\left(x+y\right)\left(y+z\right)\).

fdkrycyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyydr 

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

\(=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2\)

\(=a^2+2ab+b^2+b^2+2bc+c^2+c^2+2ca+a^2\)

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

x=67e

Ta có: a² + b² + c² = ab + bc + ca

\(\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)

\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\)

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

\(\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Rightarrow a=b=c\)(đpcm)

đây nha

  PS : nhớ k :33

                                                                                                                                                      # Aeri #