\(x\left(x-3\right)-3x+9=0\)

\(x\left(x-3\right)-3\left(x-3\right)=0\)

\(\left(x-3\right)\left(x-3\right)=0\)

\(\left(x-3\right)^2=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

Vậy x = 3

(x-3)² = 0 

:   x-3  = 0 

x = 3

học tốt

Giơ hai tay

trước

a)\(x^2-y^2+7x+7y\)

\(=\left(x-y\right).\left(x+y\right)+7.\left(x+y\right)\)

\(=\left(x+y\right).\left(x-y+7\right)\)

\(b,x^2+5x+4\)

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

\(=x.\left(x+1\right)+4.\left(x+1\right)\)

\(=\left(x+4\right).\left(x+1\right)\)

\(c,x^3-9x^2\)

\(=x^2.\left(x-9\right)\)

\(d,x^3+x^2+2x\)

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

\(e,3x^2+3y^2-6xy-1^2\)

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

\(=\left(3x-3y-1\right).\left(3x-3y+1\right)\)

cho sửa câu cuối cái nha :>

\(=3.\left(x-y\right)^2-1^{^2}\)

\(=3x^2-3y^2-1^2\)

đến đây tịt r :< sory 

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

\(A=2\left(x^2-2x+\frac{3}{2}\right)\)

\(A=2\left(x^2-2\cdot x\cdot1+1^2+\frac{1}{2}\right)\)

\(A=2\left[\left(x-1\right)^2+\frac{1}{2}\right]\)

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

Ta có \(\left(x-1\right)^2\ge0\forall x\Rightarrow2\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow2\left(x-1\right)^2+1\ge1\forall x\)

\(\Rightarrow A>0\forall x\)

ta có: A = 2x² - 4x + 3 = x² + x² - 2x - 2x + 1 + 1 + 1

A = (x² - 2x +1) + (x² -2x+1) + 1

A = (x-1)² + (x-1)²  +1

A = 2.(x-1)²  + 1

mà \(2.\left(x-1\right)^2\ge0\Rightarrow2.\left(x-1\right)^2+1\ge1.\)

=> A = 2.(x-1)² + 1 > 0 (đpcm)

...

ctv bị lạc trôi à, hay sao mak làm kiểu ý z bài náy cm mak đâu phải tìm GTNN, GTLN

:))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))chịu thôi khó mãi thôi chỉ cho câu D là được rồi 

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