tìm y nữa 

mình viết thiếu

(Phần a mình lấy vế phải bằng 0 nha ^^)

a,

\(\left(5x-1\right)^2-\left(5x-4\right)\left(5x+4\right)+7=0\\ \Leftrightarrow25x^2-10x+1-\left(25x^2-16\right)+7=0\\ \Leftrightarrow25x^2-10x+1-25x^2+16+7=0\\ \Leftrightarrow-10x+24=0\\ \Leftrightarrow x=2,4\)

b,

\(5x^2+4xy+4y^2+4x+1=0\left(1\right)\\ \Leftrightarrow4x^2+4x+1+x^2+4xy+4y^2=0\\ \Leftrightarrow\left(2x+1\right)^2+\left(x+2y\right)^2=0\left(1a\right)\)

Do \(VT\ge0\) với \(\forall x,y\in R\) nên:

\(\left(1a\right)\Leftrightarrow\left\{{}\begin{matrix}2x+1=0\\x+2y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\frac{1}{2}\\y=\frac{1}{4}\end{matrix}\right.\)

c,

\(\left(x+2\right)^3-x\left(x-1\right)\left(x+1\right)=6x^2+21\\ \Leftrightarrow x^3+6x^2+12x+8-x\left(x^2-1\right)-6x^2-21=0\\ \Leftrightarrow x^3+12x+8-x^3+x-21=0\\ \Leftrightarrow13x-13=0\\ \Leftrightarrow x=1\)

Chúc bạn học tốt nha.

\(b)5x^2 + 4xy + 4y^2 + 4x + 1 = 0\)

\(\Leftrightarrow\) \(4x^2 + 4x + 1 + x^2 + 4xy + 4y^2 = 0\)

\(\Leftrightarrow\)\((2x + 1)^2 + (x + 2y)^2 = 0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+1=0\\x+2y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\frac{1}{2}\\y=\frac{1}{4}\end{matrix}\right.\)

\(c)(x+2)^3-x(x-1)(x+1)=6x^2+21\)

\(\Leftrightarrow x^3+6x^2+12x+8-x\left(x^2-1\right)=6x^2+21\\ \Leftrightarrow13x+8=21\\ \Leftrightarrow13x=21-8\\ \Leftrightarrow13x=13\\ \Leftrightarrow x=1\)

Bài 1: 

a: \(\Leftrightarrow x^2-5x+6< =0\)

=>(x-2)(x-3)<=0

=>2<=x<=3

b: \(\Leftrightarrow\left(x-6\right)^2< =0\)

=>x=6

c: \(\Leftrightarrow x^2-2x+1>=0\)

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

hay \(x\in R\)

\(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow8x+16-5x^2-10x+4x^2+4x-8x-8+2x^2-8=0\)

\(\Leftrightarrow x^2-6x=0\Leftrightarrow x\left(x-6\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x-6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}}\)

                                    Vậy S = { 0, 6}

\(a,A=4x+2+2x-2-5x-6x^2-44x+2+2x-2-5x-6x^2-4\)

\(=\left(4x+2x-5x-44x+2x-5x\right)-\left(6x^2+6x^2\right)+\left(2-2+2-2-4\right)\)

\(=-46x-12x^2-4\)

Thay \(x=7373\) vào biểu thức A, ta có :

\(-46.7373-12.7373^2-4\)

\(-652672710\)

a) (x+2)(x+3)-(x-2)(x+5)=0

  \(x^2+3x+2x+6-x^2-5x+2x+10=0\) 

\(2x+16=0\) 

\(2x=-16\) 

\(x=-8\) 

Vậy......

b) (8-5x)(x+2)+4(x-2)(x+1)+2(x-2)(x+2)=0

  \(8x+16-5x^2-10x+4x^2+4x-8x-8+2x^2+4x-4x-8=0\) 

  \(-6x+x^2=0\) 

 \(x\left(-6+x\right)=0\) 

=> x=0   hoặc  -6+x=0  <=>x=6

Vậy \(x\in\left\{0;6\right\}\)

a) \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+2\right)x+\left(x+2\right).3-\left(x+5\right)x+\left(x+5\right).2=0\)

\(\Leftrightarrow x^2+2x+3x+6-x^2+5x+2x+10=0\)

\(\Leftrightarrow12x+16=0\)

\(\Leftrightarrow12x=-16\)

\(\Leftrightarrow x=\frac{-4}{3}\)

Vậy...

Bài tui sai rồi 2x ghi nhầm 12x