f: Ta có: \(16x^2-9\left(x+1\right)^2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\)

a: \(P=\dfrac{x^2-2x-3+3-2x}{x-4}=\dfrac{x^2-4x}{x-4}=x\)

1, x= 2

2, x = 4

**** bạn mình trước nhé

trieu dang sai ket qua vi chua doi dau

b: =>4x^2+8x-8x^2+5x-10=0

=>-4x^2+13x-10=0

=>x=2 hoặc x=5/4

c: =>2x^2-5x+6x-15=2x^2+8x

=>x-15=8x

=>-7x=15

=>x=-15/7

d: =>3x^2+15x-2x-10-3x^2-12x=5

=>x-10=5

=>x=15

e: =>x^2-3x+2x^2+2x=3x^2-12

=>-x=-12

=>x=12

- Gửi lẻ câu hỏi ra nha bạn 2 3 câu 1 lần thôi .

a) (x-3)²-4=0

⇒ (x-3)²=4

⇒ hoặc x-3=2⇒x=5

hoặc x-3=-2⇒x=1

\(a.3\left(x^2-2x+1\right)-3x^2+15x-2=0\)

\(3x^2-6x+3-3x^2+15x-2=0\)

\(9x+1=0\)

\(x=-\frac{1}{9}\)

\(b.4x^2-12x+9=0\)

\(4x^2-6x-6x+9=0\)

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

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

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

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

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

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

\(\left(x-8\right)\left(3x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-8=0\\3x+2=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=8\\x=-\frac{2}{3}\end{cases}}\)

1)a)3(2x-1)(3x-1)-(2x-3)(9x-1)=0

<=>18x²-15x+1-18x²+29x-3=0

<=>14x-2=0

<=>14x=2

<=>x=1/7

b)4(x+1)²+(2x-1)²-8(x-1)(x+1)=11

<=>4x²+8x+4+4x²-4x+1-8x²+8=11

<=>4x+13=11

<=>4x=11-13

<=>4x=-2

<=>x=-1/2

c)Sai đề phải là dấu - chứ không phải +

(x-3)(x²+3x+9)-x(x-2)(x+2)=1

<=>x³-27-x³+4x=1

<=>4x=1+27

<=>4x=28

<=>x=7

2)a)(2x-3y)(2x+3y)-4(x-y)²-8xy

=4x²-9y²-4x²+8xy-4y²-8xy

=-13y²

b)(x-2)³-x(x+1)(x-1)+6x(x-3)

=x³-6x²+12x+8-x³+x+6x²-18x

=8-5x

c)(x-2)(x²-2x+4)(x+2)(x²+2x+4)

=(x-2)(x²+2x+4)(x+2)(x²-2x+4)

=(x³-8)(x³+8)

=x⁶-64

Nguyễn Diệu Thảo sap c hk **** cho  Moon Light

\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

hay x=1