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

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

c) \(x^2-9=0\Rightarrow\left(x-3\right)\left(x+3\right)=0\)

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

d) \(\Rightarrow\left(2x-4\right)\left(2x+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

\(a,3\cdot x-15=x+35\)

\(\Rightarrow3x-x=35+15\)

\(\Rightarrow 2x=50\)

\(\Rightarrow x = 50:2\)

\(\Rightarrow x= 25\)

\(b,(8x-16)(x-5)=0\)

\(+, TH1: 8x-16=0\)

\(\Rightarrow8x=16\)

\(\Rightarrow x = 16:8\)

\(\Rightarrow x=2\)

\(+,TH2: x-5=0\)

\(\Rightarrow x =5\)

\(c,x(x+1)=2+4+6+8+10+...+2500\)  \(^{\left(1\right)}\)

Đặt \(A=2+4+6+8+10+...+2500\)

Số các số hạng của \(A\) là: \(\left(2500-2\right):2+1=1250\left(số\right)\)

Tổng \(A\) bằng: \(\left(2500+2\right)\cdot1250:2=1563750\)

Thay \(A=1563750\) vào \(^{\left(1\right)}\), ta được:

\(x\left(x+1\right)=1563750\)

\(\Rightarrow x\left(x+1\right)=1250\cdot1251\)

\(\Rightarrow x =1250\)

#\(Toru\)

Lời giải:
a. Đề có cả x,y. Bạn xem lại

b. 

PT $\Leftrightarrow 5x(x-3)-2(x-3)=0$

$\Leftrightarrow (x-3)(5x-2)=0$

$\Leftrightarrow x-3=0$ hoặc $5x-2=0$

$\Leftrightarrow x=3$ hoặc $x=\frac{2}{5}$

c.

PT $\Leftrightarrow (7x-2)(x-4)=0$

$\Leftrightarrow 7x-2=0$ hoặc $x-4=0$

$\Leftrightarrow x=\frac{2}{7}$ hoặc $x=4$

d. Đề thiếu.

a: Ta có: \(5\left(4x-1\right)+2\left(1-3x\right)-6\left(x+5\right)=10\)

\(\Leftrightarrow20x-5+2-6x-6x-30=10\)

\(\Leftrightarrow8x=43\)

hay \(x=\dfrac{43}{8}\)

b: ta có: \(2x\left(x+1\right)+3\left(x-1\right)\left(x+1\right)-5x\left(x+1\right)+6x^2=0\)

\(\Leftrightarrow2x^2+2x+3x^2-3-5x^2-5x+6x^2=0\)

\(\Leftrightarrow6x^2-3x-3=0\)

\(\Leftrightarrow2x^2-x-1=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)

c: Ta có: \(4\left(x-1\right)\left(x+5\right)-\left(x+5\right)\left(x+2\right)-3\left(x-1\right)\left(x+2\right)=0\)

\(\Leftrightarrow4\left(x^2+4x-5\right)-\left(x^2+7x+10\right)-3\left(x^2+x-2\right)=0\)

\(\Leftrightarrow4x^2+16x-20-x^2-7x-10-3x^2-3x+6=0\)

\(\Leftrightarrow6x=24\)

hay x=4

d: Ta có: \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-5+1=-4\)

hay \(x=\dfrac{2}{7}\)

\(a,x-\dfrac{3}{4}=\dfrac{1}{7}\\ x=\dfrac{1}{7}+\dfrac{3}{4}\\ x=\dfrac{25}{28}\\ b,x+\dfrac{7}{5}=\dfrac{9}{8}.\dfrac{4}{27}\\ x+\dfrac{7}{5}=\dfrac{1}{6}\\ x=\dfrac{1}{6}-\dfrac{7}{5}\\ x=\dfrac{-37}{30}\\ c,\dfrac{2}{5}-\dfrac{3}{7}=\dfrac{x}{70}\\ \dfrac{-1}{35}=\dfrac{x}{70}\\ \dfrac{-2}{70}=\dfrac{x}{70}\\ x=-2\\ d,\dfrac{2}{9}-\dfrac{7}{8}.x=1\\ \dfrac{7}{8}.x=\dfrac{2}{9}-1\\ \dfrac{7}{8}.x=\dfrac{-7}{9}\\ x=\dfrac{-7}{9}:\dfrac{7}{8}\\ x=\dfrac{-8}{9}\)

\(a,x-\dfrac{3}{4}=\dfrac{1}{7}\)

\(\Rightarrow x=\dfrac{1}{7}+\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{25}{28}\)

\(b,x+\dfrac{7}{5}=\dfrac{9}{8}.\dfrac{4}{27}\)

\(\Rightarrow x+\dfrac{7}{5}=\dfrac{1}{6}\)

\(\Rightarrow x=\dfrac{1}{6}-\dfrac{7}{5}\)

\(\Rightarrow x=-\dfrac{37}{30}\)

\(c,\dfrac{2}{5}-\dfrac{3}{7}=\dfrac{x}{70}\)

\(\Rightarrow\dfrac{-1}{35}=\dfrac{x}{70}\)

\(\Rightarrow35x=-70\)

\(\Rightarrow x=-2\)

\(d,\dfrac{2}{9}-\dfrac{7}{8}.x=1\)

\(\Rightarrow\dfrac{7}{8}x=\dfrac{2}{9}-1\)

\(\Rightarrow\dfrac{7}{8}x=-\dfrac{7}{9}\)

\(\Rightarrow x=-\dfrac{8}{9}\)

`a)x-3/4=1/7`

`=>x=1/7+3/4`

`=>x=4/28+21/28`

`=>x=25/28`

`b)x+7/5=9/8 . 4/27`

`=>x+7/5=(9.4)/(8.27)`

`=>x+7/5=1/6`

`=>x=1/6+7/5`

`=>x=5/30+42/30`

`=>x=47/30`

`c)2/5-3/7=x/70`

`=>14/35-15/35=x/70`

`=>-1/35=x/70`

`=>-2/70=x/70`

`=>x=-2`

`d)2/9-7/8x=1`

`=>7/8x=2/9-1`

`=>7/8x=-7/9`

`=>x=-7/9:7/8`

`=>x=-7/9 . 8/7`

`=>x=-8/9`

`#lv`

a: Ta có: \(\left(2x+1\right)^2-4\left(x+2\right)^2=9\)

\(\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\)

\(\Leftrightarrow-12x=24\)

hay x=-2

b: Ta có: \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)

\(\Leftrightarrow x^2+6x+9-x^2-4x+32=1\)

\(\Leftrightarrow2x=-40\)

hay x=-20

Bài 2: 

a: =>x=0 hoặc x+3=0

=>x=0 hoặc x=-3

b: =>x-2=0 hoặc 5-x=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

\(a)x^2-9x+20=0 \\<=>(x-4)(x-5)=0 \\<=>x=4\ hoặc\ x=5 \\b)x^2-3x-18=0 \\<=>(x+3)(x-6)=0 \\<=>x=-3\ hoặc\ x=6 \\c)2x^2-9x+9=0 \\<=>(x-3)(2x-3)=0 \\<=>x=3\ hoặc\ x=\dfrac{3}{2}\)

d: \(\Leftrightarrow3x^2-6x-2x+4=0\)

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

=>x=2 hoặc x=2/3

e: \(\Leftrightarrow3x\left(x^2-2x-3\right)=0\)

=>x(x-3)(x+1)=0

hay \(x\in\left\{0;3;-1\right\}\)

f: \(\Leftrightarrow x^2-5x-2+x=0\)

\(\Leftrightarrow x^2-4x-2=0\)

\(\Leftrightarrow\left(x-2\right)^2=6\)

hay \(x\in\left\{\sqrt{6}+2;-\sqrt{6}+2\right\}\)