a: =>11(x-3)=6(x-5)

=>11x-33=6x-30

=>5x=3

=>x=3/5

b: =>(4/3-1/4x-5/12)-2x=8/5*5/3=8/3

=>-9/4x+11/12=8/3

=>-9/4x=32/12-11/12=21/12=7/4

=>x=-7/9

c: =>1/2x-1/3-2/3x-1=x

=>-1/6x-4/3=x

=>-7/6x=4/3

=>x=-4/3:7/6=-4/3*6/7=-24/21=-8/7

d: =>1-2x-3x+1=7/2

=>-5x=3/2

=>x=-3/10

1: \(\left(3x-\dfrac{1}{5}\right)^2=\left(-\dfrac{3}{25}\right)^2\)

=>3x-1/5=3/25 hoặc 3x-1/5=-3/25

=>3x=8/25 hoặc 3x=2/25

=>x=8/75 hoặc x=2/75

2: \(\left(2x-\dfrac{1}{3}\right)^2=\left(-\dfrac{2}{9}\right)^2\)

=>2x-1/3=2/9 hoặc 2x-1/3=-2/9

=>2x=5/9 hoặc 2x=1/9

=>x=5/18 hoặc x=1/18

Bài 1:

a) \(\dfrac{3x^2-5}{x^2-5x}+\dfrac{5-15x}{5x-25}\)

\(=\dfrac{3x^2-5}{x\left(x-5\right)}+\dfrac{5\left(1-3x\right)}{5\left(x-5\right)}\)

\(=\dfrac{3x^2-5}{x\left(x-5\right)}+\dfrac{1-3x}{x-5}\)

\(=\dfrac{3x^2-5}{x\left(x-5\right)}+\dfrac{x\left(1-3x\right)}{x\left(x-5\right)}\)

\(=\dfrac{3x^2-5+x\left(1-3x\right)}{x\left(x-5\right)}\)

\(=\dfrac{3x^2-5+x-3x^2}{x\left(x-5\right)}\)

\(=\dfrac{-5+x}{x\left(x-5\right)}\)

\(=\dfrac{x-5}{x\left(x-5\right)}\)

\(=\dfrac{1}{x}\)

b) \(\dfrac{4+x^3}{x-3}-\dfrac{2x+2x^2}{x-3}+\dfrac{2x-13}{x-3}\)

\(=\dfrac{\left(4+x^3\right)-\left(2x+2x^2\right)+\left(2x-13\right)}{x-3}\)

\(=\dfrac{4+x^3-2x-2x^2+2x-13}{x-3}\)

\(=\dfrac{x^3-2x^2-9}{x-3}\)

\(=\dfrac{x^3-3x^2+x^2-9}{x-3}\)

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

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

\(=x^2+x+3\)

c) \(\dfrac{2}{x-5}+\dfrac{x-25}{\left(x+5\right)\left(x-5\right)}\)

\(=\dfrac{2\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}+\dfrac{x-25}{\left(x+5\right)\left(x-5\right)}\)

\(=\dfrac{2\left(x+5\right)+x-25}{\left(x+5\right)\left(x-5\right)}\)

\(=\dfrac{2x+10+x-25}{\left(x+5\right)\left(x-5\right)}\)

\(=\dfrac{3x-15}{\left(x+5\right)\left(x-5\right)}\)

\(=\dfrac{3\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}\)

\(=\dfrac{3}{x+5}\)

d) Đề sai?

Bài 2:

\(A=2\left(x+1\right)+\left(3x+2\right)\left(3x-2\right)-9x^2\)

\(A=2x+2+9x^2-4-9x^2\)

\(A=2x-2\)

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

Thay x = 15 vào A ta được:

\(A=2\left(15-1\right)\)

\(A=2.14=28\)

d) \(\left(x+2\right)\left(x^2-2x+4\right)\)

\(=\left(x+2\right)\left(x^2-2\cdot x+2^2\right)\)

\(=x^3+2^3\)

\(=x^3+8\)

e) \(\left(\dfrac{1}{4}-\dfrac{x}{5}\right)\left(\dfrac{x^2}{25}+\dfrac{x}{20}+\dfrac{1}{16}\right)\)

\(=\left(\dfrac{1}{4}-\dfrac{1}{5}x\right)\left(\dfrac{1}{25}x^2+\dfrac{1}{5}x\cdot\dfrac{1}{4}+\dfrac{1}{16}\right)\)

\(=\left(\dfrac{1}{4}-\dfrac{1}{5}x\right)\left[\left(\dfrac{1}{5}x\right)^2+\dfrac{1}{5}x\cdot\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2\right]\)

\(=\left(\dfrac{1}{4}\right)^3-\left(\dfrac{1}{5}x\right)^3\)

\(=\dfrac{1}{64}-\dfrac{1}{125}x^3\)

\(=\dfrac{1}{64}-\dfrac{x^3}{125}\)

d: (x+2)(x^2-2x+4)

=(x+2)(x^2-x*2+2^2)

=x^3+8

e: (1/4-x/5)(1/16+x/20+x^2/25)

=(1/4-x/5)[(1/4)^2+1/4*x/5+(x/5)^2]

=1/64-x^3/125

ĐKXĐ: \(\left\{{}\begin{matrix}2x\ne0\\2\left(25-x\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne25\end{matrix}\right.\)

\(\dfrac{1}{2x}+\dfrac{1}{2\left(25-x\right)}=\dfrac{1}{12}\\ \Leftrightarrow\dfrac{25-x+x}{2x\left(25-x\right)}=\dfrac{1}{12}\\ \Leftrightarrow\dfrac{25}{-2x^2+50x}=\dfrac{1}{12}\\ \Leftrightarrow-2x^2+50x=300\\ \Leftrightarrow-2x^2+50x-300=0\\ \Leftrightarrow\left[{}\begin{matrix}x=15\left(tm\right)\\x=10\left(tm\right)\end{matrix}\right.\)

Vậy...

B

4, \(\Leftrightarrow4x+4+9\left(2x+1\right)=4x+6\left(x+1\right)+7+12x\)

\(\Leftrightarrow22x+13=22x+13\)vậy pt có vô số nghiệm 

5, \(\dfrac{2x}{3}+\dfrac{2x-1}{6}=4-\dfrac{x}{3}\Rightarrow4x+2x-1=24-2x\)

\(\Leftrightarrow8x=25\Leftrightarrow x=\dfrac{25}{8}\)

6, \(\dfrac{x-1}{2}+\dfrac{x-1}{4}=1-\dfrac{2\left(x-1\right)}{3}\Rightarrow6x-6+3x-3=12-8\left(x-1\right)\)

\(\Leftrightarrow9x-9=20-8x\Leftrightarrow17x=29\Leftrightarrow x=\dfrac{29}{17}\)

\(\dfrac{1}{x}+\dfrac{1}{x+10}=\dfrac{1}{12}\)

\(ĐK:x\ne0;-10\)

\(\Leftrightarrow\dfrac{12\left(x+10\right)+12x}{12x\left(x+10\right)}=\dfrac{x\left(x+10\right)}{12x\left(x+10\right)}\)

\(\Leftrightarrow12\left(x+10\right)+12x-x\left(x+10\right)=0\)

\(\Leftrightarrow12x+120+12x-x^2-10x=0\)

\(\Leftrightarrow-x^2+14x+120=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=20\\x=-6\end{matrix}\right.\)

\(o,\dfrac{x}{2x+6}-\dfrac{x}{2x-2}=\dfrac{3x+2}{\left(x+1\right)\left(x+3\right)}\)

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

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

\(\Leftrightarrow x^2+x-x^2-3x-6x-4=0\)

\(\Leftrightarrow-8x-4=0\)

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

\(\Leftrightarrow2x+1=0\)

\(\Leftrightarrow2x=-1\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy \(S=\left\{-\dfrac{1}{2}\right\}\)