1)6x-8=3x+1

6x-3x=1+8

3x=9

x=3

Vậy x=3

2: 12-10x=25-30x

=>20x=13

=>x=13/20

3: \(3\left(2x+3\right)-2\left(4x-5\right)=10x+21\)

=>6x+9-8x+10=10x+21

=>10x+21=-2x+19

=>12x=-2

=>x=-1/6

4: \(\Leftrightarrow25x-15-6x+12=11-5x\)

=>19x-3=11-5x

=>24x=14

=>x=7/12

5: \(\Leftrightarrow8-12x-5+10x=4-6x\)

=>4-6x=-2x+3

=>-4x=-1

=>x=1/4

6: \(\Leftrightarrow32x-24-6+9x=13-40x\)

=>41x-30=13-40x

=>81x=43

=>x=43/81

7: \(\Leftrightarrow10x-5+20x=5x-11\)

=>30x-5=5x-11

=>25x=-6

=>x=-6/25

Các bn giúp mình với mình đang cần gấp

nhiều quá bạn ơi , mk nghĩ bạn nên tách ra rồi hãy đăng lên

Bài 1:

16:

=>15-30x-8+8x=4x-9

=>-22x+7=4x-9

=>-26x=-16

=>x=8/13

15: \(\Leftrightarrow12x-6+12-4x=10-6x\)

=>8x+6=10-6x

=>14x=4

=>x=2/7

14: \(\Leftrightarrow12x-18+9-15x=8x-9\)

=>-3x-9=8x-9

=>x=0

13: \(\Leftrightarrow10x+15x-10=25-10x\)

=>25x-10=25-10x

=>35x=35

=>x=1

12: \(\Leftrightarrow15x-18-4x+10=11x-10\)

=>11x-8=11x-10(loại)

1)\(\left(4x-10\right)\left(24+5x\right)=0\)

\(\Leftrightarrow2\left(2x-5\right)\left(24+5x\right)=0\)

Vì 2≠0

nên \(\left[{}\begin{matrix}2x-5=0\\24+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\5x=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{-24}{5}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{5}{2};\frac{-24}{5}\right\}\)

2) \(0,5x\left(x-3\right)=\left(x-3\right)\left(2,5x-4\right)\)

\(\Leftrightarrow0,5x\left(x-3\right)-\left(x-3\right)\left(2,5x-4\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left[0,5x-\left(2,5x-4\right)\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(0,5x-2,5x+4\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(-2x+4\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(4-2x\right)=0\)

\(\Leftrightarrow\left(x-3\right)\cdot2\cdot\left(2-x\right)=0\)

Vì 2≠0

nên \(\left[{}\begin{matrix}x-3=0\\2-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)

Vậy: x∈{2;3}

3) \(4x^2-1=\left(2x+1\right)\left(3x-5\right)\)

\(\Leftrightarrow\left(2x+1\right)\left(2x-1\right)-\left(2x+1\right)\left(3x-5\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left[2x-1-\left(3x-5\right)\right]=0\)

\(\Leftrightarrow\left(2x+1\right)\left(2x-1-3x+5\right)=0\)

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

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

Vậy: \(x\in\left\{\frac{-1}{2};4\right\}\)

4) \(\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\)

\(\Leftrightarrow\left(2-3x\right)\left(x+11\right)-\left(3x-2\right)\left(2-5x\right)=0\)

\(\Leftrightarrow\left(2-3x\right)\left(x+11\right)+\left(2-3x\right)\left(2-5x\right)=0\)

\(\Leftrightarrow\left(2-3x\right)\left(x+11+2-5x\right)=0\)

\(\Leftrightarrow\left(2-3x\right)\left(13-4x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2-3x=0\\13-4x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\4x=13\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=\frac{13}{4}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{2}{3};\frac{13}{4}\right\}\)

+)   (5x-1). (2x+3)-3. (3x-1)=0

10x^2+15x-2x-3 - 9x+3=0

10x^2 +8x=0

2x(5x+4)=0

=> x=0 hoặc x= -4/5

+)    x^3 (2x-3)-x^2 (4x^2-6x+2)=0

2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0

-2x^4 + 3x^3-2x^2=0

x^2(-2x^2+x-2)=0

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

=> x=0 hoặc x=1

+)   x (x-1)-x^2+2x=5

x^2 -x -x^2+2x=5

x=5

+)     8 (x-2)-2 (3x-4)=25

8x - 16-6x+8=25

2x=33

x=33/2

a,\(4x\left(2x+3\right)-x\left(8x-1\right)=5\left(x+2\right)\)

\(< =>8x^2+12x-8x^2+x=5x+10\)

\(< =>13x=5x+10< =>8x=10\)

\(< =>x=\frac{10}{8}=\frac{5}{4}\)

b, \(\left(3x-5\right)\left(3x+5\right)-x\left(9x-1\right)=4\)

\(< =>9x^2-25-9x^2+x=4\)

\(< =>x=4+29=33\)

c,\(3-4x\left(25-2x\right)=8x^2+x-300\)

\(< =>3-100x+8x^2=8x^2+x-300\)

\(< =>x+100x=3+300\)

\(< =>101x=303< =>x=\frac{303}{101}=3\)

d,\(2\left(1-\frac{3x}{5}\right)-\frac{2+3x}{10}=7-\frac{3\left(2x+1\right)}{4}\)

\(< =>2-\frac{6x}{5}-\frac{2+3x}{10}=7-\frac{6x+3}{4}\)

\(< =>-\frac{24x}{20}-\frac{4+6x}{20}+\frac{30x+15}{20}=5\)

\(< =>\frac{30x-6x-24x+15-4}{20}=5\)

\(< =>\frac{11}{5}=5< =>11=25\)(vo li)

e,  \(\frac{5x+2}{6}-\frac{8x-1}{3}=\frac{4x+2}{5}-5\)

\(< =>\frac{25x+10}{30}-\frac{80x-10}{30}=\frac{24x+12}{30}-5\)

\(< =>\frac{24x+12+80x-10-25x-10}{30}=5\)

\(< =>79x-8=150\)

\(< =>79x=150+8< =>x=\frac{158}{79}\)