a) \(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}\)

\(\Leftrightarrow\)\(\left(2x+3\right)\left(10x+2\right)=\left(5x+2\right)\left(4x+5\right)\)

\(\Leftrightarrow20x^2+4x+30x+6=20x^2+25x+8x+10\)

\(\Leftrightarrow20x^2-20x^2+4x+30x-25x-8x=10-6\)

\(\Leftrightarrow x=4\)

b) \(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)

\(\Leftrightarrow\left(3x-1\right)\left(5x-34\right)=\left(40-5x\right)\left(25-3x\right)\)

\(\Leftrightarrow15x^2-102x-5x+34=1000-120x-125x+15x^2\)

\(\Leftrightarrow15x^2-15x^2-102x-5x+120x+125x=1000-34\)

\(\Leftrightarrow138x=966\)

\(\Leftrightarrow x=7\)

a ) \(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}\)

\(\left(2x+3\right).\left(10x+2\right)=\left(5x+2\right)\left(4x+5\right)\)

\(20x^2+4x+30x+6=20x^2+25x+8x+10\)

\(4x+30x-25x-8x=10-6\)

\(x=4\)

b ) \(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)

\(\Leftrightarrow\left(3x-1\right).\left(5x-34\right)=\left(40-5x\right).\left(25-3x\right)\)

\(\Leftrightarrow15x^2-102x-5x+34=1000-120x-125x+15x^2\)

\(\Leftrightarrow-102x-5x-120x-125x=1000-34\)

\(\Leftrightarrow-352x=966\)

\(\Leftrightarrow x=-\frac{483}{176}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}=\frac{2.\left(2x+3\right)-\left(4x+5\right)}{2.\left(5x+2\right)-\left(10x+2\right)}=\frac{4x+6-4x-5}{10x+4-10x-2}=\frac{1}{2}\)

Suy ra:

\(\frac{2x+3}{5x+2}=\frac{1}{2}\Rightarrow2.\left(2x+3\right)=1.\left(5x+2\right)\Rightarrow4x+6=5x+2\)

\(\Rightarrow x=4\)

a ) \(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}\)

\(\Leftrightarrow\left(2x+3\right).\left(10x+2\right)=\left(5x+2\right).\left(4x+5\right)\)

\(\Leftrightarrow20x^2+4x+30x+6=20x^2+25x+8x+10\)

\(\Leftrightarrow4x+30x-25x-8x=10-6\)

\(\Leftrightarrow1x=4\)

\(\Leftrightarrow x=4:1\)

\(\Leftrightarrow x=4\)

b ) \(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)

\(\Leftrightarrow\left(3x-1\right).\left(5x-34\right)=\left(40-5x\right).\left(25-3x\right)\)

\(\Leftrightarrow15x^2-102x-5x+34=1000-120x-125x+15x^2\)

\(\Leftrightarrow-102x-5x+120x+125x=1000-34\)

\(\Leftrightarrow138x=966\)

\(\Leftrightarrow x=966:138\)

\(\Leftrightarrow x=7\)

a. \(\left(2x+3\right)\left(10x+2\right)=\left(5x+2\right)\left(4x+5\right)\)

\(20x^2+34x+5=20x^2+33x+10\)

rút gọn \(x=5\)

mình gửi câu a tí nữa mình gửi câu b

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

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

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

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

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

\(\Leftrightarrow4x=2\)

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

Vậy: \(x=\dfrac{1}{2}\)

===========

b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

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

\(\Leftrightarrow3x=13\)

\(\Leftrightarrow x=\dfrac{13}{3}\)

Vậy: \(x=\dfrac{13}{3}\)

==========

c/  \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

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

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

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)

\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)

\(\Leftrightarrow-2x=-2\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

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

\(\Leftrightarrow-14x=-4\)

\(\Leftrightarrow x=\dfrac{2}{7}\)

Vậy: \(x=\dfrac{2}{7}\)

==========

f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

\(\Leftrightarrow-x^3=8\)

\(\Leftrightarrow x=-2\)

Vậy: \(x=-2\)

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

a: 3x=81

nên x=27

b: \(5\cdot4^x=80\)

\(\Leftrightarrow4^x=16\)

hay x=2

c: \(2^x=4^5:4^3\)

\(\Leftrightarrow2^x=2^4\)

hay x=4