a) \(PT\Leftrightarrow3x-2x=2-3\Leftrightarrow x=-1\)

Vậy: \(S=\left\{-1\right\}\)

b) \(PT\Leftrightarrow-2x+3x=-7+22\Leftrightarrow x=15\)

Vậy: \(S=\left\{15\right\}\)

c) \(PT\Leftrightarrow8x-5x=3+12\Leftrightarrow3x=15\Leftrightarrow x=5\)

Vậy: \(S=\left\{5\right\}\)

d) \(PT\Leftrightarrow x+4x-2x=12+25-1\Leftrightarrow3x=36\Leftrightarrow x=12\)

Vậy: \(S=\left\{12\right\}\)

e) \(PT\Leftrightarrow x+2x+3x-3x=19+5\Leftrightarrow3x=24\Leftrightarrow x=8\)

Vậy: \(S=\left\{8\right\}\)

a)3x-2=2x-3

=>x=-1

b)7-2x=22-3x

=>x=15

c)8x-3=5x+12

=>3x=15

=>x=5

d)x-12+4x=25+2x-1

=>3x=12

=>x=4

e)x+2x+3x-19=3x+5

=>3x=24

=>x=8

`b,2(x+1)=5x-7`

`=>2x+2=5x-7`

`=>3x=9`

`=>x=3`

`d,(10x+3)/12=1+(6+8x)/9`

`<=>(10x+3)/12=(8x+15)/9`

`<=>30x+9=32x+60`

`<=>2x=-51`

`<=>x=-51/2`

\(a,8x-3=5x+12\\ \Leftrightarrow8x-5x=12+3\\ \Leftrightarrow3x=15\\ \Leftrightarrow x=\dfrac{15}{3}=5\)

\(b,x-12+4x=25+2x-1\\ \Leftrightarrow x+4x-2x=25-1+12\\ \Leftrightarrow3x=36\\ \Leftrightarrow x=\dfrac{36}{3}=12\)

\(c,7-\left(2x+4\right)=-\left(x+4\right)\\ \Leftrightarrow7-2x-4=-x-4\\ \Leftrightarrow-2x+x=-4+4-7\\ \Leftrightarrow-x=-7\\ \Leftrightarrow x=7\)

\(d,3-4x\left(45-2x\right)=8x^2+x-300\\ \Leftrightarrow3-100x+8x^2=8x^2+x-300\\ \Leftrightarrow8x^2-8x^2-100x-x=-300-3\\ \Leftrightarrow-101x=-303\\ \Leftrightarrow x=\dfrac{-303}{-101}=3\)

Đề câu d của bạn hình như sai dấu ý

sửa lại

.d. 3 - 4x (25 - 2x) = 8x ² + x - 300

=>3-100x+8x² = 8x²+x-300

=>3-100x = x-300

=>101x=303

=>x=3

250 do ban

Trả lời:

\(\frac{x-1}{2x^2-4x}-\frac{7}{8x}=\frac{5-x}{4x^2-8x}-\frac{1}{8x-16}\)\(\left(đkxđ:x\ne0;x\ne2\right)\)

\(\Leftrightarrow\frac{x-1}{2x\left(x-2\right)}-\frac{7}{8x}=\frac{5-x}{4x\left(x-2\right)}-\frac{1}{8\left(x-2\right)}\)

\(\Leftrightarrow\frac{4\left(x-1\right)}{8x\left(x-2\right)}-\frac{7\left(x-2\right)}{8x\left(x-2\right)}=\frac{2\left(5-x\right)}{8x\left(x-2\right)}-\frac{x}{8x\left(x-2\right)}\)

\(\Rightarrow4\left(x-1\right)-7\left(x-2\right)=2\left(5-x\right)-x\)

\(\Leftrightarrow4x-4-7x+14=10-2x-x\)

\(\Leftrightarrow10-3x=10-3x\)

\(\Leftrightarrow-3x+3x=10-10\)

\(\Leftrightarrow0x=0\)( luôn thỏa mãn )

Vậy S = R với \(x\ne0;x\ne2\)

\(\left(8x+5\right)^2\left(4x+3\right)\left(2x+1\right)=9\)

\(\Leftrightarrow\left(64x^2+8x+25\right)\left(8x^2+10x+3\right)-9=0\)

Đặt \(a=8x^2+10x+3\)

\(\left(8a+1\right)a-9=0\)

\(\Leftrightarrow8a^2+a-9=0\)

\(\Leftrightarrow\left(a-1\right)\left(8a+9\right)=0\)

\(\Rightarrow\orbr{\begin{cases}a=1\\a=-\frac{9}{8}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}8x^2+10x+3=1\\8x^2+10x+3=-\frac{9}{8}\end{cases}}\)

Mà \(8x^2+10x+3>0\Rightarrow8x^2+10x+3>-\frac{9}{8}\)

\(\Rightarrow8x^2+10x+3=1\Rightarrow8x^2+10x+2=0\Rightarrow2\left(x+1\right)\left(4x+1\right)=0\Rightarrow\orbr{\begin{cases}x=-1\\x=-\frac{1}{4}\end{cases}}\)

MK K BIET !

XIN LOI NHA BAN !

a) \(\left(8x+5\right)^2\left(4x+3\right)\left(2x+1\right)=9\)

\(\Leftrightarrow\left(64x^2+8x+25\right)\left(8x^2+10x+3\right)-9=0\)

Đặt a = \(8x^2+10x+3\)

\(\left(8a+1\right)a-9=0\)

\(\Leftrightarrow8a^2+a-9=0\)

\(\Leftrightarrow\left(a-1\right)\left(8a+9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=1\\a=-\frac{9}{8}\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}8x^2+10x+3=1\\8x^2+10x+3=-\frac{9}{8}\end{cases}}\)

mà \(8x^2+10x+3=1\Rightarrow8x^2+10x+2=0\)

\(\Rightarrow2\left(x+1\right)\left(4x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=-1\\x=-0,25\end{cases}}\)

cảm ơn bạn còn mấy phần còn lại ạ