a)\(x-\frac{1}{4}=\frac{5}{8}\)

\(x=\frac{5}{8}+\frac{1}{4}\)

\(x=\frac{7}{8}\)

b)\(\frac{x}{12}=\frac{-1}{24}-\frac{1}{8}\)

\(\frac{x}{12}=\frac{-1}{24}+\frac{-1}{8}\)

\(\frac{x}{12}=\frac{-1}{6}\)

\(x=\frac{-1}{6}\times12\)

\(x=-2\)

a)x-1/4=5/8

=>x=5/8+1/4

x=7/8

b) x/12=-1/24-1/8

=>x/12=-1/6

=>x=-2

c).......... Tự làm

c)

\(\left(\frac{1}{4}-\frac{3}{10}x\right)\times\frac{1}{3}=\frac{-51}{6}+\frac{1}{4}\)

\(\left(\frac{1}{4}-\frac{3}{10}x\right)\times\frac{1}{3}=\frac{-33}{4}\)

\(\frac{1}{4}-\frac{3}{10}x=\frac{-33}{4}\div\frac{1}{3}\)

\(\frac{1}{4}-\frac{3}{10}x=\frac{-33}{4}\times3\)

\(\frac{1}{4}-\frac{3}{10}x=\frac{-99}{4}\)

\(\frac{3}{10}x=\frac{1}{4}-\left(\frac{-99}{4}\right)\)

\(\frac{3}{10}x=\frac{1}{4}+\frac{99}{4}\)

\(\frac{3}{10}x=25\)

\(x=\frac{250}{3}\)

Bài 1: 

a: \(\Leftrightarrow3^x\cdot10=810\)

\(\Leftrightarrow3^x=81\)

hay x=4

c: \(\Leftrightarrow5^x\cdot5+5^x\cdot\dfrac{1}{25}=126\)

\(\Leftrightarrow5^x\cdot\dfrac{126}{25}=126\)

\(\Leftrightarrow5^x=25\)

hay x=2

Bài 2: 

a: \(27^{11}=3^{33}\)

\(81^8=3^{32}\)

mà 33>32

nên \(27^{11}>81^8\)

c: \(625^5=\left(5^4\right)^5=5^{20}\)

\(125^7=\left(5^3\right)^7=5^{21}\)

mà 20<21

nên \(625^5< 125^7\)

a: =>6/x=x/24

=>x^2=144

=>x=12 hoặc x=-12

b: =>x(1-7/12+3/8)=5/24

=>x*19/24=5/24

=>x=5/24:19/24=5/19

c: =>(x-1/3)^2=1+3/4+1/2=9/4

=>x-1/3=3/2 hoặc x-1/3=-3/2

=>x=11/6 hoặc x=-7/6

d: =>(x-3)^2=16

=>x-3=4 hoặc x-3=-4

=>x=-1 hoặc x=7

e: =>9/x=-1/3

=>x=-27

f: =>x-1/2=0 hoặc -x/2-3=0

=>x=1/2 hoặc x=-6

a) x . 3/5 = 2/3

<=> x     = 2/3 : 3/5

<=> x     = 2/3 . 5/3 

<=> x     = 10/9

Học tốt nha! :) 

a) \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=8\)

\(\Leftrightarrow x\left(x+3\right)\left(x+1\right)\left(x+2\right)=8\)

\(\Leftrightarrow\left(x^2+3x\right)\left(x^2+3x+2\right)=8\)

Đặt \(x^2+3x=u\)

Phương trình trở thành: \(u\left(u+2\right)=8\)

\(\Leftrightarrow u^2+2u-8=0\Leftrightarrow\left(u-2\right)\left(u+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}u-2=0\\u+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}u=2\\u=-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+3x=2\\x^2+3x=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2+3x-2=0\\x^2+3x+4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\pm\frac{\sqrt{17}}{2}-1\frac{1}{2}\\x\in\varnothing\end{cases}}\)

c) \(\left(x+2\right)\left(x+3\right)\left(x-7\right)\left(x-8\right)=144\)

\(\Leftrightarrow\left(x+2\right)\left(x-7\right)\left(x+3\right)\left(x-8\right)=144\)

\(\Leftrightarrow\left(x^2-5x-14\right)\left(x^2-5x-24\right)=144\)

Đặt \(x^2-5x-14=v\)

Phương trình trở thành: \(v\left(v-10\right)=144\)

\(\Leftrightarrow v^2-10v-144=0\Leftrightarrow\left(v-18\right)\left(v+8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}v-18=0\\v+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}v=18\\v=-8\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-5x-14=18\\x^2-5x-14=-8\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm\frac{3\sqrt{17}}{2}+\frac{5}{2}\\x\in\left\{6;-1\right\}\end{cases}}\)

b) \(\left(4x+3\right)^2\left(x+1\right)\left(2x+1\right)=810\)

\(\Leftrightarrow\left(4x+3\right)^2\left(x+1\right)\left(2x+1\right)-810=0\)

\(\Leftrightarrow\left(x+3\right)\left(2x-3\right)\left(16x^2+24x+89\right)=0\)

Ta có: \(16x^2+24x+89=\left(4x+3\right)^2+80>0\)

\(\Rightarrow\orbr{\begin{cases}x+3=0\\2x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=\frac{3}{2}\end{cases}}\)