`a, 8x=64`

`=>x= 64:8`

`=> x=8`

`b, (-5)x=25`

`=>x=25:(-5)`

`=>x=-5`

`c,4x+1=21`

`=>4x=21-1`

`=>4x=20`

`=>x=20:4`

`=>x=5`

`d, (-3)x-1=8`

`=>(-3)x=8+1`

`=>(-3)x=9`

`=>x=9:(-3)`

`=>x=(-3)`

Bài 1: 

a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)

\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)

\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)

\(\Leftrightarrow-12x^2+14x+13=0\)

\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)

b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)

\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)

hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)

ai giúp mik vs

Bài 3: 

a) Ta có: \(2x-3=x+\dfrac{1}{2}\)

\(\Leftrightarrow2x-x=\dfrac{1}{2}+3\)

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

b) Ta có: \(4x-\left(x+\dfrac{1}{2}\right)=2x-\left(\dfrac{1}{2}-5\right)\)

\(\Leftrightarrow3x-\dfrac{1}{2}-2x+\dfrac{1}{2}-5=0\)

\(\Leftrightarrow x=5\)

`#3107.101107`

`1.`

`a,`

`(2x - 3)^2 = |3 - 2x|`

`=> (2x - 3)^2 = |2x - 3|`

`=>`\(\left[{}\begin{matrix}2x-3=\left(2x-3\right)^2\\2x-3=-\left(2x-3\right)^2\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x-3-\left(2x-3\right)^2=0\\2x-3+\left(2x-3\right)^2=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}\left(2x-3\right)\left(1-2x+3\right)=0\\\left(2x-3\right)\left(1+2x-3\right)=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x-3=0\\4-2x=0\\2x-2=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\\x=1\end{matrix}\right.\)

Vậy, `x \in {3/2; 2; 1}`

`b,`

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

`=>`\(\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(2x-1\right)^2=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x-1=0\\2x-1=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy, `x \in {1; 1/2}`

`c,`

`5 - x^2 = 1`

`=> x^2 = 4`

`=> x^2 = (+-2)^2`

`=> x = +-2`

Vậy, `x \in {-2; 2}`

`d,`

`x - 2\sqrt{x} = 0`

`=> x^2 - (2\sqrt{x})^2 = 0`

`=> x^2 - 4x = 0`

`=> x(x - 4) = 0`

`=>`\(\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

Vậy, `x \in {0; 4}`

`g,`

`(x - 1) + 1/7 = 0`

`=> x - 1 + 1/7 = 0`

`=> x - 6/7 = 0`

`=> x = 6/7`

Vậy, `x = 6/7.`

\(a,\dfrac{3}{7}-x=\dfrac{1}{2}x-3\)

\(\Rightarrow-x-\dfrac{1}{2}x=-3-\dfrac{3}{7}\)

\(\Rightarrow-\dfrac{3}{2}x=-\dfrac{24}{7}\)

\(\Rightarrow x=-\dfrac{24}{7}:\left(-\dfrac{3}{2}\right)\)

\(\Rightarrow x=\dfrac{16}{7}\)

\(b,5x-\dfrac{2}{3}=\dfrac{5}{3}-2x\)

\(\Rightarrow5x+2x=\dfrac{5}{3}+\dfrac{2}{3}\)

\(\Rightarrow7x=\dfrac{7}{3}\)

\(\Rightarrow x=\dfrac{7}{3}:7\)

\(\Rightarrow x=\dfrac{1}{3}\)

#Toru

a: 3/7-x=1/2x-3

=>-3/2x=-3+3/7

=>-1/2x=-1+1/7=-6/7

=>1/2x=6/7

=>x=6/7*2=12/7

b: =>5x+2x=5/3+2/3

=>7x=7/3

=>x=1/3

a: =>2x-x=-5/2-1/3

=>x=-17/6

b: =>4(x-2)²=36

=>(x-2)²=9

=>x-2=3 hoặc x-2=-3

hay x=5 hoặc x=-1

c: =>2x+1/2=5/6

=>2x=1/3

hay x=1/6

a: =>2x-x=-5/2-1/3

=>x=-17/6

b: =>4(x-2)²=36

=>(x-2)²=9

=>x-2=3 hoặc x-2=-3

hay x=5 hoặc x=-1

c: =>2x+1/2=5/6

=>2x=1/3

hay x=1/6

a: Ta có: \(\left(2x+1\right)^2-4\left(x+2\right)^2=9\)

\(\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\)

\(\Leftrightarrow-12x=24\)

hay x=-2

b: Ta có: \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)

\(\Leftrightarrow x^2+6x+9-x^2-4x+32=1\)

\(\Leftrightarrow2x=-40\)

hay x=-20

\(a,-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+1}{6}=\dfrac{8}{3}\)

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

\(\Rightarrow\dfrac{-6x+8x+3x+3+4x+2}{12}=\dfrac{8}{3}\)

\(\Rightarrow\dfrac{9x+5}{12}=\dfrac{8}{3}\)

\(\Rightarrow27x+15=96\)

\(\Rightarrow27x=81\)

\(\Rightarrow x=3\left(tm\right)\)

\(b,\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3+5-2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow2x+1=13\)

\(\Rightarrow2x=12\)

\(\Rightarrow x=6\left(tm\right)\)

#Toru

a) \(-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+2}{6}=\dfrac{8}{3}\) 

\(\Rightarrow\dfrac{-6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{4\cdot8}{12}\)

\(\Rightarrow-6x+8x+3x+3+4x+2=32\)

\(\Rightarrow9x+5=32\)

\(\Rightarrow9x=32-5\)

\(\Rightarrow9x=27\)

\(\Rightarrow x=\dfrac{27}{9}\)

\(\Rightarrow x=3\)

b) \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\) (ĐK: \(x\ne-\dfrac{1}{2}\)) 

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow2x+1=13\)

\(\Rightarrow2x=12\)

\(\Rightarrow x=\dfrac{12}{2}\)

\(\Rightarrow x=6\left(tm\right)\)

a) `4\sqrt(2x-1)>8`

`<=>\sqrt(2x-1)>2`

`<=>2x-1>4`

`<=>x>5/2`

b) `2\sqrtx-1>3`

`<=>2\sqrtx>4`

`<=>\sqrtx>2`

`<=>x>4`

a) Ta có: \(4\sqrt{2x-1}>8\)

\(\Leftrightarrow2x-1>4\)

\(\Leftrightarrow2x>5\)

hay \(x>\dfrac{5}{2}\)

b) Ta có: \(2\sqrt{x}-1>3\)

\(\Leftrightarrow\sqrt{x}>2\)

hay x>4

b: =>4x^2+8x-8x^2+5x-10=0

=>-4x^2+13x-10=0

=>x=2 hoặc x=5/4

c: =>2x^2-5x+6x-15=2x^2+8x

=>x-15=8x

=>-7x=15

=>x=-15/7

d: =>3x^2+15x-2x-10-3x^2-12x=5

=>x-10=5

=>x=15

e: =>x^2-3x+2x^2+2x=3x^2-12

=>-x=-12

=>x=12