Bài 2:

a: Ta có: \(2x+79=x+45\)

nên 2x-x=45-79

hay x=-24

b: Ta có: \(15-\left(x+7\right)=\left(x-8\right)+22\)

\(\Leftrightarrow8-x-x-14=0\)

\(\Leftrightarrow2x=22\)

hay x=11

30,31

36+1

20

88

Bài 23: Tìm x:

a) 29 < 30 < 32

b) 36 < 37 + 1 < 39

c) 30 + 30 < 64

d) 46 < 88 - 42 < 48

a, 30

b,37

c, 30

d, 88

a: Ta có: \(x\left(2-x\right)+x^2+x=7\)

\(\Leftrightarrow2x-x^2+x^2+x=7\)

\(\Leftrightarrow3x=7\)

hay \(x=\dfrac{7}{3}\)

b: Ta có: \(\left(x-4\right)^2-\left(2x+1\right)^2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)

a: Ta có: \(x^3+1=0\)

\(\Leftrightarrow x^3=-1\)

hay x=-1

b: Ta có: \(6x^2-12x-48=0\)

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

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

a) \(\dfrac{x.2}{-15}=\dfrac{-5}{3}\)

\(\dfrac{x.2}{-15}=\dfrac{25}{-15}\)

x.2=25

x=12,5

b) \(\dfrac{x-1}{-12}=\dfrac{-3}{x-1}\)

(x-1)²=-3.(-12)

(x-1)²=36 

⇒(x-1)²\(\Rightarrow\left[{}\begin{matrix}x-1=6\\x-1=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-5\end{matrix}\right.\)

a: Ta có: \(x\left(2-x\right)+\left(x^2+x\right)=7\)

\(\Leftrightarrow2x-x^2+x^2+x=7\)

\(\Leftrightarrow3x=7\)

hay \(x=\dfrac{7}{3}\)

b: Ta có: \(\left(2x+1\right)^2-x\left(4-5x\right)=17\)

\(\Leftrightarrow4x^2+4x+1-4x+5x^2=17\)

\(\Leftrightarrow9x^2=16\)

\(\Leftrightarrow x^2=\dfrac{16}{9}\)

hay \(x\in\left\{\dfrac{4}{3};-\dfrac{4}{3}\right\}\)

c: Ta có: \(\left(x-4\right)^2-\left(2x+1\right)^2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)

d: ta có: \(\dfrac{2x^3-8x^2+10x}{2x}=0\)

\(\Leftrightarrow x^2-4x+5=0\)

\(\Leftrightarrow\left(x-2\right)^2+1=0\)(vô lý)

a) x = 5/8 : 3/4

x = 5/6

b) x = 5/6 - 1/12

x = 3/4

a)x=\(\dfrac{5}{8}:\dfrac{3}{4}=\dfrac{5}{6}\)

b) x=\(\dfrac{5}{6}-\dfrac{1}{12}=\dfrac{3}{4}\)

a \(x=\dfrac{5}{8}:\dfrac{3}{4}\)

\(x=\dfrac{5}{8}\times\dfrac{4}{3}\)

\(x=\dfrac{5}{6}\)

b \(x=\dfrac{5}{6}-\dfrac{1}{12}\)

\(x=\dfrac{10}{12}-\dfrac{1}{12}\)

\(x=\dfrac{9}{12}\)

\(x=\dfrac{3}{4}\)

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

\(-\dfrac{1}{3}x=-\dfrac{2}{3}-\dfrac{1}{6}\)

\(-\dfrac{1}{3}x=-\dfrac{5}{6}\)

\(x=-\dfrac{5}{6}:\left(-\dfrac{1}{3}\right)\)

\(x=\dfrac{5}{2}\)

b) \(\dfrac{4}{x}=\dfrac{x}{16}\)

\(x^2=4.16\)

\(x^2=64\)

\(\Rightarrow x=8;x=-8\)

`a)=>(12-37/3):x+1/6=-2/3`

`=>(12-37/3):x=-5/6`

`=>(-1/3):x=-5/6`

`=>x=(-1/3):(-5/6)`

`=>x=6/15=2/5`

`b)4/x=x/16`

`=>x^2=4*16`

`=>x^2=64`

`=>x^2=(+-8)^2`

a. \(x^2-2x+2\left|x-1\right|-7=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2-2x+2\left(x-1\right)-7=0\\x^2-2x-2\left(x-1\right)-7=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2-9=0\\x^2-4x-5=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2=9\\\left(x-5\right)\left(x+1\right)=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\pm3\\x=5\\x=-1\end{matrix}\right.\)

b: Ta có: \(\left(x^2+3x+2\right)\left(x^2+7x+12\right)=24\)

\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=24\)

\(\Leftrightarrow\left(x^2+5x\right)^2+10\cdot\left(x^2+5x\right)=0\)

\(\Leftrightarrow x\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)