a) x - 7 = 5. 0 => x - 7 = 0 =>x = 7.

b) x: 3 = 47 +13 => x: 3 = 60 => x = 60.3 => x = 180.

c) x :  7 - 7 = 0 hoặc x : 12 - 12 = 0. Do đó x = 49 hoặc x = 144.

d) x : 2 = 150 - 135 => x: 2 = 15 => x = 15.2 => x = 30.

e) 100: x = 140 -120 => 100: x = 20 => x = 100:20 => x = 5.

g) x : 5 = 300 - 273 => x : 5 = 27 =>x = 27.5 => x = 135

a) x - 7 = 5. 0 => x - 7 = 0 =>x = 7.

b) x: 3 = 47 +13 => x: 3 = 60 => x = 60.3 => x = 180.

c) x :  7 - 7 = 0 hoặc x : 12 - 12 = 0. Do đó x = 49 hoặc x = 144.

d) x : 2 = 150 - 135 => x: 2 = 15 => x = 15.2 => x = 30.

e) 100: x = 140 -120 => 100: x = 20 => x = 100:20 => x = 5.

g) x : 5 = 300 - 273 => x : 5 = 27 =>x = 27.5 => x = 135

           Bài làm :

\(a\text{)}...\Leftrightarrow\orbr{\begin{cases}x\div7-7=0\\x.3-12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\div7=7\\x.3=12\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=49\\x=4\end{cases}}\)

\(b\text{)}\Leftrightarrow...\Leftrightarrow x\div2=15\Leftrightarrow x=15.2=30\)

\(c\text{)}...\Leftrightarrow100\div x=20\Leftrightarrow x=100\div20=5\)

\(d\text{)}...\Leftrightarrow x\div5=27\Leftrightarrow x=27.5=135\)

a) \(x\left(x-6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

b) \(\left(-7-x\right)\left(-x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\)

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

c) \(\left(x+3\right)\left(x-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)

d) \(\left(x-3\right)\left(x^2+12\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\text{(vô lý)}\end{matrix}\right.\)

\(\Rightarrow x=3\)

e) \(\left(x+1\right)\left(2-x\right)\ge0\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+1\ge0\\2-x\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x+1\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-1\\x\le2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}-1\le x\le2\\x\in\varnothing\end{matrix}\right.\)

\(\Rightarrow-1\le x\le2\)

f) \(\left(x-3\right)\left(x-5\right)\le0\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-3\le0\\x-5\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x-3\ge0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\\\left[{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow3\le x\le5\)

a) =>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

b => \(\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)

d) => \(\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\end{matrix}\right.\)(vô lí) => x=3

c) => \(\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)

\(a,\left(x-7\right):5=0\)

\(\Rightarrow x-7=0\)

\(\Rightarrow x=7\)

\(b,\left(x:7-7\right)\left(x:12-12\right)=0\)

\(\Rightarrow\hept{\begin{cases}x:7=7\\x:12=12\end{cases}}\)

\(\Rightarrow x=1\)

\(c,300-x:5=273\)

\(\Rightarrow x:5=27\)

\(\Rightarrow x=27.5=135\)

\(d,135+x:2=150\)

\(\Rightarrow x:2=15\)

\(\Rightarrow x=30\)

a) \(\left(x-7\right):5=0\)

\(\Rightarrow x-7=0\times5\)

\(\Rightarrow x-7=0\)

\(\Rightarrow x=0+7\)

\(\Rightarrow x=7\)

Vậy x = 7

b) \(\left(x:7-7\right)\left(x:12-12\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x:7-7=0\\x:12-12=0\end{cases}}\Rightarrow\orbr{\begin{cases}x:7=7\\x:12=12\end{cases}}\Rightarrow\orbr{\begin{cases}x=49\\x=144\end{cases}}\)

Vậy x = 49 hoặc x = 144

c) \(300-x:5=273\)

\(\Rightarrow x:5=300-273\)

\(\Rightarrow x:5=27\)

\(\Rightarrow x=27\times5\)

\(\Rightarrow x=135\)

Vậy x = 135

d) \(135+x:2=150\)

\(\Rightarrow x:2=150-135\)

\(\Rightarrow x:2=15\)

\(\Rightarrow x=15\times2\)

\(\Rightarrow x=30\)

Vậy x = 30

_Chúc bạn học tốt_

a) \(\Leftrightarrow x^2-4x-x^2+6x-9=0\\ \Leftrightarrow2x=9\\ \Leftrightarrow x=4,5\)

b) \(\Leftrightarrow x^2-3x-10=0\\ \Leftrightarrow\left(x^2+2x\right)-\left(5x+10\right)=0\\ \Leftrightarrow x\left(x+2\right)-5\left(x+2\right)=0\\ \left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

c) \(\Leftrightarrow\left(2x-3-7\right)\left(2x-3+7\right)=0\\ \Leftrightarrow\left(2x-10\right)\left(2x+4\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

d) \(\Leftrightarrow\left(2x+7\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\end{matrix}\right.\)

Bài 1.

2:

a: =>x-1=0 hoặc 3x+1=0

=>x=1 hoặc x=-1/3

b: =>x-5=0 hoặc 7-x=0

=>x=5 hoặc x=7

c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)

d: =>x=0 hoặc x^2-1=0

=>\(x\in\left\{0;1;-1\right\}\)

Bạn tách ra từng câu thoi nhe .

a: Ta có: \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{23}{7}\end{matrix}\right.\)

c: Ta có: \(\left(x-3\right)^2-4=0\)

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

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

b. 

PT $\Leftrightarrow (5x^2-2x+10)^2-(3x^2+10x-8)^2=0$

$\Leftrightarrow (5x^2-2x+10-3x^2-10x+8)(5x^2-2x+10+3x^2+10x-8)=0$

$\Leftrightarrow (2x^2-12x+18)(8x^2+8x+2)=0$

$\Leftrightarrow (x^2-6x+9)(4x^2+4x+1)=0$

$\Leftrightarrow (x-3)^2(2x+1)^2=0$

$\Leftrightarrow (x-3)(2x+1)=0$

$\Leftrightarrow x-3=0$ hoặc $2x+1=0$

$\Leftrightarrow x=3$ hoặc $x=-\frac{1}{2}$

d.

$x^2-2x=24$

$\Leftrightarrow x^2-2x-24=0$

$\Leftrightarrow (x+4)(x-6)=0$
$\Leftrightarrow x+4=0$ hoặc $x-6=0$

$\Leftrightarrow x=-4$ hoặc $x=6$