Bài 15:

a: 2x+x=45

=>3x=45

=>\(x=\frac{45}{3}=15\)

b: 2x+7x=918

=>\(x\cdot\left(7+2\right)=918\)

=>9x=918

=>\(x=\frac{918}{9}=102\)

c: 2x+3x=60+5

=>5x=65

=>\(x=\frac{65}{5}=13\)

d: \(11x+22x=33\cdot2\)

=>33x=66

=>\(x=\frac{66}{33}=2\)

Bài 14:

a: (12-x)(2-x)=0

=>\(\left[\begin{array}{l}12-x=0\\ 2-x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=12\\ x=2\end{array}\right.\)

b: (x-33)(11-x)=0

=>\(\left[\begin{array}{l}x-33=0\\ 11-x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=33\\ x=11\end{array}\right.\)

c: (21-x)(12-x)=0

=>\(\left[\begin{array}{l}21-x=0\\ 12-x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=21\\ x=12\end{array}\right.\)

d: (50-x)(x-150)=0

=>\(\left[\begin{array}{l}50-x=0\\ x-150=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=50\\ x=150\end{array}\right.\)

Bài 13:

a: (x-2)(x-3)=0

=>\(\left[\begin{array}{l}x-2=0\\ x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=3\end{array}\right.\)

b: (x-3)(x-4)=0

=>\(\left[\begin{array}{l}x-3=0\\ x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=4\end{array}\right.\)

c: (x-7)(6-x)=0

=>\(\left[\begin{array}{l}x-7=0\\ 6-x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=7\\ x=6\end{array}\right.\)

d: (x-3)(x-13)=0

=>\(\left[\begin{array}{l}x-3=0\\ x-13=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=13\end{array}\right.\)

a/

\(x^3-4x^2-\left(x-4\right)=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\\x=-1\end{matrix}\right.\)

b/

\(x^5-9x=0\)

\(\Leftrightarrow x\left(x^4-9\right)=x\left(x^2-3\right)\left(x^2+3\right)=0\)

\(\Leftrightarrow x\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\)

c/

\(\left(x^3-x^2\right)^2-4x^2+8x-4=0\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\pm\sqrt{2}\end{matrix}\right.\)

Sos

3) \(x\left(x-4\right)+\left(x-4\right)^2=0\Leftrightarrow\left(x-4\right)\left(x+x-4\right)=0\Leftrightarrow2\left(x-4\right)\left(x-2\right)=0\)

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

4x.(x+1)-8(x+1)=0

(4x-8)(x+1)=0

suy ra x=2 hoặc x=-1

a: x(x-3)+x-3=0

=>(x-3)(x+1)=0

=>\(\left[\begin{array}{l}x-3=0\\ x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=-1\end{array}\right.\)

b: \(\left(5x-4\right)^2-16^2=0\)

=>(5x-4-16)(5x-4+16)=0

=>(5x-20)(5x+12)=0

=>\(\left[\begin{array}{l}5x-20=0\\ 5x+12=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\\ x=-\frac{12}{5}\end{array}\right.\)

\(a,\Leftrightarrow\left(2-x\right)\left(x^2+4\right)>0\Leftrightarrow2-x>0\Leftrightarrow x< 2\\ b,\Leftrightarrow x+3>0\Leftrightarrow x>-3\\ c,\Leftrightarrow\left[{}\begin{matrix}x< -3\\x>4\end{matrix}\right.\)

b: \(\Leftrightarrow x+3>0\)

hay x>-3

bài 2: (x-3).(y+2) = -5

    Vì x, y \(\in\)Z   => x-3 \(\in\)Ư(-5) = {5;-5;1;-1}

Ta có bảng: 

bài 3: a(a+2)<0

TH1 : \(\orbr{\begin{cases}a< 0\\a+2>0\end{cases}}\)=>\(\orbr{\begin{cases}a< 0\\a>-2\end{cases}}\)=> -2<a<0 ( TM)

TH2: \(\orbr{\begin{cases}a>0\\a+2< 0\end{cases}}\Rightarrow\orbr{\begin{cases}a>0\\a< -2\end{cases}}\Rightarrow loại\)
 

           Vậy -2<a<0

Bài 5: \(\left(x^2-1\right)\left(x^2-4\right)< 0\)

TH 1 : \(\hept{\begin{cases}x^2-1>0\\x^2-4< 0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x^2>1\\x^2< 4\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x>1\\x< 2\end{cases}}\)\(\Rightarrow\)1 < a < 2

TH 2: \(\hept{\begin{cases}x^2-1< 0\\x^2-4>0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x^2< 1\\x^2>4\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x< 1\\x>2\end{cases}}\)\(\Rightarrow\)loại

                         Vậy 1<a<2

`@` `\text {Ans}`

`\downarrow`

`1,`

`x^2 - 9 = 0`

`<=> x^2 = 0 + 9`

`<=> x^2 = 9`

`<=> x^2 = (+-3)^2`

`<=> x = +-3`

Vậy, `S = {3; -3}`

`2,`

`25 - x^2 = 0`

`<=> x^2 = 25 - 0`

`<=> x^2 = 25`

`<=> x^2 = (+-5)^2`

`<=> x = +-5`

Vậy,` S= {5; -5}`

`3,`

`-x^2 + 36 = 0`

`<=> -x^2 = 0 - 36`

`<=> -x^2 = -36`

`<=> x^2 = 36`

`<=> x^2 = (+-6)^2`

`<=> x = +-6`

Vậy, `S= {6; -6}`

`4,`

`4x^2 - 4 = 0`

`<=> 4x^2 = 0+4`

`<=> 4x^2 = 4`

`<=> x^2 = 4 \div 4`

`<=> x^2 = 1`

`<=> x^2 = (+-1)^2`

`<=> x = +-1`

Vậy, `S= {1; -1}`

`@` `\text {Kaizuu lv uuu}`

Lớp \(8\) thì nên Vậy \(S=\left\{...\right\}\) nha em ☕

a. x( x+ 3)= 0 

⇔ x= 0 hoặc x+ 3= 0

⇔ x= 0          x = -3

b. x( 2x− 1)+ 2( 2x− 1) =0 

⇔ ( 2x− 1)(x+ 2) =0

⇔ 2x− 1 =0 hoặc  x+ 2 =0

⇔ 2x       =1          x      = -2

⇔   x       =\(\dfrac{1}{2}\)         x      = -2

a) x = 8 3 .                            b) x = − 9 20 .