`(11-x):(-5)=15`

`=> 11-x=-75`

`=> x=86`

Vậy `x = 86`

`((2x+3)^2022) . (-7x+84)=0`

`=> (2x+3)^2022 = 0` hoặc `-7x + 84 = 0`

`=> 2x+3=0` hoặc `-7x = -84`

`=> x = -3/2` hoặc `x = 12`

Vậy `x = -3/2` hoặc `x = 12`

`(x-3)(x+1) < 0`

Ta có: `x - 3 < x + 1`

nên: `x - 3 < 0` và `x + 1 > 0`

`=> x < 3 và x > -1`

`=> -1 < x < 3`

Vậy `-1 <x < 3`

#\(N\)

`a, (11-x)`\(\div\)`(-5)=15`

`11-x=15.-5`

`11-x=-75`

`x=11-(-75)`

`x=11+75=86`

`b, (2x-3)^2022.(-7x+84)=0`

`=>`\(\left\{{}\begin{matrix}\left(2x-3\right)^{2022}=0\\\left(-7x+84\right)=0\end{matrix}\right.\)

`=>` \(\left\{{}\begin{matrix}2x-3=0\\-7x+84=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2x=3+0\\-7x=0-84\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2x=3\\-7x=-84\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=3\div2\\x=-84\div-7\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=12\end{matrix}\right.\)

`c, (x-3) (x+1) <0`

`-> (x-3) < x+1`

`-> (x-3)<0 , (x+1)>0`

`-> x < 3 , x> (-1)`

`-> 3 > x > -1`

Bài 1:

a) \(\Rightarrow3x^2+3x-2x^2-4x+x+1=0\)

\(\Rightarrow x^2=-1\left(VLý\right)\Rightarrow S=\varnothing\)

b) \(\Rightarrow\left(x-2020\right)\left(2x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2}\end{matrix}\right.\)

c) \(\Rightarrow\left(x-10\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)

d) \(\Rightarrow\left(x+4\right)^2=0\Rightarrow x=-4\)

e) \(\Rightarrow\left(x+6\right)\left(x-7\right)=0\)

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

f) \(\Rightarrow\left(5x-4\right)\left(5x+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)

Bài 2:

a) \(\Rightarrow3x\left(x^2-4\right)=0\Rightarrow3x\left(x-2\right)\left(x+2\right)=0\)

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

b) \(\Rightarrow x\left(x-2\right)+5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(x+5\right)=0\)

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

a, x - 96 = (443 - x) - 15

<=>x - 96 = 443 - x - 15

<=>x + x = 443 - 15 + 96

<=>2x = 524

<=>x = 524 : 2

<=>x = 262

b, (x - 12) - 15 = (20 - 7) - (18 + x)

<=>x - 12 - 15 = 20 - 7 - 18 - x

<=>x + x = 20 - 7 - 18 + 12 + 15

<=>2x = 22

<=> x = 22 : 2

<=> x = 11  

Bài 1:

a) \(x^2-6x+15=\left(x^2-6x+9\right)+6=\left(x-3\right)^2+6\ge6\)

Dấu "=" xảy ra \(\Leftrightarrow x=3\)

b) \(3x^2-15x+4=3\left(x^2-5x+\dfrac{25}{4}\right)-\dfrac{59}{4}=3\left(x-\dfrac{5}{2}\right)^2-\dfrac{59}{4}\ge-\dfrac{59}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{5}{2}\)

Bài 2:

a) \(\Rightarrow\left(x-5\right)\left(x+5\right)+2\left(x+5\right)=0\)

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

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

c) \(\Rightarrow x^2\left(x-2\right)+7\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right)\left(x^2+7\right)=0\)

\(\Rightarrow x=2\left(do.x^2+7\ge7>0\right)\)

 a, 7x . (x-10 ) = 0

=> \(\hept{\begin{cases}7x=0\\x-10=0\end{cases}}\) => \(\hept{\begin{cases}x=0\\x=10\end{cases}}\)

b,17 . (3x-6).(2x-8) = 0

          (3x-6).(2x-8) = 0 : 17

          (3x-6).(2x-8)  = 0

=>\(\hept{\begin{cases}3x-6=0\\2x-8=0\end{cases}}\)=>\(\hept{\begin{cases}x=2\\x=4\end{cases}}\)

Nhớ nha

\(\text{a, 7x. (x-10)= 0 }\)

\(\Rightarrow\orbr{\begin{cases}7x=0\\x-10=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=10\end{cases}}}\)

\(\text{ b, 17.(3x-6) . (2x-8) =0}\)

\(\left(51x-102\right).\left(2x-8\right)=0\)

\(\Rightarrow\orbr{\begin{cases}51x-102=0\\2x-8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=4\end{cases}}}\)

a) \(7x\left(x-10\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}7x=0\\x-10=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=10\end{cases}}}\)

Vậy \(x\in\left\{0;10\right\}\)

b) \(17\left(3x-6\right)\left(2x-8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\2x-8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=6\\2x=8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=4\end{cases}}}\)

Vậy \(x\in\left\{2;4\right\}\)

a)\(7x\left(x-10\right)=0\)

\(\Rightarrow\orbr{\begin{cases}7x=0\\x-10=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=10\end{cases}}}\)

b) \(17.\left(3x-6\right).\left(2x-8\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-6=0\\2x-8=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=6\\2x=8\end{cases}\Rightarrow}\orbr{\begin{cases}x=2\\x=4\end{cases}}}\)

hok tốt!!

a, 7\(x\).(\(x\) - 10) = 0

   \(\left[{}\begin{matrix}7x=0\\x-10=0\end{matrix}\right.\)

    \(\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)

Vậy \(x\in\) {0; 10}

b, 17.(3\(x\) - 6).(2\(x\) - 18) = 0

    \(\left[{}\begin{matrix}3x-6=0\\2x-18=0\end{matrix}\right.\)

    \(\left[{}\begin{matrix}3x=6\\2x-18=0\end{matrix}\right.\)

     \(\left[{}\begin{matrix}x=6:3\\x=18:2\end{matrix}\right.\)

       \(\left[{}\begin{matrix}x=2\\x=9\end{matrix}\right.\)

c, (4 - 2\(x\)).(\(x\) - 3) = 0

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

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

    \(\left[{}\begin{matrix}x=4:2\\x=3\end{matrix}\right.\)

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

    Vậy \(x\in\) {2; 3}

a) -2 (x+6)+6 (x-10)=8

-2.x + (-2).6 + 6.x - 6.10 = 8

-2x + (-12) + 6x - 60 = 8

-2x + 6x + [60 + (-12)] = 8

4x + 48 = 8

4x = 8 - 48

4x = -40

x = -40 : 4

x = -10

vậy_____

b) ( x-12)-15=(20-7) - (18-x)

x - 12 - 15 = 13 - 18 + x

x - 27 = -5 + x

x - x = -5 + 27

0x = 22

x = 22 : 0

=> x ∈ ∅

c)2(5+3x)+x=31

2.5 + 2.3x + x = 31

10 + 6x + x = 31

6x + x = 31 - 10

7x = 21

x = 21 : 7

x = 3

vậy_____

d) -x (x+7) ( x-4) =0

=> -x = 0 hoặc x+ 7 = 0 hoặc x - 4 = 0

=> x = 0 hoặc x = 0 - 7 hoặc x = 0 + 4

=> x = 0 hoặc x = -7 hoặc x = 4

vậy_____

e) x-96 = (443 - x) - 15

x - 96 = 443 - x - 15

x - 96 = (443 - 15) - x

x - 96 = 428 - x

x + x = 428 + 96

2x  = 524

x = 524 : 2

x = 262

vậy_____

Bài 3: 

a: \(x^2-16=\left(x-4\right)\cdot\left(x+4\right)\)

b: \(x^2+2x+1-y^2=\left(x+1+y\right)\left(x+1-y\right)\)

c: \(=\left(x-y\right)^2-4=\left(x-y-2\right)\left(x-y+2\right)\)