a) \(5x-65=5.3^2 \\ 5x-65=45\\5x=45+65\\5x=110\\x=22\)

b) \(200-(2x+6)=4^3\\2x+6=200-4^3\\2x+6=136\\2x=130\\x=65\)

c) \(2(x-51)=2.2^3+20\\2(x-51)=16+20\\2(x-51)=36\\x-51=18\\x=51+18=69\)

d) \(135-5(x+4)=35\\5(x+4)=135-45\\5(x-4)=90\\x-4=18\\x=18+4=22\)

e) \((2x-4)(15-3x)=0\\2(x-2).3(5-x)=0\\(x-2)(5-x)=0\\ \left[ \begin{array}{l}x-2=0\\5-x=0\end{array} \right. \\ \left[ \begin{array}{l}x=2\\x=5\end{array} \right.\)

f) \(2^{x+1} . 2^{2014}=2^{2016} \\ (2^{x+1} . 2^{2014}):2^{2014}=2^{2016} :2^{2014} \\ 2^{x=1}=2^{2016-2014} \\2^{x+1}=2^2\\x+1=2\\x=1\)

g) \(15+(x-1)^3=43\\(x-1)^3=15-42\\(x-1)^3=-27\\(x-1)^3=(-3)^3\\x-1=-3\\x=-2\)

h) \(15-x=17+(-9)\\15-x=17-9\\15-x=8\\x=15-8\\x=7\)

i) \(|x-5|=|-7|+|-4|\\|x-5|=7+4\\|x-5|=11\\ \left[ \begin{array}{l}x-5=11\\x-5=-11\end{array} \right. \\ \left[ \begin{array}{l}x=16\\x=-6\end{array} \right.\)

k) \(|x-3|-12=-9+|-7|\\|x-3|-12=-9+7\\|x-3|-12=-2\\|x-3|=10 \\ \left[ \begin{array}{l}x-3=10\\x-3=-10\end{array} \right. \\ \left[ \begin{array}{l}x=13\\x=-7\end{array} \right.\)

\(\left(1-x\right)\left(5x+3\right)=\left(3x-7\right)\left(x-1\right)\)

\(< =>\left(1-x\right)\left(5x+3+3x-7\right)=0\)

\(< =>\left(1-x\right)\left(8x-4\right)=0\)

\(< =>\orbr{\begin{cases}1-x=0\\8x-4=0\end{cases}< =>\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}}\)

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

\(< =>\left(x-2\right)\left(x+1\right)=\left(x-2\right)\left(x+2\right)\)

\(< =>\left(x-2\right)\left(x+1-x-2\right)=0\)

\(< =>-1\left(x-2\right)=0\)

\(< =>2-x=0< =>x=2\)

\(2x^3+3x^2-32x=48\)

\(< =>x^2\left(2x+3\right)-16\left(2x+3\right)=0\)

\(< =>\left(x^2-16\right)\left(2x+3\right)=0\)

\(< =>\left(x-4\right)\left(x+4\right)\left(2x+3\right)=0\)

\(< =>\hept{\begin{cases}x=4\\x=-4\\x=-\frac{3}{2}\end{cases}}\)

1. Thực hiện phép tính bằng cách hợp lí :

a) (-46) + (-125) + 46 + 25 = [(-46)+46] + [(-125)+25]

= 0+(-100) = -100

b) 25.(-15) + 25.(-5) + (-20).75 = 25.[(-15)+(-5)] + (-20).75

= 25.(-20) + (-20).75 = (-20).(25+75) = (-20).100 = -2000

c) (-151)+(-37)+(-42)+(-63)+142 =(-151)+[(-37)+(-63)]+[(-42)+142]

= (-151) + [(-100) + 100] = -151

d)32+(-149)+(-311)+(-89)+(-51) = 32+[(-149)+(-51)] + [(-311)+(-89)]

= 32+[(-200)+(-400)] = 32+(-600) = -568

e)-65.(87-17)-87.(17-65) = (-65).87 - (-65).17 - 87.17 + 87.65

= (-65).87 + 65.17 - 87.17 + 87.65 = [(-65).87+87.65] + 65.(17-87)

= 65.(-70) = -4550

g) -43.(53-16) - 53.(16-43) = (-43).53 - (-43).16 - 53.16 + 53.43

= (-43).53 + 43.16 - 53.16 + 53.43 = [(-43).53+53.43] + 16.(43-53)

= 16.(-10) = -160

1) x - 2 = -6

x = - 6 + 2

x = -4

2) -5x-(-3)=13

-5x+3=13

-5x=13-3=10

x=10:(-5)

x=-2

3) 15-(x-7)=21

15-x+7=21

15-x=21-7=14

x=15-14=1

4) 3x+17=2

3x=2-17=-15

x=-15:3=-5

5) 45-(x-9)=-35

45-x+9=-35

45-x=-35-9=-44

x=45-(-44)=89

6) -5+x=15

x=15-(-5)

x=20

7) 2x-(-17)=15

2x+17=15

2x=15-17=-2

x=-2:2=-1

ai giúp mk câu này với

8) \(\left|x-2\right|=3\\ \Rightarrow x-2=\pm3\)

+) x - 2 = 3

x = 5

+) x - 2 = -3

x = -1

\(\Rightarrow x\in\left\{5;-1\right\}\)

Các câu 9; 10; 12 tương tự câu 8; còn câu 11 tương tự câu 3; 5

Chúc bạn học tốt!

k minh minh giai cho

\(5^2-\left(5x-3\right)=43\)

\(\Leftrightarrow25-5x+3=43\)

\(\Leftrightarrow5x=-15\)

\(\Leftrightarrow x=-3\)

5² - (5x-3) = 43 

=> 5² - 5x +3 = 43 

=> 5.5-5x = 43 -3 

=> 5.(5-x) = 40 

=> 5-x = 8 

=> x = 5-8 

=> x =( -3 )

Vậy x = (-3)

\(5^2\)\(-\left(5x-3\right)=43\)

⇒ 25 - 5x + 3 = 43

⇒ 28 - 5x = 43

⇒ 5x = -15

⇒ x = -3 

Vậy x= -3

Yêu cầu là gì v c.

`@` `\text {Ans}`

`\downarrow`

`1)`

`x^3 - 2x - 1`

`= x^3 - x^2 - x + x^2 - x - 1`

`= (x^3 - x^2 - x) + (x^2 - x - 1)`

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

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

`2)`

`x^3 +3x^2 -4`

`= x^3 + 4x^2 + 4x - x^2 - 4x - 4`

`= (x^3 + 4x^2 + 4x) - (x^2 + 4x + 4)`

`= x(x^2 + 4x + 4) - (x^2 + 4x + 4)`

`= (x-1)(x^2 + 4x + 4)`

`3)`

`x^3y^3 + x^2y^2 + 4`

`= x^3 y^3 - x^2y^2 + 2x^2y^2 + 2xy - 2xy + 4`

`= (x^3y^3 - x^2y^2 + 2xy) + (2x^2y^2 - 2xy + 4)`

`= xy(x^2y^2 - xy + 2) + 2(x^2y^2 - xy + 2)`

`= (xy+2)(x^2y^2-xy+2)`

`4)`

`x^3 + 5x^2 -9xy^2+5y^3`

Bạn xem lại đề ;-;

`5)`

` x^4 + x^3 + 6x^2 + 5x + 5`

`= x^4 + x^3 + 5x^2 + x^2 + 5x + 5`

`= (x^4 + x^3 + x^2) + (5x^2 + 5x + 5)`

`= x^2(x^2 + x + 1) + 5(x^2 + x + 1)`

`= (x^2 + 5)(x^2 + x + 1)`

a. \(158-x=-12\)

\(x=158+12\)

\(x=170\)

b. \(37+x=12\)

\(x=12-37\)

\(x=-25\)

c. \(2x-15=-47\)

\(2x=\left(-47\right)+15\)

\(x=\dfrac{\left(-32\right)}{2}\)

\(x=-16\)

d. \(\left(-5\right)^2-\left(5x-3\right)=43\)

\(25-\left(5x-3\right)=43\)

\(\left(5x-3\right)=25-43\)

\(5x=\left(-18\right)+3\)

\(x=\dfrac{\left(-15\right)}{5}=-3\)

e. \(\left|x-1\right|+\left(-5\right)=2\)

\(\left|x-1\right|=2-\left(-5\right)\)

\(\left|x-1\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=7\\x-1=-7\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=8\\x=-6\end{matrix}\right.\)

f. \(\left|x+1\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

g. \(\left|x\right|=12\)

\(\Rightarrow x=\pm12\)

Câu h ko có x sao tìm

a) \(158-x=-12.\)

\(\Rightarrow x=158-\left(-12\right).\)

\(\Rightarrow x=158+12=170.\)

Vậy..........

b) \(37+x=12.\)

\(\Rightarrow x=12-37.\)

\(\Rightarrow x=-25.\)

Vậy..........

c) \(2x-15=-47.\)

\(\Rightarrow2x=-47+15.\)

\(\Rightarrow2x=-32.\)

\(\Rightarrow x=-\dfrac{32}{2}=-16.\)

Vậy..........

d) \(\left(-5\right)^2-\left(5x-3\right)=43.\)

\(\Rightarrow25-\left(5x-3\right)=43.\)

\(\Rightarrow5x-3=25-43.\)

\(\Rightarrow5x-3=-18.\)

\(\Rightarrow5x=-18+3.\)

\(\Rightarrow5x=-15.\)

\(\Rightarrow x=-\dfrac{15}{5}=3.\)

Vậy..........

e) \(\left|x-1\right|+\left(-5\right)=2.\)

\(\Rightarrow\left|x-1\right|=2-\left(-5\right).\)

\(\Rightarrow\left|x-1\right|=2+5.\)

\(\Rightarrow\left|x-1\right|=7.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=7\\x-1=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-6\end{matrix}\right..\)

Vậy..........

f) \(\left|x+1\right|=4\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right..\)

Vậy..........

g) \(\left|x\right|=12\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-12\end{matrix}\right..\)

Vậy..........