a) x+x+x+91=-2

    x.3          =(-2)-91

    x.3          = -93

    x             =(-93):3

    x             = -31

Trl:

a) \(x+x+x+91=-2\)

\(\Rightarrow x.3+91=-2\)

\(\Rightarrow x.3=-2-91\)

\(\Rightarrow x.3=-93\)

\(\Rightarrow x=-93:3\)

\(\Rightarrow x=-31\)

b) \(-125-\left(3x+1\right)=-2.27\)

\(\Rightarrow-125-\left(3x+1\right)=-54\)

\(\Rightarrow\left(3x+1\right)=-125-\left(-54\right)\)

\(\Rightarrow\left(3x+1\right)=-125+54\)

\(\Rightarrow\left(3x+1\right)=-71\)

\(\Rightarrow3x=-71-1\)

\(\Rightarrow3x=-72\)

\(\Rightarrow x=-72:3\)

\(\Rightarrow x=-24\)

Hc tốt

n: (-8)*x+17=-23

=>\(x\cdot\left(-8\right)=-23-17=-40\)

=>\(x\cdot8=40\)

=>\(x=\dfrac{40}{8}=5\)

o: \(\left(-15\right)\cdot x=10\left(-4\right)-5\)

=>\(x\cdot\left(-15\right)=-40-5=-45\)

=>\(x\cdot15=45\)

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

p: \(\left(-3\right)\cdot x-4=2\cdot\left(-7\right)+4\)

=>\(\left(-3\right)\cdot x-4=-14+4=-10\)

=>\(x\left(-3\right)=-10+4=-6\)

=>\(3x=6\)

=>\(x=\dfrac{6}{3}=2\)

q: x+x+x+91=-2

=>\(3x+91=-2\)

=>\(3x=-2-91=-93\)

=>\(x=-\dfrac{93}{3}=-31\)

r: \(-152-\left(3x+1\right)=\left(-2\right)\cdot\left(-27\right)\)

=>\(-152-3x-1=54\)

=>\(-153-3x=54\)

=>\(3x=-153-54=-207\)

=>\(x=-\dfrac{207}{3}=-69\)

A , 3 - ( 17 - x ) = 289 -  ( 36 + 289 )

3 - 17 + x = 0 - 36

-14 + x = -36

x = -36 - ( - 14 ) = -22

B, 25 - ( x + 5 ) = -415 - ( 15 - 415 )

25 - x - 5 = 0 - 15

20 - x = -15

x = 20 - ( - 15 ) = 35

C , 34 + ( 21 - x ) = ( 3747 - 30 ) - 3746

34 + 21 - x = 1 - 30 

55 - x = -29

x = 55 - (-29 ) = 74

D , -2x -  ( x -17 ) = 34 - ( -x + 25 )

- 2x - x + 17 = 34 - 25 + x

- 3x + 17 = 9 + x

- 3x - x = 9 - 17

-4x = -8

x = -8 : ( - 4 )

x = 2

E , 17x + ( -16x - 37 ) = x + 43

17x - 16x -37 = x + 43

x - 37 = x + 43

-37 - 43 = x - x 

- 80 = 0      ( vô lý )

G , ( x + 12 ) . (x - 3 ) = 0

\(\hept{\begin{cases}x+12=0\\x-3=0\end{cases}}\)

\(\hept{\begin{cases}x=-12\\x=3\end{cases}}\)

`Answer:`

a. \(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)

\(\Leftrightarrow\left|2x-\frac{3}{4}\right|=\frac{17}{2}+\frac{7}{4}\)

\(\Leftrightarrow\left|2x-\frac{3}{4}\right|=\frac{41}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-\frac{3}{4}=\frac{41}{4}\\2x-\frac{3}{4}=-\frac{41}{4}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=\frac{41}{4}+\frac{3}{4}\\2x=-\frac{41}{4}+\frac{3}{4}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=11\\2x=-\frac{19}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=11:2\\x=-\frac{19}{2}:2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{11}{2}\\x=-\frac{19}{4}\end{cases}}\)

b. \(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)

\(\Leftrightarrow\left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\)

\(\Leftrightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)

\(\Leftrightarrow\left(x+\frac{1}{5}\right)=\left(\frac{3}{5}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{5}=\frac{3}{5}\\x+\frac{1}{5}=-\frac{3}{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}-\frac{1}{5}\\x=-\frac{3}{5}-\frac{1}{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{5}\\x=-\frac{4}{5}\end{cases}}\)

c. \(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)

\(\Leftrightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)

\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}-\left(-\frac{24}{27}\right)\)

\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)

\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)

\(\Leftrightarrow3x-\frac{7}{9}=-\frac{2}{3}\)

\(\Leftrightarrow3x=-\frac{2}{3}+\frac{7}{9}\)

\(\Leftrightarrow3x=\frac{1}{9}\)

\(\Leftrightarrow x=\frac{1}{9}:3\)

\(\Leftrightarrow x=\frac{1}{27}\)

a) \(-2011-\left(200-2011\right)\)

\(=-2011-200+2011\)

\(=\left(-2011+2011\right)-200\)

\(=0-200\)

\(=-200\)

b) \(\left(-2\right)^2-\left(-2000\right)^0+\left(-1\right)^{2018}-\left|-20\right|\)

\(=4-1+1-20\)

\(=4-20\)

\(=-16\)

Bài 1 :

\(a)-2011-(200-2011)\)

\(=-2011-(200+2011)\)

\(=(-2011+2011)-200\)

\(=0-200=-200\)

\(b)(-2)^2-(-2000)^0+(-1)^{2018}-\left|-20\right|\)

\(=4-1+1-20\)

\(=4-20=-16\)

\(c)23\cdot18-23\cdot26+(-23)\cdot2\)

\(=23\cdot(18-26)+-(23\cdot2)\)

\(=23\cdot(-8)+(-46)\)

\(=-230\)

Bài 2 : Tìm số nguyên x biết :

\(a)3x-(-5)=20\)

\(\Rightarrow3x+5=20\)

\(\Rightarrow3x=20-5\)

\(\Rightarrow3x=15\Rightarrow x=5\)

\(b)3(x+2)=-4+(-2)^3\)

\(\Rightarrow3(x+2)=-4+(-8)\)

\(\Rightarrow3(x+2)=-12\)

\(\Rightarrow x+2=-12\div3\)

\(\Rightarrow x+2=-4\)

Tự tìm x câu b, và câu c,

Bài 3 tự làm

c) \(23.18-23.26+\left(-23\right).2\)

\(=23\cdot\left(18-26-2\right)\)

\(=23.\left(-10\right)\)

\(=-230\)

a) \(3x-\left(-5\right)=20\)

\(\Leftrightarrow3x=20-5\)

\(\Leftrightarrow3x=15\)

\(\Leftrightarrow x=15\div3\)

\(\Leftrightarrow x=5\)

Vậy x = 5

a: Ta có: \(F=\left(\dfrac{2\sqrt{x}}{2\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{3x}{x-2\sqrt{x}+1}\right)\)

\(=\dfrac{2x-2\sqrt{x}+1}{\sqrt{x}\left(2\sqrt{x}-1\right)}\cdot\dfrac{x-1+3x}{\left(\sqrt{x}-1\right)^2}\)

\(=\dfrac{2x-2\sqrt{x}+1}{\sqrt{x}\left(2\sqrt{x}-1\right)}\cdot\dfrac{4x-1}{\left(\sqrt{x}-1\right)^2}\)

\(=\dfrac{\left(2x-2\sqrt{x}+1\right)\left(2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)^2}\)