a, 2.(x – 5)+7 = 77

<=> 2.(x – 5) = 70 <=> x – 5 = 35 <=> x = 40

b,  x - 1 3 - 3 5 : 3 4 + 2 . 2 3 = 14

<=> x - 1 3 - 3 + 2 4 = 14

<=>  x - 1 3 = 14 + 3 - 16 = 1

<=> x – 1 = 1 <=> x = 2

c,  1 + 2 + 2 2 + 2 3 + . . . + 2 2016 = 2 x - 1 - 1

Đặt: A = 1 + 2 + 2 2 + 2 3 + . . . + 2 2016 => 2A =  2 + 2 2 + 2 3 + . . . + 2 2017

=> 2A – A = ( 2 + 2 2 + 2 3 + . . . + 2 2017 ) – ( 1 + 2 + 2 2 + 2 3 + . . . + 2 2016 )

=> A =  2 2017 - 1

Ta có:  1 + 2 + 2 2 + 2 3 + . . . + 2 2016 = 2 x - 1 - 1 =>  2 2017 - 1 =  2 x - 1 - 1 => x = 2018

d,  5 2 x - 3 - 2 . 5 2 = 5 2 . 3

<=>  5 2 x - 3 = 5 2 . 3 + 5 2 . 2

<=>  5 2 x - 3 = 5 2 . ( 3 + 2 )

<=>  5 2 x - 3 = 5 3

<=> 2x – 3 = 3 => x = 3

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

`<=>x^2-x-x^2-4x-4=1`

`<=>-5x=5`

`<=>x=-1`

`b,(x+5)(x-3)-(x-2)^2=-1`

`<=>x^2+2x-15-x^2+4x-4+1=0`

`<=>6x-18=0`

`<=>x-3=0`

`<=>x=3`

`c,x(2x-4)-(x-2)(2x+3)=0`

`<=>2x(x-2)-(x-2)(2x+3)=0`

`<=>(x-2)(2x-2x-3)=0`

`<=>-3(x-2)=0`

`<=>x-2=0`

`<=>x=2`

`d,x(3x+2)+(x+1)^2-(2x-5)(2x+5)=-12`

`<=>3x^2+2x+x^2+2x+1-4x^2+25=-12`

`<=>4x+26=-12`

`<=>4x=-38`

`<=>x=-19/2`

a: \(\Leftrightarrow-\dfrac{23}{5}\cdot\dfrac{50}{23}< =x< =-\dfrac{12}{5}:\dfrac{7}{5}=\dfrac{-12}{7}\)

=>-10<=x<=-12/7

hay \(x\in\left\{-10;-9;-8;-7;-6;-5;-4;-3;-2\right\}\)

b: \(\Leftrightarrow-\dfrac{13}{3}\cdot\dfrac{1}{3}< =x< =-\dfrac{2}{3}\cdot\dfrac{1}{8}\)

=>-13/9<=x<=-1/12

hay \(x=-1\)

a) |x|=2005

=> x=2005 hoặc -2005

vì giá trị tuyệt đối của 1 số nguyên dương là chính số đó nên x=2005

vì giá trị tuyệt đối của 1 số nguyên âm là số đối của số đó

=> -2005 có số đối là 2005 nên x cũng có thể bằng -2005

a) x = 1005 hoặc x = -1005

b) x + 15 = 22

x = 22 - 15

x = 7

Vì x là |x| nen cung co the = -7

Nhung vi theo de bai thi x>0 nen x = 7

c) x + 12 = 25

x = 25 - 12

x = 13

Vi x la |x| nen cung co the = -13

Vi theo de thi x<0 nen x = -13

a) \(\left|x\right|=1005\)

\(\Rightarrow\) x = 1005 hoặc x = -1005

b) \(\left|x\right|\) + 15 = 22 ; x > 0

x > 0 \(\Rightarrow\)\(\left|x\right|=x\)

\(\Rightarrow\left|x\right|+15=22\)

\(\Rightarrow x+15=22\)

\(\Rightarrow x=22-15\)

\(\Rightarrow x=7\)

c) \(\left|x\right|+12=25;x< 0\)

\(x< 0\Rightarrow\left|x\right|=-x\)

\(\Rightarrow\left|x\right|+12=25\)

\(\Rightarrow-x+12=25\)

\(\Rightarrow-x=25-12\)

\(\Rightarrow-x=13\)

\(\Rightarrow x=-13\)

d) \(\left|3x-15\right|=0\)

\(\Rightarrow3x-15=0\)

\(\Rightarrow3x=0+15\)

\(\Rightarrow3x=15\)

\(\Rightarrow x=15:3\)

\(\Rightarrow x=5\)

e) \(\left|-x+2\right|=4\)

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

TH1: \(-x+2=4\)

\(-x=4-2\)

\(-x=2\)

\(\Rightarrow x=-2\)

TH2: \(-x+2=-4\)

\(-x=-4-2\)

\(-x=-6\)

\(\Rightarrow x=6\)

f) \(\left|-x+\left|31-\left(-89\right)\right|\right|=14\)

\(\Rightarrow\left|-x+\left|120\right|\right|=14\)

\(\Rightarrow\left|-x+120\right|=14\)

\(\Rightarrow-x+120=14\) hoặc \(-x+120=-14\)

TH1: \(-x+120=14\)

\(\Rightarrow-x=14-120\)

\(\Rightarrow-x=-106\)

\(\Rightarrow x=106\)

TH2: \(-x+120=-14\)

\(\Rightarrow-x=-14-120\)

\(\Rightarrow-x=-134\)

\(\Rightarrow x=134\)

Lời giải:
a.

a. $(x-1)(x+2)-(x-3)(x+1)=5x-3$

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

$\Leftrightarrow 3x+1=5x-3$

$\Leftrightarrow 4=2x$

$\Leftrightarrow x=2$

b.

$(2x-1)(x+3)-(x-2)(x+3)=3x+1$

$\Leftrightarrow (2x^2+5x-3)-(x^2-4)=3x+1$

$\Leftrightarrow x^2+5x+1=3x+1$

$\Leftrightarrow x^2+2x=0$

$\Leftrightarrow x(x+2)=0$

$\Leftrightarrow x=0$ hoặc $x=-2$

c.

$x^2(x-1)-x(x-1)(x+1)=0$

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

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

$\Leftrightarrow (x-1)(-x)=0$

$\Leftrightarrow x-1=0$ hoặc $-x=0$

$\Leftrightarrow x=1$ hoặc $x=0$

d.

$4x(x-5)-(2x-3)(2x+3)=9$

$\Leftrightarrow 4x^2-20x-(4x^2-9)=9$

$\Leftrightarrow -20x=0$

$\Leftrightarrow x=0$

a: Ta có: \(\left(x-1\right)\left(x+2\right)-\left(x-3\right)\left(x+1\right)=5x-3\)

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

\(\Leftrightarrow-2x+4=0\)

\(\Leftrightarrow2x=4\)

hay x=2

b: Ta có: \(\left(2x-1\right)\left(x+3\right)-\left(x-2\right)\left(x+2\right)=3x+1\)

\(\Leftrightarrow2x^2+6x-x-3-x^2+4-3x-1=0\)

\(\Leftrightarrow x^2+2x=0\)

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

c: Ta có: \(x^2\left(x-1\right)-x\left(x-1\right)\left(x+1\right)=0\)

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

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

d: Ta có: \(4x\left(x-5\right)-\left(2x-3\right)\left(2x+3\right)=9\)

\(\Leftrightarrow4x^2-20x-4x^2+9=9\)

hay x=0

`-1/5<=x/8<=1/4`

`=>8* -1/5<=x<=1/4*8`

`=>-8/5<=x<=2`

Mà `x in ZZ`

`=>x in {-1,0,1,2}`

Mà 

\(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\) 

\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\) 

\(\Rightarrow5x\in\left\{-5;0;5;10\right\}\) 

\(\Rightarrow x\in\left\{-1;0;1;2\right\}\)

a \(2x+2>4\\ \Leftrightarrow2\left(x+1\right)>4\\ \Leftrightarrow x+1>2\\ \Leftrightarrow x>1\)

b \(3x+2>-5\\ \Leftrightarrow3x>-7\\ \Leftrightarrow x>\dfrac{-7}{3}\)

c \(10-2x>2\\ \Leftrightarrow2\left(5-x\right)>2\\ \Leftrightarrow5-x>1\\ \Leftrightarrow-x>-4\\ \Leftrightarrow x< 4\)

d \(1-2x< 3\\ \Leftrightarrow-2x< 2\\ \Leftrightarrow2x>2\\ \Leftrightarrow x>1\)

a)2x+2>4

<=> 2x>4-2

<=>2x>2

<=>x>1

Vậy...

b)3x+2>-5

<=>3x>-5-2

<=>3x>-7

<=>x>\(\dfrac{-7}{3}\)

Vậy...

c)10-2x>2

<=>-2x>-10+2

<=>-2x>-8

<=>x<4

Vậy...

d)1-2x<3

<=>-2x<3-1

<=>-2x<2

<=>x>-1

Vậy...

e)10x+3-5\(\le\)14x+12

<=>10x-2\(\le\)14x+12

<=>10x-14x\(\le\)2+12

<=>-4x\(\le\)14

<=>x\(\ge\)\(\dfrac{-7}{2}\)

Vậy...

f)(3x-1)<2x+4

<=> 3x-2x<1+4

<=>x<5

Vậy...

1/ ta co

vi x \(\in Z\Rightarrow x\in\left\{-9;-8;..;9;10\right\}\)

Tong cac so x thoa man la

-9+(-8)+(-7)+....+9+10

=(-9+9)+(-8+8)+...+(-1+1)+0

=0+0+0+..+0+0

=0

vay tong cac so ma x thoa man la 0

2/ ta co

vi x \(\in Z\Rightarrow x\in\left\{-8;-7;..;5;6;7\right\}\)

Tong cac so ma x thoa man la

-8+(-7)+(-6)+...+6+7

=-8+0+(-7+7)+(-6+6)+(-5+5)+...+(-1+1)

=-8+0+0+0+...+0

=-8

vay tong cac gia tri ma x thoa man la -8

3/ ta co

vi x \(\in Z\Rightarrow x\in\left\{-22;-21;...;22;23\right\}\)

Tong cac gia tri ma x thoa man la

(-22)+(-21)+....+22+23

=23+0+(-21+21)+(-22+22)+...+(-1+1)

=23+0+0+0+...+0

=23

vay tong cac gia tri ma x thoa man la 23

4/ ta co :

vi |x|\(\le2\Rightarrow\left|x\right|\in\left\{1;2\right\}hay.x\in\left\{2;1;-1;-2\right\}\)

Tong cac gia tri ma x thoa man la :

2+1+(-1)+(-2)

=3+(-3)

=0

vay tong cac gia tri ma x thoa man la 0

5/ ta co

│-x│< 13 nen |x| \(\in\left\{12;11;10;..;2;1;0;-1;-2;...\right\}\)

hay x \(\in\left\{12;11;10;9;...;1;0;-12;-13;...;-1\right\}\)

Tong cac so ma x thoa man la

12+13+14+15+....+1+0+(-1)+(-2)+....+(-12)

=(-12+12)+(-13+13)+...+(-1+1)+0

=0+0+0+0+...+0+0

=0

Vay tong cac gia tri ma x thoa man la 0