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4 tháng 12 2018

a,  3 ( x + 1 ) - 2 ( 3 x - 4 ) = - 13

=> 3x + 3 - 6x + 8 = - 13

=> 6x - 3x = 3 + 8 + 13

=> 3x = 24

=> x = 8

b, 2 ( x - 3 ) - 4 ( 2 x - 1 ) = - 20

=> 2x - 6 - 8x + 4 = - 20

=> 8x - 2x = - 6 + 4 + 20

=> 6x = 18

=> x = 3

c, 2 x ( x + 3 ) = 0

=> \(\orbr{\begin{cases}2x=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-3\end{cases}}}\)

d, ( x - 1 ) ( 5 x - x ) = 0

=> \(\orbr{\begin{cases}x-1=0\\5x-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\4x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=0\end{cases}}}\)

e, ( x + 3 ) 2 ( 4 - x ) = 0

=> \(\orbr{\begin{cases}\left(x+3\right)^2=0\\4-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x+3=0\\4-x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-3\\x=4\end{cases}}}\)

a) \(3\left(x+1\right)-2\left(3x-4\right)=-13\)

\(\Leftrightarrow3x+3-6x+8=-13\)

\(\Leftrightarrow3x-6x=-13-3-8\)

\(\Leftrightarrow-3x=-24\)

\(\Leftrightarrow x=8\)

Vậy \(x=8\)

b) \(2\left(x-3\right)-4\left(2x-1\right)=-20\)

\(\Leftrightarrow2x-6-8x+4=-20\)

\(\Leftrightarrow2x-8x=-20+6-4\)

\(\Leftrightarrow-6x=-18\)

\(\Leftrightarrow x=3\)

Vậy \(x=3\)

c) \(2x\left(x+3\right)=0\)

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

Vậy \(\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)

d)\(\left(x-1\right)\left(5x-x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\5x-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\4x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}\)

Vậy \(\orbr{\begin{cases}x=1\\x=0\end{cases}}\)

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

\(\orbr{\begin{cases}\left(x+3\right)^2=0\\4-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x+3=0\\-x=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3\\x=4\end{cases}}\)

Vậy \(\orbr{\begin{cases}x=-3\\x=4\end{cases}}\)

28 tháng 8 2020

a) \(\left|2x+1\right|=\left|1-x\right|\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=1-x\\2x+1=x-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x=0\\x=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)

b) \(\left|5x-4\right|=\left|x+2\right|\)

\(\Leftrightarrow\orbr{\begin{cases}5x-4=x+2\\5x-4=-x-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x=6\\6x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{3}\end{cases}}\)

c) \(\left|2x-3\right|-\left|3x+2\right|=0\Leftrightarrow\left|2x-3\right|=\left|3x+2\right|\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3=3x+2\\2x-3=-3x-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\5x=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=-5\\x=\frac{1}{5}\end{cases}}\)

d) \(\left|2+3\right|=\left|4x-3\right|\Leftrightarrow\left|4x-3\right|=5\)

\(\Rightarrow\orbr{\begin{cases}4x-3=5\\4x-3=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x=8\\4x=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-\frac{1}{2}\end{cases}}\)

e) \(\left|\frac{5}{4}-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\Leftrightarrow\left|\frac{5}{8}x+\frac{3}{5}\right|=\frac{9}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x+\frac{3}{5}=\frac{9}{4}\\\frac{5}{8}x+\frac{3}{5}=-\frac{9}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x=\frac{33}{20}\\\frac{5}{8}x=-\frac{57}{20}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{66}{25}\\x=-\frac{114}{25}\end{cases}}\)

28 tháng 8 2020

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

\(\Leftrightarrow\orbr{\begin{cases}2x+1=-x+1\\2x+1=x-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x+x=-1+1\\2x-x=-1-1\end{cases}}\)

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

b. \(\left|5x-4\right|=\left|x+2\right|\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}4x=6\\6x=2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{3}\end{cases}}\)

c. \(\left|2x-3\right|-\left|3x+2\right|=0\)

\(\Leftrightarrow\left|2x-3\right|=\left|3x+2\right|\)

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

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

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

d, e tương tự

11 tháng 9 2017

a) <=>5x-x=57+7

<=> 4x=64

<=> x=16

b) <=>4x=28+12

<=> x=40/4

<=>10

c) th1 x=0

th2 x-3=0

<=>x=3

d) th1 2x-4=0

<=> x=2

th2 x+1=0

<=> x=-1

câu e mình ko biết làm hihi 

11 tháng 9 2017

a, 5x - 7 = x + 57

5x - x = 57 + 7

4x = 64

x = 64 : 4

x = 16

b, x + 3x - 12 = 28

4x = 28 + 12

4x = 40

x = 10

c, x(x-3) = 0

\(\Rightarrow\orbr{\begin{cases}x=0\\x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}}\)

d, (2x-4)(x+1)=0

\(\Rightarrow\orbr{\begin{cases}2x-4=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)

e, 1+2+...+x=120

=> \(\frac{x\left(x+1\right)}{2}=120\)

=> x(x+1) = 120.2

=>x(x+1) = 240

=> x(x+1) = 15.16

=> x=16

6 tháng 9 2019

e)

A = \(\frac{x+5}{x-2}\) = \(\frac{\left(x-2\right)+7}{x-2}=1+\frac{7}{x-2}\)

Muốn A nguyên thì:

=> \(\frac{7}{x-2}\) ∈ Z

=> 7 ⋮ x - 2

=> x - 2 ∈ Ư (7)

=> x - 2 ∈ { 1; 7; -1; -7 }

=> x ∈ { 3; 9; -5; 1 }

6 tháng 9 2019

a) (5x - 1)(2x - 1/3) = 0

\(\Rightarrow5x-1=0\) hoặc \(2x-\frac{1}{3}=0\)

\(\Rightarrow\left[{}\begin{matrix}5x=0+1\\2x=0+\frac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}5x=1\\2x=\frac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1:5=\frac{1}{5}\\x=\frac{1}{3}:2=\frac{1}{3}.\frac{1}{2}=\frac{1}{6}\end{matrix}\right.\)

Vậy x = 1/5 hoặc x = 1/6

2 tháng 6 2017

a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)

\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)

\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)

b) \(\dfrac{39}{7}:x=13\)

\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)

c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)

\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)

\(\dfrac{14}{5}x=34+50=84\)

\(x=\dfrac{84}{\dfrac{14}{5}}=30\)

d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)

\(\dfrac{1}{6}x=\dfrac{5}{12}\)

\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)

g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)

\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)

\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)

\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)

\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)

h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)

\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)

\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)

\(x=1\)

Mỏi tay woa bn làm nốt nha!!

20 tháng 1 2020

a) x.(x-1)=0

\(\Rightarrow\)x=0 hoặc x-1=0

\(\Rightarrow\)x=0+1

\(\Rightarrow\)x=1

vậy x=1 hoặc x=0

b) -x.(x+3)=0

\(\Rightarrow\)-x = 0 hoặc x+3 = 0

\(\Rightarrow\)x= 0-3

\(\Rightarrow\)x=-3

vậy x=0 hoặc x=-3

c) (2x-4).(x+2)=0

(2x-4)= 0

2x=0+4

2x=4

x=4:2

x=2

hoặc (x+2)=0

x= 0-2

x=-2

vậy x=2 hoặc x=-2

d) (3-x).|x+5|=0

3-x = 0

x= 3-0

x=3

hoặc |x+5|=0

x+ 5=0

x=0-5

x=-5

vậy x=3 hoặc x=-5

e) (|x|+1).( 4-2x) = 0

(|x|+1) =0

|x|= 0-1

|x|=-1

hoặc( 4-2x) = 0

2x=4-0

2x=4

x=4:2

x=2

g) x2+5x=0

x2=0

x=0

hoặc 5x=0

x= 0: 5

x=0

vậy x=0

2)

a) (x+3).(y-5)= 7

(x+3)và (y-5)\(\in\)Ư(7)=\(\left\{1;-1;7;-7\right\}\)

x+3 1 7 -1 -7
y-5 7 1 -7 -1
x -2 4 -4 -10
y 12 6 2 4

b) xy + 3x - 2y= 11

x( y+3) -2y=11

x(y-3)- 2( y+3) +6 = 11

( y+3) ( x-2) = 5

vì x,y thuộc Z \(\Leftrightarrow\)y+3 và x-2 \(\in\)Z

do đó y+3 và x-2 \(\in\)Ư ( 5)= \(\left\{1;5;-1;-5\right\}\)

y+3 1 5 -1 -5
x-2 5 1 -5 -1
y -2 2 -4 -8
x 7 3 -3 1

\(\in\)\(\in\)

c) xy + 3x - 7y= 21

x( y+3) -7y= 21

x( y+3) - 7( y+3)+21= 21

(y+3)( x-7) =0

y+3 0
x-7 0
y -3
x 7

20 tháng 1 2020

@Lâm Khả Vy siêu quá đi à.

28 tháng 10 2021
Jdjsjsnnsndndndndndnnejejekekeodkndnfndndnnfnfnfnfnndkakwowoeijrhfbd historical đó có bột all of có lúc có là do sốc xó óc đó Úc được xây ff xóc đi h Hà đó ơi clan. Gì ơi có óc cm mà đi mà số of con là clair đó Úc áo
3 tháng 7 2019

a) \(\frac{1}{3}+\frac{2}{3}:x=-7\)

=> \(\frac{2}{3}:x=-7-\frac{1}{3}\)

=> \(\frac{2}{3}:x=-\frac{22}{3}\)

=> \(x=\frac{2}{3}:\left(-\frac{22}{3}\right)\)

=> \(x=-\frac{1}{11}\)

b) \(\frac{1}{3}x+\frac{2}{5}x=0\)

=> \(\frac{11}{15}x=0\)

=> \(x=0\)

c) \(\left(2x-3\right)\left(6-2x\right)=0\)

=> \(\left(2x-3\right)\left(3-x\right).2=0\)

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

=> \(\orbr{\begin{cases}2x-3=0\\3-x=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{3}{2}\\x=3\end{cases}}\)

a) \(\frac{1}{3}+\frac{2}{3}:x=-7\)

\(\Rightarrow\frac{2}{3}.\frac{1}{x}=-7-\frac{1}{3}\)

\(\Rightarrow\frac{2}{3x}=\frac{-21-1}{3}\)

\(\Rightarrow\frac{2}{3x}=\frac{-22}{3}\)

\(\Rightarrow-22.3x=6\)

\(\Rightarrow3x=\frac{-6}{22}=\frac{-3}{11}\)

\(\Rightarrow x=\frac{-3}{11}:3=\frac{-3}{11}.\frac{1}{3}\)

\(\Rightarrow x=\frac{-1}{11}\)

b) \(\frac{1}{3}x+\frac{2}{5}x=0\)

\(\Rightarrow x.\left(\frac{1}{3}+\frac{2}{5}\right)=0\)

\(\Rightarrow x=0\)

c) \(\left(2x-3\right).\left(6-2x\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}2x=3\\2x=6\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=3\end{cases}}\)

d) \(x:\frac{3}{4}+\frac{1}{4}=\frac{-2}{3}\)

\(\Rightarrow x.\frac{4}{3}=\frac{-2}{3}-\frac{1}{4}\)

\(\Rightarrow x.\frac{4}{3}=\frac{-11}{12}\)

\(\Rightarrow x=\frac{-11}{12}:\frac{4}{3}=\frac{-11}{12}.\frac{3}{4}=\frac{-11}{16}\)

e) \(\frac{3}{4}-\left|x-\frac{2}{3}\right|=\frac{1}{2}\)

\(\Rightarrow\left|x-\frac{2}{3}\right|=\frac{3}{4}-\frac{1}{2}\)

\(\Rightarrow\left|x-\frac{2}{3}\right|=\frac{1}{4}\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{2}{3}=\frac{1}{4}\\x-\frac{2}{3}=\frac{-1}{4}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{11}{12}\\x=\frac{5}{12}\end{cases}}\)