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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) \(\left|\frac{5}{4}-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\Leftrightarrow\left|\frac{5}{8}x+\frac{3}{5}\right|=\frac{9}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x+\frac{3}{5}=\frac{9}{4}\\\frac{5}{8}x+\frac{3}{5}=-\frac{9}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x=\frac{33}{20}\\\frac{5}{8}x=-\frac{57}{20}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{66}{25}\\x=-\frac{114}{25}\end{cases}}\)
\(\left|2x+1\right|=\left|1-x\right|\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=-x+1\\2x+1=x-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x+x=-1+1\\2x-x=-1-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=0\\x=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)
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
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
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
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 }
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
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!!
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 |
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) \(\frac{3}{4}-\left|x-\frac{2}{3}\right|=\frac{1}{2}\)
\(\Rightarrow\left|x-\frac{2}{3}\right|=\frac{3}{4}-\frac{1}{2}\)
\(\Rightarrow\left|x-\frac{2}{3}\right|=\frac{1}{4}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{2}{3}=\frac{1}{4}\\x-\frac{2}{3}=\frac{-1}{4}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{11}{12}\\x=\frac{5}{12}\end{cases}}\)
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 \(\orbr{\begin{cases}x=-3\\x=4\end{cases}}\)