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Ta có : 7(x - 1) + 2x(x - 1) = 0
<=> (2x + 7)(x - 1) = 0
\(\Leftrightarrow\orbr{\begin{cases}2x+7=0\\x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=-7\\x=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{7}{2}\\x=1\end{cases}}\)
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a) |2x-1|=|-7|
<=>|2x-1|=7
<=>\(\left[\begin{matrix}2x-1=7\\-\left(2x-1\right)=7\end{matrix}\right.\)
<=>\(\left[\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
Vậy: x=-3 hoặc x=4
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( x + 5 ) . ( x + 6 ) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}x+5=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\x=-6\end{matrix}\right.\)
Vậy \(x=-5\) hoặc \(x=-6\)
8x - 9x -2x - 15 = 0
\(\Rightarrow8x-9x-2x=0+15\)
\(\Rightarrow-3x=15\)
\(\Rightarrow x=15:\left(-3\right)\)
\(\Rightarrow x=-5\)
a, \(\left(x+5\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-6\end{matrix}\right.\)
Vậy ......
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a) \(\left(x-7\right)\left(x+12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x+12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-12\end{matrix}\right.\)
Vậy: x∈{7;-12}
b) \(\left(3x-15\right)\left(6-2x\right)=0\)
⇔\(3\left(x-5\right)\cdot2\cdot\left(3-x\right)=0\)
hay \(6\left(x-5\right)\left(3-x\right)=0\)
Vì 6≠0
nên \(\left[{}\begin{matrix}x-5=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\)
Vậy: x∈{3;5}
c) \(\left(3x+9\right)\left(4y-8\right)=0\)
⇔\(3\left(x+3\right)\cdot4\left(y-2\right)=0\)
hay \(12\left(x+3\right)\left(y-2\right)=0\)
Vì 12≠0
nên \(\left\{{}\begin{matrix}x+3=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=2\end{matrix}\right.\)
Vậy: x=-3 và y=2
d) \(\left(2y-16\right)\left(8x-24\right)=0\)
⇔\(2\left(y-8\right)\cdot8\left(x-3\right)=0\)
hay 16(y-8)(x-3)=0
Vì 16≠0
nên \(\left\{{}\begin{matrix}y-8=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=8\\x=3\end{matrix}\right.\)
Vậy: y=8 và x=3
e) \(\left(22-11y\right)\left(9x-18\right)=0\)
⇔\(11\left(2-y\right)9\left(x-2\right)=0\)
hay 99(2-y)(x-2)=0
Vì 99≠0
nên \(\left\{{}\begin{matrix}2-y=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=2\end{matrix}\right.\)
Vậy: x=2 và y=2
g) \(\left(7y+14\right)\cdot\left(9x-18\right)=0\)
⇔7(y+2)*9(x-2)=0
hay 63(y+2)(x-2)=0
Vì 63≠0
nên \(\left\{{}\begin{matrix}y+2=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2\\x=2\end{matrix}\right.\)
Vậy: y=-2 và x=2
h) xy=3
⇒x,y∈Ư(3)
⇒x,y∈{1;-1;3;-3}
*Trường hợp 1:
\(\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)
*Trường hợp 2:
\(\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)
*Trường hợp 3:
\(\left\{{}\begin{matrix}x=-1\\y=-3\end{matrix}\right.\)
*Trường hợp 4:
\(\left\{{}\begin{matrix}x=-3\\y=-1\end{matrix}\right.\)
Vậy: x∈{1;-1;3;-3} và y∈{1;-1;3;-3}
i) x*y=-5
⇔x,y∈Ư(-5)
⇔x,y∈{1;-1;5;-5}
*Trường hợp 1:
\(\left\{{}\begin{matrix}x=1\\y=-5\end{matrix}\right.\)
*Trường hợp 2:
\(\left\{{}\begin{matrix}x=-1\\y=5\end{matrix}\right.\)
*Trường hợp 3:
\(\left\{{}\begin{matrix}x=-5\\y=1\end{matrix}\right.\)
*Trường hợp 4:
\(\left\{{}\begin{matrix}x=5\\y=-1\end{matrix}\right.\)
Vậy: x∈{1;5;-1;-5} và y∈{1;5;-1;-5}
k) \(\left(x+4\right)\left(y-5\right)=-3\)
⇔x+4; y-5∈Ư(-3)
⇔x+4; y-5∈{1;3;-3;-1}
*Trường hợp 1:
\(\left\{{}\begin{matrix}x+4=-1\\y-5=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=8\end{matrix}\right.\)
*Trường hợp 2:
\(\left\{{}\begin{matrix}x+4=1\\y-5=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=2\end{matrix}\right.\)
*Trường hợp 3:
\(\left\{{}\begin{matrix}x+4=3\\y-5=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=4\end{matrix}\right.\)
*Trường hợp 4:
\(\left\{{}\begin{matrix}x+4=-3\\y-5=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-7\\y=6\end{matrix}\right.\)
Vậy: x∈{-5;-3;-1;-7} và y∈{8;2;4;6}
m) (x-9)(y-5)=-1
⇔x-9; y-5∈Ư(-1)
⇔x-9; y-5∈{1;-1}
*Trường hợp 1:
\(\left\{{}\begin{matrix}x-9=1\\y-5=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=4\end{matrix}\right.\)
*Trường hợp 2:
\(\left\{{}\begin{matrix}x-9=-1\\y-5=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=6\end{matrix}\right.\)
Vậy: x∈{10;8} và y∈{4;6}
n) x+3⋮x+4
⇔x+4-1⋮x+4
⇔-1⋮x+4
hay x+4∈Ư(-1)
⇔x+4∈{1;-1}
⇔x∈{-3;-5}
Vậy: x∈{-3;-5}
p)(x-5)⋮x+2
⇔x+2-7⋮x+2
hay -7⋮x+2
⇔x+2∈Ư(-7)
⇔x+2∈{1;-1;7;-7}
hay x∈{-1;-3;5;-9}
Vậy: x∈{-1;-3;5;-9}
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\(3x.\left(x-3\right)+\left(x-3\right)=0\)
\(\left(3x+1\right).\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=3\\3x=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-\frac{1}{3}\end{cases}}\)
vậy \(x=3,x=-\frac{1}{3}\)
\(b,x^3-9x-2x^2+18=0\)
\(x.\left(x^2-9\right)-2.\left(x^2-9\right)=0\)
\(\left(x-2\right).\left(x^2-9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x^2=9\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=3,x=-3\end{cases}}\)
vậy \(x=2,x=3,x=-3\)
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a ) | 2x - 1 | = | -7 |
| 2x - 1 | = 7
+ ) 2x - 1 = 7
2x = 8
x = 4
+ ) 2x - 1 = -7
2x = -6
x = -3
Vậy .........
b ) | x + 5 | = 0
x + 5 = 0
x = -5
Vậy ............
c ) 8x - 75 = 5x + 21
8x - 5x = 21 + 75
3x = 96
x = 32
Vậy .......
d ) 9x + 25 = -2x + 58
9x + 2x = 58 - 25
11x = 33
x = 3
Vậy ........
a, \(\left|2x-1\right|=\left|-7\right|\Rightarrow\left|2x-1\right|=7\)
\(\Rightarrow2x-1\in\left\{-7;7\right\}\)
+) \(2x-1=7\Rightarrow2x=7+1\)
\(\Rightarrow2x=8\Rightarrow x=8\div2=4\)
+) \(2x-1=-7\Rightarrow2x=-7+1\)
\(\Rightarrow2x=-6\Rightarrow x=-6\div2=-3\)
Vậy \(x\in\left\{-3;4\right\}\)
b, \(\left|x+5\right|=0\Rightarrow x+5=0\)
\(\Rightarrow x=0-5\Rightarrow x=-5\)
Vậy x = -5
c, \(8x-75=5x+21\Rightarrow8x-5x=21+75\)
\(\Rightarrow\left(8-5\right)x=96\Rightarrow3x=96\)
\(\Rightarrow x=96\div3\Rightarrow x=32\)
Vậy x = 32
d, \(9x+25=-\left(2x-58\right)\Rightarrow9x+25=-2x+58\)
\(\Rightarrow9x+2x=58-25\Rightarrow\left(9+2\right)x=33\)
\(\Rightarrow11x=33\Rightarrow x=33\div11\Rightarrow x=3\)
Vậy x = 3
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1: x(y-3)=-6
\(\Leftrightarrow\left(x;y-3\right)\in\left\{\left(1;-6\right);\left(-6;1\right);\left(-1;6\right);\left(6;-1\right);\left(2;-3\right);\left(-3;2\right);\left(3;-2\right);\left(-2;3\right)\right\}\)
hay \(\left(x,y\right)\in\left\{\left(1;-3\right);\left(-6;4\right);\left(-1;9\right);\left(6;2\right);\left(2;0\right);\left(-3;5\right);\left(-2;6\right);\left(3;1\right)\right\}\)
2: (2x-1)(y+2)=10
\(\Leftrightarrow\left(2x-1;y+2\right)\in\left\{\left(1;10\right);\left(-1;-10\right);\left(5;2\right);\left(-5;-2\right)\right\}\)
hay \(\left(x,y\right)\in\left\{\left(1;8\right);\left(0;-12\right);\left(3;0\right);\left(-2;-4\right)\right\}\)
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a, Vì |2x+8| và |3y-9x| đều >= 0
=> |2x+8| + |3y-9x| >= 0
Dấu "=" xảy ra <=> 2x+8=0 và 3y-9x=0 <=> x=-4 và y=-12
Vậy x=-4 và y=-12
Tk mk nha
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a) \(5\left(x-7\right)=0\)
\(\Rightarrow x-7=0\)
\(\Rightarrow x=7\)
b) \(25\left(x-4\right)=0\)
\(\Rightarrow x-4=0\)
\(\Rightarrow x=4\)
c) \(\left(34-2x\right)\left(2x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}34-2x=0\\2x-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=34\\2x=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=17\\x=3\end{matrix}\right.\)
d) \(\left(2019-x\right)\left(3x-12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2019-x=0\\3x-12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\3x=12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2019\\x=\dfrac{12}{3}=4\end{matrix}\right.\)
e) \(57\left(9x-27\right)=0\)
\(\Rightarrow9x-27=0\)
\(\Rightarrow9\left(x-3\right)=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
a) 5.(x-7)=0⇔x-7=0⇔x=7
b) 25(x-4)=0⇔x-4=0⇔x=4
c) (34-2x).(2x-6)=0
⇔ 34-2x=0 hoặc 2x-6=0
⇔2x=34 hoặc 2x=6
⇔ x=17 hoặc x=3
d) (2019-x).(3x-12)=0
⇔ 2019-x=0 hoặc 3x-12=0
⇔ x=2019 hoặc x=4
e) 57.(9x-27)=0
⇔ 9x-27=0
⇔ x=3
f) 25+(15-x)=30
⇔ 15-x=5
⇔ x=10
g) 43-(24-x)=20
⇔ 24-x=23
⇔ x=1
h) 2.(x-5)-17=25
⇔ 2(x-5)=42
⇔x-5=21
⇔ x=26
i) 3(x+7)-15=27
⇔ 3(x+7)=42
⇔ x+7=14
⇔ x=7
j) 15+4(x-2)=95
⇔ 4(x-2)=80
⇔ x-2=20
⇔ x=22
k) 20-(x+14)=5
⇔ x+14=15
⇔ x=1
l) 14+3(5-x)=27
⇔ 3(5-x)=13
⇔ 5-x=13/3
⇔ x=5-13/3
⇔ x=2/3
\(\left(9x+1\right)\left(5+2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}9x+1=0\\5+2x+0\end{cases}\Leftrightarrow\orbr{\begin{cases}9x=-1\\2x=-5\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{-1}{9}\\x=\frac{-5}{2}\end{cases}}}\)
Vậy x = -1/9 hoặc x = -5/2
\(\left(9x+1\right)\left(5+2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}9x+1=0\\5+2x=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{9}\\x=-\frac{5}{2}\end{cases}}\)