(𝑥 - 1)2 = 16 giúp mình
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1)(x+1)thuộc ước của -2
ư(2)={1;2;-1;-2}
x+1 | 1 | 2 | -1 | -2 |
x | 0 | 1 | -2 | -3 |
vậy x =0;x=1;x=-2;x=-3
2)ta có : 2x+7=2(x+3)+1
2(x+3)chia hết cho x+3
=>để 2x+7chia hết cho x+3
<=>1chia hết cho x+3
=>x+3 thuộc ư(1)
u(1)={1;-1}
x+3 | 1 | -1 |
2 | -2 | -4 |
vậy x=-2;x=-4
a: \(\Leftrightarrow x\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;9;-9;12;-12;18;-18;36;-36\right\}\)
mà -3<x<30
nên \(x\in\left\{-2;-1;1;2;3;4;6;9;12;18\right\}\)
b: \(\Leftrightarrow x\in\left\{0;4;-4;8;-8;12;-12;...\right\}\)
mà -16<=x<20
nên \(x\in\left\{-16;-12;-8;-4;0;4;8;12;16\right\}\)
c: \(\Leftrightarrow x-1+4⋮x-1\)
\(\Leftrightarrow x-1\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(x\in\left\{2;0;3;-1;5;-3\right\}\)
d: \(\Leftrightarrow2x+4-5⋮x+2\)
\(\Leftrightarrow x+2\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{-1;-3;3;-7\right\}\)
\(a,\dfrac{\left(x-1\right)^2}{x^2-1}=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\\ b,\dfrac{x^2-16}{4x-x^2}=\dfrac{\left(x-4\right)\left(x+4\right)}{x\left(4-x\right)}=\dfrac{-\left(4-x\right)\left(x+4\right)}{x\left(4-x\right)}=\dfrac{-\left(x+4\right)}{x}\\ c,\dfrac{x^2+6x+9}{2x+6}=\dfrac{\left(x+3\right)^2}{2\left(x+3\right)}=\dfrac{x+3}{2}\)
\(d,\dfrac{x^2+x}{x^2+4x+3}=\dfrac{x\left(x+1\right)}{\left(x^2+x\right)+\left(3x+3\right)}=\dfrac{x\left(x+1\right)}{x\left(x+1\right)+3\left(x+1\right)}=\dfrac{x\left(x+1\right)}{\left(x+1\right)\left(x+3\right)}=\dfrac{x}{x+3}\)
\(e,\dfrac{x^2-x+1}{x^3+1}=\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x+1}\\ f,\dfrac{\left(x+y\right)^2-z^2}{x+y+z}=\dfrac{\left(x+y-z\right)\left(x+y+z\right)}{x+y+z}=x+y-z\)
\(x-1=4\Leftrightarrow x=5\)
x=5