tìm x \(\in\)z
(x - 6)2= 9
|x| = 3
|x + 5| = 15
2x - 1= 16
5x+1 =125
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H
6 tháng 8 2023
\(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\left(\text{đ}k\text{x}\text{đ}:x\ge3\right)\\ =\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\\ =\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{2\sqrt{x}-9-\left(x-9\right)-\left(2x-4\sqrt{x}+\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}-9-x+9-2x+4\sqrt{x}-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{5\sqrt{x}-3x+2}{x-5\sqrt{x}+6}\)
__
Để \(M\in Z\) thì \(x-5\sqrt{x}+6\) thuộc ước của \(5\sqrt{x}-3x+2\)
\(\Rightarrow x-5\sqrt{x}+6=-5\sqrt{x}-3x+2\\ \Leftrightarrow x-5\sqrt{x}+6+5\sqrt{x}+3x-2=0\\ \Leftrightarrow4x-4=0\\ \Leftrightarrow4x=4\\ \Leftrightarrow x=1\)
16 tháng 2 2021
b) Ta có: \(-5+\left|3x-1\right|+6=\left|-4\right|\)
\(\Leftrightarrow\left|3x+1\right|+1=4\)
\(\Leftrightarrow\left|3x+1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=3\\3x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{2}{3};-\dfrac{4}{3}\right\}\)
c) Ta có: \(\left(x-1\right)^2=\left(x-1\right)^4\)
\(\Leftrightarrow\left(x-1\right)^2-\left(x-1\right)^4=0\)
\(\Leftrightarrow\left(x-1\right)^4-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left(x-1-1\right)\left(x-1+1\right)=0\)
\(\Leftrightarrow x\cdot\left(x-1\right)^2\cdot\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;1;2\right\}\)
d) Ta có: \(5^{-1}\cdot25^x=125\)
\(\Leftrightarrow5^{-1}\cdot5^{2x}=5^3\)
\(\Leftrightarrow5^{2x-1}=5^3\)
\(\Leftrightarrow2x-1=3\)
\(\Leftrightarrow2x=4\)
hay x=2
Vậy: x=2
25 tháng 1 2018
a) 15 - x = 1 - ( -9 )
15 - x = 10
x = 15 - 10
x = 5 thuộc Z
vậy____
b) -2 | x | + 15 = 1
-2 | x | = 1 - 15
-2 | x | = - 14
| x | = -14 : (-2)
| x | = 7
\(\Rightarrow x=\pm7\) thuộc Z
vậy_____
c) -3 . | x + 4 | = -18
| x + 4 | = -18 : (-3)
| x + 4 | = 6
\(\Rightarrow\orbr{\begin{cases}x+4=6\\x+4=-6\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\in Z\\x=-10\in Z\end{cases}}\)
Vậy_____
d) ( x - 3 ) . ( x2 + 1 ) = 0
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x^2+1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\in Z\\x\in\varnothing\end{cases}}\)
vậy x = 3
e) 12 - ( 9 - 3x) = - 2 . ( x + 5 )
=> 12 - 9 + 3x = -2x + (-2).5
=> 3 + 3x = -2x + (-10)
=> 3x + 2x = (-10) - 3
=> 5x = -13
=> x = -13 : 5
=> x = \(\frac{-13}{5}\notin Z\)
=> \(x\in\varnothing\)
f) ( x - 2 )3 = - 125
\(\Rightarrow\left(x-2\right)^3=\left(-5\right)^3\)
\(\Rightarrow x-2=5\)
\(\Rightarrow x=7\in Z\)
vậy____
g) ( x + 6 ) = 81
=> x = 81 - 6
=> x = 75 thuộc Z
vậy_____
h) -12 : | x - 4 | = - 4
=> |x-4| = -12 : (-4)
=> |x-4| = 3
\(\Rightarrow\orbr{\begin{cases}x-4=3\\x-4=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\in Z\\x=1\in Z\end{cases}}\)
vậy_____
HB
25 tháng 1 2018
a) 15 - x = 1 - ( -9 )
15 - x = 10
x = 15 - 10
x = 5
b) -2 | x | + 15 = 1
-2 x = 1 -15
- 2 x = -14
x = -14 : ( -2 )
x = 7
c) -3 . | x + 4| = -18
| x + 4 | = ( - 18 ) : ( - 3 )
| x + 4 | = 6
=> \(\orbr{\begin{cases}x+4=6\\x+4=-6\end{cases}\Rightarrow\orbr{\begin{cases}x=6-4\\x=-6-4\end{cases}\Rightarrow}\orbr{\begin{cases}x=2\\x=-8\end{cases}}}\)
Vậy x = 2 hoặc -8
d) ( x - 3 ) . ( x2 + 1 ) = 0
=> \(\orbr{\begin{cases}x-3=0\\x^2+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0+3\\x^2=0-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x^2=-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=-1\end{cases}}}\)
Vậy x = 3 hoặc -1
a,
\(\left(x-6\right)^2=9\\ \Rightarrow\left[{}\begin{matrix}x-6=-3\\x-6=3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=9\end{matrix}\right.\)
b,
\(\left|x\right|=3\\ \Rightarrow\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)
c,
\(\left|x+5\right|=15\\ \Rightarrow\left[{}\begin{matrix}x+5=-15\\x+5=15\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-20\\x=10\end{matrix}\right.\)
d,
\(2^{x-1}=16\\ \Rightarrow2^{x-1}=2^4\\ \Rightarrow x-1=4\\ \Rightarrow x=5\)
e,
\(5^{x+1}=125\\ \Rightarrow5^{x+1}=5^3\\ \Rightarrow x+1=3\\ \Rightarrow x=2\)
a: Ta có: \(\left(x-6\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=3\\x-6=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=3\end{matrix}\right.\)
b: Ta có: \(\left|x\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
c: Ta có: \(\left|x+5\right|=15\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-20\end{matrix}\right.\)