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26 tháng 11 2021
\(x^2=1\Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
\(x^2=3\Rightarrow\left[{}\begin{matrix}x=-\sqrt{3}\\x=\sqrt{3}\end{matrix}\right.\)
\(x^2=5\Rightarrow\left[{}\begin{matrix}x=-\sqrt{5}\\x=\sqrt{5}\end{matrix}\right.\Rightarrow x=-\sqrt{5}\left(vì.x< 0\right)\)
\(x^2=7\Rightarrow\left[{}\begin{matrix}x=-\sqrt{7}\\x=\sqrt{7}\end{matrix}\right.\Rightarrow x=-\sqrt{7}\left(vì.x< 0\right)\)
\(x^2=9\Rightarrow\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)
\(\left(x-2\right)^2=2\Rightarrow\left[{}\begin{matrix}x-2=-\sqrt{2}\\x-2=\sqrt{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2-\sqrt{2}\\x=2+\sqrt{2}\end{matrix}\right.\)
\(\left(x-4\right)^2=4\Rightarrow\left[{}\begin{matrix}x-2=-2\\x-2=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(\left(x-6\right)^2=6\Rightarrow\left[{}\begin{matrix}x-6=-\sqrt{6}\\x-6=\sqrt{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6-\sqrt{6}\\x=6+\sqrt{6}\end{matrix}\right.\)
\(\left(x-8\right)^2=8\Rightarrow\left[{}\begin{matrix}x-8=-2\sqrt{2}\\x-8=2\sqrt{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8-2\sqrt{2}\\x=2+2\sqrt{2}\end{matrix}\right.\)
\(\left(x-10\right)^2=10\Rightarrow\left[{}\begin{matrix}x-10=-\sqrt{10}\\x-10=\sqrt{10}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=10-\sqrt{10}\\x=10+\sqrt{10}\end{matrix}\right.\)
\(\left(x-\sqrt{3}\right)^2=3\Rightarrow\left[{}\begin{matrix}x-\sqrt{3}=-\sqrt{3}\\x-\sqrt{3}=\sqrt{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=2\sqrt{3}\end{matrix}\right.\)
\(\left(x-\sqrt{5}\right)^2=5\Rightarrow\left[{}\begin{matrix}x-\sqrt{5}=-\sqrt{5}\\x-\sqrt{5}=\sqrt{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=2\sqrt{5}\end{matrix}\right.\)
KB
5 tháng 12 2017
Bài 1:\(3^{x+2}-3^x=24\Rightarrow3^x.3^2-3^x=24\Rightarrow3^x.\left(3^2-1\right)=24\Rightarrow3^x.8=24\Rightarrow3^x=3\Rightarrow x=1\)
Bài 2:a,Chọn đáp án C.x0=1
b,Chọn đáp án D\(-\sqrt{2}+\sqrt{5}\) vì \(\sqrt{5}>\sqrt{2}\Rightarrow\left|\sqrt{2}-\sqrt{5}\right|=-\left(\sqrt{2}-\sqrt{5}\right)\)
MU
1
29 tháng 5 2022
a: \(\left(x^2-3\right)\left(2x^2-\dfrac{9}{8}\right)\left(\sqrt{\left|x\right|}-\sqrt{\dfrac{5}{2}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-3=0\\2x^2-\dfrac{9}{8}=0\\\sqrt{\left|x\right|}-\sqrt{\dfrac{5}{2}}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=3\\x^2=\dfrac{9}{16}\\\left|x\right|=\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow x\in\left\{-\sqrt{3};\sqrt{3};\dfrac{3}{4};-\dfrac{3}{4};\dfrac{-5}{2};\dfrac{5}{2}\right\}\)
b: \(x-5\sqrt{x}=0\)(ĐKXĐ: x>=0)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-5\right)=0\)
=>x=0 hoặc x=25
24 tháng 11 2021
\(a,=\dfrac{\sqrt{x}-8+5}{\sqrt{x}-8}=1+\dfrac{5}{\sqrt{x}-8}\in Z\\ \Leftrightarrow\sqrt{x}-8\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{3;7;9;13\right\}\\ \Leftrightarrow x\in\left\{9;49;81;169\right\}\left(tm\right)\\ b,=\dfrac{\sqrt{x}-2+7}{\sqrt{x}-2}=1+\dfrac{7}{\sqrt{x}-2}\in Z\\ \Leftrightarrow\sqrt{x}-2\inƯ\left(7\right)=\left\{-1;1;7\right\}\left(\sqrt{x}-2>-2\right)\\ \Leftrightarrow\sqrt{x}\in\left\{1;3;9\right\}\\ \Leftrightarrow x\in\left\{1;9;81\right\}\\ c,=\dfrac{2\left(\sqrt{x}+3\right)+2}{\sqrt{x}+3}=2+\dfrac{2}{\sqrt{x}+3}\in Z\\ \Leftrightarrow\sqrt{x}+3\inƯ\left(2\right)=\varnothing\left(\sqrt{x}+3>3\right)\\ \Leftrightarrow x\in\varnothing\)
29 tháng 10 2016
\(\sqrt{\left(x-3\sqrt{5}\right)^2}+\sqrt{\left(y+3\sqrt{5}\right)^2}+\left|x+y+z\right|=0\)
\(\Leftrightarrow\left|x-3\sqrt{5}\right|+\left|y+3\sqrt{5}\right|+\left|x+y+z\right|=0\)
\(\Leftrightarrow\begin{cases}x-3\sqrt{5}=0\\y+3\sqrt{5}=0\\x+y+z=0\end{cases}\)
\(\Leftrightarrow\begin{cases}x=3\sqrt{5}\\y=-3\sqrt{5}\\z=-x-y=-3\sqrt{5}+3\sqrt{5}=0\end{cases}\)
24 tháng 12 2023
a: Sửa đề: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne9\end{matrix}\right.\)
Để A là số nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)
=>\(\sqrt{x}-3+4⋮\sqrt{x}-3\)
=>\(4⋮\sqrt{x}-3\)
=>\(\sqrt{x}-3\in\left\{1;-1;2;-2;4;-4\right\}\)
=>\(\sqrt{x}\in\left\{4;2;5;1;7;-1\right\}\)
=>\(\sqrt{x}\in\left\{4;2;5;1;7\right\}\)
=>\(x\in\left\{16;4;25;1;49\right\}\)
b: 
NV
2
FM
3 tháng 11 2018
1) ĐKXĐ: \(x\ge0\)
\(\sqrt{x}=2\sqrt{2}\Rightarrow x=8\left(tmđkxđ\right)\)
2) ĐKXĐ: \(x\ge-1\)
\(\sqrt{\frac{x+1}{2}}=\frac{\sqrt{5}}{2}\)
\(\Leftrightarrow\frac{x+1}{2}=\frac{5}{4}\)
\(\Leftrightarrow2x+2=5\Leftrightarrow x=\frac{3}{2}\left(TMĐKXĐ\right)\)
AH
3 tháng 11 2018
1,
\(\sqrt{x}=2\sqrt{2}\)
=> \(\left(\sqrt{x}\right)=\left(2\sqrt{2}\right)^2\)
=> \(x=8\)
2.
\(\sqrt{\frac{x+1}{2}}=\frac{\sqrt{5}}{2}\)
=> \(\left(\sqrt{\frac{x+1}{2}}\right)=\left(\frac{\sqrt{5}}{2}\right)^2\)
=> \(\frac{x+1}{2}=\frac{5}{4}\)
=> 4 ( x + 1 ) = 5.2
=> 4x + 4 = 10
=> 4x = 6
=. x = \(\frac{3}{2}\)
\(\sqrt{x-5}\)=\(\sqrt{4}\)
=> x- 5 = 4
x = 4+5
x = 9
bình phương 2 vế ta có
x-5=4
x=4+5
x=9