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a: \(\Leftrightarrow\dfrac{2x-3}{x-1}=4\)
=>4x-4=2x-3
=>2x=1
hay x=1/2
b: \(\Leftrightarrow\sqrt{\dfrac{2x-3}{x-1}}=2\)
=>(2x-3)=4x-4
=>4x-4=2x-3
=>2x=1
hay x=1/2(nhận)
c: \(\Leftrightarrow\sqrt{2x+3}\left(\sqrt{2x-3}-2\right)=0\)
=>2x+3=0 hoặc 2x-3=4
=>x=-3/2 hoặc x=7/2
e: \(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
=>căn (x-5)=2
=>x-5=4
hay x=9
a) \(\sqrt{3}x-\sqrt{12}=0< =>\sqrt{3}x=\sqrt{12}=>x=2\)
Vay S = { 2 }
b) \(\sqrt{2}x+\sqrt{2}=\sqrt{8}+\sqrt{18}< =>\sqrt{2}x=\sqrt{8}+\sqrt{18}-\sqrt{2}< =>\sqrt{2}x=2\sqrt{2}+3\sqrt{2}-\sqrt{2}\) <=> \(\sqrt{2}x=4\sqrt{2}=>x=4\)
Vay S = { 4 }
c) \(\sqrt{5}x^2-\sqrt{20}=0< =>\sqrt{5}x^2=\sqrt{20}< =>x^2=2=>x=\sqrt{2}\)
Vay S = {\(\sqrt{2}\) }
d) \(\sqrt{x^2+6x+9}=3x+6< =>\sqrt{\left(x+3\right)^2}=3x+6< =>x+3=3x+6< =>-2x=\) \(3=>x=-\dfrac{3}{2}\)
Vay S = { - 3/2 }
e) \(\sqrt{x^2-4x+4}-2x+5=0< =>\sqrt{\left(x-2\right)^2}-2x+5=0< =>x-2-2x+5=0\) <=> \(-x+3=0< =>-x=-3=>x=3\)
Vay S = { 3 }
F) \(\sqrt{\dfrac{2x-3}{x-1}}=2\)
<=> \(\dfrac{2x-3}{x-1}=4< =>2x-3=4x-4< =>-2x=-1=>x=\dfrac{1}{2}\)
Vay S = { 1/2 }
g) \(\dfrac{\sqrt{2x-3}}{\sqrt{x-1}}=2< =>\sqrt{\dfrac{2x-3}{x-1}}=2< =>\dfrac{2x-3}{x-1}=4< =>2x-3=4x-4< =>-2x=-1=>x=\dfrac{1}{2}\)
bạn chưa có ĐKXĐ nên chưa xét kết quả có đúng vs Đk ko, có vài câu sai kết quả
Lời giải:
a)
\(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}=\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}+\frac{\sqrt{3}}{3}-\frac{2(3-\sqrt{3})}{(3+\sqrt{3})(3-\sqrt{3})}\)
\(=\frac{2-\sqrt{3}}{4-3}+\frac{\sqrt{3}}{3}-\frac{2(3-\sqrt{3})}{3^2-3}\)
\(=2-\sqrt{3}+\frac{\sqrt{3}}{3}-\frac{3-\sqrt{3}}{3}=\frac{6-3\sqrt{3}}{3}+\frac{2\sqrt{3}-3}{3}=\frac{3-\sqrt{3}}{3}\)
b)
\(\sqrt{x-3+2\sqrt{x-4}}=\sqrt{(x-4)+2\sqrt{x-4}+1}=\sqrt{(\sqrt{x-4}+1)^2}=|\sqrt{x-4}+1|=\sqrt{x-4}+1\)
c)
\(\sqrt{2x+4\sqrt{2x-4}}=\sqrt{(2x-4)+2.2\sqrt{2x-4}+2^2}\)
\(=\sqrt{(\sqrt{2x-4}+2)^2}=|\sqrt{2x-4}+2|=\sqrt{2x-4}+2\)
d)
\(\sqrt{x-\sqrt{2x-1}}=\frac{1}{\sqrt{2}}\sqrt{2x-2\sqrt{2x-1}}=\frac{1}{\sqrt{2}}\sqrt{(2x-1)-2\sqrt{2x-1}+1}\)
\(=\frac{1}{\sqrt{2}}\sqrt{(\sqrt{2x-1}-1)^2}=\frac{|\sqrt{2x-1}-1|}{\sqrt{2}}\)
e)
\(\sqrt{x+6\sqrt{x-9}}-\sqrt{x-9}=\sqrt{(x-9)+2.3\sqrt{x-9}+3^2}-\sqrt{x-9}\)
\(=\sqrt{(\sqrt{x-9}+3)^2}-\sqrt{x-9}=|\sqrt{x-9}+3|-\sqrt{x-9}\)
\(=\sqrt{x-9}+3-\sqrt{x-9}=3\)
Lời giải:
a)
\(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}=\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}+\frac{\sqrt{3}}{3}-\frac{2(3-\sqrt{3})}{(3+\sqrt{3})(3-\sqrt{3})}\)
\(=\frac{2-\sqrt{3}}{4-3}+\frac{\sqrt{3}}{3}-\frac{2(3-\sqrt{3})}{3^2-3}\)
\(=2-\sqrt{3}+\frac{\sqrt{3}}{3}-\frac{3-\sqrt{3}}{3}=\frac{6-3\sqrt{3}}{3}+\frac{2\sqrt{3}-3}{3}=\frac{3-\sqrt{3}}{3}\)
b)
\(\sqrt{x-3+2\sqrt{x-4}}=\sqrt{(x-4)+2\sqrt{x-4}+1}=\sqrt{(\sqrt{x-4}+1)^2}=|\sqrt{x-4}+1|=\sqrt{x-4}+1\)
c)
\(\sqrt{2x+4\sqrt{2x-4}}=\sqrt{(2x-4)+2.2\sqrt{2x-4}+2^2}\)
\(=\sqrt{(\sqrt{2x-4}+2)^2}=|\sqrt{2x-4}+2|=\sqrt{2x-4}+2\)
d)
\(\sqrt{x-\sqrt{2x-1}}=\frac{1}{\sqrt{2}}\sqrt{2x-2\sqrt{2x-1}}=\frac{1}{\sqrt{2}}\sqrt{(2x-1)-2\sqrt{2x-1}+1}\)
\(=\frac{1}{\sqrt{2}}\sqrt{(\sqrt{2x-1}-1)^2}=\frac{|\sqrt{2x-1}-1|}{\sqrt{2}}\)
e)
\(\sqrt{x+6\sqrt{x-9}}-\sqrt{x-9}=\sqrt{(x-9)+2.3\sqrt{x-9}+3^2}-\sqrt{x-9}\)
\(=\sqrt{(\sqrt{x-9}+3)^2}-\sqrt{x-9}=|\sqrt{x-9}+3|-\sqrt{x-9}\)
\(=\sqrt{x-9}+3-\sqrt{x-9}=3\)
a, \(\sqrt{9x+9}-4\sqrt{\dfrac{x+1}{4}}=5\) \(x\ge-1\)
\(\Leftrightarrow3\sqrt{x+1}-2\sqrt{x+1}=5\)
\(\Leftrightarrow x+1=25\Leftrightarrow x=24\)
2) "biểu thức"=\(\sqrt{x-5}-4\sqrt{x-5}-\sqrt{x-5}=12\Leftrightarrow4\sqrt{x-5}=12\Leftrightarrow\sqrt{x-5}=3\Leftrightarrow x=14\)
Kl: x=14
3) "biểu thức"=\(4\sqrt{x-1}-3\sqrt{x-1}+\sqrt{x-1}=5\Leftrightarrow2\sqrt{x-1}=5\Leftrightarrow\sqrt{x-1}=\dfrac{5}{2}\Leftrightarrow x=\left(\dfrac{5}{2}\right)^2+1=\dfrac{29}{4}\)
Kl: x=29/4
a) \(\sqrt{\left(x-3\right)^2}=3\Leftrightarrow\left|x-3\right|=3\) \(\Leftrightarrow\left[{}\begin{matrix}x-3=3\\x-3=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\left(N\right)\\x=0\left(N\right)\end{matrix}\right.\)
b) \(\sqrt{4x^2-20x+25}+2x=5\Leftrightarrow\left|2x-5\right|+2x-5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-5\ge0\\2x-5+2x-5=0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-5\le0\\5-2x+2x-5=0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\4x-10=0\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{5}{2}\\0x=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\x=\dfrac{10}{4}\left(N\right)\end{matrix}\right.\\x\le\dfrac{5}{2}\end{matrix}\right.\) ** 10/4 = 5/2 rồi**
Kl: x \< 5/2
c) \(\sqrt{1-12x+36x^2}=5\Leftrightarrow\left|1-6x\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}1-6x=5\\1-6x=-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\left(N\right)\\x=1\left(N\right)\end{matrix}\right.\)
Kl: x=-2/3, x=1
d) Đk: x >/ 1
\(\sqrt{x+2\sqrt{x-1}}=2\Leftrightarrow\left|\sqrt{x-1}+1\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}+1=2\left(1\right)\\\sqrt{x-1}+2=-2\left(VN\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{x-1}=1\Leftrightarrow x=2\)(N)
Kl: x=2
e) Đk: x >/ 1
\(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}-1\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}\ge1\\\left|\sqrt{x-1}-1\right|=\sqrt{x-1}-1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{x-1}-1=\sqrt{x-1}-1\) (luôn đúng)
kl: x >/ 1
f) \(\sqrt{x^2-\dfrac{1}{2}x+\dfrac{1}{16}}=\dfrac{1}{4}-x\) \(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{4}\\\left|\dfrac{1}{4}-x\right|=\dfrac{1}{4}-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{4}\\\dfrac{1}{4}-x=\dfrac{1}{4}-x\end{matrix}\right.\)
(luôn đúng)
Kl: x \< 1/4
Lần sau xé nhỏ câu hỏi giùm con nha má, để nhiều thế này thất thu T_T!
a: \(=\dfrac{2x+1-x-\sqrt{x}-1}{x\sqrt{x}-1}=\dfrac{x-\sqrt{x}}{x\sqrt{x}-1}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
b: \(=\dfrac{\sqrt{x}-4+3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
c: \(=\dfrac{x\sqrt{x}+1-\left(x-1\right)\left(\sqrt{x}+1\right)}{x-1}\)
\(=\dfrac{x\sqrt{x}+1-x\sqrt{x}-x+\sqrt{x}+1}{x-1}=\dfrac{-x+\sqrt{x}+2}{x-1}\)
\(=\dfrac{-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{x-1}=\dfrac{-\sqrt{x}+2}{\sqrt{x}-1}\)
1)
ĐKXĐ: x>4
Ta có: \(\dfrac{\sqrt{x+5}}{\sqrt{x-4}}=\dfrac{\sqrt{x-2}}{\sqrt{x+3}}\)
\(\Leftrightarrow x^2+8x+15=x^2-6x+8\)
\(\Leftrightarrow8x+6x=8-15\)
\(\Leftrightarrow14x=-7\)
hay \(x=-\dfrac{1}{2}\)(loại)
2) Ta có: \(\sqrt{4x^2-9}=3\sqrt{2x-3}\)
\(\Leftrightarrow\sqrt{2x-3}\left(\sqrt{2x+3}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)