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24 tháng 11 2021
\(a,\Leftrightarrow x-1=4\Leftrightarrow x=5\\ b,\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\3x+1=4x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\left(tm\right)\end{matrix}\right.\Leftrightarrow x=4\\ c,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=3\\ \Leftrightarrow x+5=9\\ \Leftrightarrow x=4\left(tm\right)\)
\(d,\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{5}+1\right)^2}\\ \Leftrightarrow\left|x-2\right|=\sqrt{5}+1\\ \Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{5}+1\\2-x=\sqrt{5}+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}+3\\x=1-\sqrt{5}\end{matrix}\right.\)
30 tháng 10 2023
a: ĐKXĐ: \(x\in R\)
\(\sqrt{x^2-4x+4}=7\)
=>\(\sqrt{\left(x-2\right)^2}=7\)
=>|x-2|=7
=>\(\left[{}\begin{matrix}x-2=7\\x-2=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-5\end{matrix}\right.\)
b: ĐKXĐ: x>=-3
\(\sqrt{4x+12}-3\sqrt{x+3}+\dfrac{4}{3}\cdot\sqrt{9x+27}=6\)
=>\(2\sqrt{x+3}-3\sqrt{x+3}+\dfrac{4}{3}\cdot3\sqrt{x+3}=6\)
=>\(3\sqrt{x+3}=6\)
=>\(\sqrt{x+3}=2\)
=>x+3=4
=>x=1(nhận)
VC
3
27 tháng 8 2021
`sqrt{4x+20}-3sqrt{5+x}+4/3sqrt{9x+15}=6(x>=-5)`
`<=>sqrt{4(x+5)}-3sqrt{x+5}+4/3sqrt{9(x+5)}=6`
`<=>2sqrt{x+5}-3sqrt{x+5}+4sqrt{x+5}=6`
`<=>3sqrt{x+5}=6`
`<=>sqrt{x+5}=2`
`<=>x+5=4`
`<=>x=-1(tm)`
Vậy `x=-1`
5 tháng 8 2021
a) Ta có: \(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)
\(\Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)
\(\Leftrightarrow4\sqrt{x-3}=20\)
\(\Leftrightarrow x-3=25\)
hay x=28
b) Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)
\(\Leftrightarrow2\sqrt{x+2}=6\)
\(\Leftrightarrow x+2=9\)
hay x=7
NP
2
10 tháng 7 2021
Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow\sqrt{x+5}=2\)
\(\Leftrightarrow x+5=4\)
hay x=-1
MN
10 tháng 7 2021
\(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\cdot\sqrt{9x+45}=6\left(x\ge-5\right)\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+\dfrac{4}{3}\cdot3\cdot\sqrt{x+5}=6\)
\(\Leftrightarrow3\cdot\sqrt{x+5}=6\)
\(\Leftrightarrow\sqrt{x+5}=2\)
\(\Leftrightarrow x+5=2^2=4\)
\(\Leftrightarrow x=-1\left(N\right)\)
6 tháng 2 2022
a: \(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot3\sqrt{x-2}+6\cdot\dfrac{\sqrt{x-2}}{9}=-4\)
\(\Leftrightarrow\sqrt{x-2}=4\)
=>x-2=16
hay x=18
b: \(\Leftrightarrow\left|3x+2\right|=4x\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=4x\left(x>=-\dfrac{2}{3}\right)\\3x+2=-4x\left(x< -\dfrac{2}{3}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-\dfrac{2}{7}\left(nhận\right)\end{matrix}\right.\)
c: \(\Leftrightarrow3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)
\(\Leftrightarrow4\sqrt{x-2}=40\)
=>x-2=100
hay x=102
d: =>5x-6=9
hay x=3
MH
6 tháng 2 2022
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\) (đk: x≥2)
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9\left(x-2\right)}+6\sqrt{\dfrac{1}{81}\left(x-2\right)}=-4\)
\(\dfrac{1}{3}\sqrt{x-2}-2\sqrt{x-2}+\dfrac{2}{3}\sqrt{x-2}=-4\)
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{4}{3}\sqrt{x-2}=-4\)
\(-\sqrt{x-2}=-4\)
\(\sqrt{x-2}=4\)
\(\left|x-2\right|=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=16\\x-2=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=18\left(TM\right)\\x=-14\left(L\right)\end{matrix}\right.\)
VP
7 tháng 5 2018
a)\(\sqrt{4x}< =10\)
<=> 4x <= 100
<=> x <= 25
b) \(\sqrt{9x}>=3\)
<=> 9x >= 9
<=> x >= 1
c) \(\sqrt{4x^2+4x+1}=6\)
<=>\(\sqrt{\left(2x\right)^2+2\left(2x\right).1+1^2}=6\)
<=>\(\sqrt{\left(2x+1\right)^2}=6\)
<=>\(|2x+1|=6\)
<=>\(\orbr{\begin{cases}2x+1=6\\2x+1=-6\end{cases}}\)
<=>\(\orbr{\begin{cases}2x=5\\2x=-7\end{cases}}\)
<=>\(\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{-7}{2}\end{cases}}\)
d)\(\sqrt{9x-9}-2\sqrt{x-1}=6\)
<=>\(\sqrt{9\left(x-1\right)}-2\sqrt{x-1}=6\)
<=>\(3\sqrt{x-1}-2\sqrt{x-1}=6\)
<=>\(\sqrt{x-1}=6\)
<=> x - 1 = 36
<=> x = 37
f) \(\sqrt{2x+1}=\sqrt{x-1}\)
<=> 2x + 1 = x -1
<=> 2x - x = -1 -1
<=> x = -2
g)\(\sqrt{x^2-x-1}=\sqrt{x-1}\)
<=>x2 -x -1 = x -1
<=> x2 -x-x-1+1 = 0
<=> x2 - 2x + 0 = 0
<=> x(x-2) = 0
<=>\(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
<=>\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
PQ
2
28 tháng 10 2021
\(a,ĐK:x\ge-5\\ A=2\sqrt{x+5}-2\sqrt{x+5}+3\sqrt{x+5}=3\sqrt{x+5}\\ b,A=6\Leftrightarrow\sqrt{x+5}=\dfrac{6}{3}=2\\ \Leftrightarrow x+5=4\\ \Leftrightarrow x=-1\left(tm\right)\)
ĐKXĐ: x 0
√x - √(4x) + √(9x) = 6
√x - 2√x + 3√x = 6
2√x = 6
√x = 6 : 2
√x = 3
x = 9 (nhận)
Vậy x = 9