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HP
20 tháng 9 2021
\(\sqrt{x+8}-\sqrt{5x+2}+2=0\)
<=> \(\sqrt{x+8}=\sqrt{5x+2}-2\)
<=> x + 8 = \(\left(\sqrt{5x+2}-2\right)^2\)
<=> x + 8 = \(\left(\sqrt{5x+2}\right)^2-4\sqrt{5x+2}+4\)
<=> x + 8 = 5x + 2 - \(4\sqrt{5x+2}+4\)
<=> \(4\sqrt{5x+2}=5x+2+4-x-8\)
<=> \(4\sqrt{5x+2}=4x-2\)
<=> \(4\sqrt{5x+2}=2\left(2x-1\right)\)
<=> \(\sqrt{5x+2}=\dfrac{2\left(2x-1\right)}{4}\)
<=> \(\sqrt{5x+2}=\dfrac{2x-1}{2}\)
<=> 5x + 2 = \(\dfrac{\left(2x-1\right)^2}{4}\)
<=> x = \(\dfrac{\dfrac{\left(2x-1\right)^2}{4}-2}{5}\)
<=> x = -0,278.....
28 tháng 11 2019
Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
21 tháng 9 2021
\(ĐK:x\ge-4\\ \Leftrightarrow\sqrt{x+8}=\sqrt{5x+20}-2\\ \Leftrightarrow x+8=5x+20+4-4\sqrt{5x+20}\\ \Leftrightarrow4\sqrt{5x+20}=4x+16\\ \Leftrightarrow\left(\sqrt{5x+20}\right)^2=\left[4\left(x+4\right)\right]^2\\ \Leftrightarrow5x+20=16\left(x^2+8x+64\right)\\ \Leftrightarrow5x+20=16x^2+128x+1024\\ \Leftrightarrow16x^2+123x+1004=0\\ \Leftrightarrow\left(16x^2+2\cdot4x\cdot\dfrac{123}{8}+\dfrac{15129}{64}\right)+\dfrac{49127}{64}=0\\ \Leftrightarrow\left(4x+\dfrac{123}{8}\right)^2+\dfrac{49127}{64}=0\Leftrightarrow x\in\varnothing\)
17 tháng 12 2023
a: ĐKXĐ: \(x\in R\)
\(\sqrt{\left(2x+3\right)^2}=5\)
=>|2x+3|=5
=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
b: ĐKXĐ: \(x\in R\)
\(\sqrt{9\left(x-2\right)^2}=18\)
=>\(\sqrt{9}\cdot\sqrt{\left(x-2\right)^2}=18\)
=>\(3\cdot\left|x-2\right|=18\)
=>\(\left|x-2\right|=6\)
=>\(\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
c: ĐKXĐ: x>=2
\(\sqrt{9x-18}-\sqrt{4x-8}+3\sqrt{x-2}=40\)
=>\(3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)
=>\(4\sqrt{x-2}=40\)
=>\(\sqrt{x-2}=10\)
=>x-2=100
=>x=102(nhận)
d: ĐKXĐ: \(x\in R\)
\(\sqrt{4\left(x-3\right)^2}=8\)
=>\(\sqrt{\left(2x-6\right)^2}=8\)
=>|2x-6|=8
=>\(\left[{}\begin{matrix}2x-6=8\\2x-6=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=14\\2x=-2\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)
e: ĐKXĐ: \(x\in R\)
\(\sqrt{4x^2+12x+9}=5\)
=>\(\sqrt{\left(2x\right)^2+2\cdot2x\cdot3+3^2}=5\)
=>\(\sqrt{\left(2x+3\right)^2}=5\)
=>|2x+3|=5
=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
f: ĐKXĐ:x>=6/5
\(\sqrt{5x-6}-3=0\)
=>\(\sqrt{5x-6}=3\)
=>\(5x-6=3^2=9\)
=>5x=6+9=15
=>x=15/5=3(nhận)
13 tháng 7 2021
`a)sqrt{5x-2}=3(x>=2/5)`
`<=>5x-2=9`
`<=>5x=11`
`<=>x=11/5(tm)`
`b)sqrt{x^2-4x+4}-5=0`
`<=>\sqrt{(x-2)^2}=5`
`<=>|x-2|=5`
`<=>` \(\left[ \begin{array}{l}x-2=5\\x-2=-5\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=7\\x=-3\end{array} \right.\)
`c)3sqrt{4x+8}-sqrt{9x+18}+9sqrt{(x+2)/9}=sqrt{72}(x>=-2)`
`<=>6sqrt{x+2}-3sqrt{x+2}+3sqrt{x+2}=sqrt{72}`
`<=>6sqrt{x+2}=6sqrt2`
`<=>sqrt{x+2}=sqrt2`
`<=>x+2=2`
`<=>x=0(tm)`
BA
13 tháng 7 2021
\(a,ĐK:x\ge\dfrac{2}{5}\)
\(\Leftrightarrow5x-2=9\)
\(\Leftrightarrow5x=11\)
\(\Leftrightarrow x=\dfrac{11}{5}\)
\(b,\)
\(\Leftrightarrow x^2-5x+4=25\)
\(\Leftrightarrow x^2-5x-21=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5+\sqrt{109}}{2}\\x=\dfrac{5-\sqrt{109}}{2}\end{matrix}\right.\)
\(c,\)
\(\Leftrightarrow6\sqrt{x+2}-3\sqrt{x+2}+9\cdot\sqrt{\dfrac{x+2}{9}}=6\sqrt{2}\)
\(\Leftrightarrow2\sqrt{x+2}-\sqrt{x+2}+3\cdot\sqrt{\dfrac{x+2}{9}}=2\sqrt{2}\)
Đặt \(\sqrt{x+2}=a\) ta có (1)
\(2a-a+3\cdot\dfrac{a}{\sqrt{9}}=2\sqrt{2}\)
\(\Leftrightarrow a+3\cdot\dfrac{a}{3}=2\sqrt{2}\)
\(\Leftrightarrow2a=2\sqrt{2}\)
\(\Leftrightarrow a=\sqrt{2}\)
Thay \(a=\sqrt{2}\) vào (1) ta có
\(\sqrt{x+2}=\sqrt{2}\)
\(\Leftrightarrow x+2=2\)
\(\Leftrightarrow x=0\)
27 tháng 6 2018
a) \(x^2-\sqrt{2}x+\sqrt{5}x-\sqrt{10}=0\)
\(\Leftrightarrow x\left(x-\sqrt{2}\right)+\sqrt{5}\left(x-\sqrt{2}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2}\right)\left(x+\sqrt{5}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{2}=0\\x+\sqrt{5}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{5}\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
3 tháng 8 2021
a. ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow -5x-5\sqrt{x}+12\sqrt{x}+12=0$
$\Leftrightarrow -5\sqrt{x}(\sqrt{x}+1)+12(\sqrt{x}+1)=0$
$\Leftrightarrow (\sqrt{x}+1)(12-5\sqrt{x})=0$
Dễ thấy $\sqrt{x}+1>1$ với mọi $x\geq 0$ nên $12-5\sqrt{x}=0$
$\Leftrightarrow \sqrt{x}=\frac{12}{5}$
$\Leftrightarrow x=5,76$ (thỏa mãn)
AH
Akai Haruma
Giáo viên
3 tháng 8 2021
b. ĐKXĐ: $x^2\geq 5$
PT $\Leftrightarrow \frac{1}{3}\sqrt{4}.\sqrt{x^2-5}+2\sqrt{\frac{1}{9}}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$
$\Leftrightarrow \frac{2}{3}\sqrt{x^2-5}+\frac{2}{3}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$
$\Leftrightarrow -\frac{5}{3}\sqrt{x^2-5}=0$
$\Leftrightarrow \sqrt{x^2-5}=0$
$\Leftrightarrow x=\pm \sqrt{5}$
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)
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9\left(x-2\right)}+6\sqrt{\dfrac{1}{81}\left(x-2\right)}=-4\)
\(\dfrac{1}{3}\sqrt{x-2}-2\sqrt{x-2}+\dfrac{2}{3}\sqrt{x-2}=-4\)
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{4}{3}\sqrt{x-2}=-4\)
\(-\sqrt{x-2}=-4\)
\(\sqrt{x-2}=4\)
\(\left|x-2\right|=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=16\\x-2=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=18\left(TM\right)\\x=-14\left(L\right)\end{matrix}\right.\)
ĐKXĐ: \(x\ge2\)
\(\left(x^2-6x+9\right)+\left(x-2-2\sqrt{x-2}+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(\sqrt{x-2}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\\sqrt{x-2}-1=0\end{matrix}\right.\)
\(\Leftrightarrow x=3\)