\(\sqrt{x+8}-\sqrt{5x+20}+2=0\)
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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)
b. ĐKXĐ: $x^2\geq 5$
PT $\Leftrightarrow \frac{1}{3}\sqrt{4}.\sqrt{x^2-5}+2\sqrt{\frac{1}{9}}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$
$\Leftrightarrow \frac{2}{3}\sqrt{x^2-5}+\frac{2}{3}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$
$\Leftrightarrow -\frac{5}{3}\sqrt{x^2-5}=0$
$\Leftrightarrow \sqrt{x^2-5}=0$
$\Leftrightarrow x=\pm \sqrt{5}$
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: \(x\ge-4\)
\(\Leftrightarrow\sqrt{x+8}+2=\sqrt{5x+20}\)
\(\Leftrightarrow x+12+4\sqrt{x+8}=5x+20\)
\(\Leftrightarrow\sqrt{x+8}=x+2\left(x\ge-2\right)\)
\(\Leftrightarrow x+8=x^2+4x+4\)
\(\Leftrightarrow x^2+3x-4=0\Rightarrow\left[{}\begin{matrix}x=1\\x=-4\left(ktm\right)\end{matrix}\right.\)
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2:
a: =>2x^2-4x-2=x^2-x-2
=>x^2-3x=0
=>x=0(loại) hoặc x=3
b: =>(x+1)(x+4)<0
=>-4<x<-1
d: =>x^2-2x-7=-x^2+6x-4
=>2x^2-8x-3=0
=>\(x=\dfrac{4\pm\sqrt{22}}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,ĐK:x\ge-7\\ PT\Leftrightarrow\sqrt{\left(\sqrt{x+7}+1\right)^2}+\sqrt{x+7-\sqrt{x+7}-6}=4\)
Đạt \(\sqrt{x+7}=a\ge0\)
\(PT\Leftrightarrow\sqrt{\left(a+1\right)^2}+\sqrt{a^2-a-6}=4\\ \Leftrightarrow a+1+\sqrt{a^2-a-6}=4\\ \Leftrightarrow\sqrt{a^2-a-6}=3-a\\ \Leftrightarrow a^2-a-6=a^2-6a+9\\ \Leftrightarrow5a=15\Leftrightarrow a=3\\ \Leftrightarrow\sqrt{x+7}=3\\ \Leftrightarrow x+7=9\\ \Leftrightarrow x=2\left(tm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tử số của phân số đầu phải là \(\sqrt{x}+2\) chứ không phải \(\sqrt{x+2}\), vì cái \(\sqrt{x}+2\) nó mới logic để rút gọn: )
\(Q=\left(\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}^3+8}-\dfrac{x-\sqrt{x}}{\sqrt{x}^3+8}\right)\left(\dfrac{5x-10\sqrt{x}+20}{5\sqrt{x}+4}\right)\\ =\left(\dfrac{x+4\sqrt{x}+4-x+\sqrt{x}}{\sqrt{x}^3+8}\right)\left(\dfrac{5x-10\sqrt{x}+20}{5\sqrt{x}+4}\right)\\ =\dfrac{\left(5\sqrt{x}+4\right).5.\left(x-2\sqrt{x}+4\right)}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)\left(5\sqrt{x}+4\right)}\\ =\dfrac{5}{\sqrt{x}+2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ĐK: \(x\ge5\)
\(\sqrt{4x-20}+\frac{1}{3}\sqrt{9x-45}-\frac{1}{5}\sqrt{16x-80}=0\)
\(\Leftrightarrow\)\(\sqrt{4\left(x-5\right)}+\frac{1}{3}\sqrt{9\left(x-5\right)}-\frac{1}{5}\sqrt{16\left(x-5\right)}=0\)
\(\Leftrightarrow\)\(2\sqrt{x-5}+\sqrt{x-5}-\frac{4}{5}\sqrt{x-5}=0\)
\(\Leftrightarrow\)\(\frac{11}{5}\sqrt{x-5}=0\)
\(\Leftrightarrow\)\(x-5=0\)
\(\Leftrightarrow\)\(x=5\) (t/m)
Vậy
b) \(-5x+7\sqrt{x}=-12\)
\(\Leftrightarrow\)\(5x-7\sqrt{x}-12=0\)
\(\Leftrightarrow\)\(\left(\sqrt{x}+1\right)\left(5\sqrt{x}-12\right)=0\)
đến đây tự làm
c) d) e) bạn bình phương lên
f) \(VT=\sqrt{3\left(x^2+2x+1\right)+9}+\sqrt{5\left(x^4-2x^2+1\right)+25}\)
\(=\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2}\)
\(\ge\sqrt{9}+\sqrt{25}=8\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}x+1=0\\x^2-1=0\end{cases}}\)\(\Leftrightarrow\)\(x=-1\)
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{x+8}-\sqrt{5x+2}+2=0\)
<=> \(\sqrt{x+8}=\sqrt{5x+2}-2\)
<=> 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.....
\(ĐK:x\ge-4\\ \Leftrightarrow\sqrt{x+8}=\sqrt{5x+20}-2\\ \Leftrightarrow x+8=5x+20+4-4\sqrt{5x+20}\\ \Leftrightarrow4\sqrt{5x+20}=4x+16\\ \Leftrightarrow\left(\sqrt{5x+20}\right)^2=\left[4\left(x+4\right)\right]^2\\ \Leftrightarrow5x+20=16\left(x^2+8x+64\right)\\ \Leftrightarrow5x+20=16x^2+128x+1024\\ \Leftrightarrow16x^2+123x+1004=0\\ \Leftrightarrow\left(16x^2+2\cdot4x\cdot\dfrac{123}{8}+\dfrac{15129}{64}\right)+\dfrac{49127}{64}=0\\ \Leftrightarrow\left(4x+\dfrac{123}{8}\right)^2+\dfrac{49127}{64}=0\Leftrightarrow x\in\varnothing\)