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4 tháng 5 2021

đk: \(5\ge x\ge-5\)

Ta có: \(x^2=\sqrt{5-x}+\sqrt{5+x}+12\)

\(\Leftrightarrow\left(x^2-16\right)-\left(\sqrt{5-x}-3\right)-\left(\sqrt{5+x}-1\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+4\right)-\frac{5-x-9}{\sqrt{5-x}+3}-\frac{5+x-1}{\sqrt{5+x}+1}=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+4\right)+\frac{x+4}{\sqrt{5-x}+3}-\frac{x+4}{\sqrt{5+x}+1}=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-4+\frac{1}{\sqrt{5-x}+3}-\frac{1}{\sqrt{5+x}+1}\right)=0\)

Nếu x + 4 = 0 => x = -4 (tm)

Nếu \(x-4+\frac{1}{\sqrt{5-x}+3}-\frac{1}{\sqrt{5+x}+1}=0\)

TH1: \(x=4\left(tm\right)\)

TH2: \(x>4\)

Khi đó: \(\hept{\begin{cases}\sqrt{5-x}+3< 1+3=4\\\sqrt{5+x}+1>3+1=4\end{cases}}\Rightarrow\frac{1}{\sqrt{5-x}+3}>\frac{1}{4}>\frac{1}{\sqrt{5+x}+1}\)

\(\Rightarrow x-4+\frac{1}{\sqrt{5-x}+3}-\frac{1}{\sqrt{5+x}+1}>0\)

TH3: \(x< 4\)

Khi đó: \(\hept{\begin{cases}\sqrt{5-x}+3>4\\\sqrt{5+x}+1< 4\end{cases}}\Rightarrow\frac{1}{\sqrt{5-x}+3}< \frac{1}{4}< \frac{1}{\sqrt{5+x}+1}\)

\(\Rightarrow x-4+\frac{1}{\sqrt{5-x}+3}-\frac{1}{\sqrt{5+x}+1}< 0\)

Vậy tập nghiệm của PT \(S=\left\{-4;4\right\}\)

27 tháng 8 2023

giúp mình với

30 tháng 8 2020

Bài làm:

Ta có: \(\sqrt{7+\sqrt{2x}}=3+\sqrt{5}\)

\(\Leftrightarrow7+\sqrt{2x}=\left(3+\sqrt{5}\right)^2\)

\(\Leftrightarrow7+\sqrt{2x}=14+6\sqrt{5}\)

\(\Leftrightarrow\sqrt{2x}=7+6\sqrt{5}\)

\(\Leftrightarrow2x=\left(7+6\sqrt{5}\right)^2\)

\(\Leftrightarrow2x=229+84\sqrt{5}\)

\(\Rightarrow x=\frac{229+84\sqrt{5}}{2}\)

8 tháng 10 2021

\(a,ĐK:x\le\dfrac{5}{3}\\ PT\Leftrightarrow-3x+5=49\\ \Leftrightarrow x=-\dfrac{44}{3}\left(tm\right)\\ b,ĐK:x\ge-12\\ PT\Leftrightarrow\dfrac{1}{2}x+6=2\\ \Leftrightarrow\dfrac{1}{2}x=-4\\ \Leftrightarrow x=-8\left(tm\right)\\ c,ĐK:x\ge-\dfrac{1}{2}\\ PT\Leftrightarrow2x+1=13+4\sqrt{3}\\ \Leftrightarrow x=\dfrac{12+4\sqrt{3}}{2}=6+2\sqrt{3}\left(tm\right)\\ d,PT\Leftrightarrow\left|3x-1\right|=8\Leftrightarrow\left[{}\begin{matrix}3x-1=8\\1-3x=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{7}{3}\end{matrix}\right.\)

16 tháng 2 2022

\(a,A=\left(\dfrac{x+14\sqrt{x}-5}{x-25}+\dfrac{\sqrt{x}}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}-5}\)

\(\Rightarrow A=\left(\dfrac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\left(\dfrac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{x-5\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\dfrac{x+14\sqrt{x}-5+x-5\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\dfrac{2x+9\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\dfrac{2x+10\sqrt{x}-\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)

\(\Rightarrow A=\dfrac{2\sqrt{x}\left(\sqrt{x}+5\right)-\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)

\(\Rightarrow A=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)

\(\Rightarrow A=\dfrac{2\sqrt{x}-1}{\sqrt{x}+2}\)

14 tháng 7 2021

Bài 1 : 

\(a.\sqrt{x^2-1}\)

\(ĐK:\)

\(x^2-1\ge0\)

\(\Leftrightarrow x^2\ge1\)

\(\Leftrightarrow\left[{}\begin{matrix}x\le-1\\x\ge1\end{matrix}\right.\)

Bài 2 : 

\(2\cdot\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{48}-5\sqrt{50}\)

\(=2\cdot\left|\sqrt{2}-3\right|+4\sqrt{3}-25\sqrt{2}\)

\(=-2\cdot\left(\sqrt{2}-3\right)+4\sqrt{3}-25\sqrt{2}\)

\(=-2\sqrt{2}-6+4\sqrt{3}-25\sqrt{2}\)

\(=-27\sqrt{2}-6+4\sqrt{3}\)

14 tháng 7 2021

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27 tháng 7 2016

a) \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=1+\sqrt{2}\)

b)\(\frac{x-4}{2\left(\sqrt{x}+2\right)}\) (ĐK:x\(\ge0\))

\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{2\left(\sqrt{x}+2\right)}\)

\(=\frac{\sqrt{x}-2}{2}\)

c)\(\frac{x-5\sqrt{x}+6}{3\sqrt{x}-6}\) (ĐK:x\(\ge0;x\ne4\))

\(=\frac{x-3\sqrt{x}-2\sqrt{x}+6}{3\left(\sqrt{x}-2\right)}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)-2\left(\sqrt{x}-3\right)}{3\left(\sqrt{x}-2\right)}\)

\(=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{3\left(\sqrt{x}-2\right)}\)

\(=\frac{\sqrt{x}-3}{3}\)

27 tháng 7 2016

b) Tử \(x-4=\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\) (hằng đăngt thức số 3 )

22 tháng 8 2021

bạn viết lại đề đi

22 tháng 8 2021

\(\sqrt{x+2-4\sqrt{x-2}}+\sqrt{x+7-6\sqrt{x-2}}=5\)

dạ đây ạ