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20 tháng 1 2017
2a^4=(1-a)^2=a^2-2a+1
\(A=\frac{2a-3}{\sqrt{2\left(a^2-4a+4\right)}+2a^2}=\frac{2a-3}{\sqrt{2}!\left(a-2\right)!+2a^2}\)a> 2 không thể là nghiệm=> a<2
\(A=\frac{2a-3}{\sqrt{2}\left(2-a\right)+2a^2}=\frac{2a-3}{2a^2-\sqrt{2}a+2\sqrt{2}}=\frac{2a-3}{\sqrt{2}\left(\sqrt{2}a^2-a-1+3\right)}\)
\(A=\frac{2a-3}{\sqrt{2}\left(3\right)}\)
NT
1
28 tháng 10 2021
\(\Leftrightarrow\sqrt{x+4}\left(\sqrt{x-4}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x-4=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=8\end{matrix}\right.\)
MA
1
11 tháng 7 2018
a) \(\sqrt{x^2-16}-3\sqrt{x-4}=0\)
\(\Leftrightarrow\sqrt{x^2-16}=3\sqrt{x-4}\)
\(\Leftrightarrow\sqrt{x^2-16}=\sqrt{9x-36}\)
\(\Leftrightarrow x^2-16=9x-36\)
\(\Leftrightarrow\left(x-4\right)\left(x+4\right)-9x+36=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+4\right)-9\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
vậy ...
5 tháng 7 2021
a) Pt \(\Leftrightarrow\sqrt{\left(x-2\right)^2}=5\Leftrightarrow\left|x-2\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
Vậy...
b)Đk: \(x\ge-1\)
Pt \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}\)
\(\Leftrightarrow4\sqrt{x+1}=16\)\(\Leftrightarrow x+1=16\)\(\Leftrightarrow x=15\) (tm)
Vậy...
\(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\) (a>0)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=a+\sqrt{a}-\left(2\sqrt{a}+1\right)+1=a-\sqrt{a}\)
b) \(A=a-\sqrt{a}=a-2.\dfrac{1}{2}\sqrt{a}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
Dấu "=" xảy ra khi \(\sqrt{a}=\dfrac{1}{2}\Leftrightarrow a=\dfrac{1}{4}\left(tmđk\right)\)
Vậy \(A_{min}=-\dfrac{1}{4}\)
AT
5 tháng 7 2021
a) \(\sqrt{x^2-4x+4}=5\Rightarrow\sqrt{\left(x-2\right)^2}=5\Rightarrow\left|x-2\right|=5\)
\(\Rightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
b) \(\sqrt{16x+16}-3\sqrt{x+1}+\sqrt{4x+4}=16-\sqrt{x+1}\)
\(\Rightarrow\sqrt{16\left(x+1\right)}-3\sqrt{x+1}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Rightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Rightarrow4\sqrt{x+1}=16\Rightarrow\sqrt{x+1}=4\Rightarrow x=15\)
a) \(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=a+\sqrt{a}-2\sqrt{a}-1+1=a-\sqrt{a}\)
b) Ta có: \(a-\sqrt{a}=\left(\sqrt{a}\right)^2-2.\sqrt{a}.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)
\(=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
\(\Rightarrow A_{min}=-\dfrac{1}{4}\) khi \(a=\dfrac{1}{4}\)
BC
1
28 tháng 9 2021
\(\sqrt{x+6-4\sqrt{x+2}}-\sqrt{9-4\sqrt{5}}=0\left(đk:x\ge-2\right)\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x+2}-2\right)^2}=\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(\Leftrightarrow\left|\sqrt{x+2}-2\right|=\left|\sqrt{5}-2\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+2}-2=\sqrt{5}-2\\\sqrt{x+2}-2=2-\sqrt{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=5\\x+2=21-8\sqrt{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=19-8\sqrt{5}\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{3;19-8\sqrt{5}\right\}\)
21 tháng 7 2018
ĐKXĐ:\(\hept{\begin{cases}x-2>0\\y-1>0\\z-5>0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x>2\\y>1\\z>5\end{cases}}\)
pt\(\Leftrightarrow\frac{4}{\sqrt{x-2}}+\frac{1}{\sqrt{y-1}}+\frac{25}{\sqrt{z-5}}+\sqrt{x-2}+\sqrt{y-1}+\sqrt{z-5}=16\)
Áp dụng BĐT Cauchy:
\(\frac{4}{\sqrt{x-2}}+\sqrt{x-2}+\frac{1}{\sqrt{y-1}}+\sqrt{y-1}+\frac{25}{\sqrt{z-5}}+\sqrt{z-5}\)
\(\ge2\sqrt{\frac{4}{\sqrt{x-2}}.\sqrt{x-2}}+2\sqrt{\frac{1}{\sqrt{y-1}}.\sqrt{y-1}}+2\sqrt{\frac{25}{\sqrt{z-5}}.\sqrt{z-5}}\)
\(=2\sqrt{4}+2\sqrt{1}+2\sqrt{25}=2.2+2.1+2.5\)
\(=4+2+10=16\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}x-2=4\\y-1=1\\z-5=25\end{cases}}\Leftrightarrow\hept{\begin{cases}x=6\\y=2\\z=30\end{cases}}\)
28 tháng 9 2021
1) \(\Leftrightarrow\sqrt{\left(x+5\right)^2}=4\)
\(\Leftrightarrow\left|x+5\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=4\\x+5=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-9\end{matrix}\right.\)
2) \(ĐK:x\ge2\)
\(\Leftrightarrow\sqrt{x-2}=2\)
\(\Leftrightarrow x-2=4\Leftrightarrow x=6\left(tm\right)\)
3) \(\Leftrightarrow\left(x^2-x+4\right)-\sqrt{x^2-x+4}+\dfrac{1}{4}=\dfrac{9}{4}\)
\(\Leftrightarrow\left(\sqrt{x^2-x+4}-\dfrac{1}{2}\right)^2=\dfrac{9}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}-\dfrac{1}{2}=\dfrac{3}{2}\\\sqrt{x^2-x+4}-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}=2\\\sqrt{x^2-x+4}=-1\left(VLý\right)\end{matrix}\right.\)
\(\Leftrightarrow x^2-x+4=4\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
4) \(ĐK:x\ge0\)
\(\Leftrightarrow3\sqrt{x}-3=\sqrt{x}+2\)
\(\Leftrightarrow\sqrt{x}=\dfrac{5}{2}\Leftrightarrow x=\dfrac{25}{4}\left(tm\right)\)
NA
12 tháng 8 2021
a,ĐK: x\(\ge\)1
⇔\(\sqrt{x-1-2\sqrt{x-1}+1}\)=\(\sqrt{2}\)
⇔\(\sqrt{\left(\sqrt{x-1}-1\right)^2}\)=\(\sqrt{2}\)
⇔\(\left|\sqrt{x-1}-1\right|\)=\(\sqrt{2}\)
TH1:\(\sqrt{x-1}\)-1≥0⇒\(\left|\sqrt{x-1}-1\right|\)=\(\sqrt{x-1}\)-1 bn tự giải ra nha
TH2:\(\sqrt{x-1}\)-1<0⇒\(\left|\sqrt{x-1}-1\right|\)=1-\(\sqrt{x-1}\) bn tự lm nha