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ĐKXĐ : \(4\le x\le6\)
Xét \(VP^2=6-x+x-4+2\sqrt{\left(6-x\right)\left(x-4\right)}=2+2\sqrt{\left(6-x\right)\left(x-4\right)}\)
Áp dụng bđt Cauchy ta có : \(2+2\sqrt{\left(6-x\right)\left(x-4\right)}\le2+6-x+x-4=4\)
\(\Rightarrow VP\le2\forall x\)(1)
Xét \(VT=x^2-10x+27=\left(x^2-10x+25\right)+2=\left(x-5\right)^2+2\ge2\forall x\)(2)
Từ (1);(2) \(\Rightarrow VT\ge2\ge VP\)
Dấu "=" xảy ra \(\hept{\begin{cases}6-x=x-4\\\left(x-5\right)^2=0\end{cases}\Rightarrow x=5\left(TMĐKXĐ\right)}\)
Vậy nghiệm pt là x = 5
Cau 1:
a: \(A=\dfrac{\left(\sqrt{a}-2\right)\left(a+2\sqrt{a}+4\right)+2\sqrt{a}\left(\sqrt{a}-2\right)}{a-4}\)
\(=\dfrac{\left(\sqrt{a}-2\right)\left(a+4\sqrt{a}+4\right)}{a-4}=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}=\sqrt{a}+2\)
c: \(=\dfrac{\left|c+1\right|}{\left|c\right|-1}\)
TH1: c>0
\(C=\dfrac{c+1}{c-1}\)
TH2: c<0
\(C=\dfrac{\left|c+1\right|}{-\left(c+1\right)}=\pm1\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}x-3>=0\\5-x>=0\end{matrix}\right.\)
=>3<=x<=5
\(\sqrt{x-3}+\sqrt{5-x}=2\)
=>\(\sqrt{x-3}-1+\sqrt{5-x}-1=0\)
=>\(\dfrac{x-3-1}{\sqrt{x-3}+1}+\dfrac{5-x-1}{\sqrt{5-x}+1}=0\)
=>\(\left(x-4\right)\left(\dfrac{1}{\sqrt{x-3}+1}-\dfrac{1}{\sqrt{5-x}+1}\right)=0\)
=>x-4=0
=>x=4
a, \(\sqrt{2}x-\sqrt{6}=0\Leftrightarrow\sqrt{2}x=\sqrt{6}\Leftrightarrow x=\sqrt{3}\)
b, \(\frac{x^2}{\sqrt{3}}-\sqrt{12}=0\Leftrightarrow\frac{x^2}{\sqrt{3}}=\sqrt{12}\Leftrightarrow x^2=\sqrt{12}.\sqrt{3}\Leftrightarrow x^2=\sqrt{36}\Leftrightarrow x=36\)
c, \(\sqrt{3}x+\sqrt{3}=\sqrt{12}+\sqrt{27}\Leftrightarrow\sqrt{3}x=\sqrt{12}+\sqrt{27}-\sqrt{3}\)
\(\Leftrightarrow\sqrt{3}x=2\sqrt{3}+3\sqrt{3}-\sqrt{3}\Leftrightarrow\sqrt{3}x=4\sqrt{3}\Leftrightarrow x=4\)
a.
\(\sqrt{4x^2+4x+1}-\sqrt{25x^2+10x+1}=0\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}-\sqrt{\left(5x+1\right)^2}=0\)
\(\Leftrightarrow2x+1-\left(5x+1\right)=0\)
\(\Leftrightarrow-3x=0\Leftrightarrow x=0\)
b.
\(\sqrt{x^4-16x^2+64}=\sqrt{25x^2+10x+1}\)
\(\Leftrightarrow\sqrt{\left(x^2-8\right)^2}=\sqrt{\left(5x+1\right)^2}\)
\(\Leftrightarrow x^2-8=5x+1\)
\(\Leftrightarrow x^2-5x+\dfrac{25}{4}=\dfrac{61}{4}\)
\(\Leftrightarrow\left(x-\dfrac{5}{2}\right)^2=\dfrac{61}{4}\)
............................
tương tự ..
c: \(\Leftrightarrow\sqrt{x-5}\left(\sqrt{x+5}-1\right)=0\)
=>x-5=0 hoặc x+5=1
=>x=-4 hoặc x=5
d: \(\Leftrightarrow\sqrt{2x+3}\left(\sqrt{2x-3}-2\right)=0\)
=>2x+3=0 hoặc 2x-3=4
=>x=7/2 hoặc x=-3/2
e: \(\Leftrightarrow\sqrt{x-2}\left(1-3\sqrt{x+2}\right)=0\)
=>x-2=0 hoặc 3 căn x+2=1
=>x=2 hoặc x+2=1/9
=>x=-17/9 hoặc x=2
a: ĐKXĐ: x>=0
b: \(\Leftrightarrow\dfrac{2\sqrt{2}-2\sqrt{2-\sqrt{x}}+\sqrt{2x}-\sqrt{x\left(2-\sqrt{x}\right)}+2\sqrt{2}+2\sqrt{2+\sqrt{x}}-\sqrt{2x}-\sqrt{x\left(2+\sqrt{x}\right)}}{2-2+\sqrt{x}}=\sqrt{2}\)
\(\Leftrightarrow4\sqrt{2}-2\sqrt{x\left(\sqrt{x}+2\right)}=\sqrt{2x}\)
\(\Leftrightarrow\sqrt{4x\left(\sqrt{x}+2\right)}=4\sqrt{2}-\sqrt{2x}\)
\(\Leftrightarrow4x\left(\sqrt{x}+2\right)=32-16\sqrt{x}+2x\)
\(\Leftrightarrow4x\sqrt{x}+8x-32+16\sqrt{x}-2x=0\)
=>\(x\in\left\{0;1.2996\right\}\)