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a:
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=3\)
=>|x-3|=3
=>x-3=3 hoặc x-3=-3
=>x=0 hoặc x=6
b: \(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}=2\)
=>\(\sqrt{\left(\sqrt{x-1}+1\right)^2}=2\)
=>\(\left|\sqrt{x-1}+1\right|=2\)
=>\(\left[{}\begin{matrix}\sqrt{x-1}+1=2\\\sqrt{x-1}+1=-2\left(loại\right)\end{matrix}\right.\Leftrightarrow\sqrt{x-1}=1\)
=>x-1=1
=>x=2
c:
ĐKXĐ: x>4/5
PT \(\Leftrightarrow\sqrt{\dfrac{5x-4}{x+2}}=2\)
=>\(\dfrac{5x-4}{x+2}=4\)
=>5x-4=4x+8
=>x=12(nhận)
d: ĐKXĐ: x-4>=0 và x+1>=0
=>x>=4
PT =>\(\left(\sqrt{x-4}+\sqrt{x+1}\right)^2=5^2=25\)
=>\(x-4+x+1+2\sqrt{\left(x-4\right)\left(x+1\right)}=25\)
=>\(\sqrt{4\left(x^2-3x-4\right)}=25-2x+3=28-2x\)
=>\(\sqrt{x^2-3x-4}=14-x\)
=>x<=14 và x^2-3x-4=(14-x)^2=x^2-28x+196
=>x<=14 và -3x-4=-28x+196
=>x<=14 và 25x=200
=>x=8(nhận)
a) \(\sqrt{x^2-6x+9}=3\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=3\)
\(\Leftrightarrow\left|x-3\right|=3 \)
TH1: \(\left|x-3\right|=x-3\) với \(x\ge3\)
Pt trở thành:
\(x-3=3\) (ĐK: \(x\ge3\))
\(\Leftrightarrow x=3+3\)
\(\Leftrightarrow x=6\left(tm\right)\)
TH2: \(\left|x-3\right|=-\left(x-3\right)\) với \(x< 3\)
Pt trở thành:
\(-\left(x-3\right)=3\) (ĐK: \(x< 3\))
\(\Leftrightarrow x-3=-3\)
\(\Leftrightarrow x=-3+3\)
\(\Leftrightarrow x=0\left(tm\right)\)
b) \(\sqrt{x+2\sqrt{x-1}}=2\) (ĐK: \(x\ge1\))
\(\Leftrightarrow x+2\sqrt{x-1}=4\)
\(\Leftrightarrow2\sqrt{x-1}=4-x\)
\(\Leftrightarrow4\left(x-1\right)=16-8x+x^2\)
\(\Leftrightarrow4x-4=16-8x+x^2\)
\(\Leftrightarrow x^2-12x+20=0\)
\(\Leftrightarrow\left(x-10\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=10\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
c) \(\dfrac{\sqrt{5x-4}}{\sqrt{x+2}}=2\) (ĐK: \(x\ge\dfrac{4}{5}\))
\(\Leftrightarrow\dfrac{5x-4}{x+2}=4\)
\(\Leftrightarrow5x-4=4x+8\)
\(\Leftrightarrow x=12\left(tm\right)\)
\(\sqrt{x+6-4\sqrt{x+2}}-\sqrt{9-4\sqrt{5}}=0\left(đk:x\ge-2\right)\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x+2}-2\right)^2}=\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(\Leftrightarrow\left|\sqrt{x+2}-2\right|=\left|\sqrt{5}-2\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+2}-2=\sqrt{5}-2\\\sqrt{x+2}-2=2-\sqrt{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=5\\x+2=21-8\sqrt{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=19-8\sqrt{5}\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{3;19-8\sqrt{5}\right\}\)
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) \(ĐK:x\ge0\)
\(\Leftrightarrow3\sqrt{x}-3=\sqrt{x}+2\)
\(\Leftrightarrow\sqrt{x}=\dfrac{5}{2}\Leftrightarrow x=\dfrac{25}{4}\left(tm\right)\)
Lời giải:
a/ ĐKXĐ: $x\geq 0; y\geq 1$
PT $\Leftrightarrow (x-2\sqrt{x}+1)+[(y-1)-4\sqrt{y-1}+4]=0$
$\Leftrightarrow (\sqrt{x}-1)^2+(\sqrt{y-1}-2)^2=0$
Vì $(\sqrt{x}-1)^2\geq 0; (\sqrt{y-1}-2)^2\geq 0$ với mọi $x,y$ thuộc đkxđ
Do đó để tổng của chúng bằng $0$ thì:
$\sqrt{x}-1=\sqrt{y-1}-2=0$
$\Leftrightarrow x=1; y=5$
b. ĐKXĐ: $x\geq 0; y\geq 1; z\geq 2$
PT $\Leftrightarrow 2\sqrt{x}+2\sqrt{y-1}+2\sqrt{z-2}=x+y+z$
$\Leftrightarrow (x-2\sqrt{x}+1)+[(y-1)-2\sqrt{y-1}+1]+[(z-2)-2\sqrt{z-2}+1]=0$
$\Leftrightarrow (\sqrt{x}-1)^2+(\sqrt{y-1}-1)^2+(\sqrt{z-2}-1)^2=0$
$\Rightarrow \sqrt{x}-1=\sqrt{y-1}-1=\sqrt{z-2}-1=0$
$\Leftrightarrow x=1; y=2; z=3$
ĐKXĐ \(2\le x\le4\).Đặt A=\(\sqrt[4]{\left(x-2\right)\left(4-x\right)}+\sqrt[4]{x-2}+\sqrt[4]{4-x}+6x\sqrt{3x}\)
Do x\(\ge2>0\)nên ADBĐT CAUCHY ta được:
\(\sqrt[4]{1\cdot1\cdot\left(x-2\right)\left(4-x\right)}\le\frac{1+1+x-2+4-x}{4}=1\)
\(\sqrt[4]{x-2}\le\frac{1+1+1+x-2}{4}=\frac{1}{4}\)
\(\sqrt[4]{4-x}\le\frac{1+1+1+4-x}{4}=\frac{7}{4}\)
\(6x\sqrt{3x}=2\sqrt{27x^3}\le x^3+27\)
_Do đó A\(\le1+\frac{1}{4}+\frac{7}{4}+x^3+27=x^3+30\)
Dấu = xảy ra \(\Leftrightarrow x=3\)(thỏa mãn ĐKXĐ)
=4 nha ko phải là = 44
382+28=