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B4
a) \(\frac{9}{\sqrt{3}}=\frac{9\cdot\sqrt{3}}{\sqrt{3}\cdot\sqrt{3}}=\frac{9\sqrt{3}}{3}=3\sqrt{3}\)
b)\(\frac{3}{\sqrt{5}-\sqrt{2}}=\frac{3\left(\sqrt{5}+\sqrt{2}\right)}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}=\frac{3\left(\sqrt{5}+\sqrt{2}\right)}{3}=\sqrt{5}+\sqrt{2}\)
c)\(\frac{\sqrt{2}+1}{\sqrt{2}-1}=\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\frac{\left(\sqrt{2}+1\right)^2}{1}=\left(\sqrt{2}+1\right)^2\)
d)\(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{1}=14\)
B3
a)\(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\) \(đk:x\ge1\)
\(\frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\sqrt{x-1}\cdot\left(\frac{1}{2}-\frac{9}{2}+3\right)=-17\)
\(\sqrt{x-1}\cdot\left(-1\right)=-17\)
\(\sqrt{x-1}=17\)
\(\left[{}\begin{matrix}x-1=289\left(tm\right)\\x-1=-289\left(ktm\right)\end{matrix}\right.\)
\(x=290\left(tm\right)\)
1,
\(D=\frac{1}{\sqrt{h+2\sqrt{h-1}}}+\frac{1}{\sqrt{h-2\sqrt{h-1}}}\)
\(=\frac{1}{\sqrt{h-1+2\sqrt{h-1}+1}}+\frac{1}{\sqrt{h-1-2\sqrt{h-1}+1}}\)
\(=\frac{1}{\sqrt{h-1}+1}+\frac{1}{\sqrt{h-1}-1}\)
\(=\frac{\sqrt{h-1}-1+\sqrt{h-1}+1}{h-1-1}\)
\(=\frac{2\sqrt{h-1}}{h-2}\)
Thay \(h=3\)vào D ta có:
\(D=\frac{2\sqrt{3-1}}{3-2}=2\sqrt{2}\)
Vậy với \(h=3\)thì \(D=2\sqrt{2}\)
2,
a, \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)(ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=1\Leftrightarrow x=2\left(TM\right)\)
Vậy PT có nghiệm là \(x=2\)
b, \(\sqrt{9x^2+18}+2\sqrt{x^2+2}-\sqrt{25x^2+50}+3=0\)(ĐK: \(-\sqrt{2}\le x\le\sqrt{2}\))
\(\Leftrightarrow3\sqrt{x^2+2}+2\sqrt{x^2+2}-5\sqrt{x^2+2}=-3\)
\(\Leftrightarrow0=-3\)(vô lí)
Vậy PT đã cho vô nghiệm.
\(A=4\sqrt{\dfrac{25x}{4}}-\dfrac{8}{3}\sqrt{\dfrac{9x}{4}}-\dfrac{4}{3x}\sqrt{\dfrac{9x^3}{64}}\)
\(A=4\left|\dfrac{5\sqrt{x}}{2}\right|-\dfrac{8}{3}\left|\dfrac{3\sqrt{x}}{2}\right|-\dfrac{4}{3x}\left|\dfrac{3x\sqrt{x}}{8}\right|\)
Vì \(x>0\) nên:
\(A=10\sqrt{x}-4\sqrt{x}-\dfrac{1}{2}\sqrt{x}=\dfrac{11\sqrt{x}}{2}\)
a/\(\sqrt{x^2-2x}=\sqrt{2-3x}\left(đk:x\le0\right)
\)
\(\Leftrightarrow x^2-2x=2-3x\)
\(\Leftrightarrow x^2+x-2=0\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(KTM\right)\\x=-2\left(TM\right)\end{matrix}\right.\)
Vậy x=-2 là nghiệm của PT
b/\(\sqrt{x-3}-2\sqrt{x^2-9}=0\left(đk:x\ge3\right)\)
\(\Leftrightarrow\sqrt{x-3}\left(1-2\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\1=2\sqrt{x+3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(TM\right)\\4x+12=1\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=3\\x=-\frac{11}{4}\left(KTM\right)\end{matrix}\right.\)
Vậy x=3
Lời giải:
$A=4.\sqrt{\frac{25}{4}}.\sqrt{x}-\frac{8}{3}.\sqrt{\frac{9}{4}}.\sqrt{x}-\frac{4}{3x}.\sqrt{\frac{9}{64}}.\sqrt{x^3}$
$=10\frac{x}-4\sqrt{x}-\frac{1}{2x}.x\sqrt{x}=10x-4x-\frac{1}{2}\sqrt{x}$
$=\frac{11}{2}\sqrt{x}$