căn x nhân 3 = căn x + 3 nhân 1
help me
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Lời giải:
a.
\(5+\sqrt{3}+5\sqrt{3}+3=(5+5\sqrt{3})+(\sqrt{3}+3)\)
\(=5(1+\sqrt{3})+\sqrt{3}(1+\sqrt{3})=(1+\sqrt{3})(5+\sqrt{3})\)
b.
\(\sqrt{x}+\sqrt{y}+\sqrt{xy}+1=(\sqrt{x}+\sqrt{xy})+(\sqrt{y}+1)\)
\(=\sqrt{x}(1+\sqrt{y})+(\sqrt{y}+1)=(\sqrt{y}+1)(\sqrt{x}+1)\)
c.
$x-4\sqrt{x}+3=(x-\sqrt{x})-(3\sqrt{x}-3)$
$=\sqrt{x}(\sqrt{x}-1)-3(\sqrt{x}-1)$
$=(\sqrt{x}-1)(\sqrt{x}-3)$
\(\sqrt{9.\left(x-1\right)^2}-12=0\)
=> 3.(x - 1) - 12 = 0
=> 3x - 15 = 0
=> 3x = 15
=> x = 5
b) \(\sqrt{4.\left(3-x\right)}=16\) (ĐKXĐ: x ≤ 3)
\(\Rightarrow\sqrt{3-x}=8\)
=> 3 - x = 64
=> x = -61
\(a,=\sqrt{xy}\left(\sqrt{x}-1\right)+\left(\sqrt{x}-1\right)=\left(\sqrt{xy}+1\right)\left(\sqrt{x}-1\right)\\ b,=\sqrt{xy}\left(\sqrt{x}+1\right)+\left(\sqrt{x}+1\right)=\left(\sqrt{x}+1\right)\left(\sqrt{xy}+1\right)\)
d: Ta có: \(\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}\)
\(=\dfrac{\sqrt{12+2\sqrt{11}}-\sqrt{12-2\sqrt{11}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{11}+1-\sqrt{11}+1}{\sqrt{2}}\)
\(=\sqrt{2}\)
d: Ta có: \(\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}\)
\(=\dfrac{\sqrt{12+2\sqrt{11}}-\sqrt{12-2\sqrt{11}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{11}+1-\sqrt{11}+1}{\sqrt{2}}\)
\(=\sqrt{2}\)
Bài làm
\(3\sqrt{x}=\sqrt{x}+3\Leftrightarrow3\sqrt{x}-\sqrt{x}=3\Leftrightarrow2\sqrt{x}=3\)
\(\Leftrightarrow\sqrt{x}=\frac{3}{2}\Leftrightarrow x=\frac{9}{4}\)
\(3\sqrt{x}=3+\sqrt{x}\)
\(\Leftrightarrow3\sqrt{x}-\sqrt{x}-3=0\)
\(\Leftrightarrow\left(3\sqrt{x}-\sqrt{x}\right)-3=0\)
\(\Leftrightarrow2\sqrt{x}-2=1\)
\(\Leftrightarrow2\left(\sqrt{x}-1\right)=1\)
\(\Leftrightarrow\sqrt{x}-1=\frac{1}{2}\)
\(\Leftrightarrow\sqrt{x}=1+\frac{1}{2}\Leftrightarrow\sqrt{x}=\frac{3}{2}\)
\(\Leftrightarrow x=\left(\frac{3}{2}\right)^2=\frac{9}{4}\)
Vậy \(x=\frac{9}{4}\)
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