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27 tháng 8 2018
a) \(\frac{x\sqrt[3]{y}+\sqrt[3]{x^2y^2}}{\sqrt[3]{x^2y^2}+y\sqrt[3]{x}}\)
\(=\frac{\sqrt[3]{x^2y}\left(\sqrt[3]{x}+\sqrt[3]{y}\right)}{\sqrt[3]{xy^2}\left(\sqrt[3]{x}+\sqrt[3]{y}\right)}=\sqrt[3]{\frac{x^2y}{xy^2}}=\sqrt[3]{\frac{x}{y}}\)
b) \(\frac{\sqrt[3]{54}-2\sqrt[3]{16}}{\sqrt[3]{54}+2\sqrt[3]{16}}\)
\(=\frac{\sqrt[3]{27.2}-2\sqrt[3]{8.2}}{\sqrt[3]{27.2}+2\sqrt[3]{8.2}}\)
\(=\frac{3\sqrt[3]{2}-4\sqrt[3]{2}}{3\sqrt[3]{2}+4\sqrt[3]{2}}=\frac{-\sqrt[3]{2}}{7\sqrt[3]{2}}=-\frac{1}{7}\)
BN
0
AH
Akai Haruma
Giáo viên
16 tháng 7 2020
k) ĐK: $x^2\geq 5$
PT $\Leftrightarrow 2\sqrt{x^2-5}-\frac{1}{3}\sqrt{x^2-5}+\frac{3}{4}\sqrt{x^2-5}-\frac{5}{12}\sqrt{x^2-5}=4$
$\Leftrightarrow 2\sqrt{x^2-5}=4$
$\Leftrightarrow \sqrt{x^2-5}=2$
$\Rightarrow x^2-5=4$
$\Leftrightarrow x^2=9\Rightarrow x=\pm 3$ (đều thỏa mãn)
l) ĐKXĐ: $x\geq -1$
PT $\Leftrightarrow 2\sqrt{x+1}+3\sqrt{x+1}-\sqrt{x+1}=4$
$\Leftrightarrow 4\sqrt{x+1}=4$
$\Leftrightarrow \sqrt{x+1}=1$
$\Rightarrow x+1=1$
$\Rightarrow x=0$
m)
ĐKXĐ: $x\geq -1$
PT $\Leftrightarrow 4\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}+3\sqrt{x+1}$
$\Leftrightarrow 6\sqrt{x+1}=16+2\sqrt{x+1}$
$\Leftrightarrow 4\sqrt{x+1}=16$
$\Leftrightarrow \sqrt{x+1}=4$
$\Rightarrow x=15$ (thỏa mãn)
AH
Akai Haruma
Giáo viên
16 tháng 7 2020
h)
ĐKXĐ: $x\geq -5$
PT $\Leftrightarrow \sqrt{x+5}=6$
$\Rightarrow x+5=36\Rightarrow x=31$ (thỏa mãn)
i) ĐKXĐ: $x\geq 5$
PT \(\Leftrightarrow \sqrt{x-5}+4\sqrt{x-5}-\sqrt{x-5}=12\)
\(\Leftrightarrow 4\sqrt{x-5}=12\Leftrightarrow \sqrt{x-5}=3\Rightarrow x-5=9\Rightarrow x=14\) (thỏa mãn)
j)
ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow 3\sqrt{2x}+\sqrt{2x}-6\sqrt{2x}+4=0$
$\Leftrightarrow -2\sqrt{2x}+4=0$
$\Leftrightarrow \sqrt{2x}=2$
$\Rightarrow x=2$ (thỏa mãn)
KN
13 tháng 3 2020
\(B=\frac{x}{x-16}+\frac{2}{\sqrt{x}-4}+\frac{2}{\sqrt{x}+4}\)
\(=\frac{x}{x-16}+\frac{2\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}+\frac{2\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\)
\(=\frac{x}{x-16}+\frac{2\sqrt{x}+8}{x-16}+\frac{2\sqrt{x}-8}{x-16}\)
\(=\frac{x+4\sqrt{x}}{x-16}=\frac{\sqrt{x}\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\frac{\sqrt{x}}{\sqrt{x}-4}\)
\(A=2\sqrt{12}-\sqrt{75}+\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=2\sqrt{12}-\sqrt{75}+\left(2-\sqrt{3}\right)\)(vì \(\sqrt{3}< \sqrt{4}=2\))
\(\Rightarrow\frac{1}{2}A=\sqrt{12}-\frac{\sqrt{75}}{2}+1-\frac{\sqrt{3}}{2}\)
\(=\sqrt{12}+1-\frac{\sqrt{3}\left(1+5\right)}{2}=\sqrt{12}-3\sqrt{3}+1\)
\(=\sqrt{3}+1\)
\(B-\frac{1}{2}A=0\Leftrightarrow\frac{\sqrt{x}}{\sqrt{x}-4}=\sqrt{3}+1\)
\(\Leftrightarrow\sqrt{x}=\left(\sqrt{3}+1\right)\left(\sqrt{x}-4\right)\)
\(\Leftrightarrow\sqrt{x}=\sqrt{3x}+\sqrt{x}-4\sqrt{x}-4\)
\(\Leftrightarrow\sqrt{3x}-4\sqrt{x}-4=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{3}-4\right)=4\Leftrightarrow\sqrt{x}=\frac{4}{\sqrt{3}-4}\)
\(\Rightarrow x=\left(\frac{4}{\sqrt{3}-4}\right)^2=\frac{304+128\sqrt{3}}{-173}\)
KN
13 tháng 3 2020
Mù mịt quá, sửa từ dòng 7 từ dưới lên
\(=-\sqrt{3}+1\)
\(B-\frac{1}{2}A=0\Leftrightarrow\frac{\sqrt{x}}{\sqrt{x}-4}=-\sqrt{3}+1\)
\(\Leftrightarrow\sqrt{x}=\left(\sqrt{x}-4\right)\left(1-\sqrt{3}\right)\)
\(\Leftrightarrow\sqrt{x}=\sqrt{x}-4-\sqrt{3x}+4\sqrt{3}\)
\(\Leftrightarrow-4-\sqrt{3x}+4\sqrt{3}=0\)
\(\Leftrightarrow\sqrt{3x}=4\sqrt{3}-4\)
\(\Leftrightarrow\sqrt{x}=\frac{4\left(\sqrt{3}-1\right)}{\sqrt{3}}\)
\(\Leftrightarrow x=\frac{64-32\sqrt{3}}{3}\)
31 tháng 10 2017
\(\left[\frac{\sqrt{x}}{\sqrt{x}+4}+\frac{4}{\sqrt{x}-4}\right]:\frac{x+16}{\sqrt{x}+2}\)
\(=\left[\frac{\sqrt{x}\left(\sqrt{x}-4\right)}{x-16}+\frac{4\left(\sqrt{x}+4\right)}{x-16}\right]:\frac{x+16}{\sqrt{x}+2}\)
\(=\left[\frac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}\right].\frac{\sqrt{x}+2}{x+16}\)
\(=\frac{x+16}{x-16}.\frac{\sqrt{x}+2}{x+16}\)
\(=\frac{\sqrt{x}+2}{x-16}\)
25 tháng 10 2020
Bài làm
Rút gọn
\(\left(\frac{1}{\sqrt{x}-1}-\frac{\sqrt{x}}{1-x}\right)\cdot\frac{x-\sqrt{x}}{2\sqrt{x}+1}\)
\(=\left(\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}\right)\cdot\frac{\sqrt{x}(\sqrt{x}-1)}{2\sqrt{x}+1}\)
\(=\left(\frac{\sqrt{x}+1}{(\sqrt{x}-1)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\frac{\sqrt{x}(\sqrt{x}-1)}{2\sqrt{x}+1}\)
\(=\frac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}+1}\)
\(=\frac{\sqrt{x}}{\sqrt{x}+1}\)
Tính:
\(\frac{3-\sqrt{3}}{\sqrt{3}+2}+\frac{\sqrt{3}}{\sqrt{3}-2}+\frac{21}{\sqrt{3}}\)
\(=\frac{3-\sqrt{3}}{\sqrt{3}+2}+\frac{\sqrt{3}}{\sqrt{3}-2}+\frac{7\sqrt{3}\cdot\sqrt{3}}{\sqrt{3}}\)
\(=\frac{3-\sqrt{3}}{\sqrt{3}+2}+\frac{\sqrt{3}}{\sqrt{3}-2}+7\sqrt{3}\)
\(=\frac{\left(3-\sqrt{3}\right)\left(\sqrt{3}-2\right)}{\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)}+\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}+7\sqrt{3}\)
\(=\frac{3\sqrt{3}-3-6+2\sqrt{3}}{\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)}+\frac{3+2\sqrt{3}}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}+7\sqrt{3}\)
\(=\frac{3\sqrt{3}-3-6+2\sqrt{3}+3+2\sqrt{3}}{3-4}+7\sqrt{3}\)
\(=\frac{7\sqrt{3}-6}{-1}+7\sqrt{3}\)
\(=6-7\sqrt{3}+7\sqrt{3}\)
\(=6\)
25 tháng 10 2020
Bài làm
\(\sqrt{42-10\sqrt{17}}+\sqrt{\left(\sqrt{17}-\sqrt{16}\right)^2}\)
\(=\sqrt{42-10\sqrt{17}}+\left|\sqrt{17}-\sqrt{16}\right|\)
\(=\sqrt{25-10\sqrt{17}+17}+\sqrt{17}-\sqrt{16}\)
\(=\sqrt{\left(5-\sqrt{17}\right)^2}+\sqrt{17}-\sqrt{16}\)
\(=\left|5-\sqrt{17}\right|+\sqrt{17}-\sqrt{16}\)
\(=5-\sqrt{17}+\sqrt{17}-\sqrt{16}\)
\(=5-4\)
\(=1\)