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22 tháng 10 2023
\(a)C=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ =\left(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ =\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ =\dfrac{x-\sqrt{x}+2\sqrt{x}-2-x-\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ =\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ =\dfrac{2}{x-1}\)
\(b)x=\dfrac{\sqrt{7}}{\sqrt{1+\sqrt{7}}-1}-\dfrac{\sqrt{7}}{\sqrt{1+\sqrt{7}}+1}\\ =\dfrac{\sqrt{7}\left(\sqrt{1+\sqrt{7}}+1\right)}{1+\sqrt{7}-1}-\dfrac{\sqrt{7}\left(\sqrt{1+\sqrt{7}}-1\right)}{1+\sqrt{7}-1}\\ =\sqrt{1+\sqrt{7}}+1-\sqrt{1+\sqrt{7}}+1\\ =2\)
Thay \(x=2\) vào \(C\)
\(\dfrac{2}{2-1}=\dfrac{2}{1}=2\)
17 tháng 8 2021
a: ĐKXĐ: \(x\ge\dfrac{2}{3}\)
b: Ta có: \(3\sqrt{2}-4\sqrt{18}+2\sqrt{32}-\sqrt{50}\)
\(=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}\)
\(=-6\sqrt{2}\)
1 tháng 9 2021
a: Ta có: \(2\sqrt{28}+3\sqrt{63}-3\sqrt{\dfrac{112}{9}}-\sqrt{\dfrac{196}{7}}\)
\(=4\sqrt{7}+9\sqrt{7}-4\sqrt{7}-2\sqrt{7}\)
\(=7\sqrt{7}\)
b: Ta có: \(\sqrt{8+2\sqrt{7}}-\sqrt{12-\sqrt{140}}-\sqrt{5}\)
\(=\sqrt{7}+1-\sqrt{7}+\sqrt{5}-\sqrt{5}\)
=1
13 tháng 5 2021
a)0,6.a
b)\(a^2\).(a-3)
c)36.(a-1)
d)\(\dfrac{1.a^2}{a-b}\).(a-b)
TT
3
15 tháng 9 2021
a) \(A=\dfrac{x+\sqrt{xy}}{y+\sqrt{xy}}=\dfrac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{\sqrt{x}}{\sqrt{y}}\)
b) \(B=\dfrac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}=\dfrac{\sqrt{a}\left(1+\sqrt{ab}\right)-\sqrt{b}\left(1+\sqrt{ab}\right)}{\left(\sqrt{ab}-1\right)\left(1+\sqrt{ab}\right)}=\dfrac{\left(1+\sqrt{ab}\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}-1}=\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{ab}-1}\)
15 tháng 9 2021
c) \(C=\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}=\dfrac{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}+x\right)}{1+\sqrt{x}}=1-\sqrt{x}+x\)
d) \(D=\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x+2\sqrt{xy}-y=x-\sqrt{xy}+y-x+2\sqrt{xy}-y=\sqrt{xy}\)
e) \(\dfrac{x+4\sqrt{x}+4}{\sqrt{x}+2}+\dfrac{4-x}{2-\sqrt{x}}=\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}+2}+\dfrac{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}{2-\sqrt{x}}=\sqrt{x}+2+2+\sqrt{x}=2\sqrt{x}+4\)
MM
1
\(2\sqrt{3a}-\sqrt{75a}+a\sqrt{\dfrac{13,5}{2a}}-\dfrac{2}{5}\sqrt{300a^3}=-A.\sqrt{3a}\\ \Leftrightarrow2\sqrt{3a}-\sqrt{5^2.3a}+\sqrt{\dfrac{13,5a^2}{2a}}-\sqrt{\left(\dfrac{2}{5}\right)^2.300a^3}=-A.\sqrt{3a}\\ \Leftrightarrow2\sqrt{3a}-5\sqrt{3a}+\sqrt{6,75a}-\sqrt{48a^3}=-A\sqrt{3a}\\ \Leftrightarrow2\sqrt{3a}-5\sqrt{3a}+\sqrt{\left(1,5\right)^2.3a}-\sqrt{4^2.3a}=-A\sqrt{3a}\\ \Leftrightarrow2\sqrt{3a}-5\sqrt{3a}+1,5\sqrt{3a}-4\sqrt{3a}=-A\sqrt{3a}\\ \Leftrightarrow-5,5\sqrt{3a}=-A\sqrt{3a}\\ Vậy:A=\dfrac{-5,5\sqrt{3a}}{-\sqrt{3a}}=5,5\)
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