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18 tháng 7 2016

\(A=\frac{a^5\times a^{-3}\times a}{a^2\times a^{-4}\times a^3}==\frac{a^{-3}}{a^{-4}}\times a=\frac{a^4\times a}{a^3}=a^2\)

\(P=\dfrac{3a-12\sqrt{a}+a+4\sqrt{a}-4a-8}{\left(\sqrt{a}+4\right)\left(\sqrt{a}-4\right)}:\dfrac{\sqrt{a}+4-2\sqrt{a}-5}{\left(\sqrt{a}+4\right)}\)

\(=\dfrac{-8\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+4\right)\left(\sqrt{a}-4\right)}\cdot\dfrac{\sqrt{a}+4}{-\left(\sqrt{a}+1\right)}=\dfrac{8}{\sqrt{a}-4}\)

30 tháng 6 2016

P= (\(\frac{3\sqrt{a}}{\sqrt{a}+4}+\frac{\sqrt{a}}{\sqrt{a}-4}+\frac{4\left(a+2\right)}{16-a}\)):\(\left(1-\frac{2\sqrt{a}+5}{\sqrt{a}-4}\right)\)

=\(\left(\frac{3\sqrt{a}\left(\sqrt{a}-4\right)}{a-16}+\frac{\sqrt{a}\left(\sqrt{a}+4\right)}{a-16}-\frac{4a+8}{a-16}\right):\left(\frac{\sqrt{a}-4-2\sqrt{a}-5}{\sqrt{a}-4}\right)\)

\(\left(\frac{3a-12\sqrt{a}+a+4\sqrt{a}-4a-8}{a-16}\right):\left(\frac{-\sqrt{a}-9}{\sqrt{a}-4}\right)\)

=\(\left(\frac{-8\sqrt{a}-8}{a-16}\right).\left(\frac{\sqrt{a}-4}{-\sqrt{a}-9}\right)=\frac{8\sqrt{a}+8}{\left(\sqrt{a}+4\right).\left(\sqrt{a}+9\right)}=\frac{8\sqrt{a}+8}{a+13\sqrt{a}+36}\)

5 tháng 7 2019

\(P=\frac{\sqrt{a}+3}{\sqrt{a}-2}-\frac{\sqrt{a}-1}{\sqrt{a}+2}+\frac{4\sqrt{a}-4}{4-a}\)

\(=\frac{\left(\sqrt{a}+3\right)\left(\sqrt{a}+2\right)}{a-4}-\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-4}-\frac{4\sqrt{a}-4}{a-4}\)

\(=\frac{a+5\sqrt{a}+6-\left(a-3\sqrt{a}+2\right)-\left(4\sqrt{a}-4\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)

\(=\frac{a+5\sqrt{a}+6-a+3\sqrt{a}-2-4\sqrt{a}+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)

\(=\frac{4\sqrt{a}+8}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\frac{4\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\frac{4}{\sqrt{a}-2}\)

Ai giải giúp mấy bài toán vsBài 1:A=\(\sqrt{\frac{1}{\text{√}2+1}-\frac{\text{√}8-\text{√}10}{2-\text{√}5}}\)B=\(\frac{5\text{√}5}{\text{√}5+2}+\frac{\text{√}5}{\text{√}5-1}-\frac{3\text{√}5}{3+\text{√}5}\)Bài 2 rút gọn biểu thứcA=\(\left(\frac{x+\sqrt[]{xy}}{\text{√}x+\text{√}y}-2\right):\frac{1}{\text{√}x+2}\) với x :y >0B=\(\left(\frac{a}{a-2\text{√}a}+\frac{a}{\text{√}a-2}\right):\frac{\text{√}a+1}{a-4\text{√}a+4}\)Bài 3 cho biểu...
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Ai giải giúp mấy bài toán vs

Bài 1:

A=\(\sqrt{\frac{1}{\text{√}2+1}-\frac{\text{√}8-\text{√}10}{2-\text{√}5}}\)

B=\(\frac{5\text{√}5}{\text{√}5+2}+\frac{\text{√}5}{\text{√}5-1}-\frac{3\text{√}5}{3+\text{√}5}\)

Bài 2 rút gọn biểu thức

A=\(\left(\frac{x+\sqrt[]{xy}}{\text{√}x+\text{√}y}-2\right):\frac{1}{\text{√}x+2}\) với x :y >0

B=\(\left(\frac{a}{a-2\text{√}a}+\frac{a}{\text{√}a-2}\right):\frac{\text{√}a+1}{a-4\text{√}a+4}\)

Bài 3 cho biểu thức

P=\(\left(\frac{x-2}{x+2\text{√}x}+\frac{1}{\text{√}x+2}\right)\frac{\text{√}x+1}{\text{√}x-1}\)

a)Rút gọn P

b)tìm x để P=\(\text{√}x+\frac{5}{2}\)

bài 4 rút gọn biểu thức 

A=\(\frac{1}{x+\text{√}x}+\frac{2\text{√}x}{x-1}-\frac{1}{x-\text{√}x}\)

B=\(\left(\frac{x}{x+3\text{√}x}+\frac{1}{\text{√}x+3}\right):\left(1-\frac{2}{\text{√}x}+\frac{6}{x+3\text{√}x}\right)\)

Bài 5

A=\(\left(\frac{2}{\text{√}x-3}-\frac{1}{\text{√}x+3}-\frac{x}{\text{√}x\left(x-9\right)}\right):\text{(√}x+3-\frac{x}{\text{√}x-3}\)

a)rút gọn A

b)tìm gtri x để A= -1/4

AI GIẢI GIÙM MÌNH ĐI MÌNH TẠ ƠN

0
18 tháng 8 2023

a) \(a^{\dfrac{1}{3}}\cdot a^{\dfrac{1}{2}}\cdot a^{\dfrac{7}{6}}=a^{\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{7}{6}}=a^2\)

b) \(a^{\dfrac{2}{3}}\cdot a^{\dfrac{1}{4}}:a^{\dfrac{1}{6}}=a^{\dfrac{2}{3}+\dfrac{1}{4}-\dfrac{1}{6}}=a^{\dfrac{3}{4}}\)

c) \(\left(\dfrac{3}{2}a^{-\dfrac{3}{2}}\cdot b^{-\dfrac{1}{2}}\right)\left(-\dfrac{1}{3}a^{\dfrac{1}{2}}b^{\dfrac{2}{3}}\right)=\left(\dfrac{3}{2}\cdot-\dfrac{1}{3}\right)\left(a^{-\dfrac{3}{2}}\cdot a^{\dfrac{1}{2}}\right)\left(b^{-\dfrac{1}{2}}\cdot b^{\dfrac{2}{3}}\right)\)

\(=-\dfrac{1}{2}a^{-1}b^{-\dfrac{1}{3}}\)