\(A=\dfrac{8^7\cdot9^{21}}{18^5\cdot6^7\cdot27^5}=\dfrac{2^{21}\cdot3^{42}}{2^{12}\cdot3^{32}}=2^9\cdot3^{10}\)

\(B=10\cdot\dfrac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}=10\cdot\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)

\(=10\cdot\dfrac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\cdot5}\)

\(=10\cdot2^2\cdot\dfrac{1}{5}=8\)

a) \(\sqrt{6+2\sqrt{5}}=\sqrt{5+2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}+1\right)}=\sqrt{5}+1\)

b) \(\sqrt{8-2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}=\sqrt{\left(\sqrt{7}-1\right)^2}=\sqrt{7}-1\left(do\sqrt{7}>1\right)\)

c) \(\sqrt{9+4\sqrt{5}}=\sqrt{5+2\cdot\sqrt{5}\cdot2+4}=\sqrt{\left(\sqrt{5}+2\right)^2}=2+\sqrt{5}\)

d) \(\sqrt{7-4\sqrt{3}}=\sqrt{4-2\cdot2\cdot\sqrt{3}+3}=\sqrt{\left(2-\sqrt{3}\right)^2}=2-\sqrt{3}\left(do2>\sqrt{3}\right)\)

e) \(\sqrt{8+2\sqrt{15}}=\sqrt{5+2\sqrt{3}\sqrt{5}+3}=\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}=\sqrt{3}+\sqrt{5}\)

g) \(\sqrt{9+4\sqrt{2}}=\sqrt{8+2.\sqrt{8}.1+1}=\sqrt{\left(\sqrt{8}+1\right)^2}=\sqrt{8}+1=2\sqrt{2}+1\) 

Bài 1:

a: Ta có: \(P=\left(\dfrac{\sqrt{x}}{x\sqrt{x}-1}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)

\(=\dfrac{\sqrt{x}+x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{\sqrt{x}+1}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

lỗi ;-;

lỗi

lỗi

1,        84/98

2,        20/24

làm lời giải ra hộ mình nha 

minh sẽ k cho

bài 1: 84/98

bài 2: 20/24

= 67 / 16 nha em 

ht

`3/16+4=3/16+64/16=67/16`

\(\dfrac{3}{16}+4=\dfrac{3}{16}+\dfrac{64}{16}=\dfrac{67}{16}\)

a) \(7^3.7^5=7^{3+5}=7^8\)

b)\(5^6.5^4=5^{6+4}=5^{10}\)

phân số ? 

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