\(\dfrac{102}{80}\)=\(\dfrac{51}{40}\)

\(\dfrac{102}{80}=\dfrac{102:2}{80:2}=\dfrac{51}{40}\)

\(\dfrac{70}{80}-\dfrac{2}{16}-\dfrac{3}{24}\)
\(=\dfrac{7}{8}-\dfrac{1}{8}-\dfrac{1}{8}=\dfrac{5}{8}\)

`70/80-2/16-3/24`

`=7/8-1/8-1/8`

`=5/8`

a) Số:

\(\dfrac{12}{18}\) = \(\dfrac{6}{9}\)  = \(\dfrac{2}{3}\) 

b) Rút gọn các phân số:
\(\dfrac{12}{48}\) = \(\dfrac{1}{4}\)

\(\dfrac{80}{100}\)= \(\dfrac{4}{5}\) 

\(\dfrac{75}{125}\) = \(\dfrac{3}{5}\) 

\(A=\dfrac{17^{100}+17^{96}+17^{92}+....+17^4+1}{17^{102}+17^{100}+17^{98}+....+17^2+1}\)

Gọi \(17^{100}+17^{96}+17^{92}+....+17^4+1\) là B

\(B=17^{100}+17^{96}+17^{92}+....+17^4+1\\ 17^4\cdot B=17^{104}+17^{100}+17^{96}+......+17^8+17^4\\ 17^4\cdot B-B=\left(17^{104}+17^{100}+17^{96}+......+17^8+17^4\right)-\left(17^{100}+17^{96}+17^{92}+....+17^4+1\right)\\ B\cdot\left(17^4-1\right)=17^{104}-1\\ B=\dfrac{17^{104}-1}{17^4-1}\)

Gọi \(17^{102}+17^{100}+17^{98}+....+17^2+1\) là C

\(C=17^{102}+17^{100}+17^{98}+....+17^2+1\\ C\cdot17^2=17^{104}+17^{102}+17^{100}+17^{98}+....+17^2\\ C\cdot17^2-C=\left(17^{104}+17^{102}+17^{100}+17^{98}+....+17^2\right)-\left(17^{102}+17^{100}+17^{98}+....+17^2+1\right)\\ C\cdot\left(17^2-1\right)=17^{104}-1\\ C=\dfrac{17^{104}-1}{17^2-1}\)

=>

 \(A=B:C\\ A=\dfrac{17^{104}-1}{17^4-1}:\dfrac{17^{104}-1}{17^2-1}\\ A=\dfrac{17^2-1}{17^4-1}\)

cảm ơn bạn

giúp mik đi nhé 

mik đang cần gấp nèeee

a,-7/12

b,5/14

c,-3/8

d,-1

\(\dfrac{5-2\sqrt{5}}{\sqrt{5}}-\left(5\sqrt{5}-3\right)+\sqrt{80}\\ =\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}}-5\sqrt{5}+3+4\sqrt{5}\\ =\sqrt{5}-2-5\sqrt{5}+3+4\sqrt{5}\\ =\sqrt{5}\left(1-5+4\right)-2+3\\ =0+1\\ =1\)

Lời giải:
a.

$=2\sqrt{5}-9\sqrt{5}-2\sqrt{5}=(2-9-2)\sqrt{5}=-9\sqrt{5}$

b.

$=36\sqrt{6}-2\sqrt{6}+6\sqrt{6}=(36-2+6)\sqrt{6}=40\sqrt{6}$

b) \(\dfrac{6\cdot9-2\cdot17}{63\cdot3-119}\)

\(=\dfrac{2\left(3\cdot9-17\right)}{7\cdot\left(3\cdot9-17\right)}\)

\(=\dfrac{2}{7}\)

a: \(\sqrt{20}+\sqrt{80}-\sqrt{45}\)

\(=2\sqrt5+4\sqrt5-3\sqrt5\)

\(=6\sqrt5-3\sqrt5=3\sqrt5\)

b: \(4\cdot\sqrt{\frac29}+\sqrt2+\sqrt{\frac{1}{18}}\)

\(=4\cdot\frac{\sqrt2}{3}+\sqrt2+\sqrt{\frac{2}{36}}\)

\(=\frac43\sqrt2+\sqrt2+\frac16\sqrt2=\sqrt2\left(\frac43+1+\frac16\right)=\sqrt2\left(\frac86+\frac16+1\right)=\frac{15}{6}\cdot\sqrt2=\frac52\sqrt2\)

c: \(\frac{1}{\sqrt3-1}-\frac{1}{\sqrt3+1}\)

\(=\frac{\sqrt3+1-\left(\sqrt3-1\right)}{\left(\sqrt3+1\right)\left(\sqrt3-1\right)}\)

\(=\frac{\sqrt3+1-\sqrt3+1}{2}=\frac22=1\)

d: \(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}+1}+1\)

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

\(=\frac{\sqrt{x}+1-\sqrt{x}+1-x+1}{x-1}=\frac{-x+3}{x-1}\)

e: \(\sqrt{x}-2+\frac{10-x}{\sqrt{x}+2}\)

\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+10-x}{\sqrt{x}+2}\)

\(=\frac{x-4+10-x}{x-4}=\frac{6}{x-4}\)

g: \(\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{4-x}\)

\(=\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}+\frac{\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{\sqrt{x}-2-2\left(\sqrt{x}+2\right)+\sqrt{x}}{x-4}=\frac{2\sqrt{x}-2-2\sqrt{x}-4}{x-4}=\frac{-6}{x-4}\)