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\(A=\frac{8}{9}\cdot\frac{15}{16}\cdot\frac{24}{25}\cdot...\cdot\frac{360}{361}\cdot\frac{399}{400}\)
\(A=\frac{2\cdot4\cdot3\cdot5\cdot4\cdot6\cdot...\cdot18\cdot20\cdot19\cdot21}{3\cdot3\cdot4\cdot4\cdot5\cdot5\cdot...\cdot19\cdot19\cdot20\cdot20}\)
\(A=\frac{2\cdot21}{3\cdot20}\)
\(A=\frac{7}{10}\)
\(B=\frac{9}{8}\cdot\frac{16}{15}\cdot\frac{25}{24}\cdot...\cdot\frac{441}{440}\cdot\frac{484}{483}\)
\(B=\frac{3\cdot3\cdot4\cdot4\cdot5\cdot5\cdot...\cdot21\cdot21\cdot22\cdot22}{2\cdot4\cdot3\cdot5\cdot4\cdot6\cdot...\cdot20\cdot22\cdot21\cdot23}\)
\(B=\frac{3\cdot22}{2\cdot23}=\frac{33}{23}\)
\(C=\frac{17}{23}.\left(\frac{7}{61}+\frac{28}{61}+\frac{26}{61}\right)\)
\(C=\frac{17}{23}\cdot1=\frac{17}{23}\)
a) \(-98 + 8 -(-98) -22 = 8-22=-14\)
b) \(-(-98)-61+12+61=-(-98) + 12 = 110\)
c) \(61-(-25) +7-8+(-25)=61-1=60\)
d) \((-25)-24-(-98)+24+(-98)=-25\)
a) \(x+8+x-22\)
Tại X= -98
\(-98+8-\left(-98\right)-22=-98+8-98-22=8-22=-14\)
b) \(-x-a+12+a\)
\(\Leftrightarrow-\left(-98\right)-61+12+61=12+98=110\)
c) \(a-m+7-8+m\)
\(\Leftrightarrow61-\left(-25\right)+7-8+\left(-25\right)=61+25+7-8-25=61-1=60\)d) \(m-24-x+24+x\)
\(\Leftrightarrow-25-24-\left(-98\right)+24+\left(-98\right)=-25+98-98=-25\)
a/ \(\left(x-35\right)-120=0\)
\(\Leftrightarrow x-35=0+120\)
\(\Leftrightarrow x-35=120\)
\(\Leftrightarrow x=120+35\)
\(\Leftrightarrow x=155\)
Vậy ...
b/ \(124+\left(188-x\right)=217\)
\(\Leftrightarrow188-x=217-124\)
\(\Leftrightarrow188-x=93\)
\(\Leftrightarrow x=95\)
Vậy ...
b/ \(156-\left(x+61\right)=82\)
\(\Leftrightarrow x+61=74\)
\(\Leftrightarrow x=13\)
Vậy ..
\(a)\frac{(-5)^{60}.30^5}{15^5.5^{61}}=\frac{(5.2.3)^5}{(5.3)^5.5}=\frac{5^5.2^5.3^5}{5^5.3^5.5} =\frac{2^5}{5}=\frac{32}{5}\)
\(b) \frac{(-3)^{10}.15^5}{25^3.(-9)^7}=\frac{(-3)^{10} .(3.5)^5}{(5^2)^3.[(-3).3]^7}=\frac{(-3)^{10}.3^5.5^5}{5^6.(-3)^7.3^7}=\frac{(-3)^3}{5.3^2}=\frac{-3}{5}\)
~ Hok tốt a~
\(\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+...+\dfrac{3}{59\cdot61}\)(đã sửa)
Đặt \(A=\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+....+\dfrac{3}{59\cdot61}\)
\(\dfrac{2}{3}A=\dfrac{2}{3}\left(\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+....+\dfrac{3}{59\cdot61}\right)\)
\(\dfrac{2}{3}A=\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{59\cdot61}\)
\(\dfrac{2}{3}A=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\)
\(\dfrac{2}{3}A=\dfrac{1}{5}-\dfrac{1}{61}\)
\(A=\dfrac{56}{305}:\dfrac{2}{3}=\dfrac{56}{305}\cdot\dfrac{3}{2}=\dfrac{84}{305}\)
\(\text{61 - ( - 25 ) + 7 - 8 + 61}\)
\(\Leftrightarrow146\)
\(\text{Bạn học lại công thức ( mình chỉ cho đáp số thôi )}\)
\(\text{Cho nhé bạn !!}\)
ta co:
2A=2(2 mu 60 +1 /2 mu 61 +1)
2A=2 mu 61 +2 / 2 mu 61 +1
2A=2 mu 61 +1+1/2 mu 61 +1
2A=1+1/2 mu 61 +1
ta co:
2B=2(2 mu 61 +1/2 mu 62 +1)
2B=2 mu 62 +2/2 mu 62+1
2B=2 mu 62 +1+1/2 mu 62 +1
2B=1+1/2 mu 62 +1
mà 1+1/2 mu 61+1>1+1/2 mu 62 +1 nen 2A >2B
vậy A>B
nhớ k đúng cho mk nha
Ta có:
2.A=2 mủ 61 +2/2 mủ 61 +1=1+(2/2 mủ 61 +1)
2.B=2 mủ 62 + 2 /2 mủ 62 +1=1+(2/2 mủ 62 + 1)
vì ... nên 2.A >2.B.Vậy A>B
* Cách 1 :
Ta có :
\(5A=\frac{5^{61}+5}{5^{61}+1}=\frac{5^{61}+1+4}{5^{61}+1}=\frac{5^{61}+1}{5^{61}+1}+\frac{4}{5^{61}+1}=1+\frac{4}{5^{61}+1}\)
\(5B=\frac{5^{62}+5}{5^{62}+1}=\frac{5^{62}+1+4}{5^{62}+1}=\frac{5^{62}+1}{5^{62}+1}+\frac{4}{5^{62}+1}=1+\frac{4}{5^{62}+1}\)
Vì \(\frac{4}{5^{61}+1}>\frac{4}{5^{62}+1}\) nên \(1+\frac{4}{5^{61}+1}>1+\frac{4}{5^{62}+1}\)
\(\Rightarrow\)\(5A>5B\) hay \(A>B\)
Vậy \(A>B\)
Chúc bạn học tốt ~
\(61-\left(-25\right)+7-8+61\)
\(=\left(61+25\right)+7-8+61\)
\(=86+7-8+61\)
\(=93-8+61\)
\(=85+61=146\)
61- (-25)+7-8+61
= (61 +61) +25+7-8
=122+25+7-8
=146