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\(B=\frac{2}{8}+\frac{2}{24}+\frac{2}{48}+...+\frac{2}{18\cdot20}\)
\(B=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{18\cdot20}\)
\(B=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{18}-\frac{1}{20}\)
\(B=\frac{1}{2}-\frac{1}{20}\)
\(B=\frac{9}{20}\)
=))
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(A=\frac{1}{2}+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+...+\left(\frac{1}{9}-\frac{1}{9}\right)-\frac{1}{10}\)
\(A=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
a) \(\left(2x+1\right)^3=125\)
\(\Rightarrow\left(2x+1\right)^3=5^3\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=5-1\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=4:2\)
\(\Rightarrow x=2\)
Vậy x = 2
b) \(\left(x-5\right)^4=\left(x-5\right)^6\)
\(\Rightarrow\left(x-5\right)^4-\left(x-5\right)^6=0\)
\(\Rightarrow\left(x-5\right)^4\left[1-\left(x-5\right)^2\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\1-\left(x-5\right)^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2=1\end{cases}}\)
TH 1 : \(\left(x-5\right)^4=0\Rightarrow x-5=0\Rightarrow x=5\)
TH 2 : \(\left(x-5\right)^2=1\Rightarrow\orbr{\begin{cases}x-5=1\\x-5=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)
Vậy \(x\in\left\{5;6;4\right\}\)
c) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Rightarrow\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2=1\end{cases}}\)
TH 1 : \(\left(2x-15\right)^3=0\Rightarrow2x-15=0\Rightarrow2x=15\Rightarrow x=\frac{15}{2}\)
TH 2 : \(\left(2x-15\right)^2=1\Rightarrow\orbr{\begin{cases}2x-15=1\\2x-15=-1\end{cases}}\Rightarrow\orbr{\begin{cases}2x=16\\2x=14\end{cases}}\Rightarrow\orbr{\begin{cases}x=8\\x=7\end{cases}}\)
Vậy \(x\in\left\{\frac{15}{2};8;7\right\}\)
_Chúc bạn học tốt_
A=1999/2000
B=199/200
C=511/512
hok tốt
Đáp án
mình lười trình bày cách làm lém, để đáp án thui nha
A = \(\frac{1999}{2000}\)
B = \(\frac{199}{200}\)
C = \(\frac{511}{512}\)
a) 1,28:0,125+1,28×0,25+1,28×1,75
=1,28x(8+0,25+1,75)
=1,28x10=12,8
b) bạn nên viết rõ ra
a)13x3x32,27+67,63x39
=39x32,27+67,63x39
=39x(32,27+67,63)
=39x100
=3900
b,= 1- [ 1/2 x 1/3 x1/4 x..... x 1/100 ]
=1/2 x 2/3 x 3/4 x .......x 99/100
= 1x2x3x......x99 / 2x3x4x...... x100 [ rút gọn ]
= 1/100
155,9:4,5=34 dư 29
87,5:1,75=50
52:1,6=32,5
45,6:32=1,425