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\(B=2+2^2+2^3+2^4+...+2^{99}+2^{100}=2\left(1+2^2+2^3+2^4\right)+...+2^{96}\left(1+2^2+2^3+2^4\right)=2.31+2^6.31+...+2^{96}.31=31\left(2+2^6+...+2^{96}\right)⋮31\)
280 - x.9 = 450
x.9 = 280 - 450
x.9 = -170
x= -170/9
Gọi số học sinh khối 6 là x
Theo đề, ta có: \(x-3\in BC\left(8;12;15\right)\)
\(\Leftrightarrow x-3\in\left\{120;240;360;...\right\}\)
\(\Leftrightarrow x\in\left\{123;243;363\right\}\)
mà 200<=x<=300
nên x=243
Gọi số học sinh khối 6 là a
a + 3 \(⋮8;12;15\)
\(\Rightarrow\) \(a+3\in BC\left(8;12;15\right)\)
8 = 2 . 3
12 = 22 . 3
15 = 3 . 5
\(\Rightarrow\) BCNN (8; 12; 15) = 22 . 3 . 5 = 60
Mà 203 < a + 3 < 303 học sinh
\(\Rightarrow\) a + 3 \(\in\) {240; 300}
\(\Rightarrow\) a \(\in\) {237; 207}
11:
a: góc xOy=42 độ
góc B=47 độ
mà 42<47
nên góc xOy<góc B
b: góc mAn=47 độ
góc B=47 độ
mà 47=47
nên góc mAn=góc B
Bài 2:
\(10M=\dfrac{10^{12}+10}{10^{12}+1}=1+\dfrac{9}{10^{12}+1}\)
\(10N=\dfrac{10^{11}+10}{10^{11}+1}=1+\dfrac{9}{10^{11}+1}\)
Ta có: \(10^{12}+1>10^{11}+1\)
=>\(\dfrac{9}{10^{12}+1}< \dfrac{9}{10^{11}+1}\)
=>\(\dfrac{9}{10^{12}+1}+1< \dfrac{9}{10^{11}+1}+1\)
=>10M<10N
=>M<N
Bài 1:
\(A=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)\cdot...\cdot\left(1-\dfrac{1}{900}\right)\)
\(=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{30}\right)\cdot\left(1+\dfrac{1}{2}\right)\cdot\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1+\dfrac{1}{30}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{29}{30}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{31}{30}\)
\(=\dfrac{1}{30}\cdot\dfrac{31}{2}=\dfrac{31}{60}\)
a) \(95.\left(-13\right)+\left(-13\right).15-13.\left(-10\right)\)
\(=13.\left(-95\right)+13.\left(-15\right)-13.\left(-10\right)\)
\(=13.\left[\left(-95\right)+\left(-15\right)-\left(-10\right)\right]\)
\(=13.\left(-90\right)\)
\(=-1170\)
Bài 1:
Ta có: \(S=\dfrac{1}{5\cdot9}+\dfrac{1}{9\cdot13}+\dfrac{1}{13\cdot17}+...+\dfrac{1}{41\cdot45}\)
\(=\dfrac{1}{4}\left(\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+\dfrac{4}{13\cdot17}+...+\dfrac{4}{41\cdot45}\right)\)
\(=\dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{17}+...+\dfrac{1}{41}-\dfrac{1}{45}\right)\)
\(=\dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{45}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{8}{45}=\dfrac{2}{45}\)
MTC:140
\(\dfrac{2}{4}=\dfrac{2.35}{4.35}=\dfrac{70}{140}\) \(\dfrac{4}{5}=\dfrac{4.28}{5.28}=\dfrac{112}{140}\) \(\dfrac{8}{7}=\dfrac{8.20}{7.20}=\dfrac{160}{140}\)
câu hỏi là j pạn