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Thay m=3, n=-3, ta được:
-10 * 3 + 5 * (-3) - 3 * (-3)
= -30 - 15 + 9
= -36
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{2}\left(x-2\right)+\dfrac{1}{3}\left(2-x\right)=x\\ \Leftrightarrow\dfrac{1}{2}\left(x-2\right)-\dfrac{1}{3}\left(x-2\right)=x\\ \Leftrightarrow\left(x-2\right).\left(\dfrac{1}{2}-\dfrac{1}{3}\right)=x\\ \Leftrightarrow\left(x-2\right).\left(\dfrac{3-2}{6}\right)=x\\ \Leftrightarrow\left(x-2\right).\dfrac{1}{6}=x\\ \Leftrightarrow\dfrac{1}{6}x-\dfrac{1}{3}-x=0\\ \Leftrightarrow\left(\dfrac{1}{6}-1\right)x=\dfrac{1}{3}\\ \Leftrightarrow\left(\dfrac{1-6}{6}\right)x=\dfrac{1}{3}\\ \Leftrightarrow\dfrac{-5}{6}x=\dfrac{1}{3}\\ \Leftrightarrow x=\dfrac{1}{3}:\left(-\dfrac{5}{6}\right)\\ \Leftrightarrow x=-\dfrac{2}{5}\)
Vậy \(x=-\dfrac{2}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Do A có 30 số hạng, ta nhóm 3 số thành 1 nhóm nên vừa đủ 10 nhóm và không dư số nào.
A = 2 + 2^2 + 2^3 + 2^4 + ... + 2^30
= (2+2^2+2^3)+(2^4+2^5+2^6)+...+(2^28+2^29+2^30)
= 2(1+2+2^2)+2^4(1+2+2^2)+...+2^28(1+2+2^2)
= 2.7 + 2^4 .7 + ... + 2^28 .7
= 7(2+2^4+...+2^28) chia hết cho7 (DPCM)
![](https://rs.olm.vn/images/avt/0.png?1311)
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:
\(A=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)\cdot...\cdot\left(1-\dfrac{1}{900}\right)\)
\(=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{30}\right)\cdot\left(1+\dfrac{1}{2}\right)\cdot\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1+\dfrac{1}{30}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{29}{30}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{31}{30}\)
\(=\dfrac{1}{30}\cdot\dfrac{31}{2}=\dfrac{31}{60}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}$
$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{20-19}{19.20}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}$
$=1-\frac{1}{20}=\frac{19}{20}$
![](https://rs.olm.vn/images/avt/0.png?1311)
428=22.107
422=2.211
115=5.23
180=22.32.5
160=25.5
190=2.5.9
250=2.53
350=2.52.7
324=22.34
364=22.7.13
270=2.33.5
290=2.5.29
120=23.3.5
150=2.3.52
160=25.5
\(428=2^2\cdot107\)
\(422=2\cdot211\)
\(115=5\cdot23\)
\(180=2^2\cdot3^2\cdot5\)
\(160=2^5\cdot5\)
\(190=2\cdot5\cdot19\)
\(250=2\cdot5^3\)
\(350=2\cdot5^2\cdot7\)
\(324=2^2\cdot3^4\)
\(364=2^2\cdot7\cdot13\)
\(270=3^3\cdot2\cdot5\)
\(290=2\cdot5\cdot29\)
\(120=2^3\cdot3\cdot5\)
\(150=5^2\cdot2\cdot3\)
\(160=2^5\cdot5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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}
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c)\(\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)....\left(1+\dfrac{1}{2020}\right)\left(1+\dfrac{1}{2021}\right)\)
\(=\left(\dfrac{1.2}{1.2}+\dfrac{1}{2}\right)\left(\dfrac{1.3}{1.3}+\dfrac{1}{3}\right)...\left(\dfrac{1.2021}{1.2021}+\dfrac{1}{2021}\right)\)
\(=\dfrac{3}{1.2}\cdot\dfrac{4}{1.3}\cdot\cdot\cdot\cdot\dfrac{2022}{1.2021}\)
\(=\dfrac{3.4.5...2022}{\left(1.1.1....1\right)\left(2.3.4...2021\right)}\)
\(=\)\(\dfrac{3.4.5...2022}{2.3.4...2021}\)
\(=\dfrac{2022}{2}=1011\)
\(d\))\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)....\left(1-\dfrac{1}{199}\right)\left(1-\dfrac{1}{200}\right)\)
\(=\left(\dfrac{2}{1.2}-\dfrac{1}{1.2}\right)\left(\dfrac{3}{1.3}-\dfrac{1}{1.3}\right)....\left(\dfrac{200}{1.200}-\dfrac{1}{1.200}\right)\)
\(=\dfrac{1.2.3....199}{\left(1.1.1....1\right).\left(2.3.4....200\right)}\)
\(=\dfrac{1.2.3...199}{2.3.4...200}\)
Nếu mik làm sai mong bạn thông cảm
\(1\frac{1}{3}\times50\%+\left(\frac{8}{15}-\frac{19}{30}\right)\)
= \(\frac{4}{3}\times\frac{1}{2}+\left(\frac{16}{30}-\frac{19}{30}\right)\)
= \(\frac{2}{3}+\left(-\frac{1}{10}\right)\)
= \(\frac{17}{30}\)