2 | + | 2 | + … + | 2 | = | |
3 . 5 | 5 . 7 | 155 . 157 |
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(542\cdot7+542\cdot2+542\)
\(=542\left(7+2+1\right)\)
\(=542\cdot10=5420\)
b: \(\left(126\cdot35+157\cdot55\right)\cdot\left(132-6\cdot22\right)\)
\(=\left(126\cdot35+157\cdot55\right)\left(132-132\right)\)
\(=0\cdot\left(126\cdot35+157\cdot55\right)\)
=0
c: \(897\cdot99+897\)
\(=897\left(99+1\right)\)
\(=897\cdot100=89700\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 542 x 7 + 542 x 2 + 542
= 542 x 7 + 542 x 2 + 542 x 1
= 542 x (7 + 2 + 1)
= 542 x 10
= 5420
b) (126 x 35 + 157 x 55) x (132 - 6 x 22)
= (126 x 35 + 157 x 55) x (132 - 132)
= (126 x 35 + 157 x 55) x 0 (Số bất kỳ nào nhân với 0 vẫn = 0)
= 0
c) 897 x 99 + 897
= 897 x 99 + 897 x 1
= 897 x (99 + 1)
= 897 x 100
= 89700
a) \(542.7+542.2+542\)
\(=542.\left(2+7+1\right)\)
\(=542.10=5420\)
b) \(\left(126.35+157.55\right).\left(132-6.22\right)\)
\(=13045.0=0\)
c) \(897.99+897\)
\(=897.\left(99+1\right)\)
\(=897.100=89700\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1: Tính
a) Ta có: \(\left(-25\right)\cdot68+\left(-34\right)\cdot\left(-250\right)\)
\(=-25\cdot68+\left(-340\right)\cdot\left(-25\right)\)
\(=-25\cdot\left(68-340\right)\)
\(=-25\cdot\left(-272\right)\)
\(=6800\)
b) Ta có: \(1999+\left(-2000\right)+2001+\left(-2002\right)\)
\(=1999-2000+2001-2002\)
\(=-1-1=-2\)
c) Ta có: \(515+\left[72+\left(-515\right)+\left(-32\right)\right]\)
\(=515+72-515-32\)
\(=40\)
d) Ta có: \(\left(2736-75\right)-2736+175\)
\(=2736-75-2736+175\)
\(=100\)
e) Ta có: \(-2020-\left(157-2020\right)-\left(-257\right)\)
\(=-2020-157+2020+257\)
\(=100\)
Bài 2: Tìm x
a) Ta có: \(x-\left|-2\right|=\left|-18\right|\)
\(\Leftrightarrow x-2=18\)
hay x=20
Vậy: x=20
b) Ta có: \(2x-\left|+14\right|=\left|-14\right|\)
\(\Leftrightarrow2x-14=14\)
\(\Leftrightarrow2x=28\)
hay x=14
Vậy: x=14
c) Ta có: \(\left|x+4\right|+5=20-\left(-12-7\right)\)
\(\Leftrightarrow\left|x+4\right|+5=20+12+7\)
\(\Leftrightarrow\left|x+4\right|=39-5=34\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=34\\x+4=-34\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=30\\x=-38\end{matrix}\right.\)
Vậy: x∈{30;-38}
d) Ta có: \(15-\left|2-x\right|=\left(-2\right)^2\)
\(\Leftrightarrow15-\left|2-x\right|=4\)
\(\Leftrightarrow\left|2-x\right|=11\)
\(\Leftrightarrow\left[{}\begin{matrix}2-x=11\\2-x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=13\end{matrix}\right.\)
Vậy: x∈{-9;13}
e) Ta có: \(\left|15-x\right|+\left|-25\right|=\left|-55\right|\)
\(\Leftrightarrow\left|15-x\right|+25=55\)
\(\Leftrightarrow\left|15-x\right|=30\)
\(\Leftrightarrow\left[{}\begin{matrix}15-x=30\\15-x=-30\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-15\\x=45\end{matrix}\right.\)
Vậy: x∈{-15;45}
g) Ta có: \(\left|17-\left(-4\right)\right|+\left|-24-\left(-5\right)\right|=\left|-x+3\right|\)
\(\Leftrightarrow\left|17+4\right|+\left|-24+5\right|=\left|3-x\right|\)
\(\Leftrightarrow\left|3-x\right|=40\)
\(\Leftrightarrow\left[{}\begin{matrix}3-x=40\\3-x=-40\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-37\\x=43\end{matrix}\right.\)
Vậy: x∈{-37;43}
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
+A= 1+2+2^2 +...+2^196
A= (1+2)+(2^2 +2^3) +...+(2^195 +2^196)
A= 1.3+2^2 .3+...+2^195 .3
A= 3(1+...+2^195)=> A chia hết cho 3
A=1+2+2^2+...+2^195+2^196
A= (1+2+2^2)+...+(2^194 +2^195 +2^196)
A= 1.7 +...+2^194 .7
A=7(1+...+2^194)=> A chia hết cho 7
+ta có : 3^1993 luôn luôn lẻ ;2^157 luôn luôn chan
=> 3^1993 - 2^157 là 1 số lẻ
=> ko chia hết cho 2
![](https://rs.olm.vn/images/avt/0.png?1311)
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![](https://rs.olm.vn/images/avt/0.png?1311)
1: \(=75\left(27+25-2\right)=75\cdot50=3750\)
2: \(=15\left(23+37\right)+55=15\cdot60+55=955\)
3: \(=36\cdot14+36\cdot17+36\cdot69\)
\(=36\cdot100=3600\)
4: \(=200\cdot\left(32+68\right)=200\cdot100=20000\)
\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{155.157}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{155}-\frac{1}{157}\)
\(=\frac{1}{3}-\frac{1}{157}=\frac{154}{471}\)