Lời giải:

$A=(99-97)+(95-93)+(91-89)+...+(7-5)+(8-1)$

$=\underbrace{2+2+2+...+2}_{24}+7$

$=2\times 24+7$

$=55$

\(a,=72\cdot\left(-200\right)=-14400\\ b,=\left(125\cdot8\right)\left[\left(-5\right)\left(-20\right)\right]=1000\cdot100=1000000\)

= -14400

= 1000000

a)72.(-50).(-2).(-2)

= 72.[(-50).(-2)].(-2)

= 72. 100.(-2)

=7200.(-2)=-14400

b)=125.(-5).(-20).8

=(125.8).[(-5).(-20)]

=1000.100=100000

Xét khai triển:

\(\left(1+x\right)^{90}=C_{90}^0+C_{90}^1x+C_{90}^2x^2+...+C_{90}^{90}x^{90}\)

Thay \(x=2\) ta được:

\(3^{90}=C_{90}^0+2C_{90}^1+2^2C_{90}^2+...+2^{90}C_{90}^{90}\)

Vậy \(B=3^{90}\)

\(X=\left(a+b\right)^n=\sum\limits^n_{k=0}C^k_n.a^k.b^{n-k}\)

\(\Rightarrow\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\)

\(\Rightarrow A=\sum\limits^{90}_{k=2}C^k_{90}.2^k=...\)

Hoặc có thể làm như vầy: \(A=X-C^0_{90}.2^0-C^1_{90}.2=3^{90}-1-90.2=...\)

a: \(=27\left(77+15\right)+27\cdot8\)

\(=27\cdot92+27\cdot8=27\cdot100=2700\)

b: \(=75\cdot89+75\cdot9+2\cdot75=75\left(89+9+2\right)=75\cdot100=7500\)

c: \(=0.25\left(42.9-11.7+0.8\right)=0.25\cdot32=8\)

d: 738 phút=12,3 giờ

b: =100x1000=100000

a)=\(\left[72.\left(-2\right)\right].\left[-50.\left(-2\right)\right]\)

=-144.100

=-14400

b)=(125.8).(-5).(-20)

=1000.100

=100000

c)=\(\left[125.\left(-4\right)\right]\).(-3).(-32)

=-500.96

=-48000

d)=-492.125

=-60500

Bài 1: 

\(101\cdot125+101\cdot25-101\cdot50\)

\(=101\cdot\left(125+25-50\right)\)

\(=101\cdot100\)

\(=10100\)

Bài 2: 

\(76\cdot115+56\cdot24+59\cdot24\)
\(=76\cdot115+24\cdot\left(56+59\right)\)

\(=76\cdot115+24\cdot115\)

\(=115\cdot\left(76+24\right)\)

\(=115\cdot100\)

\(=11500\)

5:

a: =>15x=165

=>x=11

b: =>x(x+1)/2=55

=>x^2+x=110

=>x=10

4: =>7x=56

=>x=8

Bài 1:

101•125+101•25+101•50

= 101•(125+25-50)

=101•100

=10100

Bài 2:

76•115+56•24+59•24

= 76•115+24•(56+59)

= 76•115+24•115

= 115•(76+24)

= 115•100

= 11500

A = (1 -1/2) + (1 - 1/6) + (1 - 1/12)  + (1 - 1/20 ) + ...+ (1 - 1/ 90)

= (1+1+1+1+1+1+1+1+1) - ( 1/2 - 1/6 - 1/12 - 1/ 20 - ...- 1/90)\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)\(=9-\left(1-\frac{1}{10}\right)=\frac{81}{10}\)

dap an dung la 81/10

\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)

\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)

\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)

\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)

\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)

\(=9-\left(1-\frac{1}{10}\right)\)

\(=9-\frac{9}{10}=\frac{81}{10}\)

\(9-A=1-\frac{1}{2}+1-\frac{5}{6}+1-\frac{11}{12}+1-\frac{19}{20}+...+1-\frac{89}{90}\)

\(\Leftrightarrow9-A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)

\(\Leftrightarrow9-A=\frac{2-1}{1\cdot2}+\frac{3-2}{2\cdot3}+\frac{4-3}{3\cdot4}+...+\frac{10-9}{9\cdot10}\)

\(\Leftrightarrow9-A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=1-\frac{1}{10}=\frac{9}{10}\)

\(\Leftrightarrow A=9-\frac{9}{10}=\frac{81}{10}\)