(1981 x 1982 - 990) : (1980 x 1982 + 992)

=(1980 x 1982+1982 -990) : (1980 x 1982 +992)

=(1980 x 1982 + 992) : ( 1980 x 1982 + 992)

=1

B=[(45.79+45.21)]:90-5^2]:5+2^3                                  B=[(45.79+45.21):90-25]:5+8                                      B=[(45.(79+21):65]:13                                                  B=[(45.100):65]:13                                                        B=[4500:65]:13                                                           B=4500:65:13                                                       

B=[(45.79+45.21) :90-5²]:5+2³

  ={[45.(79+21)]:90-25}:5+8

  ={[45.100]:90-25}:5+8

  ={4500:90-25}:5+8

  ={50-25}:5+8

  =25:5+8

  =5+8=13

M = 2²⁰¹⁰-(2²⁰⁰⁹ + 2²⁰⁰⁸+....+2¹+2⁰

Đặt A =( 2²⁰⁰⁹+2²⁰⁰⁸+...2¹ +2⁰)

Suy ra 2A = 2²⁰¹⁰+2²⁰⁰⁹+2²⁰⁰⁸+...2²+2

Suy ra 2A-A = ( 2²⁰¹⁰+2²⁰⁰⁹+2²⁰⁰⁸+...+2²+2) - (2²⁰⁰⁹+ 2²⁰⁰⁸+...+2¹+2⁰)

Suy ra A= 2²⁰¹⁰-2⁰

Suy ra M = 2²⁰¹⁰-A=2^{2010 -} 2²⁰¹⁰+2⁰=1

Vậy M=1

Đúng nha

tui biet cach lam rui

hay qua

Ta có công thức tổng quát như sau:

\(A=n^k+n^{k+1}+n^{k+2}+...+n^{k+x}\Rightarrow A=\dfrac{n^{k+x+1}-n^k}{n-1}\)

Áp dụng ta có:

\(A=1+4+4^2+...+4^6=\dfrac{4^7-1}{3}\) 

\(\Rightarrow B-3A=4^7-3\cdot\dfrac{4^7-1}{3}=1\)

______

\(A=2^0+2^1+...+2^{2008}=2^{2009}-1\)

\(\Rightarrow B-A=2^{2009}-2^{2009}+1=1\)

_____

\(A=1+3+3^2+....+3^{2006}=\dfrac{3^{2007}-1}{2}\)

\(\Rightarrow B-2A=3^{2007}-2\cdot\dfrac{3^{2007}-1}{2}=1\)

\(A=1+2+2^2+...+2^{2017}\)

\(\Rightarrow A=\dfrac{2^{2017+1}-1}{2-1}\)

\(\Rightarrow A=2^{2018}-1\)

mà \(B=2^{2018}\)

\(\Rightarrow A-B=2^{2018}-1-2^{2018}\)

\(\Rightarrow A-B=-1\)

\(2A=2+2^2+2^3+...+2^{2018}\)

\(\Rightarrow A=2A-A=2^{2018}-1\)

\(\Rightarrow A-B=2^{2018}-1-2^{2018}=-1\)

\(a.x-143=57\)

\(x=200\)

\(b.\left(8x-12\right):4=3^3\)

\(8x-12=27.4\)

\(8x-12=108\)

\(8x=120\)

\(x=15\)

\(d.10+2x=4^2\)

\(2x=16-10\)

\(2x=6\)

\(x=3\)

a) 716-(x-143)=659

<=>x-143=57

<=>x=57+143

<=>x=200

b) [(8x-12):4].3³=3⁶

<=>(8x-12):4=3³

<=>8x-12=27.4

<=>8x-12=108

<=>8x=120

<=>x=5

c) -2<|x|≤1, x thuộc Z

Vì -2<|x|≤1

=>|x| thuộc {-1;0;1}

+)Với |x|=-1=>Vô lí

+)Với |x|=0=>x=0(tm)

+)Với |x|=1=>x=-1;1(tm) 

Vậy x thuộc {-1;1;0}

d) 10+2x=4⁵:4³

<=>10+2x=16

<=>2x=6

<=>x=3

e) 4^x+1+4⁰=65

<=>4^x+1+1=65

<=>4^x+1=64

<=>4^x+1=4³

<=>x+1=3

<=>x=2

g) 96-2(x+1)=-42

<=>2(x+1)=96+42

<=>2(x+1)=138

<=>x+1=69

<=>x=68

h) 4x-20=2⁵:2²

<=>4x-20=2³

<=>4x=8+20

<=>4x=28

<=>x=7

k) 8x-75=5x+21

<=>8x-5x=75+21

<=>3x=96

<=>x=32

i) [(8x-14):2-2].31=341

<=>(8x-14):2-2=11

<=>(8x-14):2=13

<=>8x-14=26

<=>8x=40

<=>x=5

Đặt A = 2⁰ + 2¹ + 2² + 2³ + ... + 2¹⁰

Ta có : 2A = 2¹ + 2² + 2³ + ... + 2¹¹

2A - A = (2¹ + 2² + 2³ + ... + 2¹¹) - (2⁰ + 2¹ + 2² + 2³ + ... + 2¹⁰)

A = 2¹¹ - 2⁰ = 2048 - 1 = 2047

2⁰ + 2¹ + 2² + 2³ + ....... + 2¹⁰

= 1 + 2¹ + 2² + 2³ + ........... + 2¹⁰

Ta biết từ 1 đến 10 có tổng là 55

=> 1 + 2⁵⁵ 

2⁰+2¹+2²+2³+2⁴+2⁵+2⁶+2⁷+2⁸+2⁹+2¹⁰

=(2¹+2⁹)+(2²+2⁸)+(2³+2⁷)+(2⁴+2⁶)+(2⁵+2⁰)

=2¹⁰+2¹⁰+2¹⁰+2¹⁰+2⁵+1

=2⁴⁵+1

=3⁴⁵

\(a,\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6=\left(\left(\frac{3}{7}\right)^2\right)^{10}.\frac{3}{7}:\left(\frac{9}{49}\right)^6=\left(\frac{9}{49}\right)^{10}.\frac{3}{7}:\left(\frac{9}{49}\right)^6\)

\(=\left(\left(\frac{9}{49}\right)^{10}:\left(\frac{9}{49}\right)^6\right).\frac{3}{7}=\left(\frac{9}{49}\right)^{10-6}.\frac{3}{7}=\left(\frac{9}{49}\right)^4.\frac{3}{7}=\left(\left(\frac{3}{7}\right)^2\right)^4.\frac{3}{7}\)

\(=\left(\frac{3}{2}\right)^8.\frac{3}{7}=\left(\frac{3}{2}\right)^9\)

\(b,3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2=3-1+\left(\frac{1}{2}\right)^2.\frac{1}{2}=2+\left(\frac{1}{2}\right)^3=2+\frac{1}{6}=2\frac{1}{6}\)

\(c,\left(0,125\right)^3.51=\left(\frac{1}{8}\right)^3.51=\frac{1^3}{8^3}.51=\frac{1}{512}.51=\frac{51}{512}\)