23 + 3( 1/9)0 - 2 -2nhân 4 + [(-2)2:1/2] nhân 8
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AH
Akai Haruma
Giáo viên
5 tháng 2
Lời giải:
$A=9+2.3^2+2.3^3+2.3^4+...+2.3^{2023}$
$A-9=2(3^2+3^3+3^4+...+3^{2023})$
$3(A-9)=2(3^3+3^4+3^5+...+3^{2024})$
$\Rightarrow 3(A-9)-(A-9)=2(3^{2024}-3^2)$
$2(A-9)=2.3^{2024}-18$
$\Rightarrow 2A-18=2.3^{2024}-18$
$\Rightarrow A=3^{2024}\vdots 3^{2023}$ (đpcm)
31 tháng 1 2020
2 x 5 + 5 x 7 + 9 x 3
= 5 x ( 2 + 7 ) + 9 x 3
= 5 x 9 + 9 x 3
= 9 x ( 5 + 3 )
= 9 x 8
= 72
31 tháng 1 2020
2x5+5x9+9x3 15:5+27:5+8:5
=9x5+(2+3) =(15+8+27):5
=9x5+5 =50:5
=9x5+5x1 =10
=5x(9+1) 9+9x3+18:2x6
=5x10 =1x9+9x3+9x6
=50 =9x(1+6+3)
=9x10
=90
12 tháng 6 2017
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)
\(\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{23}{45}\)
\(\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{23}{45}\)
\(\frac{22}{45}.x=\frac{23}{45}\)
\(x=\frac{23}{45}:\frac{22}{45}\)
\(x=\frac{23}{22}\)
3 tháng 10 2018
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)
\(\Rightarrow\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).\frac{x}{2}=\frac{23}{45}\)
\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{46}{45}\)
\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{46}{45}\)
\(\Rightarrow\frac{22}{45}.x=\frac{46}{45}\Rightarrow x=\frac{46}{22}=\frac{23}{11}\)
KV
25 tháng 9 2023
2) -3(4 - 7) + 5(-3 + 2)
= -3.4 + 3.7 - 5.3 + 5.2
= -12 + 21 -15 + 10
= 31 - 27
= 4
4) -5(2 - 7) + 4(2 - 5)
= -5.2 + 5.7 + 4.2 - 4.5
= -10 + 35 + 8 - 20
= 38 - 30
= 8
11 tháng 6 2018
a) \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}\)
\(=1-\frac{1}{4}\)
\(=\frac{3}{4}\)
b) \(\frac{4}{7}\times\frac{8}{9}+\frac{4}{7}\times\frac{1}{9}\)
\(=\frac{4}{7}\times\left(\frac{8}{9}+\frac{1}{9}\right)\)
\(=\frac{4}{7}\times1\)
\(=\frac{4}{7}\)
c) \(\frac{13}{6}\times\frac{3}{8}-\frac{3}{8}\times\frac{7}{6}\)
\(=\frac{3}{8}\times\left(\frac{13}{6}-\frac{7}{6}\right)\)
\(=\frac{3}{8}\times1\)
\(=\frac{3}{8}\)
d) \(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{9x10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
\(2^3+3\cdot\left(\dfrac{1}{9}\right)^0-2^{-2}\cdot4+\left[\left(-2\right)^2:\dfrac{1}{2}\right]\cdot8\)
\(=8+3\cdot1-\dfrac{1}{4}\cdot4+\left(4:\dfrac{1}{2}\right)\cdot8\)
\(=8+3-\dfrac{4}{4}+4\cdot2\cdot8\)
\(=11-1+8\cdot8\)
\(=10+64\)
\(=74\)
\(=8+3-2^{-2}\cdot2^2+\left[4\cdot2\right]\cdot8\)
=11-1+8*8
=64+10=74