Câu 1: 180 = 22 x 32 x 5
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\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
2.
a, x-13=-46
=>x=(-46)+13
=>x=33
b, 4x-6=22
=>4x=22+6
=>4x=28
=>x=28:4
=>x=7
3.
a, 32=25
48=24.3
=>ƯCLN(32,48)=24=16
16=24
72=23.32
=>ƯCLN(16,72)=23=8
b,
24=23.3
60=22.3.5
=>BCNN(24,60)=23.3.5=120
72=23.32
180=22.32.5
=>BCNN(72,180)=23.32.5=360
98775 - 32 x 85
=98775 -2720
=96055
67500 - 24 x 236
= 67500 -5664
=61836
568 + 101598 : 287
= 568 +354
=922
6875 + 980 -180
=7855 -180
=7675
\(\dfrac{2}{5}+\dfrac{3}{10}-\dfrac{1}{2}\)
\(=\dfrac{7}{10}-\dfrac{1}{2}\)
= \(\dfrac{1}{5}\)
\(\dfrac{8}{11}+\dfrac{8}{33}x\dfrac{3}{4}\)
\(=\dfrac{8}{11}+\dfrac{2}{11}\)
\(=\dfrac{10}{11}\)
\(\dfrac{7}{9}x\dfrac{3}{14}:\dfrac{5}{8}\)
\(=\dfrac{1}{6}:\dfrac{5}{8}\)
\(=\dfrac{1}{6}x\dfrac{8}{5}\)
\(=\dfrac{8}{30}\)
\(=\dfrac{4}{15}\)
\(\dfrac{5}{12}-\dfrac{7}{32}:\dfrac{21}{16}\)
\(=\dfrac{5}{12}-\dfrac{7}{32}x\dfrac{16}{21}\)
\(=\dfrac{5}{12}-\dfrac{1}{6}\)
\(=\dfrac{5}{12}-\dfrac{2}{12}\)
\(=\dfrac{3}{12}=\dfrac{1}{4}\)
2^2.(x+3^2)-5=55
4.(x+9)=55+5=60
(x+9)=60:4=15
x=15-9=6
vậy x=6
180 : ( 18 x 5 )
= 180 : 90
= 2
b) 240 : ( 2 x 3 x 4 )
= 240 : 24
= 10
c) 4400 : ( 22 x 35 )
= 4400 : 770
= \(\dfrac{40}{7}\)
\(22.\left(x+32\right)-5=55.179^0\\ 22.\left(x+32\right)-5=55.1\\ 22.\left(x+32\right)-5=55\\ 22.\left(x+32\right)=55+5\\ 22.\left(x+32\right)=60\\ x+32=60:22\\ x+32=\dfrac{30}{11}\\ x=\dfrac{30}{11}-32\\ x=\dfrac{-322}{11}\)
\(22.\left(x+32\right)-5=55.179^0\)
=>22.(x+32)-5=55.1
=>22.(x+32)-5=55
=>22.(x+32)=55+5
=>22.(x+32)=60
=>x+32=60:22
=>x+32=\(\dfrac{30}{11}\)
=>x=\(\dfrac{30}{11}-32\)
\(=>x=-\dfrac{322}{11}\)
Vậy............
đề bài là gì vậy ah?