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Bài 2:
a: =>x/7=1/21
=>x=1/3
c: =>x(3x-2)=0
=>x=0 hoặc x=2/3
Bài1:
a: \(=\left(-\dfrac{7}{3}\right)^{3-2}=\dfrac{-7}{3}\)
b: \(=\left(-\dfrac{4}{9}\right)^{1-3}=\left(-\dfrac{4}{9}\right)^{-2}=\dfrac{81}{16}\)
c: \(=\left(\dfrac{1}{5}\right)^{10-7}=\left(\dfrac{1}{5}\right)^3=\dfrac{1}{125}\)
A=33. \(\left(1-\frac{2}{3}\right)\left(1-\frac{2}{5}\right)...\left(1-\frac{2}{99}\right)\)
A=33.\(\frac{1}{3}.\frac{3}{5}....\frac{97}{99}\)
A=33.\(\frac{1}{99}\)
A=\(\frac{33}{99}=\frac{1}{3}\)
1.
a) (—7/3)3:(—7/3)2=(—7/3)3–2=—7/3
b) (—4/9):(—4/9)3= (—4/9)1–3=(—4/9)—2=81/16
c) (1/5)10:(1/5)7=(1/5)10–7=(1/5)3=1/125
2.
a) —x/7 =1/—21
==> —x.(—21)=7.1
==> —x.(—21)=7
==> —x=7:(—21)
==> —x=—1/3
==> x=1/3
b) 4 2/5 . 0,5–1 3/7= 22/5 . 1/2 —10/7= 22.1/5.2–10/7= 11/5 —10/7= 77/35 — 50/35= 27/35
c) 3x2–2x=0
==> x3(3–2)=0
x3.1=0
x3=0:1
x3=0
==> x=0
c) 9x2–1=0
9x2=0+1
9x2=1
x2=1:9
x2=1/9
x2=12/32 hoặc x2=(—1/3)2
Vậy x=1/3 hoặc x=—1/3
1) ( \(\frac{55}{3}\): 15 + \(\frac{26}{3}\) . \(\frac{7}{2}\)) : [(\(\frac{37}{3}\) + \(\frac{62}{7}\)) . \(\frac{7}{18}\)] : \(\frac{-1704}{445}\)
= ( \(\frac{55}{3}\). \(\frac{1}{15}\) + \(\frac{91}{3}\)) : [ \(\frac{445}{21}\) . \(\frac{7}{18}\)] . \(\frac{-445}{1704}\)
= ( \(\frac{11}{9}\)+ \(\frac{91}{3}\)) : \(\frac{445}{54}\). \(\frac{-445}{1704}\) = \(\frac{284}{9}\). \(\frac{54}{445}\). \(\frac{-445}{1704}\)= \(\frac{284}{9}\). (\(\frac{54}{445}\). \(\frac{-445}{1704}\))
= \(\frac{284}{8}\). \(\frac{-9}{284}\)
= \(\frac{-9}{8}\)
a, Ta có: \(\frac{1}{2^2}>\frac{1}{2.3};\frac{1}{3^2}>\frac{1}{3.4};...;\frac{1}{10^2}>\frac{1}{10.11}\)
\(\Rightarrow S>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)
Vậy S > 9/22
b, Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{10^2}< \frac{1}{9.10}\)
\(\Rightarrow S>\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}=1-\frac{1}{10}=\frac{9}{10}\)
Vậy S > 9/10
\(S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(2S=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
\(2S-S=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)+\left(1+\frac{1}{2}+...+\frac{1}{2^{10}}\right)\)
\(2S-S=S=2-\frac{1}{2^{10}}\)
\(S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(2S=2\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(2S=3+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)
\(S=2S-S\)
\(S=3+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(S=3+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}-1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{10}}\)
\(S=2-\frac{1}{2^{10}}\)