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\(a)\) Ta có :
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(A=1-\frac{1}{2^{100}}< 1\)
Vậy \(A< 1\)
Chúc bạn học tốt ~
\(2.THPT\)
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(A=9\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=9\left(1-\frac{1}{100}\right)\)
\(A=9.\frac{99}{100}\)
\(A=\frac{891}{100}\)
\(B=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{93.95}\)
\(B=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)
\(B=\frac{1}{5}-\frac{1}{95}\)
\(B=\frac{18}{95}\)
\(D=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(D=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(D=\frac{1}{2}-\frac{1}{28}\)
\(D=\frac{13}{28}\)
\(a,\left(4\frac{1}{2}-\frac{2}{5}x\right):1\frac{3}{4}=\frac{11}{14}\)
\(\Rightarrow\left(\frac{9}{2}-\frac{2}{5}x\right):\frac{7}{4}=\frac{11}{4}\)
\(\Rightarrow\left(\frac{9}{2}-\frac{2}{5}x\right)=\frac{11}{4}\cdot\frac{7}{4}\)
\(\Rightarrow\left(\frac{9}{2}-\frac{2}{5}x\right)=\frac{77}{16}\)
\(\Rightarrow\frac{9}{2}-\frac{2}{5}x=\frac{77}{16}\)
\(\Rightarrow-\frac{2}{5}x=\frac{77}{16}-\frac{9}{2}\)
\(\Rightarrow-\frac{2}{5}x=\frac{5}{16}\)
\(\Rightarrow x=\frac{5}{16}:\left(-\frac{2}{5}\right)\)
\(\Rightarrow x=-\frac{25}{32}\)
\(b,\frac{2}{3}\cdot x-\frac{2}{5}x=\frac{9}{3}\)
\(\Rightarrow x\left(\frac{2}{3}-\frac{2}{5}\right)=\frac{8}{3}\)
\(\Rightarrow x\cdot\frac{4}{15}=\frac{8}{3}\)
\(\Rightarrow x=\frac{8}{3}:\frac{4}{15}\)
\(\Rightarrow x=10\)
\(c,\frac{-2}{3}|x|+1\frac{1}{2}=\frac{2}{5}\)
\(\Rightarrow\frac{-2}{3}|x|+\frac{3}{2}=\frac{2}{5}\)
\(\Rightarrow\frac{-2}{3}|x|=\frac{2}{5}-\frac{3}{2}\)
\(\Rightarrow\frac{-2}{3}|x|=-\frac{11}{10}\)
\(\Rightarrow|x|=\frac{-11}{10}:\frac{-2}{3}\)
\(\Rightarrow|x|=\frac{33}{20}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{33}{20}\\x=-\frac{33}{20}\end{cases}}\)
\(d,|2x-\frac{1}{3}|+\frac{1}{6}=\frac{3}{4}\)
\(\Rightarrow|2x-\frac{1}{3}|=\frac{3}{4}-\frac{1}{6}\)
\(\Rightarrow|2x-\frac{1}{3}|=\frac{7}{12}\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{3}=\frac{7}{12}\\2x-\frac{1}{3}=-\frac{7}{12}\end{cases}\Rightarrow\orbr{\begin{cases}2x=\frac{11}{12}\\2x=-\frac{1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{11}{24}\\x=-\frac{1}{8}\end{cases}}}\)
a) \(22\frac{1}{2}\cdot\frac{7}{9}+50\%-1,25\)
\(=\frac{45}{2}\cdot\frac{7}{9}+\frac{50}{100}-\frac{125}{100}\)
\(=\frac{5}{2}\cdot\frac{7}{1}+\frac{1}{2}-\frac{5}{4}\)
\(=\frac{35}{2}+\frac{1}{2}-\frac{5}{4}=18-\frac{5}{4}=\frac{67}{4}\)
b) \(1,4\cdot\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right):2\frac{1}{5}\)
\(=\frac{7}{5}\cdot\frac{15}{49}-\frac{22}{15}:\frac{11}{15}\)
\(=\frac{1}{1}\cdot\frac{3}{7}-\frac{22}{15}\cdot\frac{15}{11}\)
\(=\frac{3}{7}-2=\frac{3-14}{7}=\frac{-11}{7}\)
c) \(\left(-\frac{1}{2}\right)^2-\frac{7}{16}:\frac{7}{4}+75\%\)
\(=\frac{1}{4}-\frac{7}{16}\cdot\frac{4}{7}+\frac{75}{100}\)
\(=\frac{1}{4}-\frac{1}{4}+\frac{3}{4}=\frac{3}{4}\)
Bài 2 Bạn tự làm nhé
1.a,\(22\frac{1}{2}.\frac{7}{9}+50\%-1,25\)
\(=\frac{45}{2}.\frac{7}{9}+\frac{1}{2}-\frac{5}{4}\)
\(=\frac{35}{2}+\frac{1}{2}-\frac{5}{4}\)
\(=\frac{67}{4}\)
b,Các phép tính khác làm tương tự
Đổi các số ra hết thành phân số,có ngoặc thì lm ngoặc trc,Xoq đến nhân chia trước dồi mới cộng trừ
c,tương tự
2.
a,\(1\frac{3}{5}+\frac{7}{12}\div x=\frac{-9}{4}\)
\(\frac{8}{5}+\frac{7}{12}\div x=\frac{-9}{4}\)
\(\frac{7}{12}\div x=\frac{-77}{20}\)
Đến đây dễ bạn tự làm
b,\(\left(2\frac{4}{5}.x+50\right)\div\frac{2}{3}=-51\)
\(\left(\frac{14}{5}x+50\right)\div\frac{2}{3}=-51\)
\(\frac{14}{5}x+50=-34\)
\(\frac{14}{5}x=-84\)
Tự làm tiếp
c,\(\left|\frac{3}{4}.x-\frac{1}{2}\right|=\frac{1}{4}\)\(\Rightarrow\left|\frac{3}{4}x-\frac{1}{2}\right|=\varnothing\)
a, \(\frac{1}{2}-\frac{3}{5}x=4-\frac{1}{3}x\)
<=> \(\frac{1}{2}-\frac{3}{5}x+\frac{1}{3}x=4\)
<=>\(\frac{1}{2}-x.\left(\frac{3}{5}-\frac{1}{3}\right)=4\)
<=>\(\frac{1}{2}-\frac{4}{15}x=4\)
<=>\(\frac{4}{15}x=\frac{1}{2}-4\)
<=>\(\frac{4}{15}x=\frac{-7}{2}\)
<=> x = \(\frac{-7}{2}:\frac{4}{15}\)
<=> x = \(\frac{-7}{2}.\frac{15}{4}\)
<=> x = \(\frac{-105}{8}\)
b,\(\left(x^2-5\right).x^2=0\)
<=> \(x^2-5=0:x^2\)
<=>\(x^2-5=0\)
<=> \(x^2=5\)
<=> x = 5:x
c, 2 . I x - \(\frac{1}{2}\)I = \(\frac{-1}{3}+5\frac{1}{3}\)
<=>2 . I x - \(\frac{1}{2}\)I = \(\frac{-1}{3}+\frac{5}{3}\)
<=>2 . I x - \(\frac{1}{2}\)I = \(\frac{4}{3}\)
<=> I x - \(\frac{1}{2}\)I = \(\frac{4}{3}:2\)
<=> I x - \(\frac{1}{2}\)I = \(\frac{4}{3}.\frac{1}{2}\)
<=> I x - \(\frac{1}{2}\)I = \(\frac{2}{3}\)
=> x - \(\frac{1}{2}\)= \(\frac{2}{3}\)hoặc x - \(\frac{1}{2}\)= \(\frac{-2}{3}\)
TH1: x -\(\frac{1}{2}\) = \(\frac{2}{3}\)
<=> x = \(\frac{2}{3}\)+ \(\frac{1}{2}\)
<=> x = \(\frac{7}{6}\)
TH2: x - \(\frac{1}{2}\)= \(\frac{-2}{3}\)
<=> x = \(\frac{-2}{3}\)+ \(\frac{1}{2}\)
<=> x = \(\frac{-1}{6}\)
d) I 2x - 3 I - x = 6
=> 2x - 3 - x = 6 hoặc 2x - 3 - x = - 6
TH1:2x - 3 - x = 6
<=> x - 3 = 6
<=> x = 6 + 3
<=> x = 9
TH2: 2x - 3 - x = - 6
<=> x - 3 = -6
<=> x = - 6 + 3
<=> x = - 3
+ I 2x - 3 I