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Bài 1:
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + (\(\frac45\) - \(\frac{3}{17}\) + \(\frac13\)) - \(\frac17\) + (- \(\frac{14}{30}\))
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + \(\frac45\) - \(\frac{3}{17}\) + \(\frac13\) - \(\frac17\) - \(\frac{14}{30}\)
A = (\(\frac15\) + \(\frac45\)) + (\(\frac{3}{17}\) - \(\frac{3}{17}\)) - (\(\frac43-\frac13\)) - \(\frac{30}{210}\) - \(\frac{98}{210}\)
A = 1 + 0 - 1 - (\(\frac{30}{210}+\frac{98}{210}\))
A = 1 - 1 - \(\frac{228}{210}\)
A = 0 - \(\frac{128}{210}\)
A = - \(\frac{64}{105}\)
Bài 2:
B= (\(\frac58\) - \(\frac{4}{12}\) + \(\frac32\)) - (\(\frac58\) + \(\frac{9}{13}\)) - (\(\frac{-3}{2}\)) + \(\frac{7}{-15}\)
B = \(\frac58\) - \(\frac{4}{12}\) + \(\frac32\) - \(\frac58\) - \(\frac{9}{13}\) + \(\frac32\) - \(\frac{7}{15}\)
B = (\(\frac58\) - \(\frac58\)) + (\(\frac32\) + \(\frac32\)) - (\(\frac13\) + \(\frac{9}{13}\) + \(\frac{7}{15}\))
B = 0 + 3 - (\(\frac{65}{195}\) + \(\frac{135}{195}\) + \(\frac{91}{195}\))
B = 3 - (\(\frac{200}{195}\) + \(\frac{91}{195}\))
B = 3 - \(\frac{97}{65}\)
B = \(\frac{195}{65}\) - \(\frac{97}{65}\)
B = \(\frac{98}{65}\)
a: \(A=\left(\dfrac{15}{34}+\dfrac{9}{34}-1-\dfrac{15}{17}\right)+\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)
\(=\left(\dfrac{12}{17}-1-\dfrac{15}{17}\right)+1\)
\(=\dfrac{-20}{17}+1=\dfrac{-3}{17}\)
b: \(B=\dfrac{-5}{3}\cdot16\dfrac{2}{7}-\dfrac{-5}{3}\cdot28\dfrac{2}{7}\)
\(=\dfrac{-5}{3}\left(16+\dfrac{2}{7}-28-\dfrac{2}{7}\right)=\dfrac{-5}{3}\cdot\left(-12\right)=20\)
c: \(C=25\cdot\dfrac{-1}{27}+\dfrac{1}{5}-2\cdot\dfrac{1}{4}-\dfrac{1}{2}\)
\(=\dfrac{-25}{27}+\dfrac{1}{5}-1\)
\(=\dfrac{-125+27-135}{135}=\dfrac{-233}{135}\)
1. sai dấu nhé
2.a, \(\frac{45^{10}.5^{20}}{75^{15}}=\frac{\left(3^2.5\right)^{10}.5^{20}}{\left(5^2.3\right)^{15}}=\frac{3^{20}.5^{30}}{5^{30}.3^{15}}=3^5=243\)
b, \(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(\frac{4}{5}\right)^5}{\left(\frac{2}{5}\right)^6}=\frac{\left(\frac{2}{5}\cdot2\right)^5}{\left(\frac{2}{5}\right)^6}=\frac{\left(\frac{2}{5}\right)^5\cdot2^5}{\left(\frac{2}{5}\right)^5\cdot\frac{2}{5}}=2^5\div\frac{2}{5}=32\cdot\frac{5}{2}=80\)
c, \(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}=\frac{2^{15}.3^2}{2^{15}}=3^2=9\)
câu 1:
ta có \(3=\frac{2\cdot3}{2}\)
\(6=\frac{3\cdot4}{2}\)
... \(45=\frac{9\cdot10}{2}\)
\(\frac{2n+1}{\frac{n\left(n+1\right)}{2}}=\frac{2\left(2n+1\right)}{n\left(n+1\right)}\)
mà \(\frac{2n+1}{n\left(n+1\right)}=\frac{1}{n}+\frac{1}{n+1}\)
=> \(\frac{2\left(2n+1\right)}{n\left(n+1\right)}=2\left(\frac{1}{n}+\frac{1}{n+1}\right)\)
thay vào lại biểu thức ta có:
ta có: \(\) \(M=2-\frac{2\cdot5}{2\cdot3}+\frac{2\cdot7}{3\cdot4}-\frac{2\cdot9}{4\cdot5}+\cdots+\frac{2\cdot19}{9\cdot10}\)
\(M=2\left\lbrack1-\left(\frac12+\frac13\right)+\left(\frac13+\frac14\right)-\left(\frac14+\frac15\right)+\cdots+\left(\frac18+\frac19\right)-\left(\frac19+\frac{1}{10}\right)\right\rbrack\) \(M=2\left\lbrack1-\frac12-\frac{1}{10}\right\rbrack\)
\(M=2\cdot\frac{4}{10}=\frac45\)
câu 2:
\(\Leftrightarrow3A=-1+\frac13-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+\cdots+\frac{1}{3^{99}}\)
\(\Rightarrow3A+A=\left(-1+\frac13-\frac{1}{3^2}+\cdots+\frac{1}{3^{99}}\right)+\left(-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots+\frac{1}{3^{100}}\right)\)
\(4A=-1+\frac{1}{3^{100}}\)
=> \(A=\frac{-1+\frac{1}{3^{100}}}{4}\)
vì \(\frac{1}{3^{100}}<\frac13\)
=> \(-1+\frac{1}{3^{100}}<-1+\frac13=-\frac23<0\)
=> A<0
=> \(\left\vert A\right\vert=-\left(\frac{-1+\frac{1}{3^{100}}}{4}\right)=\frac{1-\frac{1}{3^{100}}}{4}\)
nhân cả hai vế với 4
\(4\left\vert A\right\vert=1-\frac{1}{3^{100}}\)
\(B=4\left\vert A\right\vert=\frac{1}{3^{100}}\)
\(B=\left(1-\frac{1}{3^{100}}\right)+\frac{1}{3^{100}}\)
\(B=1\)
vậy B=1
Câu 1.
M = 2 - 5/3 + 7/6 - 9/10 + 11/15 - 13/21 + 15/28 - 17/36 + 19/45
M = 6/5
Vì quy đồng và rút gọn các phân số ta được M = 6/5
Câu 2.
A = -1/3 + 1/3^2 - 1/3^3 + ... + 1/3^100
A = -1/4.(1 - 1/3^100)
|A| = 1/4.(1 - 1/3^100)
B = 4|A| + 1/3^100
B = 1 - 1/3^100 + 1/3^100 = 1
Vậy B = 1