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1) \(x+\dfrac{30}{100}x=-1,31\)
\(\Leftrightarrow x+\dfrac{3}{10}x=-\dfrac{131}{100}\)
\(\Leftrightarrow100x+30x=-131\)
\(\Leftrightarrow130x=-131\)
\(\Leftrightarrow x=-\dfrac{131}{130}\)
Vậy \(x=-\dfrac{131}{130}\)
b) \(\left(4,5-2x\right)\cdot\left(-1\dfrac{4}{7}\right)=\dfrac{11}{4}\)
\(\Leftrightarrow\left(\dfrac{9}{2}-2x\right)\cdot\left(-\dfrac{4}{7}\right)=\dfrac{11}{4}\)
\(\Leftrightarrow-\dfrac{18}{7}+\dfrac{8}{7}x=\dfrac{11}{4}\)
\(\Leftrightarrow-72+32x=77\)
\(\Leftrightarrow32x=77+72\)
\(\Leftrightarrow32x=149\)
\(\Leftrightarrow x=\dfrac{149}{32}\)
Vậy \(x=\dfrac{149}{32}\)
219 - 7 . ( x+1) = 100
7 . ( x + 1 ) = 219 - 100 = 119
x+ 1= 119:7= 17
x = 17-1=16
(3x - 6 ). 3100 = 3103 2018x-3 = 2018
3x - 6 = 3103 : 3100 2018x-3 = 20181
3x - 6 = 33 =27 Do x-3=1 cho nên suy ra x=4
3x = 27 + 6
3x = 33
x = 33:3
x = 11
\(a,A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2017}}+\dfrac{1}{2^{2018}}\)
\(3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2016}}+\dfrac{1}{3^{2017}}\)
\(3A-A=1-\dfrac{1}{3^{2018}}\)
\(A=\dfrac{\left(1-\dfrac{1}{3^{2018}}\right)}{2}\)
\(b,B=1+5+5^2+5^3+...+5^{100}\)
\(5B=5+5^2+5^3+5^4+...+5^{100}+5^{101}\)
\(5B-B=1-5^{101}\)
\(B=\dfrac{\left(1-5^{101}\right)}{4}\)
A=1+2+3+4+5+...+99+100
A=\(\dfrac{100.\left(100+1\right)}{2}\)=5050
Vậy A=5050
B=\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\)
B=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
B=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
B=\(1-\dfrac{1}{100}\)=\(\dfrac{99}{100}\)
Vậy B=\(\dfrac{99}{100}\)
Chuẩn ồi còn j
= 144
=1
= 0
= 400
\(12^2=144\)
\(1^{100}=1\)
\(100^1=100\)
\(0^{2018}=0\)
\(20^2=400\)