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Giải:
\(\dfrac{1}{2}-\dfrac{43}{100}+\left(-\dfrac{1}{3}\right)-\dfrac{1}{6}\)
\(=-\dfrac{43}{100}+\dfrac{1}{2}+\left(-\dfrac{1}{3}\right)-\dfrac{1}{6}\)
\(=-\dfrac{43}{100}+\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\)
\(=-\dfrac{43}{100}+\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(=-\dfrac{43}{100}+\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\)
\(=-\dfrac{43}{100}+0=-\dfrac{43}{100}\)
Vậy ...
\(\dfrac{1}{2}-\dfrac{43}{101}+\dfrac{-1}{3}-\dfrac{1}{6}\\ =\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)-\dfrac{43}{101}=0-\dfrac{43}{101}=-\dfrac{43}{101}\)
có liền luôn nè
1/2 -43/101+(-1/3)-1/6
= -43/101+(-1/3)-1/6+1/2
=-43/101+0
=-43/101
\(A=\frac{-5}{12}+\frac{4}{37}+\frac{17}{12}-\frac{41}{37}=(\frac{-5}{12}+\frac{17}{12})+(\frac{4}{37}-\frac{41}{37})=\frac{12}{12}+\frac{-37}{37}=1+(-1)=0\)
\(B=\frac{1}{2}-\frac{43}{101}+\frac{-1}{3}-\frac{1}{6}=\frac{-43}{101}+(\frac{1}{2}+\frac{-1}{3}-\frac{1}{6})=\frac{-43}{101}+(\frac{3}{6}+\frac{-2}{6}-\frac{1}{6})=\frac{-43}{101}+0=\frac{-43}{101}\)
\(A=\frac{-5}{12}+\frac{4}{37}+\frac{17}{12}-\frac{41}{37}.\)
\(A=\left(\frac{-5}{12}+\frac{17}{12}\right)-\left(\frac{41}{37}-\frac{4}{37}\right)\)
\(A=1-1=0\)
\(B=\frac{1}{2}-\frac{43}{101}+\left(\frac{-1}{3}\right)-\frac{1}{6}\)
\(B=\left(\frac{1}{2}+\left(\frac{-1}{3}\right)-\frac{1}{6}\right)-\frac{43}{101}\)
\(A=0-\frac{43}{101}=\frac{-43}{101}\)
\(C=\frac{-5}{6}\cdot\frac{12}{-7}\cdot-\frac{21}{15}\)
\(C=\frac{-5}{2.3}\cdot\frac{3.2.2}{-7}\cdot\frac{3.\left(-7\right)}{3.5}\)
\(C=\frac{-2}{1}=-2\)
\(\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{10.7}+\frac{1}{14.9}+...+\frac{1}{198.101}\)
= \(2.\left(\frac{1}{2.6}+\frac{1}{6.10}+\frac{1}{10.14}+\frac{1}{14.18}+...+\frac{1}{198.202}\right)\)
= \(2.\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+\frac{1}{14}-\frac{1}{18}+...+\frac{1}{198}-\frac{1}{202}\right)\)
= \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{202}\right)\)
= \(\frac{1}{2}.\frac{50}{101}\)
= \(\frac{25}{101}\)
1, Tìm x biết: a, 6x 1-6x=1080
b, 6x-1 6x=42 2, So sánh: E=7.(8 82 83 ....... 8100) 8 và G=8101 3, Chứng tỏ: a, 4343-1717 chia hết cho 10 b, 3636-910 chia hết cho 45
c, 2 10 2 11 2 12 7 210 211 2127 có giá trị là số tự nhiên
d, 8 10 − 8 9 − 8 8 55 810−89−8855 có giá trị là số tự nhiên
hi
Bài 1:
a: \(\Leftrightarrow6^x\left(6-1\right)=1080\)
=>6x=216
=>x=3
b: \(\Leftrightarrow6^x\left(\dfrac{1}{6}+1\right)=42\)
=>6x=36
=>x=2
Câu 3:
c: \(=\dfrac{2^{10}\left(1+2+2^2\right)}{7}=2^{10}\) là số tự nhiên
d: \(=\dfrac{8^8\left(8^2-8-1\right)}{55}=8^8\) là số tự nhiên
\(\frac{1}{2}-\frac{43}{101}+\frac{-1}{3}-\frac{1}{6}\)
\(=\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)+\left(-\frac{43}{101}\right)\)
\(=\frac{3-2-1}{6}+\frac{-43}{101}\)
\(=0+\frac{-43}{101}\)
\(=-\frac{43}{101}\)
1/2-1/3=1/6 suy ra tổng trên = -43/101