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\(S=5\times\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}\right)\)
\(=5\times\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}\right)\)
\(=5\times\left(1-\frac{1}{16}\right)\)
\(=5\times\frac{15}{16}=\frac{75}{16}\)
Vậy \(S=\frac{75}{16}\)
a) \(\frac{2}{3}.\left(\frac{1}{3}+\frac{2}{5}\right)=\frac{2}{3}.\frac{11}{15}=\frac{22}{45}\)
b) 24+25+26+...+99+100
= (100+24).77:2
= 124.77:2
= 4774
c) \(\frac{16.17-5}{16.16+11}=\frac{16.16+(16-5)}{16.16+11}=\frac{16.16+11}{16.16+11}=1\)
d) \(\frac{5}{80}+\frac{5}{90}+\frac{5}{150}+\frac{5}{210}+\frac{5}{270}\)
\(=\frac{1}{16}+\frac{1}{18}+\frac{1}{30}+\frac{1}{42}+\frac{1}{54}\)
\(=\frac{2929}{15120}\)
a)=\(\frac{2}{3}.\left(\frac{5}{15}+\frac{6}{15}\right)\)
= \(\frac{2}{3}.\frac{11}{15}=\frac{22}{45}\)
d) = 5.(\(\frac{1}{80}+\frac{1}{90}+\frac{1}{150}+\frac{1}{210}+\frac{1}{270}\))
= 50........
kq là tự tính:)))))
a) \(A=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+......+\frac{1}{2017.2022}\)
\(5A=5.\left(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+.....+\frac{1}{2017.2022}\right)\)
\(5A=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+......+\frac{5}{2017.2022}\)
\(5A=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+........+\frac{1}{2017}-\frac{1}{2022}\)
\(5A=1-\frac{1}{2022}\)
\(5A=\frac{2022}{2022}-\frac{1}{2022}\)
\(5A=\frac{2021}{2022}\)
\(A=\frac{2021}{2022}\div5\)
\(A=\frac{20201}{10110}\)
TL:
\(\frac{5}{6}=\frac{1}{2}+\frac{1}{3}\)
@@@@@@@@@@
HT
a) A = 1/3 - 1/7 + 1/7 - 1/11 +......+1/107 - 1/111
A = 1/3 - 1/111
A = ..............Bạn tự tính nhé!
b) B = 2.(3/15.18 + 3/18.21 +........+3/87.90)
B = 2.(1/15 - 1/18 + 1/18 - 1/21 +........+1/87 - 1/90)
B = 2.(1/15 - 1/90)
B = 2.5/90
B =......Tự tính nhé!
C ; D làm tương tự nhé!
c) \(\left(2x-3\right).\left(6-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{3}{2};3\right\}\)
e) \(2\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)
\(\Leftrightarrow2\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}\)
\(\Leftrightarrow\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}:2=\frac{7}{4}.\frac{1}{2}=\frac{7}{8}\)
\(\Rightarrow\left[{}\begin{matrix}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=\left(-\frac{7}{8}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{29}{12}\\x=\frac{-13}{12}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{29}{12};\frac{-13}{12}\right\}\)
Mấy bài này ko quá khó, tải MathPhoto trong đt về nó tự lm
mấy bài này quen lắm h các bn mới hok ak
đã gọi thử trí thì ko cần học