tìm x
(4.27 + x) : 2.4 =5.6
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4 tháng 7 2019
\(=\left(2.\left(2^3\right)^4.\left(3^3\right)^2+2^2.\left(2.3\right)^9\right):\left(2^{12}.\left(3^2\right)^3.31\right)\)
\(=\left(2^{13}.3^6+2^{11}.3^9\right):\left(2^{12}.3^6.31\right)\)
\(=\left[2^{11}.3^6\left(2^2+3^3\right)\right]:\left(2^{12}.3^6.31\right)\)
\(=\frac{2^{11}.3^6.31}{2^{12}.3^6.31}=\frac{1}{2}\)
4 tháng 7 2019
Đưa về phân số:
\(=\frac{2.8^4.27^2+4.6^9}{2^{12}.9^3.31}\)
\(=\frac{2.\left(2^3\right)^4.\left(3^3\right)^2+2^2.\left(2.3\right)^9}{2^{12}.\left(3^2\right)^3.31}\)
\(=\frac{2.2^{3.4}.3^{3.2}+2^2.2^9.3^9}{2^{12}.3^{2.3}.31}\)
\(=\frac{2.2^{12}.3^6+2^{2+9}.3^9}{2^{12}.3^6.31}\)
\(=\frac{2^{1+12}.3^6+2^{11}.3^9}{2^{12}.3^6.31}\)
\(=\frac{2^{13}.3^6+2^{11}.3^9}{2^{12}.3^6.31}\)
\(=\frac{2^{11}.3^6\left(2^2+3^3\right)}{2^{12}.3^6.31}\)
\(=\frac{2^{11}.3^6.31}{2^{12}.3^6.31}=\frac{1}{2}\)
Em hiểu hơn ko?
6 tháng 5 2019
a, Ta có : \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
=> \(\frac{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}}{\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}}=1\)
=> đpcm
Study well ! >_<
14 tháng 4 2018
(5,28+x):24=5,6
5,28+x=5,6×24
5,28+x=134,4
x=134,4-5,28
x=129,12
(4,27+x)=5,6x2,4
(4,27+x)=13,44
x=13,44-4,27
x=9,17
tk mình nha!
bằng 9,17 nha