2\(\dfrac{1}{547}.\dfrac{3}{211}-\dfrac{546}{547}.\dfrac{1}{211}-\dfrac{4}{547.211}\)
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KB
26 tháng 8 2018
\(2\dfrac{1}{547}.\dfrac{3}{211}-\dfrac{546}{547}.\dfrac{1}{211}-\dfrac{4}{547.211}\)
\(=\left(2+\dfrac{1}{547}\right).3.\dfrac{1}{211}-\left(1-\dfrac{1}{547}\right).\dfrac{1}{211}-4.\dfrac{1}{547}.\dfrac{1}{211}\)
Đặt \(a=\dfrac{1}{547};b=\dfrac{1}{211}\)
Thay \(a=\dfrac{1}{547};b=\dfrac{1}{211}\) vào biểu thức trên , ta được :
\(\left(2+a\right).3b-\left(1-a\right)b-4ab\)
\(=6b+3ab-b+ab-4ab\)
\(=5b\)
\(=5.\dfrac{1}{211}\)
\(=\dfrac{5}{211}\)
Vậy g/t biểu thức trên là : \(\dfrac{5}{211}\)
21 tháng 5 2022
a: \(=\dfrac{\left(2\cdot547+1\right)\cdot3}{547\cdot211}-\dfrac{546}{547\cdot211}-\dfrac{4}{547\cdot211}\)
\(=\dfrac{2735}{547\cdot211}=\dfrac{5}{211}\)
b: x=7 nên x+1=8
\(x^{15}-8x^{14}+8x^{13}-8x^{12}+...-8x^2+8x-5\)
\(=x^{15}-x^{14}\left(x+1\right)+x^{13}\left(x+1\right)-x^{12}\left(x+1\right)+...-x^2\left(x+1\right)+x\left(x+1\right)-5\)
\(=x^{15}-x^{15}-x^{14}+x^{14}-...-x^3-x^2+x^2+x-5\)
=x-5=7-5=2
TM
22 tháng 9 2020
=>\(B=\left(2\cdot\frac{1}{547}\cdot\frac{3}{211}\right)-\left(\frac{546}{547}\cdot\frac{1}{211}\right)-\left(\frac{4}{547\cdot211}\right)\)
=>\(B=\frac{6}{547\cdot211}-\frac{546}{547\cdot211}-\frac{4}{547\cdot211}\)
=>\(B=\frac{6}{115417}-\frac{546}{115417}-\frac{4}{115417}\)
=>\(B=\frac{-544}{115417}\)
BH
2
HT
27 tháng 7 2015
M = \(2.\frac{1}{547}.\frac{3}{211}-\frac{546}{547}.\frac{1}{211}-\frac{1}{247.211}\)
M = \(\frac{6}{547.211}-\frac{546}{547.211}-\frac{4}{547.211}\)
M = \(\frac{-544}{547.211}\)
27 tháng 7 2015
\(M=2.\frac{1}{547}.\frac{3}{211}-\frac{546}{547}.\frac{1}{211}-\frac{4}{547.211}\)
\(M=\left(2.\frac{1}{547}.\frac{3}{211}\right)-\left(\frac{546}{547}.\frac{1}{211}\right)-\left(\frac{4}{547.211}\right)\)
\(M=\frac{6}{547.211}-\frac{546}{547.211}-\frac{4}{547.211}\)
\(M=\frac{6}{115417}-\frac{546}{115417}-\frac{4}{115417}\)
\(M=\frac{-540}{115417}-\frac{4}{115417}\)
\(M=\frac{-544}{115417}\)
PT
0
6 tháng 2 2022
a: \(M=\dfrac{631}{315}\cdot\dfrac{1}{651}-\dfrac{1}{105}\cdot\dfrac{2603}{651}-\dfrac{4}{315\cdot651}+\dfrac{4}{105}\)
\(=\dfrac{1}{315\cdot651}\cdot\left(631-4\right)-\dfrac{1}{105}\left(\dfrac{2603}{651}-4\right)\)
\(=\dfrac{1}{105}\cdot\dfrac{1}{1953}\cdot627+\dfrac{1}{105\cdot651}\)
\(=\dfrac{1}{105\cdot651}\left(\dfrac{1}{3}\cdot627+1\right)=\dfrac{1}{105\cdot651}\cdot210=\dfrac{2}{651}\)
b: \(N=\dfrac{1095}{547}\cdot\dfrac{3}{211}-\dfrac{546}{547\cdot211}-\dfrac{4}{547\cdot211}\)
\(=\dfrac{1}{547\cdot211}\left(1095\cdot3-546-4\right)\)
\(=\dfrac{1}{547\cdot211}\cdot2735=\dfrac{5}{211}\)
- Bằng \(\dfrac{5}{211}\)
gọi 1/547=a, 1/211=b ta đc:
\(2a.3b-546a.b-4ab< =>6ab-546ab-4ab=-544ab=\dfrac{-544}{547.211}\)