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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 : \(\dfrac{5}{211}\)

28 tháng 7 2019

A = (x - 1) (x2 - 2x + 1) + 4x(x + 1)(x - 1) - 3(1 - x)(x2 + x + 1)

= (x - 1) (x2 - 2x + 1) + 4x(x + 1)(x - 1) + 3(x - 1)(x2 + x + 1)

= (x - 1) [x2 - 2x + 1 + 3(x2 + x + 1) + 4x(x + 1)]

= (x - 1) (x2 - 2x + 1 +3x2 + 3x + 3 + 4x2 + 4x)

= (x - 1) (8x2 + 5x + 4)

Vậy A = (x - 1) (8x2 + 5x + 4)

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: \(N=\dfrac{1095}{547}\cdot\dfrac{3}{211}-\dfrac{546}{547\cdot211}-\dfrac{4}{547\cdot211}\)

\(=\dfrac{1}{547\cdot211}\left(1095\cdot3-546-4\right)\)

\(=\dfrac{1}{547\cdot211}\cdot2735=\dfrac{5}{211}\)

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

Sửa đề:\(\dfrac{x+1}{2022}+\dfrac{x+2}{2021}-\dfrac{x+3}{2020}-\dfrac{x+4}{2019}=0\)

\(\Leftrightarrow\left(\dfrac{x+1}{2022}+1\right)+\left(\dfrac{x+2}{2021}+1\right)-\left(\dfrac{x+3}{2020}+1\right)-\left(\dfrac{x+4}{2019}+1\right)=0\)

=>x+2023=0

=>x=-2023

26 tháng 12 2017