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Đặt S= \(2\dfrac{1}{315}.\dfrac{1}{651}-\dfrac{1}{105}.3\dfrac{650}{651}-\dfrac{4}{315.651}+\dfrac{4}{105}\)
= \(\left(2+\dfrac{1}{315}\right).\dfrac{1}{651}-\dfrac{3}{315}.\left(3+\dfrac{651-1}{651}\right)-\dfrac{4}{315.651}+\dfrac{12}{315}\)
= \(\left(2+\dfrac{1}{315}\right).\dfrac{1}{651}-\dfrac{3}{315}.\left(3+1-\dfrac{1}{651}\right)-\dfrac{4}{315.651}+\dfrac{12}{315}\)
Đặt \(\dfrac{1}{315}=a,\dfrac{1}{651}=b\)
\(\Rightarrow S=\left(2+a\right).b-3a.\left(4-b\right)-4ab+12a\)
\(=2b+ab-12a+3ab-4ab+12a\)
\(=2b=\dfrac{2}{651}\)
Đặt \(\dfrac{1}{315}=x,\dfrac{1}{651}=y\)
\(\Rightarrow A=\left(2+x\right)y-3x\left(4-y\right)-4xy+12x\)
\(=2y+xy-12x+3xy-4xy+12x\)
\(=2y\)
Thay \(y=\dfrac{1}{651}\Rightarrow A=\dfrac{2}{651}\)
Vậy...
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
A=(2+1/315).1/651-3/315.[3+(651-1)/651]-4.1/315.1/651+12/315
A=(2+1/315).1/651-3.1/315.(3+1-1/651)-4.1/135.1/651+12.1/315
Đặt 1/315=a;1/651=b ta có:
A=(2+a)b-3a(4-b)-4ab+12a
A=2b+ab-12a+3ab-4ab+12a
A=2b=2.1/315=2/315
Đặt \(\dfrac{1}{315}=a;\dfrac{1}{651}=b;\) thay vào \(A\) đc:
\(A=(2+a).b-\left(3+1-b\right).3a-4ab+12a\)
\(=\)\(2b+ab-12a+3ab-4ab+12a\)
\(=2b=\dfrac{2}{651}\)
A=(2+1/315).1/651-3/315.[3+(651-1)/651]-4.1/315.1/651+12/315
A=(2+1/315).1/651-3.1/315.(3+1-1/651)-4.1/135.1/651+12.1/315
Đặt 1/315=a;1/651=b ta có:
A=(2+a)b-3a(4-b)-4ab+12a
A=2b+ab-12a+3ab-4ab+12a
A=2b=2.1/315=2/315
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Đặt :
\(\dfrac{1}{315}=a;\dfrac{1}{651}=b\) thay vào A ta được :
\(A=\left(2+a\right)b-\left(3+1-b\right).3a-4ab+12a\)
\(\Leftrightarrow A=2b+ab-12a+3ab-4ab+12a\)
\(\Leftrightarrow A=2b\)
Thay \(b=\dfrac{1}{651}\) ta dc :
\(A=\dfrac{2}{651}\)
Đặt a=315; b=651
\(A=2\frac{1}{315}\cdot\frac{1}{651}-\frac{1}{105}\cdot3\frac{650}{651}-\frac{4}{315\cdot651}+\frac{4}{105}\)
\(=\left(2+\frac{1}{a}\right)\cdot\frac{1}{b}-\frac{3}{a}\cdot\left(3+\frac{b-1}{b}\right)-\frac{4}{ab}+\frac{12}{a}\)
\(=\frac{2a+1}{a}\cdot\frac{1}{b}-\frac{3}{a}\cdot\frac{3b+b-1}{b}-\frac{4}{ab}+\frac{12b}{ab}\)
\(=\frac{2a+1-3\left(4b-1\right)-4+12b}{ab}=\frac{2a-3-12b+3+12b}{ab}=\frac{2a}{ab}=\frac{2}{b}\)
=2/651
Đặt a=315; b=651
\(A=2\frac{1}{315}\cdot\frac{1}{651}-\frac{1}{105}\cdot3\frac{650}{651}-\frac{4}{315\cdot651}+\frac{4}{105}\)
\(=\left(2+\frac{1}{a}\right)\cdot\frac{1}{b}-\frac{3}{a}\cdot\left(3+\frac{b-1}{b}\right)-\frac{4}{ab}+\frac{12}{a}\)
\(=\frac{2a+1}{a}\cdot\frac{1}{b}-\frac{3}{a}\cdot\frac{3b+b-1}{b}-\frac{4}{ab}+\frac{12b}{ab}\)
\(=\frac{2a+1-3\left(4b-1\right)-4+12b}{ab}=\frac{2a-3-12b+3+12b}{ab}=\frac{2a}{ab}=\frac{2}{b}\)
=2/651
Đặt a=315; b=651
\(A=2\frac{1}{315}\cdot\frac{1}{651}-\frac{1}{105}\cdot3\frac{650}{651}-\frac{4}{315\cdot651}+\frac{4}{105}\)
\(=\left(2+\frac{1}{a}\right)\cdot\frac{1}{b}-\frac{3}{a}\cdot\left(3+\frac{b-1}{b}\right)-\frac{4}{ab}+\frac{12}{a}\)
\(=\frac{2a+1}{a}\cdot\frac{1}{b}-\frac{3}{a}\cdot\frac{3b+b-1}{b}-\frac{4}{ab}+\frac{12b}{ab}\)
\(=\frac{2a+1-3\left(4b-1\right)-4+12b}{ab}=\frac{2a-3-12b+3+12b}{ab}=\frac{2a}{ab}=\frac{2}{b}\)
=2/651
Đặt a=315; b=651
\(A=2\frac{1}{315}\cdot\frac{1}{651}-\frac{1}{105}\cdot3\frac{650}{651}-\frac{4}{315\cdot651}+\frac{4}{105}\)
\(=\left(2+\frac{1}{a}\right)\cdot\frac{1}{b}-\frac{3}{a}\cdot\left(3+\frac{b-1}{b}\right)-\frac{4}{ab}+\frac{12}{a}\)
\(=\frac{2a+1}{a}\cdot\frac{1}{b}-\frac{3}{a}\cdot\frac{3b+b-1}{b}-\frac{4}{ab}+\frac{12b}{ab}\)
\(=\frac{2a+1-3\left(4b-1\right)-4+12b}{ab}=\frac{2a-3-12b+3+12b}{ab}=\frac{2a}{ab}=\frac{2}{b}\)
=2/651
Đặt a=315; b=651
\(A=2\frac{1}{315}\cdot\frac{1}{651}-\frac{1}{105}\cdot3\frac{650}{651}-\frac{4}{315\cdot651}+\frac{4}{105}\)
\(=\left(2+\frac{1}{a}\right)\cdot\frac{1}{b}-\frac{3}{a}\cdot\left(3+\frac{b-1}{b}\right)-\frac{4}{ab}+\frac{12}{a}\)
\(=\frac{2a+1}{a}\cdot\frac{1}{b}-\frac{3}{a}\cdot\frac{3b+b-1}{b}-\frac{4}{ab}+\frac{12b}{ab}\)
\(=\frac{2a+1-3\left(4b-1\right)-4+12b}{ab}=\frac{2a-3-12b+3+12b}{ab}=\frac{2a}{ab}=\frac{2}{b}\)
=2/651