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Đặt 3759=a; 5741=b
Theo đề, ta có: \(E=4\dfrac{7}{b}\cdot\dfrac{1}{a}-\dfrac{4}{a}\cdot\left(1+\dfrac{2}{b}\right)+\dfrac{1}{a}+\dfrac{1}{ab}\)
\(=\dfrac{4b+7}{b}\cdot\dfrac{1}{a}-\dfrac{4}{a}\cdot\dfrac{b+2}{b}+\dfrac{b+1}{ab}\)
\(=\dfrac{4b+7-4b-8+b+1}{ab}=\dfrac{b}{ab}=\dfrac{1}{a}=\dfrac{1}{3759}\)
\(xy\left(x-y\right)+yz\left(y-z\right)+xz\left(z-x\right)\\ =xy\left(x-y\right)+yz\left[-\left(x-y\right)-\left(z-x\right)\right]+xz\left(z-x\right)\\ =xy\left(x-y\right)-yz\left(x-y\right)-yz\left(z-x\right)+xz\left(z-x\right)\\ =\left(x-y\right)\left(xy-yz\right)+\left(z-x\right)\left(xz-yz\right)\\ =y\left(x-y\right)\left(x-z\right)+z\left(z-x\right)\left(x-y\right)\\ =\left(x-y\right)\left(x-z\right)\left(y-z\right)\)
Đặt a=4453, b=1997
Ta có: \(F=5\dfrac{6}{a}\cdot\dfrac{1}{b}-\dfrac{2}{b}\cdot2\dfrac{3}{a}\)
\(=\dfrac{5a+6}{a}\cdot\dfrac{1}{b}-\dfrac{2}{b}\cdot\dfrac{2a+3}{a}\)+
\(=\dfrac{5a+6-4a-6}{ab}\)
\(=\dfrac{1}{b}\)
\(=\dfrac{1}{1997}\)
Đặt 4453=a; 1997=b
\(A=\left(5+\dfrac{6}{a}\right)\cdot\dfrac{1}{b}-\dfrac{2}{b}\cdot\left(2+\dfrac{3}{a}\right)\)
\(=\dfrac{5a+6}{a}\cdot\dfrac{1}{b}-\dfrac{2}{b}\cdot\dfrac{2a+3}{a}\)
\(=\dfrac{5a+6-4a-6}{ab}=\dfrac{a}{ab}=\dfrac{1}{b}=\dfrac{1}{1997}\)
Bài 3
a,26/100+0,009+41/100+0,24
0,26+0,09+0,41+0,24
(0,26+0,24)+(0,09+0,41)
0,5+0,5
=1
b,9+1/4+6+2/7+7+3/5+8+2/3+2/5+1/3+5/7+3/4
(9+6+7+8)+(2/7+5/7)+(1/4+3/4)+(3/5+2/5)+(2/3+1/3)
30+1+1+1+1
=34
Bài 4,5 khó quá mik ko bít lamf^^))
Bài 4: a, \(\dfrac{2008}{2009}\) < 1; \(\dfrac{10}{9}\) > 1
\(\dfrac{2008}{2009}\) < \(\dfrac{10}{9}\)
b, \(\dfrac{1}{a+1}\) và \(\dfrac{1}{a-1}\)
Ta có: a + 1 > a - 1 ⇒ \(\dfrac{1}{a+1}\) < \(\dfrac{1}{a-1}\)
2: \(=\dfrac{203}{60}\cdot\dfrac{81}{1225}=\dfrac{783}{3500}\)
