Question

[Phillip]

Bài 1: 
Thực hiện phép tính: K = (\(\dfrac{1}{225}-\dfrac{1}{1^{^{ }2}}\)) (\(\dfrac{1}{225}-\dfrac{1}{2^{^{ }2}}\)) ... (\(\dfrac{1}{225}-\dfrac{1}{100^{^{ }2}}\))
Bài 2:
Cho hai phân số \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh rằng:

a) \(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)           b)\(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)           c) \(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)            d)\(\left(\dfrac{a+b}{c+d^{ }}\right)^2\)= \(\dfrac{a^2+b^2}{c^2+d^2}\)

Bài 3:

Tìm các số nguyên a,b thỏa mãn:
a)\(\dfrac{a}{b}=\dfrac{126}{135}\) và BCNN(a,b) = 1890      b) \(\dfrac{a}{b}=\dfrac{60}{108}\) và UCLN(a,b) = 180


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Answer 1

[Nguyễn Lê Phước Thịnh]

Bài 1:

\(K=\left(\dfrac{1}{225}-\dfrac{1}{1^2}\right)\left(\dfrac{1}{225}-\dfrac{1}{2^2}\right)\cdot...\cdot\left(\dfrac{1}{225}-\dfrac{1}{100^2}\right)\)

\(=\left(\dfrac{1}{225}-\dfrac{1}{15^2}\right)\cdot\left(\dfrac{1}{225}-\dfrac{1}{1^2}\right)\left(\dfrac{1}{225}-\dfrac{1}{2^2}\right)\cdot...\cdot\left(\dfrac{1}{225}-\dfrac{1}{100^2}\right)\)

\(=\left(\dfrac{1}{225}-\dfrac{1}{225}\right)\cdot A=0\)

Bài 2:

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

=>a=bk; c=dk

a: \(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\)

\(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\)

Do đó: \(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)

b: \(\dfrac{a+b}{a-b}=\dfrac{bk+b}{bk-b}=\dfrac{b\left(k+1\right)}{b\left(k-1\right)}=\dfrac{k+1}{k-1}\)

\(\dfrac{c+d}{c-d}=\dfrac{dk+d}{dk-d}=\dfrac{d\left(k+1\right)}{d\left(k-1\right)}=\dfrac{k+1}{k-1}\)

Do đó: \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)

c: \(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2}{d^2}\)

\(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b^2}{d^2}\)

Do đó: \(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)

d: \(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\left(\dfrac{b\left(k+1\right)}{d\left(k+1\right)}\right)^2=\left(\dfrac{b}{d}\right)^2\)

\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\dfrac{b^2}{d^2}\)

Do đó: \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)