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Ta có VP:
\(\dfrac{2}{\sqrt{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}}\)
Thay \(1=ab+bc+ca\)
\(=\dfrac{2}{\sqrt{\left(ab+bc+ca+a^2\right)\left(ab+bc+ca+b^2\right)\left(ab+bc+ca+c^2\right)}}\)
\(=\dfrac{2}{\sqrt{\left[b\left(a+c\right)+a\left(a+c\right)\right]\left[a\left(b+c\right)+b\left(b+c\right)\right]\left[b\left(a+c\right)+c\left(a+c\right)\right]}}\)
\(=\dfrac{2}{\sqrt{\left(a+c\right)\left(a+b\right)\left(a+b\right)\left(b+c\right)\left(b+c\right)\left(a+c\right)}}\)
\(=\dfrac{2}{\sqrt{\left[\left(a+c\right)\left(a+b\right)\left(b+c\right)\right]^2}}\)
\(=\dfrac{2}{\left(a+c\right)\left(a+b\right)\left(b+c\right)}\)
_____________
Ta có VT:
\(\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}+\dfrac{c}{1+c^2}\)
Thay \(1=ab+ac+bc\)
\(=\dfrac{a}{ab+ac+bc+a^2}+\dfrac{b}{ab+ac+bc+b^2}+\dfrac{c}{ab+ac+bc+c^2}\)
\(=\dfrac{a}{a\left(a+b\right)+c\left(a+b\right)}+\dfrac{b}{b\left(b+c\right)+a\left(b+c\right)}+\dfrac{c}{c\left(b+c\right)+a\left(b+c\right)}\)
\(=\dfrac{a}{\left(a+c\right)\left(a+b\right)}+\dfrac{b}{\left(a+b\right)\left(b+c\right)}+\dfrac{c}{\left(a+c\right)\left(b+c\right)}\)
\(=\dfrac{a\left(b+c\right)}{\left(a+c\right)\left(b+c\right)\left(a+b\right)}+\dfrac{b\left(a+c\right)}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}+\dfrac{c\left(a+b\right)}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)
\(=\dfrac{ab+ac+ab+bc+ac+bc}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)
\(=\dfrac{2ab+2ac+2bc}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)
\(=\dfrac{2\cdot\left(ab+ac+bc\right)}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)
\(=\dfrac{2}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\left(ab+ac+bc=1\right)\)
Mà: \(VP=VT=\dfrac{2}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)
\(\Rightarrow\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}+\dfrac{c}{1+c^2}=\dfrac{2}{\sqrt{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}}\left(dpcm\right)\)
Theo đề bài ta có :
\(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\) ( 1 )
Theo tính chất dãy tỉ số bằng nhau ta có :
\(k=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)
\(k^2=\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\) ( 2 )
Mà từ ( 1 ) = > \(k^2=\dfrac{a}{c}.\dfrac{b}{d}=\dfrac{ab}{cd}\) ( 3 )
Từ ( 2 ) , ( 3 )
= > \(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\) ( đpcm )
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk,c=dk\)
Ta có: \(\dfrac{ab}{cd}=\dfrac{bkb}{dkd}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}=\dfrac{b}{d}\left(1\right)\)
\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{bk-b}{dk-d}=\dfrac{b\left(k-1\right)}{d\left(k-1\right)}=\dfrac{b}{d}\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
\(\Rightarrow\dfrac{ab}{cd}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}\\ \dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{b^2\left(k-1\right)^2}{d^2\left(k-1\right)^2}=\dfrac{b^2}{d^2}\\ \Rightarrow\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
Giải:
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow k=\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a-c}{b-d}\) ( tính chất dãy tỉ số bằng nhau )
\(\Rightarrow k^2=\left(\dfrac{a-c}{b-d}\right)^2=\dfrac{\left(a-c\right)^2}{\left(b-d\right)^2}\) (1)
và \(k^2=\dfrac{a}{b}.\dfrac{c}{d}=\dfrac{ac}{bd}\) (2)
Từ (1), (2) \(\Rightarrow\dfrac{\left(a-c\right)^2}{\left(b-d\right)^2}=\dfrac{ab}{cd}\left(đpcm\right)\)
Vậy...
Đề sai rồi bạn ạ
Phải là : Cho\(\dfrac{a}{b}=\dfrac{c}{d}\) với c≠±1. Chứng minh rằng \(\dfrac{\left(a-c\right)^2}{\left(b-d\right)^2}=\dfrac{ac}{bd}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)Suy ra: \(\dfrac{\left(a-c\right)^2}{\left(b-d\right)^2}=\dfrac{\left(bk-dk\right)^2}{\left(b-d\right)^2}=\dfrac{\left[k\left(b-d\right)\right]^2}{\left(b-d\right)^2}\)=k2 (1)
\(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=\dfrac{k^2.bd}{bd}=k^2\) (2)
Từ (1) và (2) \(\Rightarrow\dfrac{\left(a-c\right)^2}{\left(b-d\right)^2}=\dfrac{ac}{bd}\)
Bài 1:
Áp dụng t.c của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\\ =\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a^3}{b^3}=\dfrac{a.b.c}{b.c.d}=\dfrac{a}{d}\left(dpcm\right)\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
a, Ta có: \(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{bk.b}{dk.d}=\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}\)
\(\Rightarrow\dfrac{b^2.k}{d^2.k}=\dfrac{\left[b.\left(k+1\right)\right]^2}{\left[d.\left(k+1\right)\right]^2}\Rightarrow\dfrac{b^2}{d^2}=\dfrac{b^2}{d^2}\) \(\Rightarrow\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
b, Ta có:\(\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{bk.b}{dk.d}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}\)
\(\Rightarrow\dfrac{b^2}{d^2}=\dfrac{b^2.k^2+b^2}{d^2.k^2+d^2}\Rightarrow\dfrac{b^2}{d^2}=\dfrac{b^2.\left(k^2+1\right)}{d^2.\left(k^2+1\right)}\)
\(\Rightarrow\dfrac{b^2}{d^2}=\dfrac{b^2}{d^2}\Rightarrow\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}\)
CHÚC BẠN HỌC TỐT!!
\(\dfrac{a}{b}=\dfrac{c}{d}\)=>\(\dfrac{a}{c}=\dfrac{b}{d}\)( áp dụng tỉ lệ thức )
Ta đặt:
\(\dfrac{a}{c}=\dfrac{b}{d}=k\) => a=ck ; b=dk
a) \(\dfrac{ab}{cd}=\dfrac{ck.dk}{cd}=\dfrac{k^2.\left(c.d\right)}{c.d}=k^2\) (1)
\(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(ck+dk\right)^2}{\left(c+d\right)^2}=\dfrac{k^2.\left(c+d\right)^2}{\left(c+d\right)^2}=k^2\) (2)
Từ (1) và (2) suy ra \(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
b) \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(ck\right)^2+\left(dk\right)^2}{c^2+d^2}=\dfrac{c^2k^2+d^2k^2}{c^2+d^2}=\dfrac{k^2.\left(c^2+d^2\right)}{c^2+d^2}=k^2\) (3)
Từ (1) và (3) suy ra \(\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}\)
