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ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{ac}{bd}\) (*)
mà \(\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}\)
Từ (*) \(\Rightarrow\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\left(đpcm\right)\)
\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}\Rightarrow\frac{q^2}{4}=\frac{b^2}{9}=\frac{2c^2}{32}=\frac{a^2-b^2+2c^2}{4-9+32}=\frac{108}{27}=4\)
=> \(\frac{a^2}{4}=4\Rightarrow a^2=4.4=16\Rightarrow a=+-4\)
=>\(\frac{b^2}{9}=4\Rightarrow b^2=4.9=36\Rightarrow b=+-6\)
=>\(\frac{2c^2}{32}=4\Rightarrow c^2=4.32:2=64\Rightarrow c=+-8\)
Câu 2 :
Ta có : \(\frac{a}{b}=\frac{c}{d}\) \(\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}=\frac{a-b}{c-d}\)
\(\Rightarrow\frac{a+b}{a-b}=\frac{c+d}{c-d}\)
đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
suy ra:\(\frac{ac}{bd}=\frac{bk.dk}{bd}=k.k=k^2\)
\(\frac{a^2+c^2}{b^2+d^2}=\frac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\frac{b^2k^2+d^2k^2}{b^2+d^2}=\frac{k^2.\left(b^2+d^2\right)}{b^2+d^2}=k^2\)
vậy \(\frac{ab}{bd}=\frac{a^2+c^2}{b^2+d^2}\)
Ta có:\(\frac{a}{b}=\frac{c}{d}=>\frac{a}{b}.\frac{c}{d}=\frac{c}{d}.\frac{c}{d}=>\frac{ac}{bd}=\frac{c^2}{d^2}\)
\(\frac{c}{d}=\frac{a}{b}=>\frac{a}{b}.\frac{c}{d}=\frac{a}{b}.\frac{a}{b}=>\frac{ac}{bd}=\frac{a^2}{b^2}\)
=>\(\frac{ac}{bd}=\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}\)
=>\(\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\)
a/b = c/d = K
a = bK
c = dK
ac/bd = bkdk/bd = k2
a2+c2/b2+d2 = (bK)2+(dK)2/b2+d2 = b2K2+d2K2/d2+b2 = K2(b2+d2)/b2+d2 = K2
Đăng từng bài thoy nha pn!!!
Bài 1:
Có : 2009 = 2008 + 1 = x + 1
Thay 2009 = x + 1 vào biểu thức trên,ta có :
x\(^5\)- 2009x\(^4\)+ 2009x\(^3\)- 2009x\(^2\)+ 2009x - 2010
= x\(^5\)- (x + 1)x\(^4\)+ (x + 1)x\(^3\)- (x +1)x\(^2\)+ (x + 1) x - (x + 1 + 1)
= x\(^5\)- x\(^5\)- x\(^4\)+ x\(^4\)- x\(^3\)+ x\(^3\)- x\(^2\)+ x\(^2\)+ x - x -1 - 1
= -2
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Khi đó : \(\frac{ac}{a^2+c^2}=\frac{bk.dk}{\left(bk\right)^2+\left(dk^2\right)}=\frac{k^2.bd}{k^2\left(b^2+d^2\right)}=\frac{bd}{b^2+d^2}\)
\(\Rightarrow\frac{ac}{a^2+c^2}=\frac{bd}{b^2+d^2}\left(đ\text{pcm}\right)\)