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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 \(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\) (*)
a) Từ (*)suy ra:
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\dfrac{b^2.k^2+b^2}{d^2.k^2+d^2}=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}\)\(=\dfrac{b^2}{d^2}\) (1)
\(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\dfrac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\dfrac{b^2.\left(k+1\right)^2}{d^2.\left(k+1\right)^2}=\dfrac{b^2}{d^2}\)(2)
Từ (1) và (2) suy ra: \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\) (đpcm)
b) Tương tự câu a nhé bạn!
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)\(\Rightarrow\) a = bk ; c = dk
\(\Rightarrow\)\(\dfrac{4a^2+4c^2}{4b^2+4d^2}\)=\(\dfrac{4\left(bk\right)^2+4\left(dk\right)^2}{4b^2+4d^2}\)
=\(\dfrac{4b^2k^2+4d^2k^2}{4b^2+4d^2}\)=\(\dfrac{k^2\left(4b^2+4d^2\right)}{4b^2+4d^2}\)= k2 (1)
\(\Rightarrow\)\(\dfrac{\left(a-c\right)^2}{\left(b-d\right)^2}\)=\(\dfrac{\left(bk-dk\right)^2}{\left(b-d\right)^2}\)=\(\dfrac{[k\left(b-d\right)]^2}{\left(b-d\right)^2}\)
=\(\dfrac{k^2\left(b-d\right)^2}{\left(b-d\right)^2}\)= k2 (2)
Từ (1) và (2), suy ra:
\(\dfrac{4a^2+4c^2}{4b^2+4d^2}=\dfrac{\left(a-c\right)^2}{\left(b-d\right)^2}\) (đpcm)
Đặt
\(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\left(a+2c\right)\left(b+d\right)=\left(bk+2dk\right)\left(b+d\right)\)
\(=bk\left(b+d\right)+2dk\left(b+d\right)\)
\(=b^2k+bdk+2bdk+2d^2k\)
\(=b^2k+3bdk+2d^2k\)
\(\Rightarrow\left(a+c\right)\left(b+2d\right)=\left(bk+dk\right)\left(b+2d\right)\)
\(=bk\left(b+2d\right)+dk\left(b+2d\right)\)
\(=b^2k+2bdk+bdk+2d^2k\)
\(=b^2k+3bdk+2d^2k\)
\(VT=VP\)\(\Rightarrowđpcm\)
Vì \(\dfrac{a}{b}=\dfrac{c}{d}\) (theo đề bài)
Áp dụng tính chất dãy tỉ số bằng nhau
\(\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}=\dfrac{a^2+c^2}{b^2+d^2}\)
Vậy \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}.\)
\(\dfrac{a}{b}=\dfrac{c}{d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\)
\(\Rightarrow\left(\dfrac{a}{b}\right)^2=\left(\dfrac{c}{d}\right)^2=\left(\dfrac{a+c}{b+d}\right)^2=\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}=\dfrac{a^2+c^2}{b^2+d^2}\)
\(\Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\)
Gọi \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=k\(\Rightarrow\)\(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có:\(\dfrac{ab}{cd}\)=\(\dfrac{bk.b}{dk.d}\)=\(\dfrac{b^2.k}{d^2.k}\)=\(\dfrac{b^2}{d^2}\)(vì k khác 0) 1
\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)=\(\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}\)=\(\dfrac{\left[b.\left(k-1\right)\right]^2}{\left[d.\left(k-1\right)\right]^2}\)=\(\dfrac{b^2}{d^2}\)(vì k-1 khác 0) 2
Từ 1 và 2:
\(\Rightarrow\)\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)=\(\dfrac{ab}{cd}\)
Vậy \(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)=\(\dfrac{ab}{cd}\)(điều cần chứng minh)
Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\d=ck\end{matrix}\right.\)
Ta có :
\(VT=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-dk\right)^2}{\left(dk-d\right)^2}=\dfrac{b^2-\left(k-1\right)^2}{d^2-\left(k-1\right)^2}=\dfrac{b^2}{d^2}\)\(\left(1\right)\)
\(VP=\dfrac{ab}{cd}=\dfrac{bk.b}{dk.d}=\dfrac{b^2}{d^2}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
a) \(\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
\(\Rightarrow ad=cb\)
=> \(ad+bd=bc+bd\)
\(\Rightarrow d\left(a+b\right)=b\left(c+d\right)\)
\(\Rightarrow\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
Đặt:
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\dfrac{a+b}{b}=\dfrac{bk+b}{b}=\dfrac{b\left(k+1\right)}{b}=k+1\)
\(\Rightarrow\dfrac{c+d}{d}=\dfrac{dk+d}{d}=\dfrac{d\left(k+1\right)}{d}=k+1\)
\(\Rightarrow\dfrac{a+b}{b}=\dfrac{c+d}{d}\rightarrowđpcm\)
\(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b^2}{d^2}\)
\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{\left[b\left(k-1\right)\right]^2}{\left[d\left(k-1\right)\right]^2}=\dfrac{b^2}{d^2}\)
\(\Rightarrow\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\rightarrowđpcm\)
Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\)
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{2c}{2d}=\dfrac{a+2c}{b+2d}\)
\(\Rightarrow\dfrac{a+c}{b+d}=\dfrac{a+2c}{b+2d}\)
\(\Rightarrow\left(a+2c\right)\left(b+d\right)=\left(a+c\right)\left(b+2d\right)\left(đpcm\right)\)
Vậy...
Vì \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)
\(\Rightarrow\left[{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\) (!)
Thay (!) vào đề bài:
VT = \(c\left(k+2\right).d\left(k+1\right)\left(1\right)\)
\(VP=c\left(k+1\right).d\left(k+2\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow VT=VP\)
hay \(\left(a+2c\right)\left(b+d\right)=\left(a+c\right)\left(b+2d\right)\).
vì vai trò của a,b,c,d như nhau, giả sử \(a\ge b\ge c\ge d\)
áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{\left|a-b\right|}{2}=\dfrac{\left|b-c\right|}{23}=\dfrac{\left|c-d\right|}{32}=\dfrac{\left|d-a\right|}{223}\)
=\(\dfrac{a-b+b-c+c-d-\left(-d+a\right)}{-166}=0\)
\(\Rightarrow a+b=0\Rightarrow a=b\) (1)
\(b-c=0\Rightarrow b=c\) (2)
\(c-d=0\Rightarrow c=d\) (3)
từ (1),(2) và (3) suy ra: a=b=c=d