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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 \(\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}\)
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\Rightarrow\dfrac{a.b}{c.d}=\dfrac{a+b}{c+d}.\dfrac{a+b}{c+b}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
Ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\Rightarrow\dfrac{a.b}{c.d}=\dfrac{a+b}{c+d}.\dfrac{a+b}{c+d}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
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!
Bài 1:
Áp dụng t.c của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\\ =\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a^3}{b^3}=\dfrac{a.b.c}{b.c.d}=\dfrac{a}{d}\left(dpcm\right)\)
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{c}=\dfrac{b}{d}\)
Ta đặt: \(\dfrac{a}{c}=\dfrac{b}{d}=k\) => a=ck ; b=dk
a) \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{\left(bk\right)^2-\left(dk\right)^2}{c^2-d^2}=\dfrac{b^2k^2-d^2k^2}{c^2-d^2}=\dfrac{k^2\left(b^2-d^2\right)}{b^2-d^2}=k^2\)(1)
\(\dfrac{ab}{cd}=\dfrac{ck.dk}{cd}=\dfrac{k^2\left(c.d\right)}{cd}=k^2\) (2)
Từ (1) và (2) => \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{ab}{cd}\)
b) \(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(ck-dk\right)^2}{\left(c-d\right)^2}=\dfrac{k^2\left(c-d\right)^2}{\left(c-d\right)^2}=k^2\) (3)
Từ (2) và (3) => \(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{ab}{cd}\). Chúc bạn học tốt 
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow a=bk;c=dk\)
Thay vào ta có:
\(\dfrac{ab}{cd}=\dfrac{bk.b}{dk.d}=\dfrac{b^2\cdot k}{d^2\cdot k}=\dfrac{b^2}{d^2}\left(1\right)\)
\(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}\)
\(=\dfrac{\left[b.\left(k+1\right)\right]^2}{\left[d.\left(k+1\right)\right]^2}\)
\(=\left(\dfrac{b}{d}\right)^2=\dfrac{b^2}{d^2}\left(2\right)\)
Từ (1) và (2) suy ra: đpcm
Gia su \(\dfrac{a}{b}=\dfrac{c}{d}=k\)=> a=bk; c=dk
The vao ta co:
\(\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}\)<=>\(\dfrac{b^2\cdot k}{d^2\cdot k}=\dfrac{b^2\cdot k^2-b^2}{d^2\cdot k^2-d^2}\)<=>\(\dfrac{b^2}{d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}\)
=>\(\dfrac{b^2}{d^2}=\dfrac{b^2}{d^2}\)
b,
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{b}{d}=\dfrac{a}{c}=\dfrac{b+a}{d+c}\\ \Rightarrow\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
c,
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
ta có: \(a=bk;c=dk\)
\(\Rightarrow\dfrac{2a+3c}{2b+3d}=\dfrac{2bk+3dk}{2b+3d}=\dfrac{k^2.\left(2b+3d\right)}{2b+3d}=k^2\\ \Rightarrow\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k^2.\left(2b-3d\right)}{2b-3d}=k^2\\ \Rightarrow\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
d,
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
ta có:\(a=bk;c=dk\)
\(\Rightarrow\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\\ \Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{k^2.\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\\ \Rightarrow\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
e,
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
Ta có:\(a=bk;c=dk\)
\(\Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{k^2.\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\\ \Rightarrow\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{k^2.\left(b-d\right)^2}{\left(b-d\right)^2}=k^2\\ \Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{a^2-c^2}{b^2-d^2}\)
f,
(để hôm sau lm nha, mỏi tay quá)
a, \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=> \(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{a+b}{c+d}\)=\(\dfrac{a-b}{c-d}\)(1)
\(\dfrac{a+b}{c+d}\)=\(\dfrac{a-b}{c-d}\)=> \(\dfrac{a+b}{a-b}\)=\(\dfrac{c+d}{c-d}\)
Còn các phần còn lại làm giống thế
Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\d=ck\end{matrix}\right.\)
Ta có :
\(VT=\dfrac{ab}{cd}=\dfrac{bk.b}{dk.d}=\dfrac{b^2}{d^2}\left(1\right)\)
\(VP=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\dfrac{b^2+\left(k+1\right)^2}{d^2+\left(k+1\right)^2}=\dfrac{b^2}{d^2}\)\(\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
Gọi \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=k\(\Rightarrow\)\(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có:\(\dfrac{ab}{cd}\)=\(\dfrac{bk.b}{dk.d}\)=\(\dfrac{b^2.k}{d^2.k}\)=\(\dfrac{b^2}{d^2}\)=\(\left(\dfrac{b}{d}\right)^2\)(vì k khác 0) 1
\(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)=\(\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}\)=\(\dfrac{\left[b.\left(k+1\right)\right]^2}{\left[d.\left(k+1\right)\right]^2}\)\(\left(\dfrac{b}{d}\right)^2\)=(vì k+1 khác 0) 2
Từ 1 và 2:
\(\Rightarrow\)\(\dfrac{ab}{cd}\)=\(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
Vậy \(\dfrac{ab}{cd}\)=\(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)(điều cần chứng minh)
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 \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
hình như đề sai đó bạn
bạn sửa hộ mik \(\left(\dfrac{a^2+b^2}{c^2+d^2}\right)^2\) thành\(\dfrac{a^2+b^2}{c^2+d^2}\)nha!!