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 \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)

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

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:

\(\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}\)<=>\(\dfrac{b^2\cdot k}{d^2\cdot k}=\dfrac{b^2\cdot k^2-b^2}{d^2\cdot k^2-d^2}\)<=>\(\dfrac{b^2}{d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}\)

=>\(\dfrac{b^2}{d^2}=\dfrac{b^2}{d^2}\)

hình như đề sai đó bạn

theo bài ra ta có:

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

áp dụng tính chất dảy tỉ số bằng nhau ta có:

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

Bài 1:

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)

Khi đó: \(\left\{\begin{matrix} \frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b(2k+5)}{b(3k-4)}=\frac{2k+5}{3k-4}\\ \frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{d(2k+5)}{d(3k-4)}=\frac{2k+5}{3k-4}\end{matrix}\right.\)

\(\Rightarrow \frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)

Ta có đpcm.

Bài 2:

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)

Khi đó: \(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)

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

Do đó: \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}(=\frac{b^2}{d^2})\) . Ta có đpcm.

Vào đây: Câu hỏi của nguyen lan anh - Toán lớp 7 | Học trực tuyến

Có: a/b=c/d. Áp dụng T/c tỉ lệ thức, ta có:

a/c=b/d . Đặt a/c=b/d=k=> a=ck;b=dk

Rồi cứ thế thay vào (a) và (b) thì sẽ ra

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 \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)

Bài 1:

Áp dụng t.c của dãy tỉ số bằng nhau, ta có:

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\\ =\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a^3}{b^3}=\dfrac{a.b.c}{b.c.d}=\dfrac{a}{d}\left(dpcm\right)\)

Bài 1:

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

a, Ta có: \(\dfrac{a+c}{c}=\dfrac{bk+dk}{dk}=\dfrac{\left(b+d\right)k}{dk}=\dfrac{b+d}{d}\)

\(\Rightarrowđpcm\)

b, Ta có: \(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=\dfrac{k\left(b+d\right)}{b+d}=k\) (1)

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

Từ (1), (2) \(\Rightarrowđpcm\)

c, Ta có: \(\dfrac{a-c}{a}=\dfrac{bk-dk}{bk}=\dfrac{k\left(b-d\right)}{bk}=\dfrac{b-d}{b}\)

\(\Rightarrowđpcm\)

d, Ta có: \(\dfrac{3a+5b}{2a-7b}=\dfrac{3bk+5b}{2bk-7b}=\dfrac{b\left(3k+5\right)}{b\left(2k-7\right)}=\dfrac{3k+5}{2k-7}\)(1)

\(\dfrac{3c+5d}{2c-7d}=\dfrac{3dk+5d}{2dk-7d}=\dfrac{d\left(3k+5\right)}{d\left(2k-7\right)}=\dfrac{3k+5}{2k-7}\) (2)

Từ (1), (2) \(\Rightarrowđpcm\)

e, Sai đề

f, \(\left(\dfrac{a-b}{c-d}\right)^{2012}=\left(\dfrac{bk-b}{dk-d}\right)^{2012}=\left[\dfrac{b\left(k-1\right)}{d\left(k-1\right)}\right]^{2012}=\dfrac{b^{2012}}{d^{2012}}\)(1)

\(\dfrac{a^{2012}+b^{2012}}{c^{2012}+d^{2012}}=\dfrac{b^{2012}k^{2012}+b^{2012}}{d^{2012}k^{2012}+d^{2012}}=\dfrac{b^{2012}\left(k^{2012}+1\right)}{d^{2012}\left(k^{2012}+1\right)}=\dfrac{b^{2012}}{d^{2012}}\) (2)

Từ (1), (2) \(\Rightarrowđpcm\)

Bài 2:

a, \(\dfrac{3}{x-4}=\dfrac{x+4}{3}\)

\(\Rightarrow\left(x-4\right)\left(x+4\right)=9\)

\(\Rightarrow x^2-16=9\)

\(\Rightarrow x^2=25\)

\(\Rightarrow x=\pm5\)

Vậy \(x=\pm5\)

b, \(\dfrac{x+2}{2}=\dfrac{1}{1-x}\)

\(\Rightarrow\left(x+2\right)\left(1-x\right)=2\)

\(\Rightarrow x-x^2+2-2x=2\)

\(\Rightarrow-x-x^2=0\)

\(\Rightarrow x\left(-1-x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\-1-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

Vậy x = 0 hoặc x = -1

c, \(\dfrac{x+7}{x+4}=\dfrac{x-1}{x-2}\)

\(\Rightarrow\left(x+7\right)\left(x-2\right)=\left(x+4\right)\left(x-1\right)\)

\(\Rightarrow x^2-2x+7x-14=x^2-x+4x-4\)

\(\Rightarrow x^2+5x-14=x^2+3x-4\)

\(\Rightarrow2x=10\)

\(\Rightarrow x=5\)

Vậy x = 5

Bài 3:

\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\)

\(\Rightarrow12x-4y=3x+3y\)

\(\Rightarrow9x=7y\)

\(\Rightarrow\dfrac{x}{y}=\dfrac{7}{9}\)

Vậy \(\dfrac{x}{y}=\dfrac{7}{9}\)