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

a) áp dụng tính chất của dãy tỉ số bằng nhau ta có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-c^2}{b^2-d^2}\)

Do \(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}\)=> đpcm 

b)  áp dụng tính chất của dãy tỉ số bằng nhau ta có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{ab}{cd}=\left(\frac{a-c}{b-d}\right)^2\)=> đpcm

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

thanks

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

Giải:

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)

Đặt \(\frac{a}{c}=\frac{b}{d}=k\)

a, Ta có: \(k^2=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\)  (1)

\(k^2=\frac{a}{c}.\frac{b}{d}=\frac{ab}{cd}\)  (2)

Từ (1), (2) \(\Rightarrowđpcm\)

b, Ta có: \(k=\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)

\(\Rightarrow k^2=\left(\frac{a-b}{c-d}\right)^2=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)  (1)

\(k^2=\frac{a}{c}.\frac{b}{d}=\frac{ab}{cd}\)  (2)

Từ (1), (2) \(\Rightarrowđpcm\)

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

     a)Thay vào \(\frac{a^2-b^2}{c^2-d^2}\) ta được:

\(\Rightarrow\frac{a^2-b^2}{c^2-d^2}\Rightarrow\frac{b^2k^2-b^2}{d^2k^2-d^2}\Rightarrow\frac{b^2}{d^2}\Rightarrow\frac{b.b}{d.d}\left(1\right)\)

                Ta có:\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow a=b;c=d\left(2\right)\)

                           Từ (1) và (2) suy ra:\(\frac{a^2-b^2}{c^2-d^2}=\frac{ab}{cd}\)

1, \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{3a}{3c}=\frac{b}{d}=\frac{3a+b}{3c+d}\Rightarrow\frac{a}{c}=\frac{3a+b}{3c+d}\Rightarrow\frac{a}{3a+b}=\frac{c}{3c+d}\)

2, a, Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{a}{c}=\frac{a}{c}\cdot\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{ab}{cd}\)

\(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{b}{d}\cdot\frac{b}{d}\Rightarrow\frac{ab}{cd}=\frac{b^2}{d^2}\)

\(\Rightarrow\frac{ab}{cd}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\)

b, Ta có: \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{a-b}{c-d}\cdot\frac{a-b}{c-d}\Rightarrow\frac{ab}{cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)

a) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{ab}{cd}=\frac{a.a}{c.c}=\frac{b.b}{c.d}=\frac{a^2-b^2}{c^2-d^2}\)

b) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{ab}{cd}=\frac{a}{c}.\frac{b}{d}=\frac{a-b}{c-d}.\frac{a-b}{c-d}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)

Có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)

=> \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)(Tính chất dãy tỉ số bằng nhau)

=> \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)(Đpcm)

Có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)(Tính chất dãy tỉ số bằng nhau)

=> \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{ab}{cd}\)

=> \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{ab}{cd}\)(Đpcm)