Ta có:

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

a) \(\dfrac{3a+5c}{3b+5d}=\dfrac{3\cdot bk+5\cdot dk}{3b+5d}=\dfrac{k\left(3b+5d\right)}{3b+5d}=k\) (1)

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

Từ (1) và (2) \(\Rightarrow\dfrac{3a+5c}{3b+5d}=\dfrac{a-2c}{b-2d}\left(dpcm\right)\)

b) \(\dfrac{a^2-b^2}{ab}=\dfrac{\left(bk\right)^2-b^2}{bk\cdot b}=\dfrac{b^2k^2-b^2}{b^2k}=\dfrac{b^2\left(k-1\right)}{b^2k}=\dfrac{k-1}{k}\)(1)

\(\dfrac{c^2-d^2}{cd}=\dfrac{\left(dk\right)^2-d^2}{dk\cdot d}=\dfrac{d^2k^2-d^2}{d^2k}=\dfrac{d^2\left(k-1\right)}{d^2k}=\dfrac{k-1}{k}\) (2)

Từ (1) và (2) \(\Rightarrow\dfrac{a^2-b^2}{ab}=\dfrac{c^2-d^2}{cd}\left(dpcm\right)\)

c) \(\left(\dfrac{a+b}{c+d}\right)^3=\left(\dfrac{bk+b}{dk+d}\right)^3=\dfrac{b^3\left(k+1\right)^3}{d^3\left(k+1\right)^3}=\dfrac{b^3}{d^3}\) (1)

\(\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{\left(bk\right)^3+b^3}{\left(dk\right)^3+d^3}=\dfrac{b^3k^3+b^3}{d^3k^3+d^3}=\dfrac{b^3\left(k^3+1\right)}{d^3\left(k^3+1\right)}=\dfrac{b^3}{d^3}\) (2)

Từ (1) và (2) \(\Rightarrow\left(\dfrac{a+b}{c+d}\right)^3=\dfrac{a^3+b^3}{c^3+d^3}\left(dpcm\right)\)

Cứu mình với mình đang cần gấp!~

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

Có thể dùng cách khác:v

a)\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}=t\)(với t là 1 số thực bất kì thỏa mãn)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{c}.\dfrac{b}{d}=\dfrac{ab}{cd}=t^2\\\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2-b^2}{c^2-d^2}=t^2\end{matrix}\right.\Rightarrowđpcm\)

Tương tự:v

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk,d=ck\)

a) \(\dfrac{a^2-b^2}{ab}=\dfrac{b^2k^2-b^2}{bk.b}=\dfrac{b^2\left(k^2-1\right)}{b^2.k}=\dfrac{k^2-1}{k}\) (1)

\(\dfrac{c^2-d^2}{cd}=\dfrac{d^2k^2-d^2}{dk.d}=\dfrac{d^2\left(k^2-1\right)}{d^2k}=\dfrac{k^2-1}{k}\) (2)

Tử (1) và (2) \(\Rightarrow\dfrac{a^2-b^2}{ab}=\dfrac{c^2-d^2}{cd}\)

b) \(\dfrac{\left(a+b\right)^2}{a^2+b^2}=\dfrac{\left(bk+b\right)^2}{b^2k^2+b^2}=\dfrac{\left[b\left(k+1\right)\right]^2}{b^2\left(k^2+1\right)}\)

\(=\dfrac{b^2\left(k+1\right)^2}{b^2\left(k^2+1\right)}=\dfrac{\left(k+1\right)^2}{k^2+1}\) (1)

\(\dfrac{\left(c+d\right)^2}{c^2+d^2}=\dfrac{\left(dk+d\right)^2}{d^2k^2+d^2}=\dfrac{\left[d\left(k+1\right)\right]^2}{d^2\left(k^2+1\right)}\)

\(=\dfrac{d^2\left(k+1\right)^2}{d^2\left(k^2+1\right)}=\dfrac{\left(k+1\right)^2}{k^2+1}\) (2)

Từ (1) và (2) \(\Rightarrow\dfrac{\left(a+b\right)^2}{a^2+b^2}=\dfrac{\left(c+d\right)^2}{c^2+d^2}\)

Chúc bạn học tốt ♥v♥

\(\dfrac{a}{b}=\dfrac{c}{d}=k\)

=>a=bk và c=dk

ta có \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b^2}{d^2}\)\(\dfrac{ab}{cd}=\dfrac{bk.b}{bk.d}=\dfrac{b^2}{d^2}\)

=>\(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\) (cùng =\(\dfrac{b^2}{d^2}\) ) (đpcm)

\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}\)

Đặt: \(\dfrac{a}{c}=\dfrac{b}{d}=t\)

a) \(\left\{{}\begin{matrix}\dfrac{ab}{cd}=t^2\\\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2-b^2}{c^2-d^2}\end{matrix}\right.\Rightarrowđpcm\)

b) \(\left\{{}\begin{matrix}\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\Leftrightarrow\left(\dfrac{a+b}{c+d}\right)^2=t^2\\\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2+b^2}{c^2+d^2}=t^2\end{matrix}\right.\Rightarrowđpcm\)

Ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)

Áp dụng tính chất DTSBN :

\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\)

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

 thank

a)đặt \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=k\(\Rightarrow\)a=bk, c=dk
\(\dfrac{2a+3b}{2a-3b}=\dfrac{2bk+3b}{2bk-3b}=\dfrac{b\left(2k+3\right)}{b\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\) (1)
\(\dfrac{2c+3d}{2c-3d}=\dfrac{2dk+3d}{2dk-3d}=\dfrac{d\left(2k+3\right)}{d\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\) (2)
từ (1),(2)\(\Rightarrow\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\)

b)ta có:
\(\dfrac{ab}{cd}=\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2-b^2}{c^2-d^2}\)
câu c bn tự giải nhé dễ mak ahihihichúc bn hc tốt

d: Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có: \(\dfrac{3c^2+5a^2}{3d^2+5b^2}=\dfrac{3\cdot\left(dk\right)^2+5\cdot\left(bk\right)^2}{3d^2+5b^2}=k^2\)

\(\dfrac{c^2}{d^2}=\dfrac{\left(dk\right)^2}{d^2}=k^2\)

Do đó: \(\dfrac{3c^2+5a^2}{3d^2+5b^2}=\dfrac{c^2}{d^2}\)

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