Cho :\(\dfrac{a}{b}=\dfrac{c}{d}CMR:\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}v\text{à}\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
1. Cho tỉ lệ thức \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\). CMR:
a) \(\dfrac{3a+5c}{3b+5d}\) = \(\dfrac{a-2c}{b-2d}\).
b) \(\dfrac{a^2-b^2}{ab}\) = \(\dfrac{c^2-d^2}{cd}\).
c) \(\dfrac{\left(a+b\right)^2}{a^2+b^2}\) = \(\dfrac{\left(c+d\right)^2}{c^2+d^2}\).
d) \(\left(\dfrac{a+b}{c+d}\right)^3\) = \(\dfrac{a^3+b^3}{c^3+d^3}\).
Gíup mình với cảm ơn các bạn rất nhiều!!!!!!!!!
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) \(\Rightarrow\left(\dfrac{a+b}{c+d}\right)^3=\dfrac{a^3+b^3}{c^3+d^3}\left(dpcm\right)\)
giúp mình câu d) luôn nha phong
cảm ơn phong nha
Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\). CMR : \(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\) và \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
\(\dfrac{a}{b}=\dfrac{c}{d}\\ \Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\\ \Rightarrow\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2-b^2}{c^2-d^2}\\ \dfrac{a^2}{c^2}=\dfrac{a}{c}.\dfrac{a}{c}=\dfrac{a}{c}.\dfrac{b}{d}=\dfrac{ab}{cd}\\ \Rightarrow\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)
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
Cho \(\dfrac{a}{b}=\dfrac{c}{d}.CMR:\)
a, \(\dfrac{a^2-b^2}{ab}=\dfrac{c^2-d^2}{cd}\)
b, \(\dfrac{\left(a+b\right)^2}{a^2+b^2}=\dfrac{\left(c+d\right)^2}{c^2+d^2}\)
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♥
Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\)
CMR \(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\) và \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>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) \(\left\{{}\begin{matrix}\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\Leftrightarrow\left(\dfrac{a+b}{c+d}\right)^2=t^2\\\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2+b^2}{c^2+d^2}=t^2\end{matrix}\right.\Rightarrowđpcm\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k:\)
Nên a=kb, c=kd
Do đó: \(\dfrac{ab}{cd}=\dfrac{kbb}{kdd}=\dfrac{kb^2}{kd^2}=\dfrac{b^2}{d^2}\)
Và: \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{\left(kb\right)^2-b^2}{\left(kd\right)^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b^2}{d^2}\)
Vậy \(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)
Ta có: \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(kb+b\right)^2}{\left(kd+d\right)^2}=\dfrac{\left[\left(k+1\right)b\right]^2}{\left[\left(k+1\right)d\right]^2}=\dfrac{\left(k+1\right)^2.b^2}{\left(k+1\right)^2.d^2}=\dfrac{b^2}{d^2} \)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(kb\right)^2+b^2}{\left(kd\right)^2+d^2}=\dfrac{\left(k^2+1\right)b^2}{\left(k^2+1\right)d^2}=\dfrac{b^2}{d^2}\)
Vậy \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
cho\(\dfrac{a}{b}=\dfrac{c}{d}\)với c\(\ne\) \(\pm\)1. CMR \(\dfrac{ab}{cd}=\dfrac{\left(a-c\right)^2}{\left(b-d\right)^2}\)
cho \(\dfrac{a}{b}=\dfrac{c}{d}\)CMR
\(\left(\dfrac{a-b}{c-d}\right)^2=\dfrac{ab}{cd}\)
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)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}.CMR\)
a, \(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\)
b, \(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)
c, \(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{a^2+b^2}{c^2+d^2}\)
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
cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\)(b\(\ne\)0;d\(\ne\)0)
c)\(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
d)\(\dfrac{3c^2+5a^2}{3d^2+5b^2}=\dfrac{c^2}{d^2}\)
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: \(\dfrac{3c^2+5a^2}{3d^2+5b^2}=\dfrac{c^2}{d^2}\)
Cho \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\). Cmr
a, \(\dfrac{a^2+b^2}{c^2+d^{2^{ }}}\)=\(\dfrac{ab}{cd}\)
b, \(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)=\(\dfrac{ab}{cd}\)
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