Cho TLT \(\frac{a}{b}=\frac{c}{d}\) hãy CTR
a) \(\frac{a}{b}=\frac{a+c}{b+d}\)
b) \(\frac{a-b}{b}=\frac{c-d}{d}\)
c) \(\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\)
1.Cho TLT \(\frac{a}{b}=\frac{c}{d}.CMR:\left(\frac{a+b}{c+d}\right)=\frac{a^2+b^2}{c^2+d^2}\)
Cho \(\frac{a}{b}=\frac{c}{d}.CM\)
a)\(\frac{a}{a-b}=\frac{c}{c-d}\)
b)\(\frac{a}{b}=\frac{a+c}{b+d}\)
c)\(\frac{a}{3a+b}=\frac{c}{3c+d}\)
d)\(\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\)
e)\(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)
Cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\).Hãy suy ra :
a/ \(\frac{a+c}{b+d}=\frac{a-c}{b-d}\)
b/\(\frac{2a+3c}{2d+3d}=\frac{2a-3c}{2b-3d}\)
c/ \(\frac{a^2+c^2}{b^2+d^2}=\frac{ac}{bd}\)
a/ do \(\frac{a}{b}\) = \(\frac{c}{d}\) = \(\frac{a+c}{b+d}\)=\(\frac{a-c}{b-d}\)(điều phải suy ra)
bạn viết sai đề bài b nhé phân số đầu là \(\frac{2a+3c}{2b+3d}\)
b/ đặt \(\frac{a}{b}\)= \(\frac{c}{d}\) là K
a=Kb;c=Kd
ta có:\(\frac{2a+3c}{2b+3d}\)= \(\frac{2Kb+3Kd}{2b+3d}\) = \(\frac{k\left(2b+3d\right)}{2b+3d}\) = K (1)
\(\frac{2a-3c}{2b-3d}\) = \(\frac{2Kb-3Kd}{2b-3d}\) = \(\frac{k\left(2b-3d\right)}{2b-3d}\) =K (2)
từ (!) và (2) suy ra \(\frac{2a+3c}{2b+3d}\) = \(\frac{2a-3c}{2b-3d}\)
Cho \(\frac{a}{b}=\frac{c}{d}CMR:\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{d^2-bd}\)
Ta có :
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\left(\frac{a}{b}\right)^2=\left(\frac{c}{d}\right)^2=\frac{ac}{bd}=\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+ac}{b^2+bd}=\frac{c^2-ac}{d^2-bd}\)
\(\Rightarrow\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{d^2-bd}\) (đpcm)
Cho \(\frac{a}{b}=\frac{c}{d}\) chứng minh:
1/ \(\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{d^2-bd}\)
2/ \(\frac{3a^2+c^2}{3b^2+d^2}=\frac{\left(a+c\right)^2}{\left(b+d\right)^2}\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)\(\Rightarrow a=bk;c=dk\)
1)Xét \(VT=\frac{\left(bk\right)^2+bkdk}{\left(dk\right)^2-bkdk}=\frac{b^2k^2+bdk^2}{d^2k^2-bdk^2}=\frac{k^2\left(b^2+bd\right)}{k^2\left(d^2-bd\right)}=\frac{b^2+bd}{d^2-bd}=VP\)
Suy ra Đpcm
2)Xét \(VT=\frac{3\left(bk\right)^2+\left(dk\right)^2}{3b^2+d^2}=\frac{3b^2k^2+d^2k^2}{3b^2+d^2}=\frac{k^2\left(3b^2+d^2\right)}{3b^2+d^2}=k^2\left(1\right)\)
Xét \(VP=\frac{\left(bk+dk\right)^2}{\left(b+d\right)^2}=\frac{k^2\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\left(2\right)\)
Từ (1) và (2) suy ra Đpcm
Cho \(\frac{a}{b}=\frac{c}{d}\).CMR :
a/ \(\frac{a+c}{b+d}=\frac{a-c}{b-d}\)
b/\(\frac{2a+3c}{2d+3d}=\frac{2a-3c}{2b-3d}\)
c/\(\frac{a^2+c^2}{b^2+d^2}=\frac{ac^2}{bd}\)
Cho \(\frac{a}{b}=\frac{c}{d}CMR:\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{d^2-bd}\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
\(\Rightarrow\frac{a^2+ac}{c^2-ac}=\frac{\left(bk\right)^2+\left(bk\right)\left(dk\right)}{\left(dk\right)^2-\left(bk\right)\left(dk\right)}=\frac{k^2\left(b^2+bd\right)}{k^2\left(d^2-bd\right)}=\frac{b^2+bd}{d^2-bd}\) (đpcm)
Vậy \(\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{d^2-bd}\)
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Konasuba
Cho \(\frac{a}{b}=\frac{c}{d}CMR:\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{d^2-bd}\)
giúp gấp vs mấy bn:
Tìm a,b,c ϵ Q
a)
\(\frac{a}{b}=\frac{c}{d}\left(ac\ne bd\right)Cm:\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
b)CMR nếu \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)thì\(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\)
a) Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\begin{cases}a=kb\\c=kd\end{cases}\)
=> \(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(kb\right)^2+b^2}{\left(kd\right)^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\) (1)
\(\frac{ab}{cd}=\frac{kbb}{kdd}=\frac{k.b^2}{k.d^2}=\frac{b^2}{d^2}\) (1)
Từ (1) và (2) => \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
b) Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\)
Ta có: \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=k^3\)
Mà: \(k^3=\frac{a}{d}\) => \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\)
a)Ta có:\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a}{c}\cdot\frac{b}{d}=\frac{ab}{cd}\)
\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\left(đpcm\right)\)
b)Ta có:\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
\(\Rightarrow\left(\frac{a}{b}\right)^3=\left(\frac{b}{c}\right)^3=\left(\frac{c}{d}\right)^3=\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}\)
\(\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)
Mà \(\left(\frac{a}{b}\right)^3=\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{d}=\frac{a}{d}=\frac{a^3}{b^3}\)
\(\Rightarrow\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\)