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Sai đề rồi nha bn phải là : \(\frac{a}{3a+b}=\frac{c}{3c+d}\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)\(=\frac{3a}{3c}\)
\(=\frac{3a+b}{3c+d}\)( Theo tính chất dãy tỉ số bằng nhau )
\(\Rightarrow\frac{a}{c}=\frac{3a+b}{3c+d}\)\(\Rightarrow\frac{a}{3a+b}=\frac{c}{3c+d}\)
a, Áp dụng t/c dtsbn:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)
b, Áp dụng t/c dtsbn:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{5b}{5d}=\dfrac{3a}{4c}=\dfrac{4b}{4d}=\dfrac{2a+5b}{2c+5d}=\dfrac{3a-4b}{3c-4d}\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
c, Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
Ta có \(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}\)
\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{b^2\left(k-1\right)^2}{d^2\left(k-1\right)^2}=\dfrac{b^2}{d^2}\)
Do đó \(\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
d, Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
Ta có \(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\)
Do đó \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
đặt a/b=c/d=k
suy ra a=bk;c=dk
suy ra a-b/a+b=bk-b/bk+b=b(k-1)/b(k+1)=k-1/k+1 (1)
c-d/c+d=dk-d/dk+d=d(k-1)/d(k+1)=k-1/k+1 (2)
từ 1 và 2 suy ra dpcm
Xét \(\frac{a}{b}=k;\frac{c}{d}=k\)
=> a= bk; c= dk
Thay:
\(\frac{a}{3a+b}=\frac{bk}{3.bk+b}=\frac{bk}{3.b\left(k+1\right)}=\frac{k}{3.\left(k+1\right)}\) (1)
\(\frac{c}{3c+d}=\frac{dk}{3.dk+d}=\frac{dk}{3.d\left(k+1\right)}=\frac{k}{3.\left(k+1\right)}\) (2)
Ta thấy (1)= (2)
=> \(\frac{a}{3a+b}=\frac{c}{3c+d}\) (dpcm)
theo bài ra ta có:
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{3a}{3c}\)
áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{a}{c}=\frac{b}{d}=\frac{3a}{3c}=\frac{b+3a}{d+3c}\)
=> \(\frac{a}{c}=\frac{3a+b}{3c+d}\)
=> \(\frac{a}{3a+b}=\frac{c}{3c+d}\) (đpcm)
a) \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\)
\(\Rightarrow\left(b+d\right)c=\left(a+c\right)d\)
\(\Rightarrow dpcm\)
b) \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{2a}{2b}=\dfrac{c}{d}=\dfrac{2a+c}{2b+d}=\dfrac{2a-c}{2b-d}\)
\(\Rightarrow\left(2b-d\right)\left(2a+c\right)=\left(2a-c\right)\left(2b+d\right)\)
\(\Rightarrow dpcm\)
c) \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{3c}{3d}=\dfrac{3a}{3b}=\dfrac{5c}{5d}=\dfrac{3a+5c}{3b+5d}=\dfrac{a-3c}{b-3d}\)
\(\Rightarrow\left(b-3d\right)\left(b-3d\right)=\left(3b+5d\right)\left(a-3c\right)\)
\(\Rightarrow dpcm\)
Đính chính câu c
\(\Rightarrow\left(3a+5c\right)\left(b-3d\right)=\left(3b+5d\right)\left(a-3c\right)\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)\(\Rightarrow\)\(\left\{\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
a)Xét \(VT=\frac{a}{a-b}=\frac{bk}{bk-b}=\frac{bk}{b\left(k-1\right)}=\frac{k}{k-1}\left(1\right)\)
Xét \(VP=\frac{c}{c-d}=\frac{dk}{dk-d}=\frac{dk}{d\left(k-1\right)}=\frac{k}{k-1}\left(2\right)\)
Từ (1) và (2) ta có điều phải chứng minh
b)Xét \(VT=\frac{a}{c}=\frac{bk}{dk}=\frac{b}{d}\left(1\right)\)
Xét \(VP=\frac{a+b}{c+d}=\frac{bk+b}{dk+d}=\frac{b\left(k+1\right)}{d\left(k+1\right)}=\frac{b}{d}\left(2\right)\)
Từ (1) và (2) ta có điều phải chứng minh
Ta có:a/b=c/d
<=>1 - a/b=1 - c/d
<=>a/a - a/b=c/c - c/d
<=>a/a-b=c/c-d (đpcm)
Từ \(\frac{a}{b}=\frac{c}{d}=>\frac{a}{c}=\frac{b}{d}=\frac{3a}{3c}=\frac{4b}{4d}\)
Aps dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{a}{c}=\frac{b}{d}=\frac{3a}{3c}=\frac{4b}{4d}=\frac{3a+4b}{3c+4d}\)
=>\(\frac{a}{c}=\frac{3a+4b}{3c+4d}=>\frac{3c+4d}{c}=\frac{3a+4b}{a}\)(đpcm)
a/b=c/d
=>a/c=b/d=3a/3c=4b/4d=(3a+4b)/(3c+4d) (tính chất dãy tỉ số = nhau)
có a/c=(3a+4b)/(3c+4d)
=>dpcm
a) ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}.\)
\(\Rightarrow\frac{a}{c}=\frac{a-b}{c-d}\Rightarrow\frac{a}{a-b}=\frac{c}{c-d}\)
b) ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{a^n}{c^n}=\frac{b^n}{d^n}=\frac{\left(a-b\right)^n}{\left(c-d\right)^n}\)(*)
mà \(\frac{a^n}{c^n}=\frac{b^n}{d^n}=\frac{a^n-b^n}{c^n-d^n}\)
Từ (*) \(\Rightarrow\frac{a^n-b^n}{c^n-d^n}=\frac{\left(a-b\right)^n}{\left(c-d\right)^n}\)