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a: a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{k}{k-1}\)
\(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{k}{k-1}=\dfrac{a}{a-b}\)
b: \(\dfrac{a}{b}=\dfrac{bk}{b}=k\)
\(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=k=\dfrac{a}{b}\)
c \(\dfrac{a}{3a+b}=\dfrac{bk}{3bk+b}=\dfrac{k}{3k+1}\)
\(\dfrac{c}{3c+d}=\dfrac{dk}{3dk+d}=\dfrac{k}{3k+1}=\dfrac{a}{3a+b}\)
d: \(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=k^2=\dfrac{ac}{bd}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
a/ \(VT=\dfrac{a+b}{b}=\dfrac{bk+b}{b}=\dfrac{b\left(k+1\right)}{b}=k+1=\left(1\right)\)
\(VP=\dfrac{c+d}{d}=\dfrac{dk+d}{d}=\dfrac{d\left(k+1\right)}{d}=k+1\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
b/ \(VT=\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\left(1\right)\)
\(VP=\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{a}{a-b}=\dfrac{c}{c-d}\)
c/ \(VT=\dfrac{2a-5b}{2c-5d}=\dfrac{2bk-5b}{2dk-5d}=\dfrac{b\left(2k-5\right)}{d\left(2k-5\right)}=\dfrac{b}{d}\left(1\right)\)
\(VP=\dfrac{3a+4b}{3c+4d}=\dfrac{3bk+4b}{3dk+4d}=\dfrac{b\left(3k+4\right)}{d\left(3k+4\right)}=\dfrac{b}{d}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{2a-5b}{2c-5đ}=\dfrac{3a+4b}{3c+4d}\)
d/ \(VT=\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{\left(bk\right)^2-\left(dk\right)^2}{b^2-k^2}=\dfrac{k^2\left(b^2-d^2\right)}{b^2-d^2}=k^2\left(1\right)\)
\(VP=\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{ac}{bd}\)
\(a,\)
Xét \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ad=bc\)
mà \(ad=bc\left(gt\right)\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\)
\(b,\)
\(\dfrac{a}{b}=\dfrac{c}{d}\) (Chứng minh câu a)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\)
\(\Rightarrow\dfrac{a+c}{b+d}=\dfrac{a}{b}\)
\(c,\)
Xét \(\dfrac{a}{c}=\dfrac{b}{d}\Leftrightarrow ad=bc\)
mà \(ad=bc\left(gt\right)\)
\(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
\(d,\)
\(\dfrac{a}{c}=\dfrac{b}{d}\) (Chứng minh câu c)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)
\(\Rightarrow\dfrac{a}{c}=\dfrac{a+b}{c+d}\)
\(\Rightarrow\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
\(e,\)
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{2a+b}{2c+d}\)
\(\Rightarrow\dfrac{2a+b}{2c+d}=\dfrac{a}{c}\)
b,
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{b}{d}=\dfrac{a}{c}=\dfrac{b+a}{d+c}\\ \Rightarrow\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
c,
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
ta có: \(a=bk;c=dk\)
\(\Rightarrow\dfrac{2a+3c}{2b+3d}=\dfrac{2bk+3dk}{2b+3d}=\dfrac{k^2.\left(2b+3d\right)}{2b+3d}=k^2\\ \Rightarrow\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k^2.\left(2b-3d\right)}{2b-3d}=k^2\\ \Rightarrow\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
d,
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
ta có:\(a=bk;c=dk\)
\(\Rightarrow\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\\ \Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{k^2.\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\\ \Rightarrow\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
e,
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
Ta có:\(a=bk;c=dk\)
\(\Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{k^2.\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\\ \Rightarrow\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{k^2.\left(b-d\right)^2}{\left(b-d\right)^2}=k^2\\ \Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{a^2-c^2}{b^2-d^2}\)
f,
(để hôm sau lm nha, mỏi tay quá)
a, \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=> \(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{a+b}{c+d}\)=\(\dfrac{a-b}{c-d}\)(1)
\(\dfrac{a+b}{c+d}\)=\(\dfrac{a-b}{c-d}\)=> \(\dfrac{a+b}{a-b}\)=\(\dfrac{c+d}{c-d}\)
Còn các phần còn lại làm giống thế
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}\)=k (1)
=> a=bk ,c=dk
a.Có \(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=\dfrac{k\left(b+d\right)}{b+d}=k\left(2\right)\)
Từ (1) và (2)=>\(\dfrac{a+c}{b+d}=\dfrac{a}{b}\left(=k\right)\)
b. Có \(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\)
=>\(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\left(=k^2\right)\)
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=k\)
\(\dfrac{a-c}{b-d}=\dfrac{bk-dk}{b-d}=k\)
Do đó: \(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)
b: \(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)
nên \(\dfrac{a+c}{a-c}=\dfrac{b+d}{b-d}\)
c: \(\dfrac{a}{a+c}=\dfrac{bk}{bk+dk}=\dfrac{b}{b+d}\)
Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
Lần lượt thay a và c vào các ý cần chứng minh, áp dụng theo tính chất phân phối giữa phép nhân đối với phép cộng (hay phép trừ) để tính ở mỗi vế.
Mẫu: a) Ta có : \(\dfrac{a+b}{b}=\dfrac{bk+b}{b}=\dfrac{b\left(k+1\right)}{b}=k+1\)
\(\dfrac{c+d}{d}=\dfrac{dk+d}{d}=\dfrac{d\left(k+1\right)}{d}=k+1\)
\(\Rightarrow\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
Vậy \(\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
a)\(\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
Gọi\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(a=b.k\)
\(c=d.k\)
\(\dfrac{a+b}{b}=\dfrac{bk+b}{b}=\dfrac{b.\left(k+1\right)}{b}=k+1\) (1)
\(\dfrac{c+d}{d}=\dfrac{dk+d}{d}=\dfrac{d.\left(k+1\right)}{d}=k+1\)(2)
Từ (1) và (2) \(\Rightarrow\)\(\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
b)\(\dfrac{a-b}{b}=\dfrac{c-d}{d}\)
Gọi\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(a=b.k\)
\(c=d.k\)\(\dfrac{a-b}{a}=1-\dfrac{b}{a}=1-\dfrac{b}{bk}=1-\dfrac{1}{k}\left(1\right)\)
\(\dfrac{c-d}{c}=1-\dfrac{d}{c}=1-\dfrac{d}{dk}=1-\dfrac{1}{k}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\)\(\dfrac{a-b}{b}=\dfrac{c-d}{d}\)
Ta có \(\frac{a}{b}=\frac{c}{d}=>\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}=>\frac{a}{a-b}=\frac{c}{c-d} \)
c
C (sử dụng tính chất dãy tỉ số bằng nhau)