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a.Vì \(\dfrac{a}{b}=\dfrac{c}{d}\)
=>\(\dfrac{a}{b}-1=\dfrac{c}{d}-1\)
=>\(\dfrac{a-b}{b}=\dfrac{c-d}{d}\)(đpcm)
b.Vì\(\dfrac{a}{b}=\dfrac{c}{d}\)
=>\(\dfrac{a}{c}=\dfrac{b}{d}\)
=>\(\dfrac{a}{c}-1=\dfrac{b}{d}-1\)
=>\(\dfrac{a-c}{c}=\dfrac{b-d}{d}\)(đpcm)
a)\(\dfrac{a-b}{b}\) = \(\dfrac{c-d}{d}\)
\(\dfrac{a}{b}\) = \(\dfrac{c}{d}\)
=>\(\dfrac{a}{b}\) -1= \(\dfrac{c}{d}\) -1
=> \(\dfrac{a}{b}\) - \(\dfrac{b}{b}\) = \(\dfrac{c}{d}\) - \(\dfrac{d}{d}\)
=> \(\dfrac{a-b}{b}\) = \(\dfrac{c-d}{d}\)
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) \(\Rightarrow\)\(\dfrac{a-b}{b}=\dfrac{c-d}{d}\)
Bài 2:
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{k}{k+1}\)
\(\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)
Do đó: \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
b: \(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7\cdot b^2k^2+5\cdot bk\cdot dk}{7\cdot b^2k^2-5\cdot bk\cdot dk}\)
\(=\dfrac{7b^2k^2+5bdk^2}{7b^2k^2-5bdk^2}=\dfrac{7b^2+5bd}{7b^2-5bd}\)(đpcm)
2 . a ) nếu \(\dfrac{a}{b}< \dfrac{c}{d}\)thì a.d < b.c
\(\dfrac{a}{b}\cdot\dfrac{c}{d}=\dfrac{ac}{bd}\Rightarrow\dfrac{a}{bd}< \dfrac{c}{bd}\Rightarrow a< c\)
vì a<c => a.d < b.c
=> đcpm
b) ko ghi lại đề
vì a.d<c.d => \(\dfrac{a}{bd}< \dfrac{c}{bd}\Rightarrow\dfrac{a}{b}< \dfrac{c}{d}\)( bn suy luận ngược với a nhé )
a) \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Rightarrow\dfrac{b}{a}=\dfrac{d}{c}\)
\(\Rightarrow1+\dfrac{b}{a}=1+\dfrac{d}{c}\)
\(\Rightarrow\dfrac{a+b}{a}=\dfrac{c+d}{c}\)
\(\RightarrowĐPCM\)
a, Từ \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
\(\Rightarrow ad=bc\)
\(ac-ad=ac-bc\)
\(a\left(c-d\right)=c\left(a-b\right)\)
\(\Rightarrow\dfrac{a}{a-b}=\dfrac{c}{c-d}\Rightarrow\dfrac{c-d}{c}=\dfrac{a-b}{a}\Leftrightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\)
b, Từ \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\left(1\right)\)
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{b-c}\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\Rightarrow\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)
c, Từ \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow ad=bc\)
\(\Rightarrow ad+ac=bc+ac\\ a\left(c+d\right)=c\left(a+b\right)\)
\(\Rightarrow\dfrac{a}{c}=\dfrac{a+b}{c+d}\Rightarrow\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
Đặt\(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
a) \(\dfrac{a-b}{a}=\dfrac{bk-b}{bk}=\dfrac{b\left(k-1\right)}{bk}=\dfrac{k-1}{k}\)
\(\dfrac{c-d}{c}=\dfrac{dk-d}{dk}=\dfrac{d\left(k-1\right)}{dk}=\dfrac{k-1}{k}\)
\(\Rightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\)
b) \(\dfrac{a+b}{a-b}=\dfrac{bk+b}{bk-b}=\dfrac{b\left(k+1\right)}{b\left(k-1\right)}=\dfrac{k+1}{k-1}\)
\(\dfrac{c+d}{c-d}=\dfrac{dk+d}{dk-d}=\dfrac{d\left(k+1\right)}{d\left(k-1\right)}=\dfrac{k+1}{k-1}\)
\(\Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)
c) \(\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}\)
\(\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{k}{k+1}\)
\(\Rightarrow\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
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
a, Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) ( k # 0 )
\(\Rightarrow\) \(a=b.k\)
\(c=d.k\)
Ta có: \(\dfrac{a+b}{b}=\dfrac{b.k+b}{b}=\dfrac{b.\left(k+1\right)}{b}=k+1\) (1)
\(\dfrac{c+d}{d}=\dfrac{d.k+d}{d}=\dfrac{d.\left(k+1\right)}{d}=k+1\) (2)
Từ (1) và (2) \(\Rightarrow\) \(\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
b,
, Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) ( k # 0 )
\(\Rightarrow\) \(a=b.k\)
\(c=d.k\)
Ta có: \(\dfrac{a}{a+b}=\dfrac{b.k}{b.k+b}=\dfrac{b.k}{b.\left(k+1\right)}=\dfrac{k}{k+1}\) (1)
\(\dfrac{c}{c+d}=\dfrac{d.k}{d.k+d}=\dfrac{d.k}{d.\left(k+1\right)}=\dfrac{k}{k+1}\) (2)
Từ (1) và (2) \(\Rightarrow\) \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
a) Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\)
\(\Rightarrow\frac{a^2-b^2}{c^2-d^2}=\frac{ab}{cd}\)
\(\Rightarrow\frac{a^2-b^2}{ab}=\frac{c^2-d^2}{cd}\)
b) Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có:
\(\frac{\left(a+b\right)^2}{a^2+b^2}=\frac{\left(bk+b\right)^2}{\left(bk\right)^2+b^2}=\frac{b^2\left(k+1\right)^2}{b^2\left(k^2+1\right)}=\frac{\left(k+1\right)^2}{k^2+1}\) (1)
Tương tự, ta cũng có \(\frac{\left(c+d\right)^2}{c^2+d^2}=\frac{\left(k+1\right)^2}{k^2+1}\) (2)
Từ (1), (2) suy ra \(\frac{\left(a+b\right)^2}{a^2+b^2}=\frac{\left(c+d\right)^2}{c^2+d^2}\)
Giải:
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)
\(\Rightarrow\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\left(đpcm\right)\)
Vậy...
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(\left\{{}\begin{matrix}a=b.k\\c=d.k\end{matrix}\right.\) (1)
Thay (1) vào:
\(\dfrac{a+b}{a-b}=\dfrac{b.k+b}{b.k-b}=\dfrac{b.\left(k+1\right)}{b.\left(k-1\right)}=\dfrac{k+1}{k-1}\) (2)
\(\dfrac{c+d}{c-d}=\dfrac{d.k+d}{d.k-d}=\dfrac{d.\left(k+1\right)}{d.\left(k-1\right)}=\dfrac{k+1}{k-1}\) (3)
Từ (2) và (3) =>\(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}=\dfrac{k+1}{k-1}\)
C. \(\dfrac{b}{d}=\dfrac{c}{a}\)
Chúc bạn học tốt!!