Cho \(\dfrac{a}{b}=\dfrac{c}{d}\). CM rằng
\(\dfrac{7a-11b}{21a+5b}=\dfrac{7c-11d}{4c+5d}\)
( CM bằng 2 cách)
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HD
5 tháng 11 2017
Ta có:
\(\dfrac{7a-11b}{4a+5b}=\dfrac{7c-11d}{4c+5d}\)
\(\Rightarrow\dfrac{7a-11b}{7c-11d}=\dfrac{4a+5b}{4c+5d}\)
\(\Leftrightarrow\dfrac{7a}{7c}=\dfrac{11b}{11d}=\dfrac{4a}{4c}=\dfrac{5b}{5d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Mặt khác:
\(\dfrac{a}{c}=\dfrac{b}{d}\Leftrightarrow\dfrac{a}{b}=\dfrac{c}{d}\left(đpcm\right)\)
5 tháng 11 2017
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}\)= k
Vì \(\dfrac{a}{b}=k\) = > a = bk
Vì \(\dfrac{c}{d}=k\) = > c = dk
Ta có: \(\dfrac{7a-11b}{4a+5b}=\dfrac{7.bk-11b}{4.bk+5b}=\dfrac{\left(7.11\right).b.\left(k-1\right)}{\left(4.5\right).b.\left(k+1\right)}\dfrac{\left(7.11\right).\left(k-1\right)}{\left(4.5\right).\left(k+1\right)}\)(1)
\(\dfrac{7c-11d}{4c+5d}=\dfrac{7.dk-11d}{4.dk+5d}=\dfrac{\left(7.11\right).d.\left(k-1\right)}{\left(4.5\right).d.\left(k+1\right)}=\dfrac{\left(7.11\right).\left(k-1\right)}{\left(4.5\right).\left(k+1\right)}\left(2\right)\)Từ (1) và (2) = > \(\dfrac{7a-11b}{4a+5b}=\dfrac{7c-11d}{4c+5d}\)
NA
2
AH
3 tháng 11 2018
Vì\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Đặt \(\frac{a}{c}=\frac{b}{d}\) = k
=> a = ck , b = dk
Thay a = ck , b = dk vào \(\frac{7a-11b}{4a+5b}\)ta có :
\(\frac{7a-11b}{4a+5b}=\frac{7.ck-11dk}{4ck+5dk}=\frac{k\left(7c-11d\right)}{k\left(4c+5d\right)}=\frac{7c-11d}{4c+5d}\)
Vậy \(\frac{7a-11b}{4a+5b}=\frac{7c-11d}{4c+5d}\)
NN
2
AT
5 tháng 2 2017
ta có:
\(\frac{7a-11b}{4a+5b}=\frac{7c-11d}{4c+5d}\)
\(\Rightarrow\frac{7a-11b}{7c-11d}=\frac{4a+5b}{4c+5d}\)
\(\Leftrightarrow\frac{7a}{7c}=\frac{11b}{11d}=\frac{4a}{4c}=\frac{5b}{5d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Mặt khác:
\(\frac{a}{c}=\frac{b}{d}\Leftrightarrow\frac{a}{b}=\frac{c}{d}\)
\(\Rightarrowđpcm\)
AH
Akai Haruma
Giáo viên
20 tháng 6 2019
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk,c=dk\). Khi đó ta có:
a)
\((a+c)(b-d)=(bk+dk)(b-d)=k(b+d)(b-d)\)
\((a-c)(b+d)=(bk-dk)(b+d)=k(b-d)(b+d)=k(b+d)(b-d)\)
\(\Rightarrow (a+c)(b-d)=(a-c)(b+d)\) (đpcm)
b)
\((a+c)b=(bk+dk)b=k(b+d).b=bk(b+d)\)
\((b+d).a=(b+d).bk=bk(b+d)\)
\(\Rightarrow (a+c)b=(b+d)a\)
c)
\(a(b-d)=bk(b-d)\)
\(b(a-c)=b(bk-dk)=bk(b-d)\)
\(\Rightarrow a(b-d)=b(a-c)\)
d)
\((b+d).c=(b+d).dk=dk(b+d)\)
\((a+c)d=(bk+dk)d=k(b+d)d=dk(b+d)\)
\(\Rightarrow (b+d)c=(a+c)d\)
AH
Akai Haruma
Giáo viên
20 tháng 6 2019
e)
\((b-d).c=(b-d).dk=dk(b-d)\)
\((a-c)d=(bk-dk)d=k(b-d)d=dk(b-d)\)
\(\Rightarrow (b-d)c=(a-c)d\)
f)
\((a+b)(c-d)=(bk+b)(dk-d)=b(k+1)d(k-1)=bd(k-1)(k+1)\)
\((a-b)(c+d)=(bk-b)(dk+d)=b(k-1)d(k+1)=bd(k-1)(k+1)\)
\(\Rightarrow (a+b)(c-d)=(a-b)(c+d)\)
g)
\((2a+3c)(2b-3d)=(2bk+3dk)(2b-3d)=k(2b+3d)(2b-3d)\)
\((2a-3c)(2b+3d)=(2bk-3dk)(2b+3d)=k(2b-3d)(2b+3d)\)
\(\Rightarrow (2a+3c)(2b-3d)=(2a-3c)(2b+3d)\)
h)
\((4a+3b)(4c-3d)=(4bk+3b)(4dk-3d)=b(4k+3)d(4k-3)=bd(4k+3)(4k-3)\)
\((4a-3b)(4c+3d)=(4bk-3b)(4dk+3d)=b(4k-3)d(4k+3)=bd(4k+3)(4k-3)\)
\(\Rightarrow (4a+3b)(4c-3d)=(4a-3b)(4c+3d)\)
i,k: Hoàn toàn tương tự.
11 tháng 11 2023
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(a=bk;c=dk\)
1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)
\(\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k\left(2b-3d\right)}{2b-3d}=k\)
Do đó: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)
\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)
Do đó: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
3: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
4: \(\dfrac{3a-7b}{b}=\dfrac{3bk-7b}{b}=\dfrac{b\left(3k-7\right)}{b}=3k-7\)
\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)
Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
PN
7 tháng 10 2017
1) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\) (1)
\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\) (2)
Từ (1) và (2) \(\Rightarrow\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
2) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=q\Rightarrow\left\{{}\begin{matrix}a=bq\\c=dq\end{matrix}\right.\)
Ta có: \(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bq+b}{dq+d}\right)^2=\left[\dfrac{b\left(q+1\right)}{d\left(q+1\right)}\right]^2=\dfrac{b}{d}\) (1)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bq\right)^2+b^2}{\left(dq\right)^2+d^2}=\dfrac{b^2.q^2+b^2}{d^2.q^2+d^2}=\dfrac{b^2\left(q^2+1\right)}{d^2\left(q^2+1\right)}=\dfrac{b^2}{d^2}=\dfrac{b}{d}\) (2)
Từ (1) và (2) \(\Rightarrow\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
Sửa đề:
\(\dfrac{7a-11b}{4a+5b}=\dfrac{7c-11d}{4c+5d}\)
Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{7a-11b}{4a+5b}=\dfrac{7bk-11b}{4bk+5b}=\dfrac{7k-11}{4k+5}\)
\(\dfrac{7c-11d}{4c+5d}=\dfrac{7dk-11dk}{4dk+5d}=\dfrac{7k-11}{4k+5}\)
Do đó: \(\dfrac{7a-11b}{4a+5b}=\dfrac{7c-11d}{4c+5d}\)