Cho a/b=c/d. Hãy chứng minh: A) a-c/b+d= c-d/c+d. B)2a+5b/3a-4b= 2c+5d/3c-4d C) a.b/c.d=(a-b)^2/ (c-d)^2 D)a.c/b.d=a^2+c^2/b^2+d^2 Giúp mink nha!!!!
1.Cho a/b=c/d . Chứng minh rằng
a)a-c/c=b-d/a
b)a/b=a+c/b+d
c)a+b/a-b=c+d=c-d
d)7a22+3ab/11a2-8b2
e)a/b=3a+2c/3b+2d
f)a/a+b=c/c+d
g)2a+5b/3a-4b=2c+5d/3c-4d
h)a2+c2/b2+d2=a.c/b.d
2.
a)Cho a2/b.c . Chứng minh a+b/a-b=c+a/c-a
b)Cho b2 =a.c . Chứng minh a2+b2/b2+c2=a/c
Cmr nếu a/b=c/d thì
a. a+b/a-b=c+d/c-d
b. (a+b)^2/(a-b)^2=(c+d)^2/(c-d)^2
c. 2a+5b/3a-4b=2c+5d/3c-4d
Cho a/b =c/d .Chứng minh
a. a-b/a+b = c-d/c+d
b.2a + 5b/3a + 4b = 2c - 5d/3c + 4d
a/ Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có :
\(VT=\dfrac{a-b}{a+b}=\dfrac{bk-b}{bk+b}=\dfrac{b\left(k-1\right)}{b\left(k+1\right)}=\dfrac{k-1}{k+1}\left(1\right)\)
\(VP=\dfrac{c-d}{c+d}=\dfrac{dk-d}{dk+d}=\dfrac{d\left(k-1\right)}{d\left(k+1\right)}=\dfrac{k-1}{k+1}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
b/ Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có :
\(VT=\dfrac{2a-5b}{3a+4b}=\dfrac{2bk-5b}{3bk+4b}=\dfrac{b\left(2k-5\right)}{b\left(3k+4\right)}=\dfrac{2k-5}{3k+4}\left(1\right)\)
\(VP=\dfrac{2c-5d}{3c+4d}=\dfrac{2dk-5d}{3dk+4d}=\dfrac{d\left(2k-5\right)}{d\left(3k+4\right)}=\dfrac{2k-5}{3k+4}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
a/b+c+d=b/a+c+d=c/b+a+d=d/c+b+a
P=2a+5b/3c+4d-2b+5c/3d+4a-2c+5d/3a+4b+2d+5a/3c+4b
Cho tỉ lệ thức : \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh
a) \(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{a^2-b^2}{c^2-d^2}\)
b) \(\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
\(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)
\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
b) \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)\(\Rightarrow\frac{2a}{2c}=\frac{5b}{5d}=\frac{3a}{3c}=\frac{4b}{4d}=\frac{2a+5b}{2c+5d}=\frac{3a-4b}{3c-4d}\)
\(\Rightarrow\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)
Cho a/b =c/d
Chứng minh rằng:
a, 2a+5b phần 3a-4b =2c+5d phần 3c-4d
b, a-b phần a+b=c-d phần c+d
đặt \(\frac{a}{b}=\frac{c}{d}=k\)
=> a=bk, c=dk =>\(\frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b\left(2k+5\right)}{b\left(3k-4\right)}=\frac{2k+5}{3k-4}\)(1)
=> \(\frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{2k+5}{3k-4}\) ( 2)
từ (1)( 2)=> \(\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)
câu b c/m tg tự
Bài 2 cho tỉ lệ a/b=c/d
a, ab/cd=(a-b)mũ 2/ (c-d)mũ 2
b, (a+b/c+d)= a mũ 2+5d/ c mũ 2+d mũ 2
c,a-b/a+b=c-d/c+d
d,2a+5b/3a-4b=2c+5d/3c-4d
e,2008a-2009b/2009c+2010d=2008c-2009d/2009a+2010b
giúp mk nha ai nhanh mk tick cho
chô a/b=c/d
a 2a +5b/3a-4b=2c+5d/3c-4d
2 2
b(c-d)/cd=(a-b)/ab
a. Câu hỏi của Nguyễn Ngọc Quế Anh - Toán lớp 7 - Học toán với OnlineMath
cho tỉ lệ thức a/b=c/d chứng minh 2a+5b/3a-4b=2c+5d/3c-4d
Giải:
Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{c}=\frac{b}{d}=\frac{2a}{2c}=\frac{5b}{5d}=\frac{2a+5b}{2c+5d}\)
\(\frac{a}{c}=\frac{b}{d}=\frac{3a}{3c}=\frac{4b}{4d}=\frac{3a-4b}{3c-4d}\)
\(\Rightarrow\frac{2a+5b}{2c+5d}=\frac{3a-4b}{3c-4d}\left(=\frac{a}{c}\right)\)
\(\Rightarrow\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\left(đpcm\right)\)
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