Cho a/b = c/d . CMR :
a) a/b = a+c /b+d
b) a /3a+b = c /3c+d
c) a.c /b.d = a2+c2 / b2+d2
d) a.b /c.d = (a-b)2 /(c-d)2
cho \(\frac{a}{b}=\frac{c}{d}\)chung minh rang:
\(\frac{a}{a-b}=\frac{c}{c-d}\) \(\frac{a}{b}=\frac{a+c}{b+d}\) \(\frac{a}{3a+b}=\frac{c}{3c+d}\)
\(\frac{a.b}{c.d}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\) \(\frac{a.c}{b.d}=\frac{a^2+c^2}{b^2+d^2}\)\(\frac{a.c}{b.d}=\frac{a^2-c^2}{b^2-d^2}\)
+ \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)
\(\Rightarrow\frac{a}{a-b}=\frac{c}{c-d}\)
+ \(\frac{a}{c}=\frac{3a}{3c}=\frac{b}{d}=\frac{3a+b}{3c+d}\) \(\Rightarrow\frac{a}{3a+b}=\frac{c}{3c+d}\)
+ \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{a^2}{c^2}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
\(\Rightarrow\frac{a\cdot b}{c\cdot d}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}\)
\(\Rightarrow\frac{a}{b}\cdot\frac{a}{b}=\frac{a^2+c^2}{b^2+d^2}\Rightarrow\frac{a\cdot c}{b\cdot d}=\frac{a^2+c^2}{b^2+d^2}\)
câu cuối lm tương tự
b) a.b-2.b-3.a+b
c)a.b+b.d-a.c-c.d
d)a.d-b.d-b.c+c.e+c.d+a.e
2. Hai phân số = nếu:
A. a.c = b.d B a.b = c.d C. a: c = b: d D. a.d = b.c
cho \(\frac{a}{b}\)= \(\frac{c}{d}\)chứng minh rằng:
a) \(\frac{a}{3a+b}=\frac{c}{3c+d}\)
b)\(\frac{a.c}{b.d}=\frac{a^2+c^2}{b^2+d^2}\)
c) \(\frac{a.b}{c.d}=\frac{a^2-b^2}{c^2-d^2}\)
Ta có: \(\frac{a}{b}=\frac{c}{d}\)
=> ad = bc
=> 3ac + ad = 3ac + bc
=> a(3c + d) = c(3a + b)
=> \(\frac{a}{3a+b}=\frac{c}{3a+d}\) (ĐPCM)
b) Ta có:
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
đặt \(\frac{a}{c}=k\Rightarrow\frac{b}{d}=k\)
=> a = c.k; b = d.k
=> a2 = c2.k2; b2 = d2.k2
=> \(\frac{a^2+c^2}{b^2+d^2}=\frac{\left(c^2.k^2\right)+c^2}{\left(d^2.k^2\right)+d^2}\)= \(\frac{c^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}\)=\(\frac{c^2}{d^2}=\frac{a^2}{b^2}=\frac{ac}{bd}\)
=> ĐPCM
Câu c tương tự câu b
1. Chứng minh tỉ lệ thức:
Cho \(\frac{a}{b}=\frac{c}{d}\) , chứng minh rằng :
a, \(\frac{a}{3a+b}\) \(=\frac{c}{3c+d}\)
b, \(\frac{a.c}{b.d}=\frac{a^2+c^2}{b^2+d^2}\)
c, \(\frac{a.b}{c.d}=\frac{a^2-b^2}{c^2-d^2}\)
Có \(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{ac}{bd}\)
Mà \(\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}\)
Nên \(\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\left(đpcm\right)\)
1. a) Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
=> \(\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Khi đó \(\frac{a}{3a+b}=\frac{bk}{3bk+b}=\frac{bk}{b\left(3k+1\right)}=\frac{k}{3k+1}\left(1\right)\)
\(\frac{c}{3c+d}=\frac{dk}{3dk+d}=\frac{dk}{d\left(3k+1\right)}=\frac{k}{3k+1}\left(2\right)\)
Từ (1) và (2) => \(\frac{a}{3a+b}=\frac{c}{3c+d}\)
c,
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
=> \(\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Khi đó \(\frac{ab}{cd}=\frac{b^2k}{d^2k}=\frac{b^2}{d^2}\) (3)
\(\frac{a^2-b^2}{c^2-d^2}=\frac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\frac{b^2k^2-b^2}{d^2k^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\frac{b^2}{d^2}\left(4\right)\)
Từ (3) và (4) \(\Rightarrow\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)
@@ Học tốt
Chiyuki Fujito
1) a) Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Khi đó \(\frac{a}{3a+b}=\frac{bk}{3bk+b}=\frac{bk}{b\left(3k+1\right)}=\frac{k}{3k+1}\left(1\right)\)
\(\frac{c}{3a+b}=\frac{dk}{3dk+d}=\frac{dk}{d\left(3k+1\right)}=\frac{k}{3k+1}\left(2\right)\)
\(\left(1\right)\left(2\right)\Rightarrow\frac{a}{3a+b}=\frac{c}{3c+d}\left(đpcm\right)\)
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!!!!
b) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\hept{\begin{cases}\frac{a}{c}=\frac{b}{d}=\frac{2a}{2c}=\frac{5b}{5d}=\frac{2a+5b}{2c+5d}\\\frac{a}{c}=\frac{b}{d}=\frac{3a}{3c}=\frac{4b}{4d}=\frac{3a-4b}{3c-4d}\end{cases}}\)
\(\Rightarrow\frac{2a+5b}{2c+5d}=\frac{3a-4b}{3c-4d}\)
\(\Rightarrow\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)
c) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{a.b}{c.d}\)
\(\Rightarrow\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{a.b}{c.d}\)
d) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\left(\frac{a}{b}\right)^2=\left(\frac{c}{d}\right)^2=\frac{a.c}{b.d}\)
\(\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a.c}{b.d}\)
\(\Rightarrow\frac{a^2+c^2}{b^2+d^2}=\frac{a.c}{b.d}\)
P/s: Dấu \(\Rightarrow\) dòng thứ 2 + 3 ở phần c và d đều ngang hàng nhé, đừng viết sát lề
Phần a có sai đề không vậy ? Mình cảm thấy nó không hợp lí cho lắm
Cho a/b=c/d cm rằng a)a/a-b=c/c-d
b) a/b=a+c/b+d
c) a/3a+b=c/3c+d
d)a.b/bd=a^2+c^2/b^2+d^2
E) a.b/c.d=a^2-b^2/c^2-d^2
F) a.b/c.d=(a-b)^2/(c-d)^2
Cho \(\frac{a}{b}\)=\(\frac{c}{d}\)chứng minh rằng
a)\(\frac{a}{a-b}\)=\(\frac{c}{c-d}\)
b)\(\frac{a}{b}=\frac{a+c}{b+d}\)
c)\(\frac{a}{3a+b}=\frac{c}{3c+d}\)
d)\(\frac{a.c}{b.d}=\frac{a^2+c^2}{b^2+d^2}\)
f)\(\frac{a.b}{c.d}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
a) \(\frac{a}{b}=\frac{c}{d}\)
\(\frac{a}{b}=\frac{c}{d}\)<=>\(\frac{a}{c}=\frac{b}{d}\)
áp dụng t/c dãy tỉ số = nhau :
\(\frac{a}{c}=\frac{b}{d}\)\(=\frac{a-b}{c-d}\) <=> \(\frac{a}{c}\)\(=\frac{a-b}{c-d}\)<=> \(\frac{a}{a-b}=\frac{c}{c-d}\)
mấy bài kia cũng tương tự em ạ !
gợi ý: đặt chung cho cả 4 phần a/b = c/d = k( k khác 0)
=> a=bk; c=dk
rồi thay vào các biểu thức
nnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
Cho a/b=c/d chứng minh rằng:
a)a/a-b=c/c-d
b)a/b=a+c/b+d
c) a/3a+b=c/3c+d
d)a.c/bd=a2+c2/b2+d2
e)a.b/c.d=a2-b2/c2-d2
f)a.b/cd=(a-b)2/(c-d)2
/ là phần nhé
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\) (*)
a) Từ (*) ta có:
\(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\) (1)
\(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\) (2)
Từ (1) và (2) suy ra \(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)
b) Từ (*) ta có:
\(\dfrac{a}{b}=\dfrac{bk}{b}=k\) (3)
\(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=\dfrac{k\left(b+d\right)}{b+d}=k\) (4)
Từ (3) và (4) suy ra \(\dfrac{a}{b}=\dfrac{a+c}{b+d}\)
c) Từ (*) ta có:
\(\dfrac{a}{3a+b}=\dfrac{bk}{3bk+b}=\dfrac{bk}{b\left(3k+1\right)}=\dfrac{k}{3k+1}\) (5)
\(\dfrac{c}{3c+d}=\dfrac{dk}{3dk+d}=\dfrac{dk}{d\left(3k+1\right)}=\dfrac{k}{3k+1}\) (6)
Từ (5) và (6) suy ra \(\dfrac{a}{3a+b}=\dfrac{c}{3c+d}\)
d) Từ (*) ta có:
\(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\) (7)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2.k^2+d^2.k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\) (8)
Từ (7) và (8) suy ra \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
e) Từ (*) ta có:
\(\dfrac{ab}{cd}=\dfrac{bk.b}{dk.d}=\dfrac{b^2}{d^2}=\dfrac{b}{d}\) (9)
\(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{b^2.k^2-b^2}{d^2.k^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b}{d}\) (10)
Từ (9) và (10) suy ra \(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)
f) Từ (*) ta có:
\(\dfrac{ab}{cd}=\dfrac{bk.b}{dk.d}=\dfrac{b^2}{d^2}=\dfrac{b}{d}\) (11)
\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{\left[b\left(k-1\right)\right]^2}{\left[d\left(k-1\right)\right]^2}=\dfrac{b}{d}\) (12)
Từ (11) và (12) suy ra \(\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)