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+ \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)
\(\Rightarrow\frac{a}{a-b}=\frac{c}{c-d}\)
+ \(\frac{a}{c}=\frac{3a}{3c}=\frac{b}{d}=\frac{3a+b}{3c+d}\) \(\Rightarrow\frac{a}{3a+b}=\frac{c}{3c+d}\)
+ \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{a^2}{c^2}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
\(\Rightarrow\frac{a\cdot b}{c\cdot d}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}\)
\(\Rightarrow\frac{a}{b}\cdot\frac{a}{b}=\frac{a^2+c^2}{b^2+d^2}\Rightarrow\frac{a\cdot c}{b\cdot d}=\frac{a^2+c^2}{b^2+d^2}\)
câu cuối lm tương tự
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{b}{a}=\dfrac{d}{c}\)
\(\Leftrightarrow\dfrac{b}{a}-1=\dfrac{d}{c}-1\)
\(\Leftrightarrow\dfrac{b-a}{a}=\dfrac{d-c}{c}\)
\(\Leftrightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\)
\(\Leftrightarrow\dfrac{a}{a-b}=\dfrac{c}{c-d}\)(đpcm)
\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)
\(b,Đặt:a=bk;c=dk\)
\(\frac{a}{3a+b}=\frac{bk}{3bk+b}=\frac{b.k}{b\left(3k+1\right)}=\frac{k}{3k+1};\frac{c}{3c+d}=\frac{dk}{3dk+d}=\frac{d.k}{d\left(3k+1\right)}=\frac{k}{3k+1}\)
\(\Rightarrow\frac{a}{3a+b}=\frac{c}{3c+d}\)
\(c,\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}.Đặt:a=ck;b=dk\)
\(\frac{ac}{bd}=\frac{ckc}{dkd}=\frac{c^2}{d^2}\)
\(\frac{a^2+c^2}{b^2+d^2}=\frac{c^2k^2+c^2}{d^2k^2+d^2}=\frac{c^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{c^2}{d^2}.Vậy:\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\)
\(d,Đặt:a=bk;c=dk\)
\(\frac{ab}{cd}=\frac{bkb}{dkd}=\frac{b^2}{d^2}và:\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\frac{b^2k^2-2kb^2+b^2}{d^2k^2-2kd^2+d^2}=\frac{b^2\left(k^2-2k+1\right)}{d^2\left(k^2-2k+1\right)}=\frac{b^2}{d^2}\)
\(Vậy:\frac{ab}{cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
Mình giải câu a) thôi nhé, những câu còn lại bạn làm tương tự như mình thôi
a) Đặt a/b=c/d=k
suy ra: a=kb và c=kd
a/b=kb/b=k (1)
a+c/b+d=kb+kd/b+d=k(b+d)/b+d=k (2)
Từ (1) và (2) suy ra: a/b=a+c/b+d
(những câu còn lại bạn đặt k rồi làm như mình nhé)