cho a/b =c/d cm a/3a+b=c/3c+d
cho a/b = c/d .Chứng minh
a) 3a-c/3b-d = 2a+3c/2b+3d
b) 3a-b/3a+d = 3c-a/3c+d
c) a^2 - b^2/c^2-d^2 = 2ab + b^2/2cd + d^2
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{3a-c}{3b-d}=\dfrac{3bk-dk}{3b-d}=k\)
\(\dfrac{2a+3c}{2b+3d}=\dfrac{2bk+3dk}{2b+3d}=k\)
Do đó: \(\dfrac{3a-c}{3b-d}=\dfrac{2a+3c}{2b+3d}\)
c: \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2}{d^2}\)
\(\dfrac{2ab+b^2}{2cd+d^2}=\dfrac{2\cdot bk\cdot b+b^2}{2\cdot dk\cdot d+d^2}=\dfrac{b^2}{d^2}\)
Do đó: \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{2ab+b^2}{2cd+d^2}\)
a,cho 3a-b/3a+b=3c-d/3c+d cmr a/b=c/d
b,cho a/b=c/d cmr:b^2+d^2/a^2+c^2=bd/ac
a) Ta có\(\frac{3a-b}{3a+b}=\frac{3c-d}{3c+d}\)
=> (3a - b)(3c + d) = (3a + b)(3c - d)
=> 9ac + 3ad - 3bc - bd = 9ac - 3ad + 3bc - bd
=> 3ad - 3bc = -3ad + 3bc
=> 3ad + 3ad = 3bc + 3bc
=> 6ad = 6bc
=> ad = bc
=> \(\frac{a}{b}=\frac{c}{d}\left(\text{đpcm}\right)\)
b) Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Khi đó \(\frac{b^2+d^2}{a^2+c^2}=\frac{b^2+d^2}{\left(bk\right)^2+\left(dk\right)^2}=\frac{b^2+d^2}{d^2k^2+d^2k^2}=\frac{b^2+d^2}{k^2\left(b^2+d^2\right)}=\frac{1}{k^2}\)(1);
\(\frac{bd}{ac}=\frac{bd}{bkdk}=\frac{1}{k^2}\left(2\right)\)
Từ (1)(2) => \(\frac{b^2+d^2}{a^2+c^2}=\frac{bd}{ac}\)(đpcm)
Chứng minh tỉ lệ thức sau
Cho a/b=c/d
CM: a/3a+b=c/3c+d
Xét \(\frac{a}{b}=k;\frac{c}{d}=k\)
=> a= bk; c= dk
Thay:
\(\frac{a}{3a+b}=\frac{bk}{3.bk+b}=\frac{bk}{3.b\left(k+1\right)}=\frac{k}{3.\left(k+1\right)}\) (1)
\(\frac{c}{3c+d}=\frac{dk}{3.dk+d}=\frac{dk}{3.d\left(k+1\right)}=\frac{k}{3.\left(k+1\right)}\) (2)
Ta thấy (1)= (2)
=> \(\frac{a}{3a+b}=\frac{c}{3c+d}\) (dpcm)
theo bài ra ta có:
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{3a}{3c}\)
áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{a}{c}=\frac{b}{d}=\frac{3a}{3c}=\frac{b+3a}{d+3c}\)
=> \(\frac{a}{c}=\frac{3a+b}{3c+d}\)
=> \(\frac{a}{3a+b}=\frac{c}{3c+d}\) (đpcm)
Cho a/b=c/d CMR
a, 3a/b=3c/d
b, a+b/b=c+b/a
c, 3a+b/b= 3c+d/d
a ta có a/b=c/d=>ac=bd.nhân cả 2 vế vs 3 ta được 3ac=3bd=>3a/b=3c/d
c từ ý a có 3a/b=3c/d=>3a/b+1=3c/d +1(cộng cả hai vế vs 1).sau đó quy đồng được 3a+b/b=3c+d/d
còn ý b thì hình như bạn chép sai r thì phải,đề bài đúng chắc là như thế nầy a+b/b=c+a/a.nếu đề bài như thế thì sẽ giải giông ý c bạn nha!^^
cho \(\dfrac{a}{b}\) =\(\dfrac{c}{d}\) cm rằng
a) \(\dfrac{a}{a-b}\) =\(\dfrac{c}{c-d}\) b)\(\dfrac{a}{b}\) =\(\dfrac{a+c}{b+d}\) c) \(\dfrac{a}{3a+d}\) =\(\dfrac{c}{3c+d}\) d)\(\dfrac{a.c}{b.d}\) =\(\dfrac{a^2+c^2}{b^2+c^2}\) e)\(\dfrac{a.b}{c.d}\) =\(\dfrac{a^2-b^2}{c^2-d^2}\) f)\(\dfrac{a.b}{c.d}\) =\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
mn giúp mk vs ạ! thanks
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)
Cho tỉ lệ thuận: 2a trừ 13b phần 3a trừ 7b bằng 2c + 13d phần 3c trừ 7d. CM rằng a phần b bằng c phần d và ngược lại: cho a phần b bằng c phần d suy ra 2a trừ 13b phần 3a + 7b bằng 2c+ 13b phần 3c trừ 7d
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 a/b=c/d . CMR a/a-b = c/c-d , a/3a - b = c/3c - d
Ta có:\(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\)
a/ Ta có: ad = bc => ac - ad = ac - bc => a . (a - d) = c . (a - b) => \(\frac{a}{a-b}=\frac{c}{c-d}\)
b/ Ta có: ad = bc => 3ac - ad = 3ac - bc => a . (3c - d) = c . (3a - b) => \(\frac{a}{3a-b}=\frac{c}{3c-d}\)
cho a/b = c/d. chứng minh a/3a+b = c/ 3c+d
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow\begin{cases}a=bk\\c=dk\end{cases}\)
Ta có :
\(\frac{a}{3a+b}=\frac{bk}{3bk+b}=\frac{bk}{b\left(3k+1\right)}=\frac{k}{3k+1}\left(1\right)\)
\(\frac{c}{3c+d}=\frac{dk}{3dk+d}=\frac{dk}{d\left(3k+1\right)}=\frac{k}{3k+1}\left(2\right)\)
Từ 1 và 2
=> \(\frac{a}{3a+b}=\frac{c}{3c+d}\)
Giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=b.k,c=d.k\)
Ta có:
\(\frac{a}{3a+b}=\frac{b.k}{3.b.k+b}=\frac{b.k}{b.\left(3.k+1\right)}=\frac{k}{3.k+1}\) (1)
\(\frac{c}{3c+d}=\frac{d.k}{3.d.k+d}=\frac{d.k}{d.\left(3.k+1\right)}=\frac{k}{3.k+1}\) (2)
Từ (1) và (2) suy ra \(\frac{a}{3a+b}=\frac{b}{3c+d}\)