Bạn nào giải giúp mk vs ak.M tặng ngay 3***
Chứng minh
a+3c/b+3d=a+c/b+d
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a)có:a/b=c/d=>a/c=b/d=a+b/c+d=a-b/c-d
b)có:a/b=c/d=a/b=3c/3d=a+3c/b+3d=a+c/b+d
c)có:a/b=c/d=>a/c=b/d=a+b/c+d
=>a/c=a+b/c+d=>a/a+b=c/c+d
áp dụng t/c DTSBN hết nha bạn
Cho a/b=c/d Với b/d khác +-3/2 . Chứng minh rằng:
a)2a+3c/2b+3d=2a-3c/2b-3d.
b)a^2+c^2/b^2+d^2=ac/bd
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: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
đặt a/b =c/d =k
=> a=bm , c=dm
=> 2a+3c/2b+3d =2bm+3bm/ 2b +3d = m.(2d+3d)/2d+3d =m (1)
=> 2a-3c/2d-3d=2bm-3dm /2b -3d =m.(2b-3d)/2b-3d= m (2)
Từ (1) và (2) => 2a+3c/2b+3d =2a-3c/2b-3d
câu 2 tương tự nha
Giả sử \(\frac{a+3c}{b+3d}=\frac{a+c}{b+d}\)
\(\Leftrightarrow\left(a+3c\right)\left(b+d\right)=\left(b+3d\right)\left(a+c\right)\)
\(\Leftrightarrow a\left(b+d\right)+3c\left(b+d\right)=a\left(b+3d\right)+c\left(b+3d\right)\)
\(\Leftrightarrow ab+ad+3bc+3cd=ab+3ad+bc+3cd\)
\(\Leftrightarrow2bc=2ad\)
\(\Leftrightarrow bc=ad\)
\(\Leftrightarrow\frac{a}{b}=\frac{c}{d}\)
Mình nghĩ đề phải cho \(\frac{a}{b}=\frac{c}{d}\)thì điều giả sử là đúng