Chứng tỏ a/a+c=b/b+c, biết a/b=c/d
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Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\dfrac{a}{3a+b}=\dfrac{bk}{3bk+b}=\dfrac{k}{3k+1}\)
\(\dfrac{c}{3c+d}=\dfrac{dk}{3dk+d}=\dfrac{k}{3k+1}\)
Do đó: \(\dfrac{a}{3a+b}=\dfrac{c}{3c+d}\)
(a+b+c+d)(a+d-b-c)=(a-b+c-d)(a+b-c-d)
=>(a+d)^2-(b+c)^2=(a-d)^2-(b-c)^2
=>(a+d)^2-(a-d)^2=(b+c)^2-(b-c)^2
=>(a+d-a+d)(a+d+a-d)=(b+c+b-c)(b+c-b+c)
=>4ad=4bc
=>ad=bc
=>a/c=b/d
Hai số đói nhau có tổng bằng 0
x+y=-a+b-c-d+c-b+d+a=0
Vậy x và y là 2 số đối nhau
hộ caiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
bài 2:
a. Ta có : (a-b)-(b+c)+(c-a)-(a-b-c)
= a-b-b-c+c-a-a+b+c
=-a-b+c=-(a+b-c)
a/a=1 b/b=1
1+c=1+c
easy man