Ta có :\(\left(\dfrac{a+b}{c+d}\right)^2\)= \(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)= \(\dfrac{a^2+b^2}{c^2+d^2}\)
=> \(\left(\dfrac{a+b}{c+d}\right)^2\)= \(\dfrac{a^2+b^2}{c^2+d^2}\)
cho \(\dfrac{a}{b}=\dfrac{c}{d}\).CMR:
a) \(\dfrac{a-b}{a}=\dfrac{c-d}{c}\)
b) \(\dfrac{a-b}{c-d}=\dfrac{a^2-b^2}{c^2-d^2}\)
c) \(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{a^2+b^2}{c^2+d^2}\)
d) \(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{2b^2+5bd}{7b^2-5bd}\)
Bài 1:
a) Cho a(y+z) = b(z+c) = c(x+y) Tính: \(\dfrac{y-z}{a\left(b-c\right)}=\dfrac{z-c}{b\left(c-a\right)}=\dfrac{x-y}{c\left(a-b\right)}\)
b) \(Cho\dfrac{a}{2014}=\dfrac{b}{2015}=\dfrac{c}{2016}cm:4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)
c) \(\dfrac{a}{a'}+\dfrac{b'}{b}=1\) và \(\dfrac{b}{b'}+\dfrac{c'}{c}=1\)
cm: abc+a'b'c'=0
bài 4:
a) \(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\) Tính: \(\dfrac{x}{y}\)
b) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\) Tính P = \(\dfrac{xy+yz+xz}{x^2+y^2-z^2}\)
c) \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}\)
Tính : P = \(\dfrac{a+b}{c+d}+\dfrac{c+b}{a+d}=\dfrac{c+d}{a+b}=\dfrac{a+d}{c+b}\)
d) \(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}\) Tính: \(P=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}.CMR:\)
a, \(\dfrac{a^2-b^2}{ab}=\dfrac{c^2-d^2}{cd}\)
b, \(\dfrac{\left(a+b\right)^2}{a^2+b^2}=\dfrac{\left(c+d\right)^2}{c^2+d^2}\)
cho \(\dfrac{b}{a}=\dfrac{c}{d}\)cmr:
a,\(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)
b,\(\dfrac{a}{b}=\dfrac{a+c}{b+d}\)
c,\(\dfrac{a}{3a+b}=\dfrac{c}{3c+d}\)
d,\(\dfrac{ac}{bd}=\dfrac{a^2+b^2}{b^2+d^2}\)
e,\(\dfrac{a.b}{c.d}=\dfrac{a^2-b^2}{c^2-d^2}\)
f,\(\dfrac{a.b}{c.d}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}.CMR\)
a, \(\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)
b, \(\dfrac{7a-4b}{3a+5b}=\dfrac{7c-4d}{3c+5d}\)
c, \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(c-a\right)^2}{\left(d-b\right)^2}\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\)
CMR \(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{a^2+b^2}{c^2+d^2}\)
Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\) . Chứng minh rằng ta có các tỉ lệ thức sau (giả thiết các tỉ lệ thức là có nghĩa ) :
a) \(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)
b) \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
Biết \(\dfrac{a}{b}< \dfrac{c}{d}\left(b,d>0\right)\)
CMR \(\dfrac{a}{b}=\dfrac{ab+cd}{b^2+d^2}< \dfrac{c}{d}\)
\(\dfrac{a}{b}=\dfrac{c}{d};b+d\ne0\)
Chứng tỏ rằng : \(\dfrac{3.a^2+c^2}{3.b^2+d^2}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\)
