a: Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{2a-3b}{2a+3b}=\dfrac{2bk-3b}{2bk+3b}=\dfrac{2k-3}{2k+3}\)
\(\dfrac{2c-3d}{2c+3d}=\dfrac{2dk-3d}{2dk+3d}=\dfrac{2k-3}{2k+3}\)
Do đó: \(\dfrac{2a-3b}{2a+3b}=\dfrac{2c-3d}{2c+3d}\)
b: \(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=k\)
\(\dfrac{a-c}{b-d}=\dfrac{bk-dk}{b-d}=k\)
Do đó: \(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)
1.C/m rằng
Nếu \(\dfrac{a}{b}=\dfrac{a}{c}\) thì
a,\(\dfrac{2a-3b}{2a+3b}=\dfrac{2c-3d}{3c+3d}\)
b,\(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)
c,\(\left(\dfrac{a-b}{c-d}\right)^4=\dfrac{a^4+b^4}{c^4+d^4}\)
Chứng minh \(\dfrac{a}{b} = \dfrac{c}{d}\) nếu biết
a, \(\dfrac {4a-3b}{4c-3d} = \dfrac {4a+3b}{4c+3d}\)
b, \(\dfrac {2a-3b}{2a+3b} = \dfrac {2c-3d}{2c+3d}\)
Chứng minh : \(\dfrac{a}{b}=\dfrac{c}{d}\) nếu biết :
a,\(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
b,\(\dfrac{2a-3b}{2a+3b}=\dfrac{2c-3d}{2c+3d}\)
c,\(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
d,\(\dfrac{4a-3b}{a}=\dfrac{4c-3d}{c}\)
e,\(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
Chứng minh \(\dfrac{a}{b}=\dfrac{c}{d}\) nếu biết :
a,\(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
b,\(\dfrac{2a-3b}{2a+3b}=\dfrac{2c-3d}{2c+3d}\)
c,\(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
d,\(\dfrac{4a-3b}{a}=\dfrac{4c-3d}{c}\)
e,\(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
a) Cho a,b,c,d >0 và dãy tỉ số :\(\dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}\)
Tính :P=\(\dfrac{\left(3a-2b\right)\left(3b-2c\right)\left(3c-2a\right)}{\left(3a-c\right)\left(3b-a\right)\left(3c-b\right)}\)
b)Tìm giá trị nguyên dương của x và y sao cho:\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{5}\)
hộ tui vs các chế
Cho tỉ lệ thức \(\dfrac{a}{c}=\dfrac{c}{b}\) chứng minh rằng:
a)\(\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{b-a}{a}\)
b)\(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}.CMR\)
a, \(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)
b, \(\dfrac{c}{a+c}=\dfrac{b}{b+d}\)
c, \(\dfrac{a+b}{a}=\dfrac{d}{c+d}\)
d, \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
e, \(\dfrac{4a-3b}{a}=\dfrac{4c-3d}{c}\)
f, \(\dfrac{a^2+b^2}{a^2-b^2}=\dfrac{c^2+d^2}{c^2-d^2}\)
Cho a , b, c > 0 và dãy tỉ số \(\dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}\)
Tính P= \(\dfrac{\left(3a-2b\right).\left(3b-2c\right).\left(3c-2a\right)}{\left(3a-c\right).\left(3b-a\right).\left(3c-b\right)}\)
Cho a,b,c > 0 và dãy tỉ số : \(\dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}\)
Tính M =\(\dfrac{\left(3a-2b\right).\left(3b-2c\right).\left(3c-2a\right)}{\left(3a-c\right).\left(3b-a\right).\left(3c-b\right)}\)
