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Biết \(\dfrac{a}{b}< \dfrac{c}{d}\left(b,d>0\right)\)
CMR \(\dfrac{a}{b}=\dfrac{ab+cd}{b^2+d^2}< \dfrac{c}{d}\)
biết:\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\) với a,b,c,d\(\ne\)0. CMR:
\(\dfrac{a}{b}=\dfrac{c}{d}\) hoặc \(\dfrac{a}{b}=\dfrac{d}{c}\)
Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}.CMR:\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)
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 a/b = c/d, cmr: \(\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
DX
9 tháng 8 2021 lúc 16:44
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) \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
1,tìm x,y:
a)0,2 : 1dfrac{1}{6} dfrac{2}{3}: (6x+7)
b)dfrac{a}{a+2b} dfrac{c}{c+2d}. Tính dfrac{a^2.d^2-4b^2.c^2}{abcd}
c)dfrac{a}{b} dfrac{a^2+c^2}{ab+cd}.CMR:dfrac{a}{b} dfrac{c}{d}(b,d khác 0)
Đọc tiếp
1,tìm x,y:
a)0,2 : 1\(\dfrac{1}{6}\)= \(\dfrac{2}{3}\): (6x+7)
b)\(\dfrac{a}{a+2b}\)= \(\dfrac{c}{c+2d}\). Tính \(\dfrac{a^2.d^2-4b^2.c^2}{abcd}\)
c)\(\dfrac{a}{b}\)= \(\dfrac{a^2+c^2}{ab+cd}\).CMR:\(\dfrac{a}{b}\)= \(\dfrac{c}{d}\)(b,d khác 0)
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}
Đọc tiếp
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 \(\dfrac{a}{b}=\dfrac{c}{d}\).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) \(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{2b^2+5bd}{7b^2-5bd}\)