\(ab=c\cdot c\)
nên a/c=c/b
Đặt a/c=c/b=k
=>a=ck; c=bk
\(\dfrac{a^2+c^2}{b^2+c^2}=\dfrac{c^2k^2+c^2}{b^2+b^2k^2}=\dfrac{c^2}{b^2}=k^2\)
\(\dfrac{a}{b}=\dfrac{ck}{\dfrac{c}{k}}=ck\cdot\dfrac{k}{c}=k^2\)
Do đó: \(\dfrac{a^2+c^2}{b^2+c^2}=\dfrac{a}{b}\)
Cho a,b,c,d > 0 . Và a.b.c.d = 1 . Chứng minh a.b + c.d lớn hoặc = 2
chứng minh tỉ lệ thức
\(\dfrac{a.b}{c.d}\)=\(\dfrac{a^2-b^2}{c^2-d^2}\)
Bài 1: Cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\)
Chứng minh rằng \(\frac{a.b}{c.d}=\frac{a^2-b^2}{c^2-d^2}\)
Bài 2: Chứng minh rằng nếu \(\frac{a}{b}=\frac{c}{d}\)
a) \(\frac{5a+3b}{5a-3b}=\frac{5a+3b}{5a-3b}\)
b) \(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
Cho \(\dfrac{a}{c}=\dfrac{c}{b}\) chứng minh rằng : \(\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{b-a}{a}\)
Cho c2==ab . Chứng minh rằng :
a) \(\dfrac{a^{2^{ }}+c^2}{b^{2^{ }}+c^{2^{ }}}=\dfrac{a}{b}\)
b)\(\dfrac{b^{2^{ }}-a^{2^{ }}}{a^{2^{ }}+c^2}=\dfrac{b-a}{a}\)
Bai 1: Cho a2=b.c. Chứng minh: \(\dfrac{a^2+c^2}{a^2+b^2}=\dfrac{c}{d}\)
Bài 2: Cho: \(^{\dfrac{a}{b}=\dfrac{b}{c}}\). Chứng minh: \(\dfrac{a^2-b^2}{b^2-c^2}\)=\(\dfrac{a}{c}=\dfrac{\left(a-b\right)^2}{\left(b-c\right)^2}\)
GIÚP KHẨN CẤP NHA !
cho \(\frac{a}{b}=\frac{c}{d}\)chung minh rang:
\(\frac{a}{a-b}=\frac{c}{c-d}\) \(\frac{a}{b}=\frac{a+c}{b+d}\) \(\frac{a}{3a+b}=\frac{c}{3c+d}\)
\(\frac{a.b}{c.d}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\) \(\frac{a.c}{b.d}=\frac{a^2+c^2}{b^2+d^2}\)\(\frac{a.c}{b.d}=\frac{a^2-c^2}{b^2-d^2}\)
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 a,b,c là 3 cạnh của 1 tam giác. Chứng minh rằng: 2*( a*b+ b*c+ c*a) > a^2+ b^2+ c^2.
