a/ \(\dfrac{a^2+c^2}{b^2+c^2}=\dfrac{a}{b}\)
\(Tacó\dfrac{a^2+ab}{b^2+ab}=\dfrac{a\left(a+b\right)}{b\left(b+a\right)}=\dfrac{a}{b}\) (vì \(c^2=ab\) )
Vậy....
Cho \(\dfrac{a}{c}=\dfrac{c}{b}\) chứng minh rằng : \(\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{b-a}{a}\)
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) Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) . Chứng minh rằng \(\dfrac{2a^2-3ab+5b^2}{2a^2+3ab}=\dfrac{2c^2-3cd+5d^2}{2c^2+3cd}\)
2) Cho \(\dfrac{a}{c}=\dfrac{c}{b}\). Chứng minh rằng \(\dfrac{b^2-c^2}{a^2+c^2}=\dfrac{b-a}{a}\)
3) Cho \(\dfrac{a}{b}=\dfrac{c}{d}\).Chứng minh rằng\(\dfrac{3a^6+c^6}{3b^6+d^6}=\dfrac{\left(a+c\right)^6}{\left(b+d\right)^6}\)
Cho x,y,z,a,b,c khác 0 và \(\dfrac{x^2-yz}{a}=\dfrac{y^2-xz}{b}=\dfrac{z^2-xy}{c}\).Chứng minh rằng \(\dfrac{a^2-bc}{x}=\dfrac{b^2-ac}{y}=\dfrac{c^2-ab}{z}\)
Cho \(\dfrac{a}{c}=\dfrac{c}{b}\). Chứng minh rằng \(\dfrac{a^2+c^2}{b^2+c^2}=\dfrac{a}{b}\)
a) Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) (\(a,b,c,d\ne0\)). Chứng minh rằng:
1) \(\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
2) \(\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}\)
3) \(\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{\left(a+b\right)^3}{\left(c+d\right)^3}\) \(\left(\dfrac{a}{b}=\dfrac{c}{d}\ne1\right)\)
b)Cho \(\dfrac{2a+13b}{3a-7b}=\dfrac{2c+13d}{3c-7d}\). Chứng minh rằng:\(\dfrac{a}{b}=\dfrac{c}{d}\)
c)Cho \(\dfrac{cy-bz}{x}=\dfrac{az-cx}{y}=\dfrac{bx-ay}{z}\). Chứng minh rằng: \(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) chứng minh rằng:
a) \(\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}\)
b)\(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
c)\(\dfrac{7a^2-3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)
cho a + b ≠ c ; b ≠ c; c2 = 2( ac + bc - ab ). Chứng minh rằng \(\dfrac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\dfrac{a-c}{b-c}\)
cho \(\dfrac{x^2-yz}{a}=\dfrac{y^2-zx}{b}=\dfrac{z^2-xy}{c}\). Chứng minh rằng \(\dfrac{a^2-bc}{x}=\dfrac{b^2-ca}{y}=\dfrac{c^2-ab}{z}\)
