cho a^2=bc, cmr
a/ \(\frac{c}{2a-5c}\)=\(\frac{a}{2b-5a}\)
b/\(\frac{3a-7c}{2a+5c}\)=\(\frac{3b-7a}{2b+5a}\)
c/\(\frac{2a^2-c^2}{a^2+3c^2}\)=\(\frac{2b^2-a^2}{b^2+3a^2}\)
Cho a2 = bc
CMR:
a,\(\frac{a+b}{a-b}=\frac{c+a}{c-a}\)
b,\(\frac{c}{2a-5c}=\frac{a}{2b-5a}\)
c,\(\frac{3a-7c}{2a+5c}=\frac{3b-7a}{2b+5a}\)
d,\(\frac{2a^2-c^2}{a^2+3c^2}=\frac{2b^2-a^2}{b^2+3a^2}\)
1. Cho \(\frac{a}{2b+3c}=\frac{b}{2c+3a}=\frac{c}{2a+3b}\). Chứng minh \(a=b=c\).
2. Cho \(\frac{a}{5b-2c}=\frac{b}{5c-2a}=\frac{c}{5a-2b}\). Chứng minh \(a=b=c\).
Lưu ý: Giải theo cách lớp 7
Bài 1: Cho \(\frac{2a+3b}{2c+3d}=\frac{5a+b}{5c+d}\) . Chứng minh rằng \(\left(\frac{2a+3c}{2b+3d}\right)^3=\frac{2a^3+3c^2}{2b^2+3d^2}\)
Bài 2:Tìm các số x,y biết \(\frac{x-3}{2y}=\frac{5y+6}{4}=\frac{3}{2y+2}\)
cho a/b=c/d. CMR:
a,5a-3b/3a+2b=5c-3d/3c+2d
b,2a+7b/a-2b=2c+d/c-2d
c,ac/bd=(ac)mũ 2/(bd)mũ 2
d,2a mũ 2+3c mũ 2/3b mũ 2+3d mũ 2=5a mũ 2-2c mũ 2/2b mũ 2- 2d mũ 2
cho tỉ lệ thức :
\(\frac{2a+13b}{3a-7b}=\frac{2c+13d}{3c-7d}\)
CMR \(\frac{a}{b}=\frac{c}{d}\)
cho tỉ lệ thức :
\(\frac{a}{b}=\frac{c}{d}\)
CMR
a)\(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
b)\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
Cho dãy tỉ số :\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\)
CMR \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)
1 a) 2a=3b:5b=7c và 3a +5c-7b=30
b)\(\frac{x-1}{2}=\frac{x+3}{4}=\frac{z-5}{6}\)và 5z-3x-4y=50
c)3x=4y=6z và x-3y+2z=70
d)\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\)và x+y+z=20
2 cho \(\frac{a}{b}=\frac{c}{d}\)và a;b;c;d\(\ne\)0
a)\(\frac{a}{a-b}\frac{c}{d}\)
b)\(\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\)
c)\(\frac{a}{3a+b}=\frac{c}{3c+d}\)
d)\(\frac{a^2-b^2}{c^2-d^2}=\frac{ab}{cd}\)
g)\(\frac{5a+3b}{5c+3b}=\frac{5a-3b}{5c-3d}\)
h)\(\frac{2a+3b}{2a-3d}=\frac{2c+3d}{2c-3d}\)
cho a,b,c,d thỏa mãn: \(\frac{2a+3c}{2b+3d}\)=\(\frac{3a-4c}{3b-4d}\). Tính \(\frac{4a^3d^3-b^3c^2}{4b^3c^3-a^3d^3}\)
Cho a,b,c>0 và \(\frac{2b+c-a}{a}=\frac{2c-b+a}{b}=\frac{2a+b-c}{c}\). Tính \(P=\frac{\left(3a-2b\right)\left(3b-2c\right)\left(3c-2a\right)}{\left(3a-c\right)\left(3b-a\right)\left(3c-b\right)}\)