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 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 \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)
c/m nếu a/b=c/d
a)\(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)
b)\(\frac{5a^3+3b^6}{11a^2-8b^2}=\frac{5c+3cd}{11c^2-8d^2}\)
1, chứng minh rằng nếu \(\frac{a}{b}=\frac{c}{d}\)thì
a, \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
b, \(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
2, tìm x,y,z biết \(\frac{3x}{8}=\frac{3y}{64}=\frac{3z}{216}\)và \(2x^2+2y^2+z^2=1\)
3, tìm x,y biết \(\frac{2x+1}{5}=\frac{4y-5}{9}=\frac{2x+4y-4}{7x}\)
4, cho \(\frac{a}{_a,}+\frac{b}{b^,}=1\)và \(\frac{b}{b^,}+\frac{c^,}{c}=1\)
chứng minh rằng abc+\(a^,b^,c^,=0\)
giải chi tiết nha các bạn
Chứng minh:\(\frac{5a+3b}{5c+3d}=\frac{2a-3b}{2c-3d}\) Cho a/b=c/d.
Bài 1: Tìm a, b, c biết:
\(\frac{a}{2}=\frac{b}{3};\frac{b}{5}=\frac{c}{4}\)và \(a-b+c=49\)
Bài 2: Tìm x, y, z biết:
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)và \(x=2y+3z=14\)
Bài 3: Chứng minh rằng:
Nếu \(\frac{a}{b}=\frac{c}{d}\)thì
\(a,\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-5d}\)
\(b,\frac{7a+8b}{7a-8b}=\frac{7c+7d}{7c-7d}\)
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 \(\frac{a}{b}=\frac{c}{d}\)chứng minh rằng:
a, \(\frac{a-b}{a+b}=\frac{c-d}{c+d}\)
b, \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
c, \(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
Cho \(\frac{a}{b}=\frac{c}{d}\). c/m:
a) \(\frac{a}{b}=\frac{a^2+c^2}{b^2+c^2}\)
b) \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
CMR Nếu \(\frac{a}{b}=\frac{c}{d}\)thì:
a)\(\left(\frac{a-b}{c-d}\right)^4=\frac{a^4+b^4}{c^4+d^4}\)
b)\(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
c)\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
