Bài 1 : Tìm x,y,z biết :

a) 2x = 3y ; 5y = 7z và 3x - 7y + 5z = -30 

b) 3x =5y ; 7y = 2z và x + y + z = 74 

c) x : z = \(\dfrac{2}{3}\) : \(\dfrac{1}{2}\) ; z : y = 1 : \(\dfrac{4}{7}\) và y + z = 66

d) x : y : z = 3 : 4 : 5 và \(2x^2\) + \(2y^2\) - \(3z^2\) = -100

e) \(x:y:z\) = 2 : 5 : 6 và \(2x^2\) + \(4y^2\) - \(4z^2\) = -324

f) \(\dfrac{x-1}{2}\) = \(\dfrac{y-2}{3}\) = \(\dfrac{z-3}{4}\) và \(x-2y+3z=14\)

g)\(\dfrac{x-1}{2}\) = \(\dfrac{y+3}{4}\) =\(\dfrac{z-5}{6}\) và \(5z-3x-4y=50\)

h) \(\dfrac{x}{2}=\dfrac{y}{7}\) và \(xy=56\)

i)\(\dfrac{x-y}{3}=\dfrac{x+y}{13}=\dfrac{xy}{200}\)

k) \(\dfrac{x-5}{6}=\dfrac{x+5}{18}\)

l) \(\dfrac{2x-11}{12}=\dfrac{x+5}{20}\)

Câu 1 : cho tỉ lệ thức a/b =c/d .Chứng minh :  \(\dfrac{a+2b}{a-2b}\) = \(\dfrac{c+2d}{c-2d}\)

Câu 2 : Tìm x,y,z biết : (áp dụng công thức dãy tỉ số bằng nhau)

a) 2x=3y , 5y =7z và 3x+5y-7z =30. 

b) \(\dfrac{x-1}{2}\)=\(\dfrac{y+3}{4}\)=\(\dfrac{z-5}{6}\)và 5z-3x-4y=50. 

c) \(\dfrac{1}{2}\)x =\(\dfrac{2}{3}\)y=\(\dfrac{3}{4}\)z và x-y=15. 

Tìm x,y,z biết :

1) \(x:y:z=3:5:\left(-2\right)\) và \(5x-y+3z=-16\)

2) \(\dfrac{x}{2}=\dfrac{y}{-3};\dfrac{z}{3}=\dfrac{y}{4}\) và \(x+y+z=5,2\)

3) \(2x=3y;7z=5y\) và \(3x-7y+5z=30\)

4) \(3x=4y=5z\) và \(x-\left(y+z\right)=-21\)

5) \(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\) và \(2x+3y-z=50\)

Rút gọn:
a) \(\dfrac{3\left(x-y\right)\left(x-z\right)^2}{6\left(x-y\right)\left(x-z\right)}\)
b) \(\dfrac{6x^2y^2}{8xy^5}\)
c) \(\dfrac{3x\left(1-x\right)}{2\left(x-1\right)}\)
d) \(\dfrac{9-\left(x+5\right)^2}{x^2+4x+4}\)
e) \(\dfrac{x^2-2x+1}{x^2-1}\)
f) \(\dfrac{8x-4}{8x^3-1}\)
g) \(\dfrac{x^2+5x+6}{x^2+4x+4}\)
k) \(\dfrac{20x^2-45}{\left(2x+3\right)^2}\)

Bài 1: Cho 4 số a,b,c,d thỏa mãn \(b^2=ac;c^2=bd\\ \) . Chứng minh \(\dfrac{a}{d}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\)

Bài 2 : Cho \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh

a) \(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)

b) \(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)

Bài 3 : CMR : Nếu a(y+z)=b(z+x)=c(x+y) trong đó a,b,c là các số thực khác nhau thì \(\dfrac{y-z}{a\left(b-c\right)}=\dfrac{z-x}{b\left(c-a\right)}=\dfrac{x-y}{c\left(a-b\right)}\)

Bài 4 : Cho \(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}\). Chứng minh \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\)

Bài 5 : CMR : Nếu \(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\) thì \(\dfrac{a}{x+2y+z}=\dfrac{b}{2x+y-z}=\dfrac{c}{4x-4y+z}\)

1. Cho biết x , y , z # 0 và \(\dfrac{\left(ax+by+cz\right)^2}{x^2+y^2+z^2}=a^2+b^2+c^2\) .
Chứng minh rằng : \(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)
2. Rút gọn : \(\dfrac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\) , biết rằng : x + y + z = 0
3. Cho 3x - y = 3z và 2x + y = 7z . Tính giá trị cua biểu thức :
M = \(\dfrac{x^2-2xy}{x^2+y^2}\) ( x # 0 ; y # 0 )

a,Tìm x,y,z biết: \(\dfrac{y+z+1}{x}\)=\(\dfrac{x+z+2}{y}\)=\(\dfrac{x+y-3}{z}\)=\(\dfrac{1}{x+y+z}\)

b,Cho \(\dfrac{a}{b}\)=\(\dfrac{b}{c}\)=\(\dfrac{c}{d}\). Chứng minh rằng: (\(\dfrac{a+b+c}{b+c+d}\))³=\(\dfrac{a}{d}\)

c,Cho \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\). Chứng minh rằng: \(\dfrac{5a+3b}{5c+3d}\)=\(\dfrac{5a-3b}{5c-3d}\)

d,Cho \(\dfrac{3x-2y}{4}\)=\(\dfrac{2z-4x}{3}\)=\(\dfrac{4y-3z}{2}\).Chứng minh rằng: \(\dfrac{x}{2}\)=\(\dfrac{y}{3}\)=\(\dfrac{z}{4}\)

Bài 1: 

a)  Cho a(y+z) = b(z+c) = c(x+y)  Tính: \(\dfrac{y-z}{a\left(b-c\right)}=\dfrac{z-c}{b\left(c-a\right)}=\dfrac{x-y}{c\left(a-b\right)}\)

b) \(Cho\dfrac{a}{2014}=\dfrac{b}{2015}=\dfrac{c}{2016}cm:4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)

c) \(\dfrac{a}{a'}+\dfrac{b'}{b}=1\) và \(\dfrac{b}{b'}+\dfrac{c'}{c}=1\)

cm: abc+a'b'c'=0

bài 4: 

a) \(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\)    Tính: \(\dfrac{x}{y}\)

b) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)  Tính P = \(\dfrac{xy+yz+xz}{x^2+y^2-z^2}\)

c) \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}\)

Tính : P = \(\dfrac{a+b}{c+d}+\dfrac{c+b}{a+d}=\dfrac{c+d}{a+b}=\dfrac{a+d}{c+b}\)

d) \(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}\) Tính: \(P=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)