A = \(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\)
Mà \(\dfrac{a}{b+c}=\dfrac{b}{a+c}=\dfrac{c}{a+b}\)
\(\Rightarrow\)A = a + b + c (với a, b, c \(\in\) R
A = \(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\)
Mà \(\dfrac{a}{b+c}=\dfrac{b}{a+c}=\dfrac{c}{a+b}\)
\(\Rightarrow\)A = a + b + c (với a, b, c \(\in\) R
Biết \(\dfrac{a}{b+c}=\dfrac{b}{c+a}=\dfrac{c}{a+b}\)
Tính A=\(\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{c}{a+b}\)
Cho \(\dfrac{a+b+c-d}{d}\)=\(\dfrac{b+c+d-a}{a}\)=\(\dfrac{c+d+a-b}{b}\)=\(\dfrac{d+a+b-c}{c}\), (a+b+c+d) khác 0
tính giá trị của biểu thức: P=(1+\(\dfrac{b+c}{a}\))(1+\(\dfrac{c+d}{b}\))(1+\(\dfrac{d+a}{c}\))(1+\(\dfrac{a+b}{d}\))
Bài 17: Cho a, b, c là 3 số thực khác 0, thỏa mãn điều kiện : \(a+b\ne-c\) và \(\dfrac{a+b-c}{c}\)=\(\dfrac{b+c-a}{a}\)=\(\dfrac{c+a-b}{b}\). Tính giá trị biểu thức P=\(\left(1+\dfrac{b}{a}\right)\)x\(\left(1+\dfrac{a}{c}\right)\)x\(\left(1+\dfrac{c}{b}\right)\)
Cho : \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{b+c+a}\)
Tính : M = \(\dfrac{a+b}{c+d}+\dfrac{b+c}{a+d}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}\)
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: \(P=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
Tìm 3 số a,b,c biết \(\dfrac{b+c+1}{a}=\dfrac{a+c+2}{b}=\dfrac{a+b-3}{c}=\dfrac{1}{a+b+c}\)
Câu 1 ; Ba số a;b ; c khác nhau và khác 0 thỏa mãn \(\dfrac{a}{b+c}=\dfrac{b}{b+c}=\dfrac{c}{a+b}\).
Giá trị P = \(\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{a}{c}+\dfrac{c}{a}+\dfrac{b}{c}+\dfrac{c}{b}\)
Tính P =
Câu 2 : Tìm x , biết :
\(\dfrac{1+3y}{12}=\dfrac{1+5y}{5x}=\dfrac{1+7y}{2}\)
giúp mk nha
Cho
\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)
Tính:
\(\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}\)
Cho : \(\dfrac{a}{c}=\dfrac{c}{b}.CMR:\\ a,\dfrac{a^2+c^2}{b^2+c^2}=\dfrac{a}{b}\\ b,\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{b-a}{a}\)