Cho a+b+c=1 và \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}=\dfrac{1}{3}\)
Tính S=\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\)
Cho a + b + c = 4034 và \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{2}\)
Tính \(S=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
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 \(\dfrac{a+b+c-d}{d}\)=\(\dfrac{b+c+d-a}{a}\)=\(\dfrac{c+d+a-b}{b}\)=\(\dfrac{d+a+b-c}{c}\), (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+\(\dfrac{a+b}{d}\))
Cho a+b+c=4034 vµ \(\dfrac{1}{c+b}+\dfrac{1}{a+c}+\dfrac{1}{a+b}=\dfrac{1}{2}\) .Tính \(\dfrac{a}{c+b}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\)
cho a+b+c=2016 và \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{4}\)
Tính \(A=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
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: \(P=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) Chứng minh rằng:
a, \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
b \(\dfrac{a^{2018}+c^{2018}}{b^{2018}+d^{2018}}=\dfrac{\left(a+c\right)^{2018}}{\left(b+d\right)^{2018}}\)
Cho \(\dfrac{1}{c}=\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) (với a,b,c khác 0, b khác c) chứng minh rằng \(\dfrac{a}{b}=\dfrac{a-c}{c-b}\)
