Ta có:
a/b = c/d => 2018a/2018b = 2018c/2018d = 2018a - 2018c / 2018b- 2018d
a/b = c/d => 2017a/2017b = 2017c/2017d =2017a+ 2017c/ 2017b+ 2017d
=> 2018a-2018c/2018b-2018d = 2017a+2017c/2017b+2017d (=a/b=c/d)
1, Cho a, b, c, d là 4 số khác 0 thỏa mãn \(b^2\)=ac và \(c^2\)=bd
Chứng minh rằng: \(\dfrac{2016a^3+2017b^3+2018c^3}{2016b^3+2017c^3+2018d^3}\)=\(\dfrac{a}{d}\)
Cho a/b=b/c=c/d=d/a,(a+b+c+d khac 0)
Tinh M=2019a-2018b/a+b=2018c-2017d/c+d
Cho
\(\dfrac{2016c-2017b}{2015}\)=\(\dfrac{2017a-2015c}{2016}\)=\(\dfrac{2015b-2016a}{2017}\).
Chứng minh \(\dfrac{a}{2015}\)=\(\dfrac{b}{2016}\)=\(\dfrac{c}{2017}\)
Bài 1:Tìm 3 số a,b,c biết
\(\dfrac{3a-2b}{5}=\dfrac{2c-5a}{3}=\dfrac{5b-3c}{2}\) và a+b+c= -50
Bài 2: Chứng minh rằng:Nếu các số a,b,c,d thỏa mãn:
[ab(ab-2cd)+c2.d2].[ab(ab-2)+2(ab+1)] =0
Thì a,b,c,d lập thành một tỉ lệ thức
Bài 3:Cho b2= a.c; c2=b.d (c,b,d\(\ne0\) và b+c\(\ne0\) ; b3+d3\(\ne d^3\) )
CMR \(\dfrac{a^3+b^3-c^3}{b^3+c^3-d^3}=\left(\dfrac{a+b-c}{b+c-d}\right)^3\)
Bài 4: Cho b2 = a.c (a,c\(\ne0\) )
CMR \(\dfrac{a}{c}=\left(\dfrac{2016a-2017b}{2016b-2017c}\right)^2\)
Cho \(\text{a,b,c \in R; a,b,c \ne0}\)thỏa mãn: b2 = a.c
Chứng minh rằng : \(\frac{a}{c}=\left(\frac{a+2018b}{b+2018c}\right)^2\)
Cho \(\dfrac{a-b}{c-d}=\dfrac{a+b}{c+d}\)
Cm:\(\dfrac{a}{c}=\dfrac{b}{d}\)
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 \(\dfrac{1}{c}=\dfrac{1}{2}.\left(\dfrac{1}{a}+\dfrac{1}{c}\right)vớia,b,c\)khác 0 b khác 0
cm rằng\(\dfrac{a}{b}=\dfrac{a}{c}-\dfrac{c}{d}\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\). CM rằng
\(\dfrac{7a-11b}{21a+5b}=\dfrac{7c-11d}{4c+5d}\)
( CM bằng 2 cách)
