cho : \(\dfrac{a}{b}=\dfrac{c}{d}\) (b,d khác 0, a khác 0, a khác 3c )
cm: (a+3c).(b-d)=(a-c).(b+3d)
b, CTR : C= \(^{3^{10}.199-3^9.500}\) chia hết cho 97
Cho b^2=ac;c^2=bd Với b,c,d Khác 0, 2b+3c khác 4d,b^3+c^3 khác d^3
CMR
(a+b-c/b+c-d)^3=(2a+3b-4c/2b+3d-4c)^3
Giải:
Ta có: \(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c}\)
\(c^2=bd\Rightarrow\frac{b}{c}=\frac{c}{d}\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\)
\(\Rightarrow a=bk,b=ck,c=dk\)
Ta có:
\(\left(\frac{a+b-c}{b+c-d}\right)^3=\left(\frac{bk+ck-dk}{b+c-d}\right)^3=\left[\frac{k\left(b+c-d\right)}{b+c-d}\right]^3=k^3\) (1)
\(\left(\frac{2a+3b-4c}{2b+3c-4d}\right)^2=\left(\frac{2bk+3ck-4dk}{2b+3c-4d}\right)^3=\left[\frac{k\left(2b+3c-4d\right)}{2b+3c-4d}\right]^3=k^3\) (2)
Từ (1) và (2) suy ra \(\left(\frac{a+b-c}{b+c-d}\right)^3=\left(\frac{2a+3b-4c}{2b+3c-4d}\right)^3\) ( đpcm )
Cho a+b+c+d ≠ 0 thỏa mãn:
\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{b+a+d}=\dfrac{d}{c+b+a}\)
Tính P = \(\dfrac{2a+5b}{3c+4d}+\dfrac{2b+5c}{3d+4a}+\dfrac{2c+5d}{3a+4b}+\dfrac{2d+5a}{3c+4b}\)
Cho a+b+c+d ≠ 0 và \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{b+a+d}=\dfrac{d}{c+b+a}\)
Tính giá trị biểu thức:
P = \(\dfrac{2a+5b}{3c+4d}-\dfrac{2b+5c}{3d+4a}+\dfrac{2c+5d}{3a+4b}+\dfrac{2d+5a}{3c+4b}\)
Cho các số thực a,b,c,d khác 0 thỏa mãn \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}.\)Chứng minh rằng
\(\frac{a^3+2b^3+3c^3}{b^3+2c^3+3d^3}=\left(\frac{a+2b+3c}{b+2c+3d}\right)^3=\frac{a}{d}\)
Cho tỉ lệ thức a/b. Với b/d khác +- 3/2
Cm : 1) 2a + 3c/2b + 3d = 2a - 3c /2b - 3d
2) a^2 + c^2/b^2+d^2
Cho a / 3b = b / 3c = c / 3d = d / 3a và a + b + c + d khác 0
Chứng minh rằng a = b = c =d
Theo dãy tỉ số (=) ta* có:
\(\frac{a}{3b}=\frac{b}{3c}=\frac{c}{3d}=\frac{d}{3a}=\frac{a+b+c+d}{3a+3b+3c+3d}=\frac{a+b+c+d}{3\left(a+b+c+d\right)}=\frac{1}{3}\)
=> a = 1/3 . 3b = b (1)
=> b = 1/3 . 3c = c (2)
=> c = 1/3 . 3d = d (3)
Từ(1) (2) và (3) =. a = b= c =d => ĐPCM
Cho a, b, c, d > 0. CMR \(\dfrac{a}{b+2c+3d}+\dfrac{b}{c+2d+3a}+\dfrac{c}{d+2a+3b}+\dfrac{d}{a+2b+3c}\ge\dfrac{2}{3}\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT=\dfrac{a}{b+2c+3d}+\dfrac{b}{c+2d+3a}+\dfrac{c}{d+2a+3b}+\dfrac{d}{a+2b+3c}\)
\(=\dfrac{a^2}{ab+2ac+3ad}+\dfrac{b^2}{bc+2bd+3ab}+\dfrac{c^2}{cd+2ac+3bc}+\dfrac{d^2}{ad+2bd+3cd}\)
\(\ge\dfrac{\left(a+b+c+d\right)^2}{4\left(ab+ad+bc+bd+ca+cd\right)}\ge\dfrac{\left(a+b+c+d\right)^2}{\dfrac{3}{2}\left(a+b+c+d\right)^2}=\dfrac{2}{3}\)
*Chứng minh \(4\left(ab+ad+bc+bd+ca+cd\right)\le\dfrac{3}{2}\left(a+b+c+d\right)^2\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-d\right)^2+\left(b-c\right)^2+\left(b-d\right)^2+\left(a-c\right)^2+\left(c-d\right)^2\ge0\)
Cho a/b=c/d Với b/d khác +-3/2 . Chứng minh rằng:
a)2a+3c/2b+3d=2a-3c/2b-3d.
b)a^2+c^2/b^2+d^2=ac/bd
Cho các số a,b,c,d thỏa điều kiện: a/3b=b/3c=c/3d=d/3a và a+b+c+d khác 0. C/M :a=b=c=d