Cho \(\frac{a+b}{c+d}=\frac{a-2b}{c-2d}\)
CMR: \(\frac{a}{b}=\frac{c}{d}\)
Những câu hỏi liên quan
Cho \(\frac{a+2c}{b+2d}=\frac{2a+c}{2b+d}\) .
CMR : \(\frac{a}{b}=\frac{a+c}{b+d};\frac{2a-c}{2b-d}=\frac{a-2c}{b-2d};\frac{a+2b}{a-b}=\frac{c+2d}{c-d}\)
\(\frac{a+2b}{a-2b}=\frac{c+2d}{c-2d}CMR\frac{a}{b}=\frac{c}{d}\)
Cho a, b, c, d > 0. CMR \(\frac{a^4}{a^3+2b^3}+\frac{b^4}{a^3+2b^3}+\frac{c^4}{c^3+2d^3}+\frac{d^4}{d^3+2a^3}\ge\frac{a+b+c+d}{3}\)
Cho:\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}\)
Tính: P\(\frac{2a-b}{2c-d}+\frac{2b-c}{2d-a}+\frac{2c-d}{2a-b}+\frac{2d-a}{2b-c}\)
Giúp với ai nhanh mình tick cho.
Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{b+c+d+a}=1\)
=> a = b = c = d
=> \(D=\frac{2a-a}{2a-a}+\frac{2a-a}{2a-a}+\frac{2a-a}{2a-a}+\frac{2a-a}{2a-a}\)
D = 1 + 1 + 1 + 1 = 4
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cho a,b,c,d > 0. CMR \(\frac{a^4}{a^3+2b^3}+\frac{b^4}{b^3+2c^3}+\frac{c^4}{c^3+2d^3}+\frac{d^4}{d^3+2a^3}\ge\frac{a+b+c+d}{3}\)
Cho \(a,b,c,d\ge0\).CMR \(\frac{a-b}{a+2b+c}+\frac{b-c}{b+2c+d}+\frac{c-d}{c+2d+a}+\frac{d-a}{d+2a+b}\ge0\)
@BĐT Vasile cirtoaje
Đặt cái ban đầu là A
Dầu tiên ta có
\(\text{(3a+c)(a+2b+c)+(3b+d)(b+2c+d)+(3c+a)(c+2d+a)+(3d+b)(d+2a+b)}\)
\(=4\left(a+b+c+d\right)^2\)
Ta có: \(\frac{a-b}{a+2b+c}+\frac{1}{2}=\frac{1}{2}.\frac{3a+c}{a+2b+c}=\frac{1}{2}.\frac{\left(3a+c\right)^2}{\left(3a+c\right)\left(a+2b+c\right)}\)
Tương tự ta có
\(\frac{b-c}{b+2c+d}+\frac{1}{2}=\frac{1}{2}.\frac{\left(3b+d\right)^2}{\left(3b+d\right)\left(b+2c+d\right)}\)
\(\frac{c-d}{c+2d+a}+\frac{1}{2}=\frac{1}{2}.\frac{\left(3c+a\right)^2}{\left(3c+a\right)\left(c+2d+a\right)}\)
\(\frac{d-a}{d+2a+b}+\frac{1}{2}=\frac{1}{2}.\frac{\left(3d+b\right)^2}{\left(3d+b\right)\left(d+2a+b\right)}\)
Cộng vế theo vế ta được
\(\frac{a-b}{a+2b+c}+\frac{1}{2}+\frac{b-c}{b+2c+d}+\frac{1}{2}+\frac{c-d}{c+2d+a}+\frac{1}{2}+\frac{d-a}{d+2a+b}+\frac{1}{2}=\frac{1}{2}.\frac{\left(3d+b\right)^2}{\left(3d+b\right)\left(d+2a+b\right)}+\frac{1}{2}.\frac{\left(3c+a\right)^2}{\left(3c+a\right)\left(c+2d+a\right)}+\frac{1}{2}.\frac{\left(3b+d\right)^2}{\left(3b+d\right)\left(b+2c+d\right)}+\frac{1}{2}.\frac{\left(3a+c\right)^2}{\left(3a+c\right)\left(a+2b+c\right)}\)
\(\ge\frac{1}{2}.\frac{\left(3a+c+3b+d+3c+a+3d+b\right)^2}{\left(3a+c\right)\left(a+2b+c\right)+\left(3b+d\right)\left(b+2c+d\right)+\left(3c+a\right)\left(c+2d+a\right)+\left(3d+b\right)\left(d+2a+b\right)}\)
\(=\frac{1}{2}.\frac{16\left(a+b+c+d\right)^2}{4\left(a+b+c+d\right)^2}=2\)
\(\Rightarrow A+2\ge2\)
\(\Leftrightarrow A\ge0\)
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=4(a+b+c+d)2
Ta có: a−ba+2b+c +12 =12 .3a+ca+2b+c =12 .(3a+c)2(3a+c)(a+2b+c)
Tương tự ta có
b−cb+2c+d +12 =12 .(3b+d)2(3b+d)(b+2c+d)
c−dc+2d+a +12 =12 .(3c+a)2(3c+a)(c+2d+a)
d−ad+2a+b +12 =12 .(3d+b)2(3d+b)(d+2a+b)
Cộng vế theo vế ta được
a−ba+2b+c +12 +b−cb+2c+d +12 +c−dc+2d+a +12 +d−ad+2a+b +12 =12 .(3d+b)2(3d+b)(d+2a+b) +12 .(3c+a)2(3c+a)(c+2d+a) +12 .(3b+d)2(3b+d)(b+2c+d) +12 .(3a+c)2(3a+c)(a+2b+c)
≥12 .(3a+c+3b+d+3c+a+3d+b)2(3a+c)(a+2b+c)+(3b+d)(b+2c+d)+(3c+a)(c+2d+a)+(3d+b)(d+2a+b)
=12 .16(a+b+c+d)24(a+b+c+d)2 =2
⇒A+2≥2
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Cho \(\frac{a}{b}=\frac{c}{d}\) CMR: \(\frac{4a-2b}{5a+2b}=\frac{4c-2d}{5c+2d}\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{4a}{4c}=\frac{2b}{2d}=\frac{4a-2b}{4c-2d}=\frac{5a}{5c}=\frac{2b}{2d}=\frac{5a+2b}{5c+2d}\)
Suy ra \(\frac{4a-2b}{4c-2d}=\frac{5a+2b}{5c+2d}\)Suy ra điều phải chứng minh: \(\frac{4a-2b}{5a+2b}=\frac{4c-2d}{5c+2d}\)
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25 tháng 4 2020 lúc 10:42
Cho a, b, c, d > 0. CMR: \(\frac{a^4}{a^3+2b^3}+\frac{b^4}{b^3+2c^3}+\frac{c^4}{c^3+2d^3}+\frac{d^4}{d^3+2a^3}\ge\frac{a+b+c+d}{3}\) (Dùng Cô-si )
Bạn tham khảo (hoàn toàn dùng Cô-si):
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cho \(\frac{a+b}{c+d}=\frac{a-2b}{c-2d}\)
CMR \(\frac{a}{b}=\frac{c}{d}\)
(a+b).(c-2d)=(a-2b).(c+d)
ac-2ad+bc-2bd=ac+ad-2bc-2bd
ac-ac+bc+2bc=2ad+ad+2bd-2bd
3bc=3ad
bc=ad
=> a/b=c/d
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