Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt \(\frac{a+b}{c+d}=\frac{a-2b}{c-2d}=k\)
=> \(\begin{cases}a+b=k.\left(c+d\right)=k.c+k.d\\a-2b-k.\left(c-2d\right)=k.c-k.2d\end{cases}\)
=> (a + b) - (a - 2b) = (k.c + k.d) - (k.c - k.2d)
=> a + b - a + 2b - k.c + k.d - k.c + k.2d
=> 3b = 3kd
=> b = kd
Mà a + b = k.c + k.d
=> a = k.c
=> \(\frac{a}{b}=\frac{k.c}{k.d}=\frac{c}{d}\left(đpcm\right)\)
Cách 2:
Ta có: \(\frac{a+b}{c+d}=\frac{a-2b}{c-2d}\)
=> (a + b).(c - 2d) = (c + d).(a - 2b)
=> (a + b).c - (a + b).2d = (c + d).a - (c + d).2b
=> ac + bc - 2ad - 2bd = ac + ad - 2bc - 2bd
=> ac + bc - 2ad - 2bd - ac - ad + 2bc + 2bd = 0
=> 3bc - 3ad = 0
=> 3bc = 3ad
=> bc = ad
=> \(\frac{a}{b}=\frac{c}{d}\left(đpcm\right)\)
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
\(\frac{a-b}{a+2b+c}+\frac{1}{2}+\frac{b-c}{b+2c+d}+\frac{1}{2}+\frac{c-d}{c+2d+a}+\frac{1}{2}+\frac{d-a}{d+2a+b}+\frac{1}{2}=\frac{1}{2}.\frac{\left(3d+b\right)^2}{\left(3d+b\right)\left(d+2a+b\right)}+\frac{1}{2}.\frac{\left(3c+a\right)^2}{\left(3c+a\right)\left(c+2d+a\right)}+\frac{1}{2}.\frac{\left(3b+d\right)^2}{\left(3b+d\right)\left(b+2c+d\right)}+\frac{1}{2}.\frac{\left(3a+c\right)^2}{\left(3a+c\right)\left(a+2b+c\right)}\)
\(\ge\frac{1}{2}.\frac{\left(3a+c+3b+d+3c+a+3d+b\right)^2}{\left(3a+c\right)\left(a+2b+c\right)+\left(3b+d\right)\left(b+2c+d\right)+\left(3c+a\right)\left(c+2d+a\right)+\left(3d+b\right)\left(d+2a+b\right)}\)
\(=\frac{1}{2}.\frac{16\left(a+b+c+d\right)^2}{4\left(a+b+c+d\right)^2}=2\)
\(\Rightarrow A+2\ge2\)
\(\Leftrightarrow A\ge0\)
=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
Bạn tham khảo (hoàn toàn dùng Cô-si):
Câu hỏi của Trần Anh Thơ - Toán lớp 8 | Học trực tuyến
1.
\(P=\frac{a^4}{abc}+\frac{b^4}{abc}+\frac{c^4}{abc}\ge\frac{\left(a^2+b^2+c^2\right)^2}{3abc}=\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)\left(a+b+c\right)}{3abc\left(a+b+c\right)}\)
\(P\ge\frac{\left(a^2+b^2+c^2\right).3\sqrt[3]{a^2b^2c^2}.3\sqrt[3]{abc}}{3abc\left(a+b+c\right)}=\frac{3\left(a^2+b^2+c^2\right)}{a+b+c}\)
Dấu "=" khi \(a=b=c\)
2.
\(P=\sum\frac{a^2}{ab+2ac+3ad}\ge\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}\ge\frac{\left(a+b+c+d\right)^2}{4.\frac{3}{8}\left(a+b+c+d\right)^2}=\frac{2}{3}\)
Dấu "=" khi \(a=b=c=d\)
a,Cách 1: \(\frac{a+b}{b}=\frac{c+d}{d}\)
=> (a+b)d = b(c+d)
=> ad + bd = bc + bd
=> ad = bc
=> \(\frac{a}{b}=\frac{c}{d}\)
Cách 2:
\(\frac{a+b}{b}=\frac{c+d}{d}\Rightarrow\frac{a}{b}+1=\frac{c}{d}+1\Rightarrow\frac{a}{b}=\frac{c}{d}\)
b,\(\frac{a}{a-2b}=\frac{c}{c-2d}\Rightarrow a\left(c-2d\right)=c\left(a-2b\right)\Rightarrow ac-2ad=ac-2bc\Rightarrow-2ad=-2bc\Rightarrow ad=bc\Rightarrow\frac{a}{b}=\frac{c}{d}\)