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Bạn đánh lại đề đi, Để ghi dấu mũ bạn ấn nút "x2" trên thanh công cụ, sau khi bạn gõ xong dấu mũ rồi bạn ấn lại nó để đưa về trạng thái thường
\(\frac{\left(a+b\right)2}{\left(c+d\right)2}=\frac{2a+2b}{2c+2d}\)
Vậy \(\frac{\left(a+b\right)2}{\left(c+d\right)2}=\frac{2a+2b}{2c+2d}\)
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Ta có:
a/(1+b²) = a- ab²/(1+b²) ≥ a - ab/2 (do 1+b² ≥ 2b)
Tương tự ta có:
b/(1+c²) ≥ b- bc/2
c/(1+d²) ≥ c - cd/2
d/(1+a²) ≥ d - ad/2
Cộng vế với vế ta được:
VT = a/(1+b²) + b/(1+c²) + c/(1+d²) + d/(1+a²) ≥ (a+b+c+d) - (ab+bc+cd+da)/2
VT ≥ (a+b+c+d -ab+bc+cd+da)/2 + (a+b+c+d)/2
Ta có:
ab+bc+cd+da = (a+c)(b+d) ≤ [(a+b+c+d)/2]² = 4 = a+b+c+d
=> a+b+c+d ≥ ab+bc+cd+da
=> VT ≥ (a+b+c+d)/2 =2
Dấu = khi a=b=c=d=1
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\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{b}.\dfrac{a}{b}=\dfrac{c}{d}.\dfrac{c}{d}=\dfrac{a}{b}.\dfrac{c}{d}\)
\(\Rightarrow\dfrac{ac}{bd}=\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}=\dfrac{a^2+c^2}{b^2+d^2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
đặt a/b=c/d=k=>a=bk;c=dk
=>\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\frac{\left(b\left(k+1\right)\right)^2}{\left(d\left(k+1\right)\right)^2}=\frac{b^2}{d^2}\) (1)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\frac{b^2.k^2+b^2}{d^2.k^2+d^2}=\frac{b^2.\left(k^2+1\right)}{d^2.\left(k^2+1\right)}=\frac{b^2}{d^2}\) (2)
từ (1) và (2)=>đpcm
tick nhé
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a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{3b+5d}{3a+5c}=\dfrac{3b+5d}{3bk+3dk}=\dfrac{1}{k}\)
\(\dfrac{b-2d}{a-2c}=\dfrac{b-2d}{bk-2dk}=\dfrac{1}{k}\)
=>\(\dfrac{3b+5d}{3a+5c}=\dfrac{b-2d}{a-2c}\)
b: \(\dfrac{ab}{a^2-b^2}=\dfrac{bk\cdot b}{b^2k^2-b^2}=\dfrac{k}{k^2-1}\)
\(\dfrac{cd}{c^2-d^2}=\dfrac{dk\cdot d}{d^2k^2-d^2}=\dfrac{k}{k^2-1}\)
=>ab/a^2-b^2=cd/c^2-d^2
c: \(\dfrac{a^2+b^2}{\left(a+b\right)^2}=\dfrac{b^2k^2+b^2}{\left(bk+b\right)^2}=\dfrac{k^2+1}{\left(k+1\right)^2}\)
\(\dfrac{c^2+d^2}{\left(c+d\right)^2}=\dfrac{d^2k^2+d^2}{\left(dk+d\right)^2}=\dfrac{k^2+1}{\left(k+1\right)^2}\)
=>\(\dfrac{a^2+b^2}{\left(a+b\right)^2}=\dfrac{c^2+d^2}{\left(c+d\right)^2}\)
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\(a,\dfrac{a}{c}=\dfrac{c}{b}\Leftrightarrow\dfrac{a^2}{c^2}=\dfrac{c^2}{b^2}=\dfrac{a^2+c^2}{b^2+c^2}\left(1\right)\)
Mà \(\dfrac{a}{c}=\dfrac{c}{b}\Leftrightarrow ab=c^2\Leftrightarrow\dfrac{a}{b}=\dfrac{c^2}{b^2}\left(2\right)\)
Từ \(\left(1\right)\left(2\right)\tođpcm\)
\(b,\dfrac{a}{c}=\dfrac{c}{b}\Leftrightarrow ab=c^2\)
\(\Leftrightarrow\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{\left(b-a\right)\left(b+a\right)}{a^2+ab}=\dfrac{\left(b-a\right)\left(b+a\right)}{a\left(a+b\right)}=\dfrac{b-a}{a}\left(đpcm\right)\)
Có: \(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)(Tính chất dãy tỉ số bằng nhau)
=> \(\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{\left(a+c\right)^2}{\left(b+d\right)^2}=\frac{a^2+c^2}{b^2+d^2}\)(Tính chất dãy tỉ số bằng nhau)
=> \(\frac{\left(a+c\right)^2}{\left(b+d\right)^2}=\frac{a^2+c^2}{b^2+d^2}\)
=> Đpcm
Có: \(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)(Tính chất dãy tỉ số bằng nhau)
=> \(\frac{a^n}{b^n}=\frac{c^n}{d^n}=\frac{\left(a+c\right)^n}{\left(b+d\right)^n}=\frac{a^n+c^n}{b^n+d^n}\)(Tính chất dãy tỉ số bằng nhau)
=> \(\frac{\left(a+c\right)^n}{\left(b+d\right)^n}=\frac{a^n+c^n}{b^n+d^n}\)
=> Đpcm