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[ab(ab−2cd)+c2d2].[ab(ab−2)+2(ab+1)]=0[ab(ab−2cd)+c2d2].[ab(ab−2)+2(ab+1)]=0
⇔(ab−cd)2((ab)2+2)=0⇔ab=cd.⇔(ab−cd)2((ab)2+2)=0⇔ab=cd.
![](https://rs.olm.vn/images/avt/0.png?1311)
[ab(ab - 2cd) + c2d2].[ab(ab - 2) + 2(ab + 1)] = 0
=> ab(ab - 2cd) + c2d2 = 0 hoặc ab(ab - 2) + 2(ab + 1) = 0
+) ab(ab - 2cd) + c2d2 = 0 => (ab)2 - 2(ab).(cd) + (cd)2 = 0 => (ab)2 - (ab).(cd) - (ab).(cd) + (cd)2 = 0
=> (ab - cd).(ab - cd) = 0 => (ab - cd)2 = 0 => ab - cd = 0 => ab = cd => \(\frac{a}{c}=\frac{d}{b}\) => a; b; c;d lập được thành 1 tỉ lệ thức
+) ab(ab - 2) + 2(ab + 1) = 0 => (ab)2 + 2 = 0 (Vô lí, vì (ab)2 + 2 > 0 với mọi a; b)
Vậy..................
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\left[ab\left(ab-2cd\right)+c^2d^2\right]\left[ab\left(ab-2\right)+2\left(ab+1\right)\right]\)
\(=\left(a^2b^2-2abcd+c^2d^2\right)\cdot\left(a^2b^2-2ab+2ab+2\right)\)
=\(\left(ab-cd\right)^2\left(a^2b^2+2\right)=0\)
Vif \(a^2b^2+2>0\)nên \(ab-cd=0\Leftrightarrow ab=cd\)
Suy ra 4 tỉ lên thức:
\(\orbr{\begin{cases}\frac{a}{c}=\frac{d}{b}\\\frac{b}{c}=\frac{d}{a}\end{cases} và} \orbr{\begin{cases}\frac{a}{d}=\frac{c}{b}\\\frac{b}{d}=\frac{c}{a}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\left[ab\left(ab-2cd\right)+c^2d^2\right].\left[ab\left(ab-2\right)+2\left(ab+1\right)\right]=0\)
\(\Leftrightarrow\left(a^2b^2-2acbd+c^2d^2\right).\left(a^2b^2-2ab+2ab+2\right)=0\)
\(\Leftrightarrow\left(ab-cd\right)^2.\left(a^2b^2+2\right)=0\)
Vì \(a^2b^2+2>0\forall a;b\)
\(\Leftrightarrow\left(ab-cd\right)^2=0\)
\(\Leftrightarrow ab-cd=0\)
\(\Leftrightarrow ab=cd\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
[ab(ab-2cd)+c2 d2 ] [ab(ab-2)+2(ab+1)=0<=>(a2b2-2abcd+c2d2)(a2b2-2ab+2ab+2)=0
<=>[(a2b2 - abcd)+(-abcd+c2d2)](a2b2+2)=0<=>ab(ab-cd)-cd(ab-cd)=0(vì a2b2 > 0)
<=>(ab-cd)2=0<=>ab=cd
![](https://rs.olm.vn/images/avt/0.png?1311)
cái này trong đề h.s giỏi lớp 7 nè tui làm r cả đề :)
Ta có:
\left[ab\left(ab-2cd\right)+c^2d^2\right].\left[ab\left(ab-2\right)+2\left(ab+1\right)\right]=0[ab(ab−2cd)+c2d2].[ab(ab−2)+2(ab+1)]=0
\Leftrightarrow\left(a^2b^2-2acbd+c^2d^2\right).\left(a^2b^2-2ab+2ab+2\right)=0⇔(a2b2−2acbd+c2d2).(a2b2−2ab+2ab+2)=0
\Leftrightarrow\left(ab-cd\right)^2.\left(a^2b^2+2\right)=0⇔(ab−cd)2.(a2b2+2)=0
Vì a^2b^2+2>0\forall a;ba2b2+2>0∀a;b
\Leftrightarrow\left(ab-cd\right)^2=0⇔(ab−cd)2=0
\Leftrightarrow ab-cd=0⇔ab−cd=0
\Leftrightarrow ab=cd\left(đpcm\right)⇔ab=cd(đpcm)
<=>(a2b2-2abcd+c2d2)(a^2*b^2-2ab+2ab+2)=0
<=>(ab-cd)^2.(a^2*b^2+2)=0
<=>ab-cd=0 (vì a^2*b^2+2>0 với mọi a,b)
nên a/c=b/d