Ké ạ

a(b+1)+b(a+1)=(a+1)(b+1)

<=>ab+a+ab+b=ab+a+b+1

<=>ab+a+ab+b-ab-a-b=1

<=>ab=1 (đpcm)

NQ có 2 lần số đo kìa bạn ơi

\(SA\perp\left(ABCD\right)\Rightarrow SA\perp BD\)

Lại có \(BD\perp AC\) (hai đường chéo hv)

\(\Rightarrow BD\perp\left(SAC\right)\)

\(\left\{{}\begin{matrix}SA\perp\left(ABCD\right)\Rightarrow SA\perp BC\\BC\perp AB\left(gt\right)\end{matrix}\right.\) \(\Rightarrow BC\perp\left(SAB\right)\) (1)

\(\Rightarrow\widehat{CSB}\) là góc giữa SC và (SAB)

\(SB=\sqrt{SA^2+AB^2}=\frac{2a\sqrt{3}}{3}\)

\(\Rightarrow tan\widehat{CSB}=\frac{BC}{SB}=\frac{\sqrt{3}}{2}\Rightarrow\widehat{CSB}\approx41^0\)

\(\left\{{}\begin{matrix}SA\perp\left(ABCD\right)\Rightarrow SA\perp CD\\CD\perp AD\left(gt\right)\end{matrix}\right.\) \(\Rightarrow CD\perp\left(SAD\right)\) (2)

\(\Rightarrow\widehat{SDA}\) là góc giữa (SCD) và (ABCD)

\(tan\widehat{SDA}=\frac{SA}{AD}=\frac{\sqrt{3}}{3}\Rightarrow\widehat{SDA}=30^0\)

Từ (1) \(\Rightarrow BC\perp AH\), mà \(AH\perp SB\Rightarrow AH\perp\left(SBC\right)\Rightarrow AH\perp SC\) (3)

Từ (2) \(\Rightarrow CD\perp AK\), mà \(AK\perp SD\Rightarrow AK\perp\left(SCD\right)\Rightarrow AK\perp SC\) (4)

(3);(4) \(\Rightarrow SC\perp\left(AHK\right)\) \(\Rightarrow\left(SAC\right)\perp\left(AHK\right)\)

\(\dfrac{1}{c}+b^2c=ab\left(a+b+c\right)+b^2c=ab\left(a+c\right)+b^2\left(a+c\right)=b\left(a+b\right)\left(a+c\right)\)

\(\dfrac{1}{c}+a^2c=ab\left(a+b+c\right)+a^2c=a\left(a+b\right)\left(b+c\right)\)

\(\Rightarrow\left(\dfrac{1}{c}+b^2c\right)\left(\dfrac{1}{c}+a^2c\right)=ab\left(a+b\right)^2\left(b+c\right)\left(a+c\right)\)

\(\Leftrightarrow\left(1+b^2c^2\right)\left(1+a^2c^2\right)=c^2\left(a+b\right)^2ab\left(ab+bc+ac+c^2\right)\)\(=c^2\left(a+b\right)^2\left(a^2b^2+ab^2c+a^2bc+abc^2\right)\)\(=c^2\left(a+b\right)^2\left[a^2b^2+abc\left(a+b+c\right)\right]=c^2\left(a+b\right)^2\left(a^2b^2+1\right)\)

\(\Rightarrow\dfrac{\left(1+b^2c^2\right)\left(1+a^2c^2\right)}{c^2\left(a^2b^2+1\right)}=\left(a+b\right)^2\)

\(\Leftrightarrow\sqrt{\dfrac{\left(1+b^2c^2\right)\left(1+a^2c^2\right)}{c^2+a^2b^2c^2}}=a+b\) (đpcm)