a: Ta có: \(\widehat{ICA}+\widehat{ICB}=\widehat{ACB}=90^0\)

\(\widehat{ICB}+\widehat{NCB}=\widehat{NCI}=90^0\)

Do đó: \(\widehat{ICA}=\widehat{NCB}\)

Ta có: \(\widehat{CAI}+\widehat{CBI}=90^0\)(ΔCBA vuông tại C)

\(\widehat{CBI}+\widehat{CBN}=\widehat{NBI}=90^0\)

Do đó: \(\widehat{CAI}=\widehat{CBN}\)

Xét ΔCAI và ΔCBN có

\(\widehat{CAI}=\widehat{CBN}\)

\(\widehat{ICA}=\widehat{NCB}\)

Do đó: ΔCAI~ΔCBN

b: Ta có: \(\widehat{ACM}+\widehat{ACI}=\widehat{ICM}=90^0\)

\(\widehat{ICA}+\widehat{ICB}=\widehat{ACB}=90^0\)

Do đó: \(\widehat{ACM}=\widehat{ICB}\)

Ta có: \(\widehat{CAM}+\widehat{CAB}=\widehat{BAM}=90^0\)

\(\widehat{CAB}+\widehat{CBA}=90^0\)(ΔCAB vuông tại C)

Do đó: \(\widehat{CAM}=\widehat{CBA}\)

Xét ΔCAM và ΔCBI có

\(\widehat{CAM}=\widehat{CBI}\)

\(\widehat{ACM}=\widehat{BCI}\)

Do đó: ΔCAM~ΔCBI

=>\(\dfrac{AC}{CB}=\dfrac{AM}{BI}\)

=>\(AC\cdot BI=MA\cdot BC\)

c: Xét tứ giác CIBN có \(\widehat{ICN}+\widehat{IBN}=90^0+90^0=180^0\)

nên CIBN là tứ giác nội tiếp

=>\(\widehat{CIN}=\widehat{CBN}\)

=>\(\widehat{CIN}=\widehat{BAC}\)

Ta có:\(3\left(a^2+b^2+c^2\right)=\left(a+b+c\right)^2\)

\(\Leftrightarrow3a^2+3b^2+3c^2=a^2+b^2+c^2+2ab+2bc+2ca\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Vì \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(c-a\right)^2\ge0\) với mọi \(a;b;c\inℝ\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) với mọi \(a;b;c\inℝ\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\)

\(\Rightarrow a=b=c\)

\(\Rightarrow P=a^3+a^3+c^3-3.a.a.a\)

\(\Leftrightarrow P=3a^3-3a^3\)

\(\Leftrightarrow P=0\)

Vậy ...

`556^2 - 553 . 559 `

`= 556^2 - (556 - 3) . (556 + 3) `

`= 556^2 - (556^2 - 3^2)`

`= 556^2 - 556^2 + 9`

`= 0 + 9`

= 9

`456^2 + 456 . 88 + 44^2`

`= 456^2 + 456 . 88 + 44^2`

`= 456^2 + 2 .456 . 4 + 44^2`

`= (456 + 44)^2`

`= 500^2`

`= 250000`

--------------------------------

Áp dụng các HDT sau nhé: 

`(a+b)^2 = a^2 + 2ab + b^2`

`a^2 - b^2 = (a+b)(a-b)`

  (\(x+1\)) + (\(x-1\))² 

= \(x\) + 1  + \(x^2\) - 2\(x\) + 1

= \(x^2\) - (2\(x\) - \(x\)) + (1 + 1)

= \(x^2\) - \(x\) + 2

\(\left(x+1\right)+\left(x-1\right)^2\\ =\left(x+1\right)+\left(x^2-2x+1\right)\\ =x+1+x^2-2x+1\\ =x^2+\left(x-2x\right)+\left(1+1\right)\\ =x^2-x+2\)

Đặt \(x^2-x+1=a;x+1=b\)

Phương trình sẽ trở thành: \(3a^2-2b^2=5ab\)

=>\(3a^2-5ab-2b^2=0\)

=>\(3a^2-6ab+ab-2b^2=0\)

=>3a(a-2b)+b(a-2b)=0

=>(a-2b)(3a+b)=0

=>\(\left[{}\begin{matrix}a-2b=0\\3a+b=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2-x+1-2\left(x+1\right)=0\\3\left(x^2-x+1\right)+x+1=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x^2-x+1-2x-2=0\\3x^2-3x+3+x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2-3x-1=0\\3x^2-2x+4=0\end{matrix}\right.\)

=>\(x^2-3x-1=0\)

=>\(x=\dfrac{3\pm\sqrt{13}}{2}\)

`(x+1)^2 + (x-1)^2`

`= x^2 + 2x + 1 + x^2 - 2x + 1`

`= 2x^2 + 2`

`= 2(x^2 +1)`

-----------------------------------

Áp dụng hằng đẳng thức: 

\(\left(a\pm b\right)^2=a^2\pm2ab+b^2\)

\(\left(x+1\right)^2-\left(x-1\right)^2\\ =\left[\left(x+1\right)-\left(x-1\right)\right]\left[\left(x+1\right)+\left(x-1\right)\right]\\ =\left(x+1-x+1\right)\left(x+1+x-1\right)\\ =2\cdot2x\\ =4x\)