\(\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac\)

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

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

\(=a^2\)

`a)x^2(x+4)(x-4)-(x^2+1)(x^2-1)`

`=x^2(x^2-16)-(x^2+1)(x^2-1)`

`=x^4-16x^2-(x^4-1)`

`=-16x^2+1`

`b) (a-b+c)^2-(a-c)^2-2ac+2ab`

`=a^2+b^2+c^2-2ab-2bc+2ac-(a^2-2ac+c^2)-2ac+2ab`

`=a^2+b^2+c^2-2ab-2bc+2ac-a^2+2ac-c^2-2ac+2ab`

`=b^2-2bc+2ac`

a) Ta có: \(x^2\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)

\(=x^2\left(x^2-16\right)-\left(x^4-1\right)\)

\(=x^4-16x^2-x^4+1\)

\(=-16x^2+1\)

b) Ta có: \(\left(a-b+c\right)^2-\left(a-c\right)^2-2ac+2ab\)

\(=\left(a-b+b-a+c\right)\left(a-b+c+a-c\right)-2ac+2ab\)

\(=c\left(2a-b\right)-2ac+2ab\)

\(=2ac-2bc-2ac+2ab\)

\(=2ab-2bc\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow ab+bc+ca=0\Rightarrow\left\{{}\begin{matrix}bc=-ab-ac\\ab=-bc-ac\\ac=-ab-bc\end{matrix}\right.\)

\(M=\dfrac{1}{a^2+bc-ab-ac}+\dfrac{1}{b^2+ac-ab-bc}+\dfrac{1}{c^2+ab-bc-ac}\)

\(=\dfrac{1}{a\left(a-b\right)-c\left(a-b\right)}+\dfrac{1}{b\left(b-c\right)-a\left(b-c\right)}+\dfrac{1}{c\left(c-a\right)-b\left(c-a\right)}\)

\(=\dfrac{1}{\left(a-b\right)\left(a-c\right)}-\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(a-c\right)\left(b-c\right)}\)

\(=\dfrac{b-c-\left(a-c\right)+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)

\(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}\)

\(=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}\)

\(=\frac{\left(a+b-c\right)\left(a+b+c\right)}{\left(a+c-b\right)\left(a+c+b\right)}\)

\(=\frac{a+b-c}{a+c-b}\)

Bạn sai đề nên mik sửa và làm luôn nha

 \(a^2+b^2-c^2+2ab\)

______________________

\(a^2+b^2+c^2+2ac\)

= \(a^2+b^2-c^2+2ab\) (Ở đây ta gạch a²,b²,c²,2a)

_____________________________

\(a^2+b^2+c^2+2ac\)   (Ở đây ta cũng gạch a²,b²,c²,2a)

=> Kết quả cuối của biểu thức là: \(\frac{b}{c}\)

Tíck cho mình nha

\(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}\)

\(=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}\)

\(=\frac{\left(a+b-c\right)\left(a+b+c\right)}{\left(a+c-b\right)\left(a+b+c\right)}\)

\(=\frac{a+b-c}{a+c-b}\left(a+b+c\ne0\right)\)

Ta có :

\(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}\)

\(=\frac{a^2+2ab+b^2-c^2}{a^2+2ac+c^2-b^2}\)

\(=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}\)

\(=\frac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+c+b\right)\left(a+c-b\right)}=\frac{a+b-c}{a-b+c}\)

\(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\frac{\left(a+b\right)^2-2ab-c^2+2ab}{\left(a+c\right)^2-2ac-b^2+2ac}.\)

\(=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}\)

\(=\frac{\left(a+b-c\right)\left(a+b+c\right)}{\left(a+c-b\right)\left(a+b+c\right)}\)

\(=\frac{a+b-c}{a+c-b}\)

Ta có : \(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\)

\(\Leftrightarrow2\left(ab+ac+bc\right)=0\Rightarrow ab+ac+bc=0\Rightarrow\hept{\begin{cases}ab=-ac-bc\\ac=-ab-bc\\bc=-ac-ab\end{cases}}\)

Nên \(\frac{a^2}{a^2+2bc}=\frac{a^2+ab+bc+ac}{a^2+bc-ac-ab}=\frac{\left(a+c\right)\left(a+b\right)}{\left(a-c\right)\left(a-b\right)}\)

\(\frac{b^2}{b^2+2ac}=\frac{b^2+ab+bc+ac}{b^2+ac-ab-bc}=\frac{\left(a+b\right)\left(b+c\right)}{\left(b-a\right)\left(b-c\right)}\)

\(\frac{c^2}{b^2+2ab}=\frac{c^2+ab+ac+bc}{b^2+ab-ac-bc}=\frac{\left(c+b\right)\left(c+a\right)}{\left(c-b\right)\left(c-a\right)}\)

\(P=\frac{\left(a+b\right)\left(a+c\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(a+b\right)\left(b+c\right)}{\left(b-a\right)\left(b-c\right)}+\frac{\left(c+b\right)\left(c+a\right)}{\left(c-b\right)\left(c-a\right)}\)

\(=\frac{\left(a+b\right)\left(a+c\right)\left(b-c\right)+\left(a+b\right)\left(b+c\right)\left(c-a\right)+\left(c+b\right)\left(c+a\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\frac{\left(a+b\right)\left[\left(a+c\right)\left(b-c\right)+\left(b+c\right)\left(c-a\right)\right]+\left(c+b\right)\left(c+a\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\frac{\left(a+b\right)\left(2bc-2ac\right)+\left(c+b\right)\left(c+a\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\frac{-2c\left(a+b\right)\left(a-b\right)+\left(c+b\right)\left(c+a\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\frac{\left(a-b\right)\left[-2c\left(a+b\right)+\left(b+c\right)\left(c+a\right)\right]}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\frac{\left(a-b\right)\left(-a^2+ab+c^2-bc\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)

Vậy \(P=1\)

cuối cùng P bằng 1 yên tâm mình tính rùi

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

\(\Leftrightarrow abc.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0\Leftrightarrow\hept{\begin{cases}bc=-\left(ab+ac\right)\\ab=-\left(bc+ac\right)\\ac=-\left(bc+ab\right)\end{cases}}\)

Ta có: \(a^2+2bc=a^2+bc+bc=a^2+bc+\left(-ab-ac\right)=\left(a-b\right)\left(a-c\right)\)

Tương tự \(b^2+2ac=\left(b-a\right)\left(b-c\right);c^2+2ab=\left(c-a\right)\left(c-b\right)\)

\(\Leftrightarrow N=\frac{bc}{\left(a-b\right)\left(a-c\right)}+\frac{ac}{\left(b-a\right)\left(b-c\right)}+\frac{ab}{\left(c-a\right)\left(c-b\right)}\)

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