Ta có

 \(a^2+1=a^2+ab+bc+ca=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right).\left(a+c\right)\\ Cmtt:b^2+1=\left(b+a\right).\left(b+c\right)\\ c^2+1=\left(c+a\right).\left(c+b\right)\)

Nên

 \(\dfrac{b-c}{a^2+1}+\dfrac{c-a}{b^2+1}+\dfrac{a-b}{c^2+1}\\ =\dfrac{\left(b-c\right)}{\left(a+b\right)\left(a+c\right)}+\dfrac{\left(c-a\right)}{\left(b+c\right)\left(b+a\right)}+\dfrac{\left(a-b\right)}{\left(c+a\right)\left(c+b\right)}\\ =\dfrac{\left(b-c\right)\left(b+c\right)+\left(c-a\right)\left(c+a\right)+\left(a-b\right)\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\\ =\dfrac{b^2-c^2+c^2-a^2+a^2-b^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\\ =0\)

\(\dfrac{b-c}{a^2+1}+\dfrac{c-a}{b^2+1}+\dfrac{a-b}{c^2+1}\)

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

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

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

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

\(=\dfrac{b^2-c^2+c^2-a^2+a^2-b^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\) 

?

giải j bn?

:)?

b: Phương trình hoành độ giao điểm của \(\left(d1\right),\left(d2\right)\) là:

2x=-x+3

\(\Leftrightarrow3x=3\)

hay x=1

Thay x=1 vào y=2x, ta được:

\(y=2\cdot1=2\)

Vậy: \(A\left(1;2\right)\)

Thay y=0 vào \(\left(d2\right)\), ta được:

\(-x+3=0\)

hay x=3

Vậy: \(B\left(3;0\right)\)

\(AB=\sqrt{\left(1-3\right)^2+\left(2-0\right)^2}=2\sqrt{2}\)

\(OA=\sqrt{\left(0-1\right)^2+\left(0-2\right)^2}=\sqrt{5}\)

\(OB=\sqrt{\left(0-3\right)^2}=3\)

\(P=\dfrac{AB+OA+AB}{2}=\dfrac{3+2\sqrt{2}+\sqrt{5}}{2}\)

\(S=\sqrt{P\cdot\left(P-OA\right)\left(P-OB\right)\left(P-AB\right)}=3\left(đvdt\right)\)

Câu 7:

a: =>x+5=-16

=>x=-21

b: =>20-16-(x-6)=90

=>4-(x-6)=90

=>10-x=90

=>x=-80

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

\(-2xy\left(4xy+9x^2+4y\right)=-2xy.4xy+\left(-2xy\right).9x^2+\left(-2xy\right).4y\)

\(=-8x^2y^2-18x^3y-8xy^2\)

\(x\left(x+8y\right)-y\left(8x-y^2\right)=x^2+8xy-8xy+y^3\)

\(=x^2+y^3\)

1: =11,2-11,2+4,5-4,5+4,2=4,2

2: =3,7(-3,15-6,85)

=-3,7*10=-37

3: \(=\left[-64,008\right]\cdot\dfrac{16}{5}\cdot\dfrac{10}{8}\cdot\dfrac{1}{5}\cdot8\)

=-64,008*32/5

=-409,6512

4: \(=\left[0,2\left(-30,17-9,83\right)\right]-\left[4,48+2,52\right]:0,4\)

=-8-7:0,4

=-8-17,5

=-25,5

1) = (11,2 + -11,2) + (-4,5 + 4,5) + 4,2

= 0 + 0 + 4,2

= 4,2 

2) = ( -3,15 + -6,85 ) . 3,7 

= -10 . 3,7 

= -3,7

\(\overrightarrow{AB}=\left(1;-2\right)\Rightarrow AB=\sqrt{1^2+\left(-2\right)^2}=\sqrt{5}\)

(C) tâm A đi qua B nên có bán kính \(R=AB=\sqrt{5}\)

Phương trình:

\(\left(x-1\right)^2+\left(y-1\right)^2=5\)