Đặt x^2 = a, y^2 = b, z^2 = c  =>  abc = (xyz)^2 = 1

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

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

Ta có: \(a\left(c+a-b\right)\left(a+b-c\right)=a\left[a^2-\left(b-c\right)^2\right]=a^3-ab^2-c^2a+2abc\)

Tương tự: \(b\left(a+b-c\right)\left(b+c-a\right)=b^3-bc^2-a^2b+2abc\)

                  \(c\left(b+c-a\right)\left(c+a-b\right)=c^3-ca^2-b^2c+2abc\)

Tử thức của A = \(a^3+b^3+c^3-a^2b-ab^2-b^2c-bc^2-c^2a-ca^2+6abc\)

Lại có: \(\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)=\left(b+c-a\right)\left[a^2-b^2-c^2+2bc\right]\)

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

\(\Rightarrow A=\frac{a^3+b^3+c^3-a^2b-ab^2-b^2c-bc^2-c^2a-ca^2+6abc}{-a^3-b^3-c^3+a^2b+ab^2+b^2c+bc^2+c^2a+ca^2-2abc}\)

        \(=\frac{4abc}{-a^3-b^3-c^3+a^2b+ab^2+b^2c+bc^2+c^2a+ca^2-2abc}-1\)

        \(=\frac{4}{\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)}-1\) 

       \(=\frac{4}{\left(y^2+z^2-x^2\right)\left(z^2+x^2-y^2\right)\left(x^2+y^2-z^2\right)}-1\)

4𝑥−2𝑥+1=2(12−𝑥)

2𝑥+1=2(12−𝑥)

2𝑥+1=2(−𝑥+12)

2𝑥+1=−2𝑥+24

2𝑥+1−1=−2𝑥+24−1

2𝑥=−2𝑥+23

2𝑥+2𝑥=−2𝑥+23+2𝑥

4𝑥=23

x=23/4

998989891193423894236429827289452985724896379627586439587438562735448 ko tin thì tính đi :))

(X+2)^2+(x-1)^2=2.(x+3).(x+1)

\(x^2+4x+4+x^2-2x+1=2x^2+8x+6\)

\(2x^2+2x+5=2x^2+8x+6\)

\(2x^2+2x-2x^2-8x=6-5\)

\(-6x=1\)

\(x=\frac{-1}{6}\)

vậy \(x=\frac{-1}{6}\)