\(A=\left(a+b+c\right)^3-\left(b+c-a\right)^3-\left(c+a-b\right)^3-\left(a+b-c\right)^3\)

Đặt \(x=b+c-a,y=c+a-b,z=a+b-c\).

Khi đó \(x+y+z=a+b+c\).

\(A=\left(x+y+z\right)^3-x^3-y^3-z^3\)

\(=\left(y+z\right)\left[\left(x+y+z\right)^2+x\left(x+y+z\right)+x^2\right]-\left(y+z\right)\left(y^2-yz+z^2\right)\)

\(=\left(y+z\right)\left(3x^2+3xy+3zx+3yz\right)\)

\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

\(=24abc=24\)

Đặt \(\hept{\begin{cases}a+b-c=x\\b+c-a=y\\c+a-b=n\end{cases}}\)

\(\Rightarrow x+y+n=a+b+c\)

\(\Rightarrow x+y+2b\)

\(\Rightarrow y+n=2c\)

\(\Rightarrow n+x=2a\)

Ta có:

\(A=\left(x+y+n\right)^3-y^3-n^3-x^3\)

\(=x^3+y^3+n^3+3\left(x+y\right)\left(y+n\right)\left(n+x\right)-y^3-n^3-x^3\)

\(=3\left(x+y\right)\left(y+n\right)\left(n+z\right)\)

\(=3.2b.2c.2a=24abc\)

\(\Rightarrow A=24\) (Vì đề ra \(abc=1\))

Ta có :

\(\left(ax+b\right)\left(x^2-2cx+abc\right)=x^3-4x^2+3x+\frac{9}{5}\)

\(\Leftrightarrow ax^3+2acx^2+bx^2-2bcx+ab^2c=x^3-4x^2+3x+\frac{9}{5}\)

\(\Leftrightarrow ax^3+\left(2ac+b^2\right)x^2+\left(a^2bc-2bc\right)x+ab^2c=x^3-4x^2+3x+\frac{9}{5}\)

Đồng nhất hệ số ta được :

a = 1

2ac + b² = -4

a²bc - 2bc = 3

\(ab^2c=\frac{9}{5}\)

\(\Rightarrow a=1;b=\frac{3}{5};c=5\)

Tl

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#Kirito

mệnh đè 2 sai

5(x²-9)+(2x+3)²=5x²-45+4x²+12x+9=9x²+12x-36=3(3x²+4x-12)

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

\(=x^4+3x^2y^2-2x^3y-2x^3y+y^4+2x^3y-3x^2y^2+2xy^3\)

\(=x^4+y^4\)