Ta có : 

\(\left(3x+2\right)\left(9x^2-6x+4\right)-\left(x-3\right)\left(x+3\right)\)

\(=\)\(\left(3x+2\right)\left[\left(3x\right)^2-3x.2+2^2\right]-\left(x^2-3^2\right)\)

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

\(=\)\(27x^3-x^2+8-9\)

\(=\)\(27x^3-x^2-1\)

Chúc bạn học tốt ~ 

x²(3x+2)

= x²3x+2x²

= 3x³+2x²

đề thiếu hay sai j đó

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

\(x^2-6x+9+x^2+4x+4-2x^2=3\)

\(-2x+13=3\)

\(-2x=3-13\)

\(-2x=-10\)

\(x=\frac{-10}{-2}\)

\(x=5\)

Vậy \(x=5\)

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

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

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

x^3+ y^3+ 3xy

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

=x^2 + 2xy + y^2

=(x+y)^2 =1

=> x^3+ y^3+ 3xy=1

còn câu b ai giúp m vs

\(x^4+4x^3+12\)

\(=\left(x^2\right)^2+2.2x^3+\left(2x\right)^2-4x^2+12\)

\(=\left(x^2+2x\right)^2-4x^2+12\)

Có \(\left(x^2+2x\right)^2-4x^2+12>0\)

=> Vô nghiệm

1,\(4x+5y=10\)

\(\Rightarrow x=\frac{10-5y}{4}\)

\(\Rightarrow x=\frac{8+2-4y-y}{4}\)

\(\Rightarrow x=2-y+\frac{2-y}{4}\)

Để x nguyên => 2-y=4k(k thuộc N*)

=> y = 2-4k

=> x = 2-2+4k+4k : 4

=> x = 4k+k

Vậy \(\left(x;y\right)\in\left(4k+k;2-4k\right).Với\forall k\inℕ^∗\)

Ta có A=\(\left(ab+bc+ca\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)

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

\(A=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2=a^2-ab+b^2+3ab\left(1-2ab\right)+6a^2b^2\)

=\(\left(a+b\right)^2-3ab+3ab-6a^2b^2+6a^2b^2=1\)

2) Ta có \(A=\left(a-1\right)\left(b-1\right)\left(c-1\right)=abc-ab-bc-ca+a+b+c-1=0\)

Vì \(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)

\(\Rightarrow a^2+b^2+c^2+2\left(ab+ac+bc\right)=0\)

\(\Rightarrow2\left(ab+ac+bc\right)=-1\)

\(\Rightarrow ab+ac+bc=-\frac{1}{2}\)

\(\Rightarrow\left(ab+bc+ac\right)^2=\left(-\frac{1}{2}\right)^2=\frac{1}{4}\)

\(\Rightarrow a^2b^2+b^2c^2+a^2c^2+2\left(ab^2c+a^2bc+abc^2\right)=\frac{1}{4}\)

\(\Rightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(b+a+c\right)=\frac{1}{4}\)

\(\Rightarrow a^2b^2+b^2c^2+a^2c^2=\frac{1}{4}\)

Xét \(\left(a^2+b^2+c^2\right)^2=1\)

\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=1\)

\(\Rightarrow a^4+b^4+c^4+2.\frac{1}{4}=1\)

\(\Rightarrow a^4+b^4+c^4=1-\frac{1}{2}=\frac{1}{2}\)

\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+ac+bc\right)=0\)

\(\Leftrightarrow ab+ac+bc=-\frac{1}{2}\)

\(\Leftrightarrow\left(ab+ac+bc\right)^2=\left(ab\right)^2+\left(ac\right)^2+\left(bc\right)^2+2abc\left(a+b+c\right)=\frac{1}{4}\)

\(\Rightarrow\left(ab\right)^2+\left(ac\right)^2+\left(bc\right)^2=\frac{1}{4}\)

Do đó \(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left[\left(ab\right)^2+\left(ac\right)^2+\left(bc\right)^2\right]=1\)

\(\Leftrightarrow a^4+b^4+c^4+2.\frac{1}{4}=1\Rightarrow a^4+b^4+c^4=\frac{1}{2}\)