\(A=x^2-6x+11\)

\(A=\left(x^2-6x+9\right)+2\)

\(A=\left(x-3\right)^2+2\ge2\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\left(x-3\right)^2=0\)

\(\Leftrightarrow\)\(x-3=0\)

\(\Leftrightarrow\)\(x=3\)

Vậy GTNN của \(A\) là \(2\) khi \(x=3\)

\(B=x^2-20x+101\)

\(B=\left(x^2-20x+100\right)+1\)

\(B=\left(x-10\right)^2+1\ge1\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\left(x-10\right)^2=0\)

\(\Leftrightarrow\)\(x-10=0\)

\(\Leftrightarrow\)\(x=10\)

Vậy GTNN của \(B\) là \(1\) khi \(x=10\)

Chúc bạn học tốt ~ 

\(A=x^2-6x+11\)

\(A=\left(x^2-6x+9\right)+2\)

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

Mà  \(\left(x-3\right)^2\ge0\)

\(\Rightarrow A\ge2\)

Dấu "=" xảy ra khi :  \(x-3=0\Leftrightarrow x=3\)

Vậy  \(A_{Min}=2\Leftrightarrow x=3\)

b) \(B=x^2-20x+101\)

\(B=\left(x^2-20x+100\right)+1\)

\(B=\left(x-10\right)^2+1\)

Mà  \(\left(x-10\right)^2\ge0\)

\(\Rightarrow B\ge1\)

Dấu "=" xảy ra khi :  \(x-10=0\Leftrightarrow x=10\)

Vậy  \(B_{Min}=1\Leftrightarrow x=10\)

c)  \(C=x^2-4xy+5y^2+10x-22y+28\)

\(C=\left(x^2-4xy+4y^2\right)+y^2+10x-22y+28\)

\(C=\left[\left(x-2y\right)^2+2\left(x-2y\right).5+25\right]+\)\(\left(y^2-2y+1\right)+2\)

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

Mà  \(\left(x-2y+5\right)^2\ge0\)

      \(\left(y-1\right)^2\ge0\)

\(\Rightarrow C\ge2\)

Dấu "=" xảy ra khi : 

\(\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

Vây  \(C_{Min}=2\Leftrightarrow\left(x;y\right)=\left(-3;1\right)\)

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