2.

c) \(C=2x^2+4x-6=2\left(x^2+2x+1\right)-8\)

\(=2\left(x+1\right)^2-8\ge-8\forall x\)

Dấu"=" xảy ra<=> \(2\left(x+1\right)^2=0\Leftrightarrow x=-1\)

3.

c) \(C=-3x^2-6x+9=-3\left(x^2+2x+1\right)+12\)

\(=-3\left(x+1\right)^2+12\le12\forall x\)

Dấu "=" xảy ra<=> \(-3\left(x+1\right)^2=0\Leftrightarrow x=-1\)

\(2,GTNN\)

\(A=x^2-6x+5=x^2+6x+9-4\)

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

\(A_{min}=-4\Leftrightarrow\left(x+3\right)^2=0\Rightarrow x=-3\)

\(B=4x^2-8x+7=4\left(x^2-2x+\frac{7}{4}\right)\)

\(=4\left(x^2-2x+1+\frac{3}{4}\right)=4\left(x-1\right)^2+3\ge3\)

\(\Rightarrow B_{min}=3\Leftrightarrow\left(x-1\right)^2=0\Rightarrow x=1\)

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

\(=2\left(x^2+2x+1-4\right)=2\left(x+1\right)^2-8\ge-8\)

\(\Rightarrow C_{min}=-8\Leftrightarrow\left(x+1\right)^2=0\Rightarrow x=-1\)

\(3,GTLN\)

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

\(=-\left(x^2-2x+1-4\right)=-\left(x-1\right)^2+4\le4\)

\(A_{max}=4\Leftrightarrow-\left(x-1\right)^2=0\Rightarrow x=1\)

\(B=-9x^2+6x-4=-\left[9x^2-6x+4\right]\)

\(=-\left[\left(3x\right)^2-6x+1+3\right]=-\left(3x-1\right)^2-3\)

\(B_{max}=-3\Leftrightarrow-\left(3x-1\right)^2=0\Rightarrow x=\frac{1}{3}\)

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

\(=-3\left(x^2+2x+1-4\right)=-3\left(x+1\right)^2+12\)

\(C_{max}=12\Leftrightarrow-3\left(x+1\right)^2=0\Rightarrow x=-1\)

Sửa lại đề \(x^3-6x^2-x+30=0\)

\(PT\Leftrightarrow x^3+2x^2-8x^2-16x+15x+30=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-8x+15\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-3\right)\left(x-5\right)=0\)

Đến đây tự giải nha :)

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\(3=a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}\)\(\Leftrightarrow\)\(a+b+c\le3\)

\(P=\frac{\sqrt{4\left(a+3\right)}+\sqrt{4\left(b+3\right)}+\sqrt{4\left(c+3\right)}}{2}\le\frac{a+b+c+21}{4}\le6\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Ta có : \(x^n-1⋮x-1\)

          \(x^{n+1}-1⋮x-1\)

=> \(\left(x^n-1\right)\left(x^{n+1}-1\right)⋮\left(x-1\right)^2\)(1)

Do n; n+1 là 2 số tự nhiên liên tiếp => 1 trong 2 số chia hết cho 2 

+)Th1: n chia hết cho 2 hay n chẵn => \(x^n-1⋮x^2-1\) hay \(⋮x+1\)(2)

+)Th2: n+1 chia hết cho 2 hay n+2 chẵn.CM như trên 

Mà \(x+1\), \(\left(x-1\right)^2\) ko có nhân tử chung. Từ (1),(2) suy ra \(\left(x^n-1\right)\left(x^{n+1}-1\right)⋮\left(x-1\right)^2\)\(\left(x+1\right)\)(đpcm)

Vì f(x) chia x-3 dư 7 

\(\Rightarrow f\left(x\right)=\left(x-3\right)q\left(x\right)+7\)

\(\Rightarrow f\left(3\right)=7\)

Vì f(x) chia x-2 dư 5

\(\Rightarrow f\left(x\right)=\left(x-2\right)q\left(x\right)+5\)

\(\Rightarrow f\left(2\right)=5\)

Ta có f(x) khi chia (x-2)(x-3) thì được thương là 3x và còn dư

\(\Rightarrow f\left(x\right)=\left(x-2\right)\left(x-3\right)3x+ax+b\)

\(\Rightarrow\hept{\begin{cases}f\left(2\right)=2a+b=5\\f\left(3\right)=3a+b=7\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a=2\\b=1\end{cases}}\)

Vậy \(f\left(x\right)=\left(x-2\right)\left(x-3\right)3x+2x+1\)