Ta có: \(x=71\Rightarrow\left\{{}\begin{matrix}x-1=70\\x-37=34\end{matrix}\right.\)

Thay \(x-1=70\) và \(x-37=34\) vào biểu thức, ta được:

\(A=x^5-\left(x-1\right)x^4-\left(x-1\right)x^3-\left(x-1\right)x^2-\left(x-1\right)x+x-37\)

\(\Leftrightarrow A=x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x+x-37\)

\(\Leftrightarrow A=2x-37\)

Thay \(x=71\) vào ta được:

\(A=2\cdot71-37=142-37=105\)

c) \(-6x^2-5y+3xy+10x\)

\(=3xy-6x^2-5y+10x\)

\(=3x\left(y-2x\right)-5\left(y-2x\right)\)

\(=\left(3x-5\right)\left(y-2x\right)\)

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

\(=abc-ab-bc-ca+a+b+c-1\)

\(=abc-ab-bc+b-ca+a+c-1\)

\(=ab\left(c-1\right)-b\left(c-1\right)-a\left(c-1\right)+\left(c-1\right)\)

\(=\left(ab-b-a+1\right)\left(c-1\right)\)

\(=\left[b\left(a-1\right)-\left(a-1\right)\right]\left(c-1\right)\)

\(=\left(a-1\right)\left(b-1\right)\left(c-1\right)\)

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b)\(B=9x-3x^2=-3\left(x^2-3x\right)=-3\left(x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}\right)+\dfrac{27}{4}=-3\left(x-\dfrac{3}{2}\right)^2+\dfrac{27}{4}\le\dfrac{27}{4}\)

( Vì : \(\left(x-\dfrac{3}{2}\right)^2\ge0\Rightarrow-3\left(x-\dfrac{3}{2}\right)^2\le0\))

Vậy \(MaxB=\dfrac{27}{4}\) khi \(x=\dfrac{3}{2}\)

e)\(E=5x-x^2=-x^2+5x=-x^2+2\cdot x\cdot\dfrac{5}{2}-\dfrac{25}{4}+\dfrac{25}{4}=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\)

(Vì: \(\left(x-\dfrac{5}{2}\right)^2\ge0\Rightarrow-\left(x-\dfrac{5}{2}\right)^2\le0\))

Vậy \(MaxE=\dfrac{25}{4}\) khi \(x=\dfrac{5}{2}\)

c)\(C=x^2-6x+11=x^2-6x+9+2=\left(x-3\right)^2+2\ge2\) (Vì:\(\left(x-3\right)^2\ge0\))

Vậy \(MinC=2\) khi \(x=3\)

a)Sửa đề: \(2x^2-8x-10\)

\(=2x^2-8x+8-18\)

\(=2\left(x^2-4x+4\right)-18\)

\(=2\left(x-2\right)^2-18\ge-18\) (Vì: \(\left(x-2\right)^2\ge0\Rightarrow2\left(x-2\right)^2\ge0\))

Vậy GTNN của \(2x^2-8x-10\) là \(-18\)

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