\(D=4x^2-2x+3x\left(x-5\right)=4x^2-2x+3x^2-15x=7x^2-17x=7\left(-1\right)^2-17\left(-1\right)=24\)

\(E=x^{10}-2020x^9+2020x^8-2020x^7+...+2020x^2-2020x=x^9\left(x-2019\right)-x^8\left(x-2019\right)+x^7\left(x-2019\right)-...-x^2\left(x-2019\right)+x\left(x-2019\right)-x=x^9\left(2019-2019\right)-...+x\left(2019-2019\right)-2019=-2019\)

Bài 1:

\(P=2a^2-2b^2-a^2+2ab-b^2+a^2+2ab+b^2+b^2=2a^2-b^2+4ab\\ Q=\left(2x+3\right)^2+\left(2x-3\right)^2-2\left(2x-3\right)\left(2x+3\right)\\ Q=\left(2x+3-2x+3\right)^2=9^2=81\)

Bài 2:

\(Sửa:A=x^2+2xy+y^2-4x-4y+2=\left(x+y\right)^2-4\left(x+y\right)+4-2\\ A=\left(x+y-2\right)^2-2=\left(3-2\right)^2-2=1-2=-1\)

Bài 1:

a: ĐKXĐ: x∉{-1/2;1/2}

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

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

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

\(=\frac{4x-2+4x^2+2x+2}{2}=\frac{4x^2+6x}{2}=2x^2+3x\)

b: Thay x=0 vào A, ta được:

\(A=2\cdot0^2+3\cdot0=0\)

THay x=-3 vào A, ta được:

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

=18-9

=9

Vì x=1/2 không thỏa mãn ĐKXĐ
nên khi x=1/2 thì A không có giá trị

c: A=2

=>\(2x^2+3x=2\)

=>\(2x^2+3x-2=0\)

=>\(2x^2+4x-x-2=0\)

=>(x+2)(2x-1)=0

=>x=1/2(loại) hoặc x=-2(nhận)

Bài 2: Để B là số nguyên thì \(2n^2+5n-1\) ⋮2n-1

=>\(2n^2-n+6n-3+2\) ⋮2n-1

=>2⋮2n-1

mà 2n-1 lẻ

nên 2n-1∈{1;-1}

=>2n∈{2;0}

=>n∈{1;0}

\(A=\left(x^2-4x+4\right)+4=\left(x-2\right)^2+4\ge4\)

\(minA=4\Leftrightarrow x=2\)

\(B=\left(4x^2-12x+9\right)+2=\left(2x-3\right)^2+2\ge2\)

\(minB=2\Leftrightarrow x=\dfrac{3}{2}\)

\(C=3\left(x^2+2x+1\right)-8=3\left(x+1\right)^2-8\ge-8\)

\(minC=-8\Leftrightarrow x=-1\)

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

\(maxD=-4\Leftrightarrow x=1\)

\(E=-\left(4x^2-6x+\dfrac{9}{4}\right)-\dfrac{11}{4}=-\left(2x-\dfrac{3}{2}\right)^2-\dfrac{11}{4}\le-\dfrac{11}{4}\)

\(maxA=-\dfrac{11}{4}\Leftrightarrow x=\dfrac{3}{4}\)

\(F=-2\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{55}{8}=-2\left(x-\dfrac{1}{4}\right)^2-\dfrac{55}{8}\le-\dfrac{55}{8}\)

\(maxF=-\dfrac{55}{8}\Leftrightarrow x=\dfrac{1}{4}\)

\(G=\left(x^2-4xy+4y^2\right)+\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-2y\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(maxG=\dfrac{3}{4}\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-\dfrac{1}{2}\end{matrix}\right.\)

\(H=-\left(x^2-2x+1\right)-\left(y^2+4y+4\right)+16=-\left(x-1\right)^2-\left(y+2\right)^2+16\le16\)

\(maxH=16\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

a>(8x^2y+10xy6^2-6xy):2xy=4xy+5y-3

b>(3x^2-4x).(2x-6)=6x^3-26x^2+24x