B3:\(\Rightarrow90.10^n-10^n.10^2+10^n.10-20\Rightarrow10^n.\left(90-10^2\right)+10^n.10-20\)

\(\Rightarrow10^n.\left(90-100\right)+10^n.10-20\Rightarrow-10.10^n+10^n.10-20\Rightarrow-20\)

\(A=-\left(x^2-x+5\right)=-\left(x^2-2.\frac{1}{2}x+\frac{1}{4}+\frac{19}{4}\right)=-\left[\left(x-\frac{1}{2}\right)^2+\frac{19}{4}\right]\)

\(=-\left(x-\frac{1}{2}\right)^2-\frac{19}{4}\le-\frac{19}{4}\)

Vậy \(A_{min}=-\frac{19}{4}\Leftrightarrow x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)

Cứ nói người ta ngu trong khi cứ ngồi đó,giỏi thì làm đi

$\textbf{1)}$

$8x(2x-1)-(4x-1)^2-13$

$=16x^2-8x-(16x^2-8x+1)-13$

$=-14.$

$\textbf{2)}$

$(2x-4)(x+3)-2x(x+1)$

$=2x^2+2x-12-2x^2-2x$

$=-12.$

Bài 4: 

Ta có: \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)

\(\Leftrightarrow8x-24x^2+2-6x+24x^2-60x-4x+40=-50\)

\(\Leftrightarrow-62x=-92\)

hay \(x=\dfrac{46}{31}\)

ĐKXĐ: x<>1/2

Để \(\frac{4x^3-6x^2+8x}{2x-1}\) là số nguyên thì \(4x^3-6x^2+8x\vdots2x-1\)

=>\(4x^3-2x^2-4x^2+2x+6x-3+3\vdots2x-1\)

=>3⋮2x-1

=>2x-1∈{1;-1;3;-3}

=>2x∈{2;0;4;-2}

=>x∈{1;0;2;-1}

1: \(\left(2x+1\right)\left(4x^2-2x+1\right)-8x^3+30=2x+7\)

=>\(\left(2x\right)^3+1-8x^3+30=2x+7\)

=>2x+7=31

=>2x=24

=>x=12

2: \(\left(x-2\right)\left(x+3\right)+\left(x+1\right)^2-2x^2-3x\)

\(=x^2+3x-2x-6+x^2+2x+1-2x^2-3x\)

=-6+1

=-5

c: \(-x^2+2x-2=-\left(x-1\right)^2-1\le-1\forall x\)

\(\Leftrightarrow V\ge-1\forall x\)

Dấu '=' xảy ra khi x=1

$\textbf{a)}$

$S=\dfrac{3}{2x^2+2x+3}$

$=\dfrac{3}{2\left(x+\dfrac12\right)^2+\dfrac52}.$

Vì $2\left(x+\dfrac12\right)^2+\dfrac52\ge\dfrac52$

$\Rightarrow S\le\dfrac{3}{5/2}=\dfrac65.$

Dấu ``='' khi $x=-\dfrac12.$

Vậy $\max S=\dfrac65.$

Giups mik vs

Lời giải:
$(2x-1)(8x-3)-(4x-1)^2+6x=16x^2-6x-8x+3-(16x^2-8x+1)+6x$

$=16x^2-14x+3-16x^2+8x-1+6x$

$=(16x^2-16x^2)+(-14x+8x+6x)+(3-1)=0+0+2=2$ là giá trị không phụ thuộc vào biến $x$