a) Đặt  \(A=-x^2+9x-12\)

\(-A=x^2-9x+12\)

\(-A=\left(x^2-9x+\frac{81}{4}\right)-\frac{33}{4}\)

\(-A=\left(x-\frac{9}{2}\right)^2-\frac{33}{4}\)

Mà  \(\left(x-\frac{9}{2}\right)^2\ge0\forall x\)

\(\Rightarrow-A\ge-\frac{33}{4}\Leftrightarrow A\le\frac{33}{4}\)

Dấu "=" xảy ra khi :  \(x-\frac{9}{2}=0\Leftrightarrow x=\frac{9}{2}\)

Vậy  \(A_{Max}=\frac{33}{4}\Leftrightarrow x=\frac{9}{2}\)

b) Đặt \(B=2x^2+10x-1\)

\(B=2\left(x^2+5x+\frac{25}{4}\right)-\frac{29}{4}\)

\(B=2\left(x+\frac{5}{2}\right)^2-\frac{29}{4}\)

Mà  \(\left(x+\frac{5}{2}\right)^2\ge0\forall x\Rightarrow2\left(x+\frac{5}{2}\right)^2\ge0\forall x\)

\(\Rightarrow B\ge-\frac{29}{4}\)

Dấu "=" xảy ra khi :  \(x+\frac{5}{2}=0\Leftrightarrow x=-\frac{5}{2}\)

Vậy  \(B_{Min}=-\frac{29}{4}\Leftrightarrow x=-\frac{5}{2}\)

c) Đặt  \(C=\left(2x+6\right)\left(x-1\right)\)

\(C=2x^2-2x+6x-6\)

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

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

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

Mà  \(\left(x+1\right)^2\ge0\forall x\Rightarrow2\left(x+1\right)^2\ge0\forall x\)

\(\Rightarrow C\ge-8\)

Dấu "=" xảy ra khi :  \(x+1=0\Leftrightarrow x=-1\)

Vậy  \(C_{Min}=-8\Leftrightarrow x=-1\)

d) Đặt  \(D=3x-2x^2\)

\(-2D=4x^2-6x\)

\(-2D=\left(4x^2-6x+\frac{9}{4}\right)-\frac{9}{4}\)

\(-2D=\left(2x-\frac{3}{2}\right)^2-\frac{9}{4}\)

Mà  \(\left(2x-\frac{3}{2}\right)^2\ge0\forall x\)

\(\Rightarrow-2D\ge-\frac{9}{4}\)

\(\Leftrightarrow D\le\frac{9}{8}\)

Dấu "=" xảy ra khi :  \(2x-\frac{3}{2}=0\Leftrightarrow x=\frac{3}{4}\)

Vậy  \(D_{Max}=\frac{9}{8}\Leftrightarrow x=\frac{3}{4}\)

\(\left|2x-\frac{1}{2}\right|+1=3x\)

\(\Leftrightarrow\left|2x-\frac{1}{2}\right|=3x-1\)

\(\Leftrightarrow\orbr{\begin{cases}2x-\frac{1}{2}=3x-1\\2x-\frac{1}{2}=1-3x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3x=-1+\frac{1}{2}\\2x+3x=1+\frac{1}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-x=-\frac{1}{2}\\5x=\frac{3}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{3}{10}\end{cases}}\)

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

b, \(\left(3x-1\right)^2-9x\left(x+1\right)=9x^2-6x+1-9x^2-9x=1-15x\)

c, \(\left(x-1\right)^2-x\left(x+1\right)=x^2-2x+1-x^2-x=1-3x\)

\(\left(3x-1\right)^2-9^2=13\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=\sqrt{94}\\3x-1=-\sqrt{94}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{94}+1}{3}\\x=\dfrac{-\sqrt{94}+1}{3}\end{matrix}\right.\)

Bài 1.

a) -2x( -3x + 2 ) - ( x + 2 )²

= 6x² - 4x - ( x² + 4x + 4 )

= 6x² - 4x - x² - 4x - 4

= 5x² - 8x - 4

b) ( x + 2 )( x² - 2x + 4 ) - 2( x + 1 )( 1 - x )

= x³ + 8 + 2( x + 1 )( x - 1 )

= x³ + 8 + 2( x² - 1 )

= x³ + 8 + 2x² - 2

= x³ + 2x² + 6

c) ( 2x - 1 )² - 2( 4x² - 1 ) + ( 2x + 1 )²

= 4x² - 4x + 1 - 8x² + 2 + 4x² + 4x + 1

= 4

d) x² - 3x + xy - 3y

= x( x - 3 ) + y( x - 3 )

= ( x - 3 )( x + y )

Bài 2.

a) 4x² - 4xy + y² = ( 2x - y )²

b) 9x³ - 9x²y - 4x + 4y

= 9x²( x - y ) - 4( x - y )

= ( x - y )( 9x² - 4 )

= ( x - y )( 3x - 2 )( 3x + 2 )

c) x³ + 2 + 3( x³ - 2 )

= x³ + 2 + 3x³ - 6

= 4x³ - 4

= 4( x³ - 1 )

= 4( x - 1 )( x² + x + 1 )

Bài 3.

2( x - 2 ) = x² - 4x + 4

⇔ ( x - 2 )² - 2( x - 2 ) = 0

⇔ ( x - 2 )( x - 2 - 2 ) = 0

⇔ ( x - 2 )( x - 4 ) = 0

⇔ x = 2 hoặc x = 4

Học tốt <3

x(3x-1)+(9x-5)(x-2)=3x²-x+9x(x-2)-5(x-2)=3x²-x+9x²-18x-5x+10=12x²-22x+10

a)4x²+12xy+9y²=  (2x)²+2.2x.3y+(3y)²=(2x+3y)²   

b)y²+1-2y= y²-2.y.1+1²=(y-1)²

1. = \(\dfrac{x+y}{x-y}\)
2. = \(\dfrac{x}{x+3}\)

\(a,A=\dfrac{3x+2-3x+2+3x-6}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{3x-2}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{1}{3x+2}\\ b,B=\dfrac{1}{2}+\dfrac{x}{\dfrac{x+2-x}{x+2}}=\dfrac{1}{2}+\dfrac{x}{\dfrac{2}{x+2}}=\dfrac{1}{2}+\dfrac{x\left(x+2\right)}{2}\\ B=\dfrac{1+x^2+2x}{2}=\dfrac{\left(x+1\right)^2}{2}\)