a. \(2x^3+3x^2+2x+3=2x\left(x^2+1\right)+3\left(x^2+1\right)=\left(2x+3\right)\left(x^2+1\right)\)

b. \(a^2-ab+a-b=a\left(a+1\right)-b\left(a+1\right)=\left(a-b\right)\left(a+1\right)\)

c. \(2x^2+4x+2-2y^2=2\left(x^2+2x+1-y^2\right)=2\left(x+1+y\right)\left(x+1-y\right)\)

d. \(x^4-2x^3+10x^2-20x=x\left(x^3-2x^2+10x-20\right)\)

\(==x.x\left(x^2+10\right)-2\left(x^2+10\right)=x\left(x-2\right)\left(x^2+10\right)\)

e. \(x^3+2x^2+x=x^2\left(x+1\right)+x\left(x+1\right)=\left(x^2+x\right)\left(x+1\right)\)

f. \(xy+y^2-x-y=x\left(y-1\right)+y\left(y-1\right)=\left(x+y\right)\left(y-1\right)\)

a) 2x³ + 3x² + 2x + 3

= ( 2x³ + 2x ) + ( 3x² + 3 )

= 2x( x² + 1 ) + 3( x² + 1 )

= ( x² + 1 )( 2x + 3 )

b) a² - ab + a - b

= ( a² + a ) - ( ab + b )

= a( a + 1 ) - b( a + 1 )

= ( a - b )( a + 1 )

c) 2x² + 4x + 2 - 2y²

= ( 2x² - 2y² ) + ( 4x + 2 )

= 2( x² - y² ) + 2( 2x + 1 )

= 2( x² - y² + 2x + 1 )

= 2[ ( x² + 2x + 1 ) - y² ]

= 2[ ( x + 1 )² - y² ]

= 2( x - y + 1 )( x + y + 1 )

d) x⁴ - 2x³ + 10x² - 20x

= x( x³ - 2x² + 10x - 20 )

= x[ ( x³ - 2x² ) + ( 10x - 20 ) ]

= x[ x²( x - 2 ) + 10( x - 2 ) ]

= x( x - 2 )( x² + 10 )

e) x³ + 2x² + x = x( x² + 2x + 1 ) = x( x + 1 )²

f) xy + y² - x - y

= ( xy - x ) + ( y² - y )

= x( y - 1 ) + y( y - 1 )

= ( x + y )( y - 1 )

a) ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne-2\end{cases}}\)

\(N=\frac{\left(x+2\right)^2}{x}.\left(1-\frac{x^2}{x+2}\right)-\frac{x^2+6x+4}{x}\)

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

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

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

\(N=\frac{-x^3-2x^2-2x}{x}\)

\(N=\frac{-x\left(x^2+2x+2\right)}{x}\)

\(N=-\left(x^2+2x+2\right)\)

b) \(N=-\left(x^2+2x+2\right)\)

\(\Leftrightarrow N=-\left(x^2+2x+1+1\right)\)

\(\Leftrightarrow N=-\left(x+1\right)^2-1\le-1\)

Max N = -1 \(\Leftrightarrow x=-1\)

Vậy .......................

3(x - 1)² - (x + 1)² + 2(x - 3)(x + 3) - (2x + 3)² - (5 - 20x)

= 3(x² - 2x + 1) - (x + 1)(x + 1) + 2(x² - 9) - (2x + 3)(2x + 3) - 5 + 20x

= 3x² - 6x + 3 - x(x + 1) - 1(x + 1) + 2x² - 18 - 2x(2x + 3) - 3(2x + 3) - 5 + 20x

= 3x² - 6x + 3 - x² - 2x - 1 + 2x² - 18 - 4x² - 6x - 6x - 9 - 5 + 20x

= (3x² - x² + 2x² - 4x²) + (-6x - 2x - 6x - 6x + 20x) + (3 - 1 - 18 - 9 - 5)

= -30

=> biểu thức A không phụ thuộc vào x

A = 3( x - 1 )² - ( x + 1 )² + 2( x - 3 )( x + 3 ) - ( 2x + 3 )² - ( 5 - 20x )

A = 3( x² - 2x + 1 ) - ( x² + 2x + 1 ) + 2( x² - 9 ) - ( 4x² + 12x + 9 ) - 5 + 20x 

A = 3x² - 6x + 3 - x² - 2x - 1 + 2x² - 18 - 4x² - 12x - 9 - 5 + 20x

A = ( 3x² - x² + 2x² - 4x² ) + ( -6x - 2x - 12x + 20x ) + ( 3 - 1 - 18 - 9 - 5 )

A = -30 ( đpcm )

Pt \(\Leftrightarrow\left(4x-1\right)\sqrt{x^3+1}=2\left(x^3+x\right)+1\)

Đặt \(\sqrt{x^3+1}=i\)

Ta có pt : \(\left(4x-1\right)i=2i^2+1\)

\(\Leftrightarrow2i^2+\left(4x-1\right)i+2x-1=0\Leftrightarrow\orbr{\begin{cases}i=\frac{1}{2}\\i=2x-1\end{cases}}\)

Tới đây tự blabla tiếp:))

mình nghĩ sửa đề bài là  \(\frac{\sqrt{x^2-x+6}+7\sqrt{x}-\sqrt{6\left(x^2+5x-2\right)}}{x+3-\sqrt{2\left(x^2+10\right)}}\le0\) 

a)\(A=\frac{6}{-2^2-3}\)

Ta có: \(x^2\ge0\Rightarrow2x^2+3\ge3\forall x\Rightarrow-2x^2-3\le-3\)

\(\Rightarrow A\ge-2\Rightarrow MinA=-2\)khi x=0

b) Ta có: \(x^2+2x+6=\left(x+1\right)^2+5\ge5\Rightarrow-x-2x-6\le-5\)

\(\Rightarrow B\ge\frac{-1}{5}\Rightarrow MinB=\frac{-1}{5}\)khi x=-1

c) Ta có:\(10x-x^2+3=-\left(x^2-10x+25\right)+28\le28\)\(\Rightarrow C\ge\frac{7}{28}=\frac{1}{4}\)

c) ĐKXĐ : \(x\ne4\)

Để biểu thức \(\frac{3x^3-4x^2+x-1}{x-4}\) nguyên với \(x\) nguyên thì :

\(3x^3-4x^2+x-1⋮x-4\)

\(\Leftrightarrow3x^3-12x^2+8x^2-32x+33x-132+131⋮x-4\)

\(\Leftrightarrow3x^2.\left(x-4\right)+8x.\left(x-4\right)+31.\left(x-4\right)+131⋮x-4\)

\(\Leftrightarrow131⋮x-4\)

\(\Leftrightarrow x-4\inƯ\left(131\right)\)

\(\Leftrightarrow x-4\in\left\{-1,1,131,-131\right\}\)

\(\Leftrightarrow x\in\left\{3,5,135,-127\right\}\)

d) ĐKXĐ : \(x\ne-\frac{3}{2}\)

Để biểu thức \(\frac{3x^2-x+1}{3x+2}\) nhận giá trị nguyên với \(x\) nguyên thì :

\(3x^2-x+1⋮3x+2\)

\(\Leftrightarrow3x^2+2x-3x-2+3⋮3x+2\)

\(\Leftrightarrow x.\left(3x+2\right)-\left(3x+2\right)+3⋮3x+2\)

\(\Leftrightarrow3⋮3x+2\)

\(\Leftrightarrow3x+2\inƯ\left(3\right)\)

\(\Leftrightarrow3x+2\in\left\{-1,1,-3,3\right\}\)

\(\Leftrightarrow x\in\left\{-1,-\frac{1}{3},-\frac{5}{3},\frac{1}{3}\right\}\) mà \(x\) nguyên 

\(\Rightarrow x=-1\)

a) ĐKXĐ : \(x\ne3\)

Để \(\frac{2}{x-3}\)nguyên

=> \(2⋮x-3\)

=> \(x-3\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Cả 4 giá trị đều tmđk

KL : Vậy x = { 4 ; 2 ; 5 ; 1 }

b) ĐKXĐ : \(x\ne-2\)

Để \(\frac{3}{x+2}\)nguyên

=> \(3⋮x+2\)

=> \(x+2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

Cả 4 giá trị đều tmđk

KL : Vậy x = { -1 ; -3 ; 1 ; -5 }

a, ĐKXĐ : \(x\ne3\)

 \(\frac{2}{x-3}\)có giá trị nguyên

\(\Leftrightarrow x-3\inƯ\left(2\right)\)

\(Ư\left(2\right)=\left\{\pm1;\pm2\right\}\)

 +, TH1 : \(x-3=1\Leftrightarrow x=4\left(TM\right)\)

+, TH2 : \(x-3=-1\Leftrightarrow x=2\left(TM\right)\)

+, TH3 : \(x-3=2\Leftrightarrow x=5\left(TM\right)\)

+, TH4 : \(x-3=-2\Leftrightarrow x=1\left(TM\right)\)

Vậy với \(x\in\left\{4;2;5;1\right\}\)thì \(\frac{2}{x-3}\)có giá trị nguyên