1: A=4x^2+12x+9-4x^2+4x-1-6x=10x+8

Khi x=201 thì A=10*201+8=2018

2: B=4x^2+20x+25-4x^2+12=20x+37

Khi x=1/20 thì B=1+37=38

1, \(A=\left(2x+3\right)^2-\left(2x-1\right)^2-6x\)

\(A=\left[\left(2x+3\right)+\left(2x-1\right)\right]\left[\left(2x+3\right)-\left(2x-1\right)\right]-6x\)

\(A=\left(2x+3+2x-1\right)\left(2x+3-2x+1\right)-6x\)

\(A=4\left(4x+2\right)-6x\)

\(A=16x+8-6x\)

\(A=10x+8\)

Thay \(x=201\) vào A ta có:

\(A=10\cdot201+8=2010+8=2018\)

Vậy: ....

2, \(B=\left(2x+5\right)^2-4\left(x+3\right)\left(x-3\right)\)

\(B=\left(2x+5\right)^2-4\left(x^2-9\right)\)

\(B=4x^2+20x+25-4x^2+36\)

\(B=20x+61\)

Thay \(x=\dfrac{1}{20}\) vào B ta có:

\(B=20\cdot\dfrac{1}{20}+61=1+61=62\)

Vậy: ...

chịu

`@` `\text {Ans}`

`\downarrow`

`P(x)+Q(x)-R(x)`

`= 5x^2 + 5x - 4 +2x^2 - 3x + 1 - (4x^2 - x + 3)`

`= 5x^2 + 5x - 4 + 2x^2 - 3x + 1 - 4x^2 + x - 3`

`= (5x^2 + 2x^2 - 4x^2) + (5x - 3x + x) + (-4 + 1 - 3)`

`= 3x^2 + 3x - 6`

Thay `x=-1/2`

`3*(-1/2)^2 + 3*(-1/2) - 6`

`= 3*1/4 - 3/2 - 6`

`= 3/4 - 3/2 - 6`

`= -3/4 - 6 = -27/4`

Vậy, khi `x=-1/2` thì GTr của đa thức là `-27/4`

P(x)+Q(x)-R(x)

=5x^2+5x-4+2x^2-3x+1-4x^2+x-3

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

Khi x=-1/2 thì (1) sẽ là 2*1/4+3*(-1/2)-6=1/2-3/2-6=-7

\(x=\sqrt{\dfrac{2\sqrt{3}+2-6\sqrt{3}}{2\sqrt{3}\left(2\sqrt{3}+2\right)}}=\sqrt{\dfrac{2-4\sqrt{3}}{2\sqrt{3}\left(2\sqrt{3}+2\right)}}\) ko tồn tại vì 2-4căn 3<0

a: \(A=\left(\dfrac{2x^2+2}{x^3-1}+\dfrac{x^2-x+1}{x^4+x^2+1}-\dfrac{x^2+3}{x^3-x^2+3x-3}\right):\dfrac{1}{x-1}\)

\(=\left(\dfrac{2x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{x^2-x+1}{x^4+2x^2+1-x^2}-\dfrac{x^2+3}{x^2\left(x-1\right)+3\left(x-1\right)}\right)\cdot\dfrac{x-1}{1}\)

\(=\left(\dfrac{2x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{\left(x^2-x+1\right)}{\left(x^2+1\right)^2-x^2}-\dfrac{x^2+3}{\left(x-1\right)\left(x^2+3\right)}\right)\cdot\dfrac{x-1}{1}\)

\(=\left(\dfrac{2x^2+3}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{x^2-x+1}{\left(x^2+1+x\right)\left(x^2+1-x\right)}-\dfrac{1}{x-1}\right)\cdot\dfrac{x-1}{1}\)

\(=\left(\dfrac{2x^2+3}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x^2+x+1}-\dfrac{1}{x-1}\right)\cdot\dfrac{x-1}{1}\)

\(=\dfrac{2x^2+3+x-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x-1}{1}\)

\(=\dfrac{x^2+1}{x^2+x+1}\)

b: Để A là số nguyên thì \(x^2+1⋮x^2+x+1\)

=>\(x^2+x+1-x⋮x^2+x+1\)

=>\(x⋮x^2+x+1\)

=>\(x^2+x⋮x^2+x+1\)

=>\(x^2+x+1-1⋮x^2+x+1\)

=>\(-1⋮x^2+x+1\)

=>\(x^2+x+1\in\left\{1;-1\right\}\)

=>\(x^2+x+1=1\)

=>x²+x=0

=>x(x+1)=0

=>\(x\in\left\{0;-1\right\}\)

Bài 1:

a. $2x^3+3x^2-2x=2x(x^2+3x-2)=2x[(x^2-2x)+(x-2)]$

$=2x[x(x-2)+(x-2)]=2x(x-2)(x+1)$

b.

$(x+1)(x+2)(x+3)(x+4)-24$

$=[(x+1)(x+4)][(x+2)(x+3)]-24$

$=(x^2+5x+4)(x^2+5x+6)-24$

$=a(a+2)-24$ (đặt $x^2+5x+4=a$)

$=a^2+2a-24=(a^2-4a)+(6a-24)$

$=a(a-4)+6(a-4)=(a-4)(a+6)=(x^2+5x)(x^2+5x+10)$

$=x(x+5)(x^2+5x+10)$

Bài 2:

a. ĐKXĐ: $x\neq 3; 4$

\(A=\frac{2x+1-(x+3)(x-3)+(2x-1)(x-4)}{(x-3)(x-4)}\\ =\frac{2x+1-(x^2-9)+(2x^2-9x+4)}{(x-3)(x-4)}\\ =\frac{x^2-7x+14}{(x-3)(x-4)}\)

b. $x^2+20=9x$

$\Leftrightarrow x^2-9x+20=0$

$\Leftrightarrow (x-4)(x-5)=0$

$\Rightarrow x=5$ (do $x\neq 4$)

Khi đó: $A=\frac{5^2-7.5+14}{(5-4)(5-3)}=2$

Bài 3:

$(2x^2-7x^2:13x:2):(2x-1)=(2x^2-\frac{7}{26}x):(2x-1)$

$=[x(2x-1)+\frac{19}{52}(2x-1)+\frac{19}{52}]:(2x-1)$

$=[(2x-1)(x+\frac{19}{52})+\frac{19}{52}]: (2x-1)$

$\Rightarrow$ thương là $x+\frac{19}{52}$ và thương là $\frac{19}{52}$

a. Làm gọn 1 chút xíu:

\(y=\left(x^{11}+2x^7-3x^5-6x\right)\left(3x^7+6x^2-2\right)\)

\(y'=\left(11x^{10}+14x^6-15x^4-6\right)\left(3x^7+6x^2-2\right)+\left(21x^6+12x\right)\left(x^{11}+2x^7-3x^5-6x\right)\)

b.

 \(y'=5\left(x^4-\dfrac{2}{3x}\right)^4\left(4x^3+\dfrac{2}{3x^2}\right)\Rightarrow y'\left(10\right)=5\left(10^4-\dfrac{2}{30}\right)^4\left(4.10^3+\dfrac{2}{300}\right)=?\)

c.

\(y'=\dfrac{7}{\left(x+1\right)^2}\Rightarrow y'\left(4\right)=\dfrac{7}{25}\)

a: \(=x^2-2x-3x^2+5x-4+2x^2-3x+7=3\)

b: \(=2x^3-4x^2+x-1-5+x^2-2x^3+3x^2-x=4\)

c: \(=1-x-\dfrac{3}{5}x^2-x^4+2x+6+0.6x^2+x^4-x=7\)