K hiểu 😐😐😐

\(1,\\ a,=x^3-5x\\ b,=3x^3y-6x^3y^2+9xy\\ c,=6x^2-6x-36\\ d,=x^3+2x^2y-3xy^2\\ 2,\\ a,=4x^2-25\\ b,=x^2-6x+9\\ c,=9x^2+24x+16\\ d,=x^3-6x^2y+12xy^2-8y^3\\ e,=125x^3+225x^2y+135xy^2+27y^3\\ f,=125-x^3\)

\(g,=8y^3+x^3\\ 3,\\ a,=x\left(x+2\right)\\ b,=\left(x-3\right)^2\\ c,=\left(x-y\right)\left(y+5\right)\\ d,=2x\left(y+1\right)-y\left(y+1\right)=\left(2x-y\right)\left(y+1\right)\\ e,=6x^2y^2\left(xy^2+2y-3x\right)\)

a) x³ - 64x = 0

x(x² - 64) = 0

x(x - 8)(x + 8) = 0

x = 0 hoặc x - 8 = 0 hoặc x + 8 = 0

*) x - 8 = 0

x = 8

*) x + 8 = 0

x = -8

Vậy x = -8; x = 0; x = 8

b) x³ - 4x² = -4x

x³ - 4x² + 4x = 0

x(x² - 4x + 4) = 0

x(x - 2)² = 0

x = 0 hoặc (x - 2)² = 0

*) (x - 2)² = 0

x - 2 = 0

x = 2

Vậy x = 0; x = 2

c) x² - 16 - (x - 4) = 0

(x - 4)(x + 4) - (x - 4) = 0

(x - 4)(x + 4 - 1) = 0

(x - 4)(x + 3) = 0

x - 4 = 0 hoặc x + 3 = 0

*) x - 4 = 0

x = 4

*) x + 3 = 0

x = -3

Vậy x = -3; x = 4

d) (2x + 1)² = (3 + x)²

(2x + 1)² - (3 + x)² = 0

(2x + 1 - 3 - x)(2x + 1 + 3 + x) = 0

(x - 2)(3x + 4) = 0

x - 2 = 0 hoặc 3x + 4 = 0

*) x - 2 = 0

x = 2

*) 3x + 4 = 0

3x = -4

x = -4/3

Vậy x = -4/3; x = 2

e) x³ - 6x² + 12x - 8 = 0

(x - 2)³ = 0

x - 2 = 0

x = 2

f) x³ - 7x - 6 = 0

x³ + 2x² - 2x² - 4x - 3x - 6 = 0

(x³ + 2x²) - (2x² + 4x) - (3x + 6) = 0

x²(x + 2) - 2x(x + 2) - 3(x + 2) = 0

(x + 2)(x² - 2x - 3) = 0

(x + 2)(x² + x - 3x - 3) = 0

(x + 2)[(x² + x) - (3x + 3)] = 0

(x + 2)[x(x + 1) - 3(x + 1)] = 0

(x + 2)(x + 1)(x - 3) = 0

x + 2 = 0 hoặc x + 1 = 0 hoặc x - 3 = 0

*) x + 2 = 0

x = -2

*) x + 1 = 0

x = -1

*) x - 3 = 0

x = 3

Vậy x = -1; x = -1; x = 3

a,x\(^3\)-64=0
   x\(^3\)     =64
  =>x=3
b,x\(^3\)-4x\(^2\)=-4x

   x\(^3\)-4x\(^2\)+4x=0
   x(x\(^2\)-4x+4)=0
   x(x-2)\(^2\)=)
TH1:x=0
TH2:x-2=0
     =>x=2

c,x\(^2\)-16-(x-4)=0

   (x+4)(x-4)-(x-4)=0
   (x-4)(x+4-1)=0

   (x-4)(x+3)=0
TH1:x-4=0
     =>x=4

TH2:x+3=0

    =>x=-3

d,(2x+1).2=3+x
    4x+2-3-x=0
    3x-1=0
     x=\(\dfrac{1}{3}\)

e,x\(^3\)-6x\(^2\)+12x-8=0
   (x-2)\(^3\)=0
=>x-2=0
=>x=2

f,x\(^3\)-7x+6=0
  x\(^3\)-x-6x+6=0
  x(x\(^2\)-1)-6(x-1)=0
  x(x+1)(x-1)-6(x-1)=0
  (x-1)(x\(^2\)+x-6)=0
TH1:x-1=0

  =>x=1
TH2:x\(^2\)+x-6=0

       x\(^2\)+3x-2x-6=0

       x(x+3)-2(x+3)=0

       (x+3)(x-2)=0

=>x+3=0                =>x-2=0

+>x=-3                   =>x=2

d,(2x+1)\(^2\)=(3+x)\(^2\)

4x\(^2\)+4x+1-9-6x-x\(^2\)=0
3x\(^2\)-2x-8=0
3x\(^2\)-6x+4x-8=0
3x(x-2)+4(x-2)=0

(3x+4)(x-2)=0

TH1:3x+4=0                     TH2:x-2=0
=>x=\(\dfrac{-4}{3}\)                              =>x=2

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

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

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

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

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

r) Tham khảo: https://hoc247.net/hoi-dap/toan-8/phan-tich-da-thuc-x-y-2-z-2-y-z-2-x-2-z-x-2-y-2-thanh-nhan-tu-faq343704.html

\(7,\\ a,A=x^2-4x+3+11=\left(x-2\right)^2+10\ge10\\ \text{Dấu }"="\Leftrightarrow x=2\\ b,B=-\left(4x^2-4x+1\right)+6=-\left(2x-1\right)^2+6\le6\\ \text{Dấu }"="\Leftrightarrow x=\dfrac{1}{2}\\ c,x-y=2\Leftrightarrow x=y+2\\ \Leftrightarrow B=y^2-3x^2=y^2-3\left(y+2\right)^2\\ \Leftrightarrow B=y^2-3y^2-12y-12=-4y^2-12y-12\\ \Leftrightarrow B=-\left(4y^2+12y+9\right)-3=-\left(2y+3\right)^2-3\le-3\\ \text{Dấu }"="\Leftrightarrow y=-\dfrac{3}{2}\Leftrightarrow x=\dfrac{1}{2}\)

\(8,\\ \Leftrightarrow x^3-3x^2+5x+a=\left(x-2\right)\cdot a\left(x\right)\)

Thay \(x=2\Leftrightarrow8-12+10+a=0\Leftrightarrow a=-6\)

Bài 7:

a.

$A=(x-1)(x-3)+11=x^2-4x+3+11=x^2-4x+14$

$=(x^2-4x+4)+10=(x-2)^2+10\geq 10$
Vậy gtnn của $A$ là $10$ khi $x=2$

b.

$B=5-4x^2+4x=6-(4x^2-4x+1)=6-(2x-1)^2\leq 6$

Vậy gtln của $B$ là $6$ khi $2x-1=0\Leftrightarrow x=\frac{1}{2}$

c.

$x-y=2\Rightarrow x=y+2$. Khi đó:

$B=y^2-3x^2=y^2-3(y+2)^2=y^2-(3y^2+12y+12)=-2y^2-12y-12$

$=6-2(y^2+6y+9)=6-2(y+3)^2\leq 6$

Vậy $B_{\max}=6$

Bài 8:

Đặt $f(x)=x^3-3x^2+5x+a$

Theo định lý Bê-du, để $f(x)\vdots x-2$ thì $f(2)=0$

$\Leftrightarrow 6+a=0$

$\Leftrightarrow a=-6$

Bài 8 cách khác:

$x^3-3x^2+5x+a=x^2(x-2)-x(x-2)+3(x-2)+(a+6)$

$=(x-2)(x^2-x+3)+(a+6)$

Vậy $x^3-3x^2+5x+a$ chia $x-2$ có dư là $a+6$

Để phép chia là chia hết thì số dư phải bằng $0$

Tức là $a+6=0$

$\Rightarrow a=-6$

a: \(5x^4-x^3+7x\)

\(=x\left(5x^3-x^2+7\right)\)

c: \(x^2-5x+6=\left(x-2\right)\left(x-3\right)\)

a) \(=x\left(5x^3-x^2+7\right)\)

b) \(=\left(5x+4y\right)\left(x-y\right)\)

c) \(\left(x^2-2x\right)-\left(3x-6\right)=\left(x-2\right)\left(x-3\right)\)

a) \(5x^4-x^3+7x=x\left(5x^3-x^2+7\right)\)

b) \(5x\left(x-y\right)-4y\left(y-x\right)=5x\left(x-y\right)+4y\left(x-y\right)=\left(x-y\right)\left(5x+4y\right)\)

c) \(x^2-5x+6=x^2-2x-3x+6=\left(x^2-2x\right)-\left(3x-6\right)=x\left(x-2\right)-3\left(x-2\right)=\left(x-3\right)\left(x-2\right)\)

rút gọn nha

a, (x²+1)(x-3)-(x-3)(x²+3x+9)

=(x-3)(x²+1+x²+3x+9)

(x-3)(2x²+3x+10)

tl mình nha

a) \(\left(x^2+1\right)\left(x-3\right)-\left(x-3\right)\left(x^2+3x+9\right)\)

\(=\left(x-3\right)\left[x^2+1-\left(x^2+3x+9\right)\right]\)

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

\(=\left(x-3\right)\left(-3x-8\right)\)

b) \(\left(x+2\right)^2+x\left(x+5\right)\)

\(=x^2+4x+4+x^2+5x\)

\(=2x^2+9x+4\)

c) \(\left(5x+4y\right)\left(5x-4y\right)-24x^2+15y^2\)

\(=25x^2-16y^2-24x^2+15y^2\)

\(=x^2-y^2\)

\(=\left(x+y\right)\left(x-y\right)\)