Bài 1

a) 5x²y - 20xy²

= 5xy(x - 4y)

b) 1 - 8x + 16x² - y²

= (1 - 8x + 16x²) - y²

= (1 - 4x)² - y²

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

c) 4x - 4 - x²

= -(x² - 4x + 4)

= -(x - 2)²

d) x³ - 2x² + x - xy²

= x(x² - 2x + 1 - y²)

= x[(x² - 2x+ 1) - y²]

= x[(x - 1)² - y²]

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

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

e) 27 - 3x²

= 3(9 - x²)

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

f) 2x² + 4x + 2 - 2y²

= 2(x² + 2x + 1 - y²)

= 2[(x² + 2x + 1) - y²]

= 2[(x + 1)² - y²]

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

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

Bài 2:

a: \(x^2\left(x-2023\right)+x-2023=0\)

=>\(\left(x-2023\right)\left(x^2+1\right)=0\)

mà \(x^2+1>=1>0\forall x\)

nên x-2023=0

=>x=2023

b: 

ĐKXĐ: x<>0

\(-x\left(x-4\right)+\left(2x^3-4x^2-9x\right):x=0\)

=>\(-x\left(x-4\right)+2x^2-4x-9=0\)

=>\(-x^2+4x+2x^2-4x-9=0\)

=>\(x^2-9=0\)

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

=>\(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

c: \(x^2+2x-3x-6=0\)

=>\(\left(x^2+2x\right)-\left(3x+6\right)=0\)

=>\(x\left(x+2\right)-3\left(x+2\right)=0\)

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

=>\(\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

d: 3x(x-10)-2x+20=0

=>\(3x\left(x-10\right)-\left(2x-20\right)=0\)

=>\(3x\left(x-10\right)-2\left(x-10\right)=0\)

=>\(\left(x-10\right)\left(3x-2\right)=0\)

=>\(\left[{}\begin{matrix}x-10=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=10\end{matrix}\right.\)

Câu 1:

a: \(5x^2y-20xy^2\)

\(=5xy\cdot x-5xy\cdot4y\)

\(=5xy\left(x-4y\right)\)

b: \(1-8x+16x^2-y^2\)

\(=\left(16x^2-8x+1\right)-y^2\)

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

\(=\left(4x-1-y\right)\left(4x-1+y\right)\)

c: \(4x-4-x^2\)

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

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

d: \(x^3-2x^2+x-xy^2\)

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

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

\(=x\left[\left(x-1\right)^2-y^2\right]\)

\(=x\left(x-1-y\right)\left(x-1+y\right)\)

e: \(27-3x^2\)

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

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

f: \(2x^2+4x+2-2y^2\)

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

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

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

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

Bài 2

a) x²(x - 2023) - 2023 + x = 0

x²(x - 2023) - (x - 2023) = 0

(x - 2023)(x² - 1) = 0

x - 2023 = 0 hoặc x² - 1 = 0

*) x - 2023 = 0

x = 2023

*) x² - 1 = 0

x² = 1

x = 1 hoặc x = -1

Vậy x = -1; x = 1; x = 2023

b) -x(x - 4) + (2x³ - 4x² - 9x) : x = 0

-x² + 4x + 2x² - 4x - 9 = 0

x² - 9 = 0

x² = 9

x = 3 hoặc x = -3

Vậy x = 3; x = -3

c) x² + 2x - 3x - 6 = 0

(x² + 2x) - (3x + 6) = 0

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

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

x + 2 = 0 hoặc x - 3 = 0

*) x + 2 = 0

x = -2

*) x - 3 = 0

x = 3

Vậy x = -2; x = 3

d) 3x(x - 10) - 2x + 20 = 0

3x(x - 10) - (2x - 20) = 0

3x(x - 10) - 2(x - 10) = 0

(x - 10)(3x - 2) = 0

x - 10 = 0 hoặc 3x - 2 = 0

*) x - 10 = 0

x = 10

*) 3x - 2 = 0

3x = 2

x = 2/3

Vậy x = 2/3; x = 10

b) ( 2x+3)^2 - (2x+1)(2x-1) =22

=> 4x²+12x+9-4x²+1=22

=> 12x=12

=>x=1

c) (4x+3)(4x-3) -(4x-5)^2 =16

=>16x²-9-16x²+40x-25=16

=>40x=50

=>x=4/5

a)\(\left(x-13\right)^2-4=0\\\left(x-13\right)^2=4\\ \left(x-13\right)^2=2^2\\ \Rightarrow\left\{{}\begin{matrix}x-13=2\\x-13=-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}15\\-11\end{matrix}\right.\)

vậy...

d) x^3 -9x^2 +27x-27 =-8

=>(x-3)³=-8

=>x-3=-2

=>x=1

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

=>x³+3x²+3x+1-x³-3x²=2

=>3x=1

=>x=1/3

a) ( x - 3 )² - 4 = 0

<=> ( x - 3 )² - 2² = 0

<=> ( x - 3 - 2 )( x - 3 + 2 ) = 0

<=> ( x - 5 )( x - 1 ) = 0

<=> x = 5 hoặc x = 1

b( 2x + 3 )² - ( 2x + 1 )( 2x - 1 ) = 22

<=> 4x² + 12x + 9 - ( 4x² - 1 ) = 22

<=> 4x² + 12x + 9 - 4x² + 1 = 22

<=> 12x + 10 = 22

<=> 12x = 12

<=> x = 1

c) ( 4x + 3 )( 4x - 3 ) - ( 4x - 5 )² = 16

<=> 16x² - 9 - ( 16x² - 40x + 25 ) = 16

<=> 16x² - 9 - 16x² + 40x - 25 = 16

<=> 40x - 34 = 16

<=> 40x = 50

<=> x = 50/40 = 5/4

d) x³ - 9x² + 27x - 27 = -8

<=> ( x - 3 )³ = -8

<=> ( x - 3 )³ = (-2)³

<=> x - 3 = -2

<=> x = 1 

e) ( x + 1 )³ - x²( x + 3 ) = 2

<=> x³ + 3x² + 3x + 1 - x³ - 3x² = 2

<=> 3x + 1 = 2

<=> 3x = 1

<=> x = 1/3

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

<=> x³ - 6x² + 12x - 8 - x( x² - 1 ) + 6x² = 5

<=> x³ + 12x - 8 - x³ + x = 5

<=> 13x - 8 = 5

<=> 13x = 13

<=> x = 1

a) \(\left(x-3\right)^2-4=0\)

=> \(\left(x-3\right)^2-2^2=0\)

=> \(\left(x-3-2\right)\left(x-3+2\right)=0\)

=> \(\left(x-5\right)\left(x-1\right)=0\)

=> \(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

b) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)

=> \(\left(2x+3\right)^2-\left[\left(2x\right)^2-1^2\right]=22\)

=> \(\left(2x+3\right)^2-\left(4x^2-1\right)=22\)

=> \(\left(2x\right)^2+2\cdot2x\cdot3+3^2-4x^2+1=22\)

=> \(4x^2+12x+9-4x^2+1=22\)

=> \(12x+9+1=22\)

=> \(12x+10=22\)

=> 12x = 12

=> x = 1

c) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=16\)

=> \(\left(4x\right)^2-3^2-\left[\left(4x\right)^2-2\cdot4x\cdot5+5^2\right]=16\)

=> \(16x^2-9-\left(16x^2-40x+25\right)=16\)

=> \(16x^2-9-16x^2+40x-25=16\)

=> \(-9+40x-25=16\)

=> \(40x=16+25-\left(-9\right)=16+25+9=50\)

=> x = 50/40 = 5/4

d) \(x^3-9x^2+27x-27=-8\)

=> \(x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3=8\)

=> \(\left(x-3\right)^3=-8\)

=> \(\left(x-3\right)^3=\left(-2\right)^3\)

=> x - 3  = -2 => x = 1

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

=> \(x^3+3x^2+3x+1-x^3-3x^2=2\)

=> \(3x+1=2\)

=> \(3x=1\)=> x = 1/3

f) \(\left(x-2\right)^3-x\left(x-1\right)\left(x+1\right)+6x^2=5\)

=> \(x^3-3\cdot x^2\cdot2+3\cdot x\cdot2^2-2^3-x\left(x^2-1\right)+6x^2=5\)

=> \(x^3-6x^2+12x-8-x^3+x+6x^2=5\)

=> \(\left(12x+x\right)-8=5\)

=> 13x  = 13

=> x = 1

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

=>(x+3)^2 = 4

=>(x+3)^2 = 2^2 = (-2)^2

=>x+3 = 2 hoặc -2

=> x= -1 hoặc -5

b) ( 2x + 3)² – (2x + 1)(2x – 1) = 22

<=> 4x² + 12x + 9 - 4x² + 1 = 22

<=> 12x + 10 = 22

<=> 12x = 12

<=> x = 1

Vậy x = 1

c) (4x + 3)(4x – 3) – (4x - 5)² = 16

<=> 16x² - 9 - 16x² + 40x - 25 = 16

<=> 40x - 34 = 16

<=> 40x = 50

<=> \(\frac{5}{4}\)

Vậy \(x=\frac{5}{4}\)

e) (x + 1)³ – x²(x + 3) = 2

<=> x³ + 3x² + 3x + 1 - x³ - 3x² = 2

<=> 3x + 1 = 2

<=> 3x = 2

<=> x = \(\frac{2}{3}\)

Vậy \(x=\frac{2}{3}\)

f) (x – 2)³ – x(x – 1)(x + 1) + 6x² = 5

<=> x³ - 6x² + 12x - 8 - x³ + x + 6x² = 5

<=> 13x - 8 = 5

<=> 13x = 13

<=> x = 1

Vậy x = 1

e: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)

f: Ta có: \(x^3-6x^2+12x-19=0\)

\(\Leftrightarrow x^3-6x^2+12x-8-11=0\)

\(\Leftrightarrow\left(x-2\right)^3=11\)

hay \(x=\sqrt[3]{11}+2\)

\(a,\Leftrightarrow\left(3x-7\right)\left(3x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{7}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-1-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\ c,\Leftrightarrow4x^2-7x-2-4x^2+4x+3=7\\ \Leftrightarrow-3x=6\Leftrightarrow x=-2\\ d,\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=0\\ \Leftrightarrow4x=-26\Leftrightarrow x=-\dfrac{13}{2}\\ e,\Leftrightarrow x^3+27-x^3+x-27=0\\ \Leftrightarrow x=0\\ f,\Leftrightarrow\left(4x-3\right)\left(4x-3+3x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)

a) 9x²-49=0

(3x)²-7²=0

<=> (3x-7)(3x+7)=0

th1: 3x-7=0

<=>3x=7

<=>x=\(\dfrac{7}{3}\)

th2: 3x+7=0

<=>3x=-7

<=>x=\(-\dfrac{7}{3}\)

\(b,\Rightarrow\left(x+2\right)\left(x+2-x+3\right)=0\\ \Rightarrow5\left(x+2\right)=0\\ \Rightarrow x=-2\\ c,\Rightarrow2x\left(x^2-2x+1\right)=0\\ \Rightarrow2x\left(x-1\right)^2=0\\ \Rightarrow\left[{}\begin{matrix}2x=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\ d,\Rightarrow\left(x-1-2x-1\right)\left(x-1+2x+1\right)=0\\ \Rightarrow3x\left(-x-2\right)=0\\ \Rightarrow-3x\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}-3x=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

a)thiếu dấu

b)(x+2)² -(x+2)(x-3)=0

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

(x+2)5=0

x+2=0

x=-2

c)2x³-4x²+2x=0

2x(x²-2x+1)=0

2x(x-1)²

suy ra 2 trường hợp

x=0

x-1=0=>x=1

d)(x-1)²-(2x+1)²=0

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

(x-2)3x=0

x=0

x=2

c: =>(x-1)(x+1)=0

hay \(x\in\left\{1;-1\right\}\)

plss

a,

\(=\dfrac{3}{4x}.\left(x-3\right)\left(x+3\right)\)=0

\(\left\{{}\begin{matrix}\dfrac{3}{4x}=0\\x-3=0\\x+3=0\end{matrix}\right.\)

=>\(x=\left\{3,-3\right\}\)

b,

\(x^3-16x=0\\x\left(x^2-16\right)\\ x\left(x-4\right)\left(x+4\right)\)

\(\left\{{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\)

=>\(x=\left\{-4,0,4\right\}\)

d,

\(3x^3-27x=0\\ 3x\left(x^2-9\right)=0\\ 3x\left(x-3\right)\left(x+3\right)=0\)

\(\left\{{}\begin{matrix}3x=0\\x-3=0\\x+3=0\end{matrix}\right.\)

=>\(x=\left\{-3,0,3\right\}\)

e,

\(x^2+\left(x+1\right)+2x\left(x+1\right)=0\\ x\left(x+1\right)\left(x+2\right)=0\)

\(\left\{{}\begin{matrix}x=0\\x+1=0\\x+2=0\end{matrix}\right.\)

=>\(x=\left\{-2,-1,0\right\}\)

f,

\(x\left(2x-3\right)-2\left(3-2x\right)=0\\ \left(2x-3\right)\left(x+2\right)=0\)

\(\left\{{}\begin{matrix}2x-3=0\\x+2=0\end{matrix}\right.\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)

a) 3x(4x-3)-2x(5-6x)=0

\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)

\(\Leftrightarrow24x^2-19x=0\)

\(\Leftrightarrow x\left(24x-19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{19}{24}\)

b) 5(2x-3)+4x(x-2)+2x(3-2x)=0

\(\Leftrightarrow\)10x-15+4x²-8x+6x-4x²=0

\(\Leftrightarrow8x-15=0\)

\(\Leftrightarrow8x=15\)

\(\Leftrightarrow x=\dfrac{15}{8}\)

vậy x=\(\dfrac{15}{8}\)

c)3x(2-x)+2x(x-1)=5x(x+3)

\(\Leftrightarrow6x-3x^2+2x^2-2x=5x^2+15x\\ \Leftrightarrow4x-x^2=5x^2+15x\\ \Leftrightarrow4x-x^2-5x^2-15x=0\\ \)

\(\Leftrightarrow-6x^2-11x=0\\ \Leftrightarrow-x\left(6x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\6x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-11}{6}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{-11}{6}\)