Bài 1.

\(a, (3x-4)^2\)

\(=\left(3x\right)^2-2\cdot3x\cdot4+4^2\)

\(=9x^2-24x+16\)

\(b,\left(1+4x\right)^2\)

\(=1^2+2\cdot1\cdot4x+\left(4x\right)^2\)

\(=16x^2+8x+1\)

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

\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot3+3\cdot2x\cdot3^2+3^3\)

\(=8x^3+36x^2+54x+27\)

\(d,\left(5-2x\right)^3\)

\(=5^3-3\cdot5^2\cdot2x+3\cdot5\cdot\left(2x\right)^2-\left(2x\right)^3\)

\(=125-150x+60x^2-8x^3\)

\(e,49x^2-25\)

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

\(=\left(7x-5\right)\left(7x+5\right)\)

\(f,\dfrac{1}{25}-81y^2\)

\(=\left(\dfrac{1}{5}\right)^2-\left(9y\right)^2\)

\(=\left(\dfrac{1}{5}-9y\right)\left(\dfrac{1}{5}+9y\right)\)

Bài 2.

\(a,\left(x-5\right)^2-\left(x+7\right)\left(x-7\right)=8\)

\(\Rightarrow x^2-2\cdot x\cdot5+5^2-\left(x^2-7^2\right)=8\)

\(\Rightarrow x^2-10x+25-\left(x^2-49\right)=8\)

\(\Rightarrow x^2-10x+25-x^2+49=8\)

\(\Rightarrow\left(x^2-x^2\right)-10x=8-25-49\)

\(\Rightarrow-10x=-66\)

\(\Rightarrow x=\dfrac{33}{5}\)

\(b,\left(2x+5\right)^2-4\left(x+1\right)\left(x-1\right)=10\)

\(\Rightarrow\left(2x\right)^2+2\cdot2x\cdot5+5^2-4\left(x^2-1^2\right)=10\)

\(\Rightarrow4x^2+20x+25-4x^2+4=10\)

\(\Rightarrow\left(4x^2-4x^2\right)+20x=10-25-4\)

\(\Rightarrow20x=-19\)

\(\Rightarrow x=\dfrac{-19}{20}\)

#\(Toru\)

Bài 1

a) (3x - 4)²

= (3x)² - 2.3x.4 + 4²

= 9x² - 24x + 16

b) (1 + 4x)²

= 1² + 2.1.4x + (4x)²

= 1 + 8x + 16x²

c) (2x + 3)³

= (2x)³ + 3.(2x)².3 + 3.2x.3² + 3³

= 8x³ + 36x² + 54x + 27

d) (5 - 2x)³

= 5³ - 3.5².2x + 3.5.(2x)² - (2x)³

= 125 - 150x + 60x² - 8x³

e) 49x² - 25

= (7x)² - 5²

= (7x - 5)(7x + 5)

f) 1/25 - 81y²

= (1/5)² - (9y)²

= (1/5 - 9y)(1/5 + 9y)

Bài 3.

\(a,A=x^2-6x+19\)

\(=x^2-6x+9+10\)

\(=\left(x^2-2\cdot x\cdot3+3^2\right)+10\)

\(=\left(x-3\right)^2+10\)

Ta thấy: \(\left(x-3\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-3\right)^2+10\ge10\forall x\)

Dấu \("="\) xảy ra \(\Leftrightarrow x-3=0\Leftrightarrow x=3\)

Vậy: \(Min_A=10\) khi \(x=3\)

\(b,B=-x^2+8x-20\)

\(=-x^2+8x-16-4\)

\(=-\left(x^2-8x+16\right)-4\)

\(=-\left(x^2-2\cdot x\cdot4+4^2\right)-4\)

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

Ta thấy: \(\left(x-4\right)^2\ge0\forall x\)

\(\Rightarrow-\left(x-4\right)^2\le0\forall x\)

\(\Rightarrow-\left(x-4\right)^2-4\le-4\forall x\)

Dấu \("="\) xảy ra \(\Leftrightarrow x-4=0\Leftrightarrow x=4\)

Vậy \(Max_B=-4\) khi \(x=4\)

\(c,C=4x^2+12x+100\)

\(=4x^2+12x+9+91\)

\(=\left[\left(2x\right)^2+2\cdot2x\cdot3+3^2\right]+91\)

\(=\left(2x+3\right)^2+91\)

Ta thấy: \(\left(2x+3\right)^2\ge0\forall x\)

\(\Rightarrow\left(2x+3\right)^2+91\ge91\forall x\)

Dấu \("="\) xảy ra \(\Leftrightarrow2x+3=0\Leftrightarrow x=-\dfrac{3}{2}\)

Vậy \(Min_C=91\) khi \(x=\dfrac{-3}{2}\)

\(d,D=25+4x-x^2\)

\(=-x^2+4x-4+29\)

\(=-\left(x^2-2\cdot x\cdot2+2^2\right)+29\)

\(=-\left(x-2\right)^2+29\)

Ta thấy: \(\left(x-2\right)^2\ge0\forall x\)

\(\Rightarrow-\left(x-2\right)^2\le0\forall x\)

\(\Rightarrow-\left(x-2\right)^2+29\le29\forall x\)

Dấu \("="\) xảy ra \(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy \(Max_D=29\) khi \(x=2\)

#\(Toru\)

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

bài 1:

a) x(x-2)-5y-(x-2)=(x-5y)(x-2)

b) =(2x-3-4x)(2x-3+4x)=(-2x-3)(6x-3)

bài 2 bạn tự luyện nhé

Bài 1:

Ta có: \(4-2\left(x+1\right)=2\)

\(\Leftrightarrow2\left(x+1\right)=2\)

\(\Leftrightarrow x+1=1\)

hay x=0

Bài 2: 

Ta có: \(\left|2x-3\right|-1=2\)

\(\Leftrightarrow\left|2x-3\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)

chưa biết

\(1,\\ a,=7x^3-49x^2+21x\\ b,=x^2-x-42\\ c,=x^2-16x+64\\ d,=9x^2+12x+4\\ e,=x^2-16-25+10x-x^2=10x-41\\ 2,\\ a,\Rightarrow2\left(x-7\right)=19\\ \Rightarrow x-7=\dfrac{19}{2}\Rightarrow x=\dfrac{33}{2}\\ b,\Rightarrow4x^2-20x+25-4x^2+3x-2x=50\\ \Rightarrow-19x=25\Rightarrow x=-\dfrac{25}{19}\)

Bài 3:

3: \(6x\left(x-y\right)-9y^2+9xy\)

\(=6x\left(x-y\right)+9xy-9y^2\)

\(=6x\left(x-y\right)+9y\left(x-y\right)\)

\(=\left(x-y\right)\left(6x+9y\right)\)

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

Bài 4:

a) \(3xy^2-12x\)

\(=3x\left(y^2-4\right)\)

Bài 1:

b: \(=\left(x-2y\right)\left(x+2y\right)+4\left(x+2y\right)\)

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

c: \(=\left(x+y-3\right)\left(x+y+3\right)\)