mình cần gấp nhé mọi người 13 giờ mk đi học rồi

Bài 1:

b: =x^2-10x+x-10

=(x-10)(x+1)

c: \(=2x^2-5x+2x-5=\left(2x-5\right)\left(x+1\right)\)

d: \(=3x^2+5x-3x-5=\left(3x+5\right)\left(x-1\right)\)

e: \(=\left(2x+y\right)^3\)

\(2x\left(x^2-7x-3\right)=2x^3-14x-6x\)

\(4xy^2\left(-2x^3+y^2-7xy\right)=-8x^4y^2+4xy^5-28x^2y^3\)

a) Ta có: \(\left(x^4+2x^2y^2+y^4\right):\left(x^2+y^2\right)\)

\(=\left(x^2+y^2\right)^2:\left(x^2+y^2\right)\)

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

b) Ta có: \(\left(49x^2-81y^2\right):\left(7x+9y\right)\)

\(=\frac{\left(7x+9y\right)\left(7x-9y\right)}{7x+9y}\)

\(=7x-9y\)

c) Ta có: \(\left(x^3+3x^2y+3xy^2+y^3\right):\left(x+y\right)\)

\(=\left(x+y\right)^3:\left(x+y\right)\)

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

d) Ta có: \(\left(x^3-3x^2y+3xy^2-y^3\right):\left(x^2-2xy+y^2\right)\)

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

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

e)Sửa đề: \(\left(8x^3+1\right):\left(2x+1\right)\)

Ta có: \(\left(8x^3+1\right):\left(2x+1\right)\)

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

\(=4x^2-2x+1\)

f) Ta có: \(\left(8x^3-1\right):\left(4x^2+2x+1\right)\)

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

\(=2x-1\)

a, (x⁴ + 2x²y² + y⁴) : (x² + y²)

= (x² + y²)² : (x² + y²)

= x² + y²

b, (49x² - 81y²) : (7x + 9y)

= (7x - 9y)(7x + 9y) : (7x + 9y)

= 7x - 9y

c, (x³ + 3x²y + 3xy² + y³) : (x + y)

= (x + y)³ : (x + y)

= (x + y)²

d, (x³ - 3x²y + 3xy² - y³) : (x² - 2xy + y²)

= (x - y)³ : (x - y)²

= x - y

Phần e thiếu thì phải

f, (8x³ - 1) : (4x² + 2x + 1)

= (2x - 1)(4x² + 2x + 1) : (4x² + 2x + 1)

= 2x - 1

Chúc bn học tốt!

a: =-1/5x^5y^2

b: =-9/7xy^3

c: =7/12xy^2z

d: =2x^4

e: =3/4x^5y

f: =11x^2y^5+x^6

a) x² - 2x - 4y² - 4y

= (x² - 4y²) - (2x + 4y)

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

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

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

b) x⁴ + 2x³ - 4x - 4

= (x⁴ - 4) + ( 2x³ - 4x)

= (x² - 2)(x² + 2) + 2x(x² - 2)

= (x² - 2)(x² + 2 + 2x)

c) x³ + 2x²y - x -2y

= (x³ - x) + (2x²y - 2y)

= x(x² - 1) + 2y(x² - 1)

= (x + 2y)(x² - 1)

Trả lời:

1, \(P=9x^2-7x+2=9\left(x^2-\frac{7}{9}x+\frac{2}{9}\right)=9\left[\left(x^2-2x\frac{7}{18}+\frac{49}{324}\right)+\frac{23}{324}\right]\)

\(=9\left[\left(x-\frac{7}{18}\right)^2+\frac{23}{324}\right]=9\left(x-\frac{7}{18}\right)^2+\frac{23}{36}\)

Ta có: \(9\left(x-\frac{7}{18}\right)^2\ge0\forall x\)

\(\Leftrightarrow9\left(x-\frac{7}{18}\right)^2+\frac{23}{26}\ge\frac{23}{26}\forall x\)

Dấu "=" xảy ra khi \(x-\frac{7}{18}=0\Leftrightarrow x=\frac{7}{18}\)

Vậy GTNN của P = 23/36 khi x = 7/18

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 \(x=2\)

#\(Toru\)