dây là toán lớp 8 à 

Đáp án:

a) x= 1; x = 2

b) x = - 2; x = - 1/3

Giải thích các bước giải:

1) Đề bài ko đầy đủ ở PT thứ 2 của hệ

2)

a) (x² + 2x + 2) = (x + 1)² + 1 > 0; (x² + x + 2) = (x + 1/2)² + 3/4 > 0

Đặt y = (x² + 2x + 2)/(x² + x + 2) = 1 + x/(x² + x + 2) ⇒ 2x/(x² + x + 2) = 2(y - 1)

⇒ 1/y = (x² + x + 2)/(x² + 2x + 2) = 1 - x/(x² + 2x + 2) ⇒ x/(x² + 2x + 2) = 1 - 1/y

Thay vào ta có PT theo ẩn y : (1 - 1/y) + 2(y - 1) = 7/10

⇔ 20y² - 17y - 10 = 0

⇔ (5y + 2)(4y - 5) = 0

⇔ 4y - 5 = 0 ( vì y > 0)

⇔ (x² + 2x + 2)/(x² + x + 2) = 5/4

⇔ x² - 3x + 2 = 0

⇔ (x - 1)(x - 2) = 0

⇔ x = 1; x = 2

b) Tương tự câu a

3x² - 5x + 2 # 0 ⇔ x # 1; x # 2/3 ; (3x² + x + 2) = 2x² + (x + 1/2)² + 3/4 > 0

Đặt y = (3x² - 5x + 2)/(3x² + x + 2) = 1 - 6x/(3x² + x + 2) ⇒ 13x/(x² + x + 2) = (13/6)(1 - y)

⇒ 1/y = (3x² + x + 2)/(3x² - 5x + 2) = 1 + 6x/(3x² - 5x + 2) ⇒ 2x/(x² + 2x + 2) = (1/3)(1/y - 1)

Thay vào ta có PT theo ẩn y : (1/3)(1/y - 1) + (13/6)(1 - y) = 6

⇔ 13y² - 25y - 2 = 0

⇔ (y - 2)(13y + 1) = 0

@ Với y - 2 = 0

⇔ (3x² - 5x + 2)/(3x² + x + 2) = 2

⇔ 3x² + 7x + 2 = 0

⇔ (x + 2)(3x + 1) = 0

⇔ x = - 2; x = - 1/3

@ Với 13y + 1 = 0

⇔ (3x² - 5x + 2)/(3x² + x + 2) = - 1/13

⇔ 21x² - 32x + 14 = 0 ( vô nghiệm)

kết quả: (-y)(y-x^2)

\(^{x^2-x-y^2-y}\)

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

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

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

a2+2ab+b2−ac−bc=(a+b)2−c(a+b)=(a+b)(a+b−c)a2+2ab+b2−ac−bc=(a+b)2−c(a+b)=(a+b)(a+b−c)

x2−2x+1−4y2−4y−1=(x−1)2−(2y+1)2=(x+2y)(x−2y−2)x2−2x+1−4y2−4y−1=(x−1)2−(2y+1)2=(x+2y)(x−2y−2)x2(x−1)−16(x−1)=(x−1)(x2−16)=(x−1)(x+4)(x−4)x2(x−1)−16(x−1)=(x−1)(x2−16)=(x−1)(x+4)(x−4) x2y−x3−9y+9x=x2(y−x)−9(y−x)=(y−x)(x2−9)(y−x)(x−3)(x+3)

x2−y2−2x+2y=(x+y)(x−y)−2(x−y)=(x+y−2)(x−y)x2−y2−2x+2y=(x+y)(x−y)−2(x−y)=(x+y−2)(x−y) 2x+2y−x2−xy=2(x+y)−x(x+y)=(2−x)(x+y)2x+2y−x2−xy=2(x+y)−x(x+y)=(2−x)(x+y)

3a2−6ab+3b2−12c2=3(a2−2ab+b2−4c2)=3(a−b)2−3(2c)2=3(a−b+2c)(a−b−2c)3a2−6ab+3b2−12c2=3(a2−2ab+b2−4c2)=3(a−b)2−3(2c)2=3(a−b+2c)(a−b−2c) x2−25+y2+2xy=(x+y)2−25=(x+y−5)(x+y+5)

TL
x=-4/3,

x=1/2

HT

TL

x=1/2

Hok tốt

a) \(3x^2-3xy-2x+2y\)

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

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

b) \(6x^2+3xy-2ax-ay\)

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

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

a) x^3 - 2x^2y + xy^2
= x.x^2 - x.2xy + y^2
= x.( x^2 - 2xy + y^2) ( đặt nhân tử chung)
= x(x-y)^2 ( hđt số 2)
Vậy : x^3 - 2x^2y + xy^2 = x(x-y)^2

a.x^2-x(x+2)>3x-1
<=> x^2 - x^2 - 2x > 3x - 1
<=> -2x -3x > -1
<=> -5x > -1
<=> x < 0,2
b.x -2/6 - x - 1/3 < x/2
<=> 6x/6 - 2/6 - 6x/6 -2/6 < 3x/6
<=> 6x - 2 - 6x -2 < 3x
<=> 6x - 6x - 3x < 0 + 2 + 2
<=> -3x < 4
<=> x > -4/3

TL :

x²-x-y²-y=(x²-y²)-(x+y)=(x-y)(x+y)-(x+y)=(x+y)(x-y-1)

x²-xy+x-y=x(x-y)+(x-y)=(x+1)(x-y)

ta có :