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Bài 1. Tính:
a) \(x^2\left(x-2x^3\right)\)
\(=x^3-2x^5\)
b) \(\left(x^2+1\right)\left(5-x\right)\)
\(=5x^2-x^3+5-x\)
c. \(\left(x-2\right)\left(x^2+3x-4\right)\)
\(=x^3+3x^2-4x-2x^2-6x+8\)
\(=x^3+x^2-10x+8\)
d) \(\left(x-2\right)\left(x-x^2+4\right)\)
\(=x^2-x^3+4x-2x+2x^2-8\)
\(=3x^2-x^3+2x-8\)
e) \(\left(x^2-1\right)\left(x^2+2x\right)\)
\(=x^4+2x^3-x^2-2x\)
f) \(\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)
\(=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)
\(=\left(6x^2+x-2\right)\left(3-x\right)\)
\(=18x^2+3x-6-6x^3-x^2+2x\)
\(=17x^2+5x-6-6x^3\)
g) \(\left(x+3\right)\left(x^2+3x-5\right)\)
\(=x^3+3x^2-5x+3x^2+9x-15\)
\(=x^3+6x^2+4x-15\)
h) \(\left(xy-2\right)\left(x^3-2x-6\right)\)
\(=x^4y-2x^2y-6xy-2x^3+4x+12\)
i) \(\left(5x^3-x^2+2x-3\right)\left(4x^2-x+2\right)\)
\(=20x^3-5x^4+10x^3-4x^4+x^3-2x^2+8x^3-2x^2+4x-12x^2+3x-6\)
\(=39x^3-9x^4-16x^2+7x-6\)
Bài 5: Tìm x, biết
1) \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)
\(\Leftrightarrow\left(x^2-4x+4\right)-\left(x^2-9\right)-6=0\)
\(\Leftrightarrow x^2-4x+4-x^2+9-6=0\)
\(\Leftrightarrow-4x+7=0\)
\(\Leftrightarrow-4x=-7\)
\(\Leftrightarrow x=\dfrac{-7}{-4}=\dfrac{7}{4}\)
Vậy \(x=\dfrac{7}{4}\)
2) \(4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)
\(\Leftrightarrow4\left(x^2-6x+9\right)-\left(4x^2-1\right)-10=0\)
\(\Leftrightarrow4x^2-24x+36-4x^2+1-10=0\)
\(\Leftrightarrow-24x+27=0\)
\(\Leftrightarrow-24x=-27\)
\(\Leftrightarrow x=\dfrac{-27}{-24}=\dfrac{9}{8}\)
Vậy \(x=\dfrac{9}{8}\)
4) \(\left(x-4\right)^2-\left(x-2\right)\left(x+2\right)=6\)
\(\Leftrightarrow\left(x^2-8x+16\right)-\left(x^2-4\right)-6=0\)
\(\Leftrightarrow x^2-8x+16-x^2+4-6=0\)
\(\Leftrightarrow-8x+14=0\)
\(\Leftrightarrow-8x=-14\)
\(\Leftrightarrow x=\dfrac{-14}{-8}=\dfrac{7}{4}\)
Vậy \(x=\dfrac{7}{4}\)
5) \(9\left(x+1\right)^2-\left(3x-2\right)\left(3x+2\right)=10\)
\(\Leftrightarrow9\left(x^2+2x+1\right)-\left(9x^2-4\right)-10=0\)
\(\Leftrightarrow9x^2+18x+9-9x^2+4-10=0\)
\(\Leftrightarrow18x+3=0\)
\(\Leftrightarrow18x=-3\)
\(\Leftrightarrow x=\dfrac{-3}{18}=\dfrac{-1}{6}\)
Vậy \(x=\dfrac{-1}{6}\)
Bài 2
a. (x-2y)2 =2x-4y
b. (2x^2 +3)2 =4x^2+6
c. (x-2) (x^2+2x+4) = x^3-8 (hằng đẳng thức)
d. (2x-1)3 = 6x-3
Xin lỗi mik chỉ lm ổn bài 2 thôi!
a) \(2x\left(x^2-7x-3\right)=2x.x^2-2x.7x-2x.3=2x^3-14x^2-6x\)
b) \(\left(-2x^3+y^2-7xy\right)4xy^2=\left(-2x^3\right)4xy^2+y^24xy^2-7xy.4xy^2=-8x^4y^2+4xy^4-28x^2y^3\)
c) \(\left(-5x^3\right)\left(2x^2+3x-5\right)=-5x^32x^2-5x^33x-5x^3.-5=-10x^5-15x^4+25x^3\)
d) \(\left(2x^2-xy+y^2\right)\left(-3x^3\right)=-3x^32x^2-3x^3.-xy-3x^3y^2=-6x^5+3x^4y-3x^3y^2\)
e) \(\left(x^2-2x+3\right)\left(x-4\right)=x\left(x^2-2x+3\right)-4\left(x^2-2x+3\right)=x^3-2x^2+3x-4x^2+8x-12=x^3-6x^2+11x-12\)
f) \(\left(2x^3-3x-1\right)\left(5x+2\right)=5x\left(2x^3-3x-1\right)+2\left(2x^3-3x-1\right)=10x^4-15x^2-5x+4x^3-6x-2=10x^4+4x^3-15x^2-11x-2\)
Bạn chú ý đăng lẻ câu hỏi! 1/
a/ \(=x^3-2x^5\)
b/\(=5x^2+5-x^3-x\)
c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)
d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)
e/ \(=x^4-x^2+2x^3-2x\)
f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)
Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5
a) 2x.(x2 - 7x - 3)
= 2xx2 + 2x(-7x) + 2x(-3)
= 2x2x - 2.7xx - 2.3x
= 2x3 - 14x2 - 6x
Bài 12:
1) A = x2 - 6x + 11
= (x2 - 6x + 9) + 2
= (x - 3)2 + 2
Ta có: (x - 3)2 ≥ 0 ∀ x
Dấu ''='' xảy ra khi x - 3 = 0 ⇔ x = 3
Do đó: (x - 3)2 + 2 ≥ 2
Hay A ≥ 2
Dấu ''='' xảy ra khi x = 3
Vậy Min A = 2 tại x = 3
2) B = x2 - 20x + 101
= (x2 - 20x + 100) + 1
= (x - 10)2 + 1
Ta có: (x - 10)2 ≥ 0 ∀ x
Dấu ''='' xảy ra khi x - 10 = 0 ⇔ x = 10
Do đó: (x - 10)2 + 1 ≥ 1
Hay B ≥ 1
Dấu ''='' xảy ra khi x = 10
Vậy Min B = 1 tại x = 10
Câu 1:
a) x2(x - 2x3)
= x3 - 2x5
b) (x2 + 1)(5 - x)
= - x3 + 5x2 - x + 5
c) (x - 2)(x2 + 3x - 4)
= x3 + 3x2 - 4x - 2x2 - 6x + 8
= x3 + x2 - 10x + 8
d) (x - 2)(x - x2 + 4)
= x2 - x3 + 4x - 2x + 2x2 - 8
= -x3 + 3x2 + 2x - 8
e) (x2 - 1)(x2 + 2x)
= x4 + 2x3 - x2 - 2x
f) (2x - 1)(3x + 2)(3 - x)
= (6x2 + 4x - 3x - 2)(3 - x)
= (6x2 + x - 2)(3 - x)
= 18x2 - 6x3 + 3x - x2 - 6 + 2x
= -6x3 + 17x2 + 5x - 6
g) (x + 3)(x2 + 3x - 5)
= x3 + 3x2 - 5x + 3x2 + 9x - 15
= x3 + 6x2 + 4x - 15
h) (xy - 2)(x3 - 2x - 6)
= x4y - 2x2y - 6xy - 2x3 + 4x + 12
i) (5x3 - x2 + 2x - 3)(4x2 - x + 2)
= 20x5 - 5x4 + 10x3 - 4x4 + x3 - 2x2 + 8x3 - 2x2 + 4x - 12x2 + 3x - 6
= 20x5 - 9x4 + 19x3 - 16x2 + 7x - 6
Câu hỏi của Huỳnh Thị Thu Hằng - Toán lớp 8
Bài 2:
a) (x - 2y)2
= x2 - 4xy + 4y2
b) (2x2 + 3)2
= 4x4 + 12x2 + 9
c) (x - 2)(x2 + 2x + 4)
= x3 - 8
d) (2x - 1)3
= 8x3 - 12x2 + 6x - 1
8.
\(5x^2-10xy+5y^2-20z^2\\ =5\left(x^2-2xy+y^2-z^2\right)\\ =5\left(\left(x-y\right)^2-z^2\right)\\ =5\left(x-y-z\right)\left(x-y+z\right)\\ x^2-5x+5y-y^2\\ =\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\\ =\left(x-y\right)\left(x+y-5\right)\\ 3x^2-6xy+3y^2-12z^2\\ =3\left(x^2-2xy+y^2-4z^2\right)\\ =3\left(\left(x-y\right)^2-\left(2z\right)^2\right)\\ =3\left(x-y-2z\right)\left(x-y+2z\right)\\ x^2+4x+3\\ =\left(x^2+x\right)+\left(3x+3\right)\\ =x\left(x+1\right)+3\left(x+1\right)\\ =\left(x+1\right)\left(x+3\right)\\ \left(x^2+1\right)-4x^2\\ =\left(x^2-2x+1\right)\left(x^2+2x+1\right)\\ =\left(x-1\right)^2.\left(x+1\right)^2\\ x^2-4x-5\\ =\left(x^2+x\right)-\left(5x+5\right)\\ =x\left(x+1\right)-5\left(x+1\right)\\ =\left(x+1\right)\left(x-5\right)\)
Bài 3:
(6x + 1)2 + (6x - 1)2 - 2(1 + 6x)(6x - 1)
= (6x + 1 - 6x + 1)2
= 22
= 4
Bài 4:
a) 1012
= (100 + 1)2
= 1002 + 2.100 + 1
= 10000 + 200 + 1
= 10201
b) 97.103
= (100 - 3)(100 + 3)
= 1002 - 32
= 10000 - 9
= 9991
c) 772 + 232 + 77.46
= 772 + 2.77.23 + 232
= (77 + 23)2
= 1002
= 10000
d) 1052 - 52
= (105 - 5)(105 + 5)
= 100.110
= 11000
e) A = (x - y)(x2 + xy + y2) + 2y3
= x3 - y3 + 2y3
= x3 + y3
Bài 5:
1) (x - 2)2 - (x - 3)(x + 3) = 6
⇔ x2 - 4x + 4 - x2 + 9 = 6
⇔ -4x + 13 = 6
⇔ 4x = 7
⇔ x = \(\dfrac{7}{4}\)
2) 4(x - 3)2 - (2x - 1)(2x + 1) = 10
⇔ 4(x2 - 6x + 9) - (4x2 - 1) = 10
⇔ 4x2 - 24x + 36 - 4x2 + 1 = 10
⇔ -24x + 37 = 10
⇔ 24x = 27
⇔ x = \(\dfrac{9}{8}\)
3) (x - 4)2 - (x - 2)(x + 2) = 6
⇔ x2 - 8x + 16 - x2 + 4 = 6
⇔ -8x + 20 = 6
⇔ 8x = 14
⇔ x = \(\dfrac{7}{4}\)
4) 9(x + 1)2 - (3x - 2)(3x + 2) = 10
⇔ 9(x2 + 2x + 1) - (9x2 - 4) = 10
⇔ 9x2 + 18x + 9 - 9x2 + 4 = 10
⇔ 18x + 13 = 10
⇔ 18x = -3
⇔ x = \(\dfrac{-3}{18}\)
Bài 6:
a) 1 - 2y + y2
= 12 - 2.1.y + y2
= (1 - y)2
b) (x + 1)2 - 25
= (x + 1)2 - 52
= (x + 1 + 5)(x + 1 - 5)
= (x + 6)(x - 4)
c) 1 - 4x2
= 12 - (2x)2
= (1 - 2x)(1 + 2x)
d) 8 - 27x3
= 23 - (3x)3
= (2 - 3x)(4 + 6x + 9x2)
e) 27 + 27x + 9x2 + x3
= 33 + 3.32.x + 3.3.x2 + x3
= (3 + x)3
f) 8x3 - 12x2y + 6xy2 - y3
= (2x)3 - 3.(2x)2.y + 3.2x.y2 - y3
= (2x - y)3
g) x3 + 8y3
= x3 + (2y)3
= (x + 2y)(x2 + 2xy + 4y2)
Bài 7:
a) 3x2 - 6x + 9x2
= 3x(x - 2 + 3x)
= 3x(4x - 2)
= 6x(2x - 1)
b) 10x(x - y) - 6y(y - x)
= 10x(x - y) + 6y(x - y)
= (x - y)(10x + 6y)
= 2(x - y)(5x + 3y)
c) 3x2 + 5y - 3xy - 5x
= (3x2 - 3xy) + (5y - 5x)
= 3x(x - y) + 5(y - x)
= 3x(x - y) - 5(x - y)
= (x - y)(3x - 5)
d) 3y2 - 3z2 +3x2 + 6xy
= 3(y2 - z2 + x2 + 2xy)
= 3[(x2 + 2xy + y2) - z2]
= 3[(x + y)2 - z2 ]
= 3(x + y - z)(x + y + z)
e) 16x3 + 54y3
= 2(8x3 + 27y3)
= 2[(2x)3 + (3y)3]
= 2(2x + 3y)(4x2 + 6xy + 9y2)
f) x2 - 25 - 2xy + y2
= (x2 - 2xy + y2) - 25
= (x - y)2 - 52
= (x - y - 5)(x - y + 5)
g) x5 - 3x4 + 3x3 - x2
= x2(x3 - 3x2 + 3x - 1)
= x2(x - 1)3.
Bài 7:
a) 3x2 - 6x + 9x2
= 3x(x - 2 + 3x)
= 3x(4x - 2)
b) 10x(x - y) - 6y(y - x)
= 10x(x - y) + 6y(x - y)
= (x - y)(10x + 6y)
c) 3x2 + 5y - 3xy - 5x
= (3x2 - 3xy) + (5y - 5x)
= 3x(x - y) + 5(y - x)
= 3x(x - y) - 5(x - y)
= (x - y)(3x - 5)
d) 3y2 - 3z2 + 3x2 + 6xy
= 3(y2 - z2 + x2 + 2xy)
= 3[(x + y)2 - z2 ]
= 3(x + y - z)(x + y + z)
e) 16x3 + 54y3
= 2(8x3 + 27y3)
= 2(2x + 3y)(4x2 + 6xy + 9y2)
f) x2 - 25 - 2xy + y2
= (x2 - 2xy + y2) - 25
= (x - y)2 - 52
= (x - y - 5)(x - y + 5)
g) x5 - 3x4 + 3x3 - x2
= x2(x3 - 3x2 + 3x - 1)
= x2(x - 1)3
Bài 8:
1) 5x2 - 10xy + 5y2 - 20z2
= 5(x2 - 2xy + y2 - 4z2)
= 5[(x - y)2 - (2z)2 ]
= 5(x - y - 2z)(x - y + 2z)
3) x2 - 5x + 5y - y2
= (x2 - y2) - (5x - 5y)
= (x - y)(x + y) - 5(x - y)
= (x - y)(x + y - 5)
4) 3x2 - 6xy + 3y2 - 12z2
= 3(x2 - 2xy + y2 - 4z2)
= 3[(x - y)2 - (2z)2 ]
= 3(x - y - 2z)(x - y + 2z)
5) x2 + 4x + 3
= x2 + x + 3x + 3
= x(x + 1) + 3(x + 1)
= (x + 1)(x + 3)
6) (x2 + 1)2 - 4x2
= (x2 + 1)2 - (2x)2
= (x2 - 2x + 1)(x2 + 2x + 1)
= (x - 1)2(x + 1)2
7) x2 - 4x - 5
= x2 + x - 5x - 5
= x(x + 1) - 5(x + 1)
= (x + 1)(x - 5)
Bài 7
1.
\(3x^2-6x+9x^2\\ =12x^2-6x\\ =6x\left(2x-1\right)\\ \)
2.
\(10x\left(x-y\right)-6y\left(x-y\right)\\ =10x\left(x-y\right)+6y\left(x-y\right)\\ =\left(x-y\right)\left(10x+6y\right)\)
3.
\(3x^2+5y-3xy-5x\\ =3x\left(x-y\right)-5\left(x-y\right)\\ =\left(x-y\right)\left(3x-5\right)\)
4.
\(3y^2-3z^2+3x^2+6xy\\ =3\left(x^2+2xy+y^2-z^2\right)\\ =3\left(\left(x+y\right)^2-z^2\right)\\ =3\left(x+y-z\right)\left(x+y+z\right)\)
5.
\(16x^3+54y^3\\ =2\left(8x^3+27y^3\right)\\ =2\left(\left(2x\right)^3+\left(3y\right)^3\right)\\ =2\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\\ =2\left(2x+3y\right)\left(2x-3y\right)^2\)
6.
\(x^2-25-2xy+y^2\\ =\left(x-y\right)^2-25\\ =\left(x-y-5\right)\left(x-y+5\right)\\ x^5-3x^4+3x^3-x^2\\ =x^5-x^4-2x^4+2x^3+x^3-x^2\\ =x^4\left(x-1\right)-2x^3\left(x-1\right)+x^2\left(x-1\right)\\ \\ \left(x-1\right)\left(x^4-2x^3+x^2\right)\)
câu 3 là sao z bạn