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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!
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
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
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 \(x=\dfrac{-1}{6}\)