Lesson 1: analyzing the polynomial factors.
Notes + 2 x-1
x 3 + 6x2 + 11x + 6
x 4 + 2 x 2-3
AB + ac + b2 + 2bc + c2
A3-b3 + c3 + 3abc
Lesson 2: for functions:
search conditions of x to A means.
A shortening.
Computer x to A < 1.
Post 3: prove the inequality:
For a + b + c = 0. Prove that: a3 + b3 + c3 = 3abc.
For a, b, c are the sidelengths of the triangle. Proof that:
Prove that x 5 + y5 ≥ x4y + xy4 with x, y ≠ 0 and x + y ≥ 0
Lesson 4: solve the equation:
x 2-3 x + 2 + | x-1 | = 0
Lesson 5: find the largest and smallest value (if any)
A = x 2-2 x + 5
B =-2 x 2-4 x + 1.
C =
Lesson 6: calculate the value of expression.
Know a – b = 7 feature: A = (a + 1) a2-b2 (b-1) + ab-3ab (a-b + 1)
For three numbers a, b, c is not zero catches up deals for equality:
Computer: P =
Article 7: proof that
8351634 + 8241142 divisible 26.
A = n3 + 6n2-19n-24 divisible by 6.
B = (10n-9n-1) divisible 27 with n in N *.
Article 8:
In the motorcycle race three cars depart at once. The second car in a one-hour run slower than the first car 15 km and 3 km third cars. rapidly should the destination more slowly the first car 12 minutes and the third car earlier today. No stops along the way. Calculate the speed of each car, race distance and the time each car
Những câu hỏi liên quan
Bài 1 : phân tích đa thức thành nhân tử.
3x2 + 2x – 1
x3 + 6x2 + 11x + 6
x4 + 2x2 – 3
ab + ac +b2 + 2bc + c2
a3 – b3 + c3 + 3abc
(1) (a+b+c)2a2+b2+c2+2ab+2bc+2ac(a+b+c)2a2+b2+c2+2ab+2bc+2ac(2) (a+b−c)2a2+b2+c2+2ab−2bc−2ac(a+b−c)2a2+b2+c2+2ab−2bc−2ac(3) (a−b−c)2a2+b2+c2−2ab−2ac+2bc(a−b−c)2a2+b2+c2−2ab−2ac+2bc(4) a3+b3(a+b)3−3ab(a+b)a3+b3(a+b)3−3ab(a+b)(5) a3−b3(a−b)3+3ab(a−b)a3−b3(a−b)3+3ab(a−b)(6) (a+b+c)3a3+b3+c3+3(a+b)(b+c)(c+a)(a+b+c)3a3+b3+c3+3(a+b)(b+c)(c+a)(7) a3+b3+c3−3abc(a+b+c)(a2+b2+c2−ab−bc−ac)a3+b3+c3−3abc(a+b+c)(a2+b2+c2−ab−bc−ac) (8) (a−b)3+(b−c)3+(c−a)33(a−b)(b−c)(c−a)(a−b)3+(b−c)3+(c−a)33(a−b)(b−c)(c−a)(9)...
Đọc tiếp
(1) (a+b+c)2=a2+b2+c2+2ab+2bc+2ac(a+b+c)2=a2+b2+c2+2ab+2bc+2ac
(2) (a+b−c)2=a2+b2+c2+2ab−2bc−2ac(a+b−c)2=a2+b2+c2+2ab−2bc−2ac
(3) (a−b−c)2=a2+b2+c2−2ab−2ac+2bc(a−b−c)2=a2+b2+c2−2ab−2ac+2bc
(4) a3+b3=(a+b)3−3ab(a+b)a3+b3=(a+b)3−3ab(a+b)
(5) a3−b3=(a−b)3+3ab(a−b)a3−b3=(a−b)3+3ab(a−b)
(6) (a+b+c)3=a3+b3+c3+3(a+b)(b+c)(c+a)(a+b+c)3=a3+b3+c3+3(a+b)(b+c)(c+a)
(7) a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ac)a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ac)
(8) (a−b)3+(b−c)3+(c−a)3=3(a−b)(b−c)(c−a)(a−b)3+(b−c)3+(c−a)3=3(a−b)(b−c)(c−a)
(9) (a+b)(b+c)(c+a)−8abc=a(b−c)2+b(c−a)2+c(a−b)2(a+b)(b+c)(c+a)−8abc=a(b−c)2+b(c−a)2+c(a−b)2
(10) (a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)−abc(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)−abc
(11) ab2+bc2+ca2−a2b−b2c−c2a=(a−b)3+(b−c)3+(c−a)33ab2+bc2+ca2−a2b−b2c−c2a=(a−b)3+(b−c)3+(c−a)33
(12)ab3+bc3+ca3−a3b−b3c−c3a=(a+b+c)[(a−b)3+(b−c)3+(c−a)3]3ab3+bc3+ca3−a3b−b3c−c3a=(a+b+c)[(a−b)3+(b−c)3+(c−a)3]3
Chứng minh giùm mik hằng đẳng thức kia vs
Phân tích đa thức thành nhân tử (đừng để ý đến cái tên nha các bạn)
x3 + 6x2 + 11x + 6x4 + 2x2 – 3ab + ac +b2 + 2bc + c2a3 – b3 + c3 + 3ab
a]
x^3 + 6x^2 + 11x + 6
= x^3 + x^2 + 5x^2 + 5x + 6x + 6
= x^2(x + 1) + 5x(x + 1) + 6(x + 1)
= (x + 1)(x^2 + 5x + 6)
= (x + 1)(x^2 + 2x + 3x + 6)
= (x + 1)[x(x + 2) + 3(x + 2)
= (x + 1)(x + 2)(x + 3)
b thiếu đề bài nè x^4 - 2x^2 - 3 = 0
(x^2)(x^2)-2(x^2)-3=0
(x^2)(x^2)-3(x^2)+1(x^2)-3=0
(x^2)(x^2-3) + 1(x^2-3) = 0
(x^2-3) (x^2+1)=0
c bó tay
d (a3 + b3 + c3) - 3abc
= ( (a+b+c)3 - 3ab(a+b) -3bc(b+c) -3ac(a+c) - 6abc) - 3abc
= (a+b+c)3 - 3ab(a+b) - 3bc(b+c) - 3ac(a+c) - 9abc
= (a+b+c)3
- 3ab(a+b) - 3abc
- 3bc(b+c) - 3abc
- 3ac(a+c) - 3abc
= (a+b+c)3
- 3ab(a+b+c)
- 3bc(a+b+c)
- 3ac(a+b+c)
= (a+b+c)( (a+b+c)2 - 3ab -3bc 3ac)
=(a+b+c)( a2 + b2 + c2 + 2ab +2bc + 2ca -3ab - 3bc -3ac)
=(a+b+c) (a2 + b2 + c2 - ab - bc -ac)
ý d hình như đề sai
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chị ui kết bạn với em đi em hết lượt kết bạn rùi
em học lớp 5
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1. x3 + 6x2 + 11x + 6
= x3 + 3x2 + 3x2 + 9x + 2x + 6
= x2(x + 3) + 3x(x + 3) + 2(x + 3)
= (x + 3)(x2 + 3x + 2)
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Xem thêm câu trả lời
phân tích đa thức thành nhân tử.
1) 3x^2 + 2x – 1
2) x^3 + 6x^2 + 11x + 6
3) x^4 + 2x^2 – 3
4) ab + ac +b^2 + 2bc + c^2
5) a^3 – b^3 + c^3 + 3abc
hộ mk cái
1) \(3x^2+2x-1\)
\(=3x^2+3x-x-1\)
\(=3x\left(x+1\right)-\left(x+1\right)\)
\(=\left(x+1\right)\left(3x-1\right)\)
2) \(x^3+6x^2+11x+6\)
\(=x^3+x^2+5x^2+5x+6x+6\)
\(=x^2\left(x+1\right)+5x\left(x+1\right)+6\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+5x+6\right)\)
\(=\left(x+1\right)\left(x+2x+3x+6\right)\)
\(=\left(x+1\right)\left(x+2\right)\left(x+3\right)\)
3) \(x^4+2x^2-3\)
\(=\left(x^2+1\right)^2-4\)
\(=\left(x^2+1-2\right)\left(x^2+1+2\right)\)
\(=\left(x^2-1\right)\left(x^2+3\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+3\right)\)
4) \(ab+ac+b^2+2bc+c^2\)
\(=a\left(b+c\right)+\left(b+c\right)^2\)
\(=\left(b+c\right)\left(a+b+c\right)\)
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1, \(3x^2+2x-1\)
\(=3x^2+3x-x-1\)
\(=3x\left(x+1\right)-\left(x+1\right)\)
\(=\left(x+1\right)\left(3x-1\right)\)
2, \(x^3+6x^2+11x+6\)
\(=\left(x^3+3x^2\right)+\left(3x^2+9x\right)+\left(2x+6\right)\)
\(=x^2\left(x+3\right)+3x\left(x+3\right)+2\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2+3x+2\right)\)
\(=\left(x+3\right)\left(x+1\right)\left(x+2\right)\)
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2. Chứng minh rằng:
a. a3+ b3 = (a + b)3 - 3ab (a + b)
b. a3+ b3 + c3 - 3abc = (a + b + c) (a2 + b2 c2 - ab - bc - ca)
a )
`VP= (a+b)^3-3ab(a+b)`
`=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2`
`=a^3+b^3 =VT (đpcm)`
b)
b) Ta có
`VT=a3+b3+c3−3abc`
`=(a+b)3−3ab(a+b)+c3−3abc`
`=[(a+b)3+c3]−3ab(a+b+c)`
`=(a+b+c)[(a+b)2+c2−c(a+b)]−3ab(a+b+c)`
`=(a+b+c)(a2+b2+2ab+c2−ac−bc−3ab)`
`=(a+b+c)(a2+b2+c2−ab−bc−ca)=VP`
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a) Ta có:
`VP= (a+b)^3-3ab(a+b)`
`=a^3 + b^3+3ab ( a + b )- 3ab ( a + b )`
`=a^3 + b^3=VT(dpcm)`
b) Ta có
`VT=a^3+b^3+c^3−3abc`
`=(a+b)^3−3ab(a+b)+c^3−3abc`
`=[(a+b)^3+c^3]−3ab(a+b+c)`
`=(a+b+c)[(a+b)^2+c^2−c(a+b)]−3ab(a+b+c)`
`=(a+b+c)(a^2+b^2+2ab+c^2−ac−bc−3ab)`
`=(a+b+c)(a^2+b^2+c^2−ab−bc−ca)=VP`
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4x2-1
X(x+y)-6x-6y
X2-2xy+y2-z2
A2+2+2a+2ab+b2-ac-bc
9x2-1 phần 4
X2-2x-4y2-4y
9(x-y)2-4(x+y)2
(3x-2y)2-(2x-3y)2
9(x-y)2-4(x+y)2
\(4x^2-1=\left(2x-1\right)\left(2x+1\right)\)
\(x\left(x+y\right)-6x-6y=\left(x+y\right)\left(x-6\right)\)
\(x^2-2xy+y^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
\(9x^2-\dfrac{1}{4}=\left(3x-\dfrac{1}{2}\right)\left(3x+\dfrac{1}{2}\right)\)
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Suppose that the polynomial f(x) = x5 - x4 - 4x3 + 2x2 + 4x + 1 has 5 solutions x1; x2; x3; x4; x5. The other polynomial k(x) = x2 - 4. Find the value of P = k(x1) x k(x2) x k(x3) x k(x4) x k(x5)
thầy mình bảo phân tích cách này thành nhân tử rồi nhớ nghiệm và máy tính mà bấm chứ chắc cái này cao siêu quá chưa đến lượt bọn mình giải đâu
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help
Question 7:
Bottle A contains 15% syrup. Bottle B contains 40% syrup. When these 2 bottles of syrup are mixed, the syrup content is 30% and the total volume is 600ml. How much syrup is in the bottle A at first?
Let ABCD be a trapezoid with bases AB, CD and O be the intersection of AC and BD. If the areas of triangle OAB, triangle OCD are 16cm2, 40cm2respectively and M is the midpoint of BD, then the area of the triangle AMD is .........cm2.
Question 14:
Suppose that the polynomial...
Đọc tiếp
help
Question 7:Bottle A contains 15% syrup. Bottle B contains 40% syrup. When these 2 bottles of syrup are mixed, the syrup content is 30% and the total volume is 600ml. How much syrup is in the bottle A at first?
Let ABCD be a trapezoid with bases AB, CD and O be the intersection of AC and BD. If the areas of triangle OAB, triangle OCD are 16cm2, 40cm2respectively and M is the midpoint of BD, then the area of the triangle AMD is .........cm2.
Question 14:Suppose that the polynomial f(x) = x5 - x4 - 4x3 + 2x2 + 4x + 1 has 5 solutions x1; x2; x3; x4; x5. The other polynomial k(x) = x2 - 4.
Find the value of P = k(x1) x k(x2) x k(x3) x k(x4) x k(x5)
CMR :1,a2+b2=<a+b>2-2ab
2,a3+b3=<a+b>3-3ab.<a+b>
3,a3-b3=<a-b>3+3ab.<a+b>
Cho :a+b=1
Tính :A=a3+b3+3ab
2
Ta có:
VP=(a+b)3−3ab(a+b)VP=(a+b)3-3ab(a+b)
=a3+b3+3ab(a+b)−3ab(a+b)=a3+b3+3ab(a+b)-3ab(a+b)
=a3+b3=VT(dpcm)
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1, \(VT=a^2+b^2=a^2+b^2+2ab-2ab=\left(a+b\right)^2-2ab=VP\left(đpcm\right)\)
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