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
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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
chị ui kết bạn với em đi em hết lượt kết bạn rùi
em học lớp 5
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) \(ab+ac+b^2+2bc+c^2\)
\(=a\left(b+c\right)+\left(b+c\right)^2\)
\(=\left(b+c\right)\left(a+b+c\right)\)
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, \(x^3+6x^2+11x+6\)
\(=\left(x^3+3x^2\right)+\left(3x^2+9x\right)+\left(2x+6\right)\)
\(=x^2\left(x+3\right)+3x\left(x+3\right)+2\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2+3x+2\right)\)
\(=\left(x+3\right)\left(x+1\right)\left(x+2\right)\)
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`
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`
\(4x^2-1=\left(2x-1\right)\left(2x+1\right)\)
\(x\left(x+y\right)-6x-6y=\left(x+y\right)\left(x-6\right)\)
\(x^2-2xy+y^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
\(9x^2-\dfrac{1}{4}=\left(3x-\dfrac{1}{2}\right)\left(3x+\dfrac{1}{2}\right)\)
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)
1, \(VT=a^2+b^2=a^2+b^2+2ab-2ab=\left(a+b\right)^2-2ab=VP\left(đpcm\right)\)
