Cho đa thức f(x)= -3x^4 - 5x^2 + 13x^4 - 7x + 5x^3 - 10-x^2 + 7x - 2
Chứng tỏ rằng f(-1) + f(1) + 14 = 0
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Q
5 tháng 7 2017
Mấy bài kia phá tung tóe rồi rút gọn hết sức xong thay x vào, làm câu c thôi nhé:
c) \(C=x^{14}-10x^{13}+10x^{12}-10x^{11}+...+10x^2-10x+10\)
riêng câu này ta thay x = 9 vào luôn, vậy ta có:
\(C=9^{14}-10\cdot9^{13}+10\cdot9^{12}-10\cdot9^{11}+...+10\cdot9^2-10\cdot9+10\)
\(=9^{14}-\left(9+1\right)\cdot9^{13}+\left(9+1\right)\cdot9^{12}-\left(9+1\right)\cdot9^{11}+...+\left(9+1\right)\cdot9^2-\left(9+1\right)\cdot9+10\)
\(=9^{14}-9^{14}-9^{13}+9^{13}+9^{12}-9^{12}-9^{11}+...+9^3+9^2-9^2-9+10\)
\(=-9+10\)
\(=1\)
7 tháng 7 2018
a)f(x)+g(x)=\(x^5-4x^4-2x^2-7-2x^5+6x^4-2x^2+6.\)
=\(-x^5+2x^4-4x^2-1\)
f(x)-g(x)=\(x^5-4x^4-2x^2-7+2x^5-6x^4+2x^2-6\)
=\(3x^5-10x^4-13\)
b)f(x)+g(x)=\(5x^4+7x^3-6x^2+3x-7-4x^4+2x^3-5x^2+4x+5\)
=\(x^4+9x^3-11x^2+7x-2\)
f(x)-g(x)=\(5x^4+7x^3-6x^2+3x-7+4x^4-2x^3+5x^2-4x-5\)
=\(9x^4+5x^3-x^2-x-12\)
AK
7 tháng 7 2018
a )
\(f\left(x\right)+g\left(x\right)=x^5-4x^4-2x^2-7+-2x^5+6x^4-2x^2+6\)
\(\Rightarrow f\left(x\right)+g\left(x\right)=\left(x^5-2x^5\right)+\left(6x^4-4x^4\right)-\left(2x^2+2x^2\right)+\left(6-7\right)\)
\(\Rightarrow f\left(x\right)+g\left(x\right)=-x^5+2x^4-4x^2-1\)
\(f\left(x\right)-g\left(x\right)=x^5-4x^4-2x^2-7-\left(-2x^5+6x^4-2x^2+6\right)\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=x^5-4x^4-2x^2-7+2x^5-6x^4+2x^2-6\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=\left(x^5+2x^5\right)-\left(4x^4+6x^4\right)+\left(2x^2-2x^2\right)-\left(6+7\right)\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=3x^5-10x^4-13\)
XO
4 tháng 7 2021
1) x2 -7x + 10 = x2 - 2x - 5x + 10 = x(x - 2) - 5(x - 2) = (x - 5)(x - 2)
2) x2 + 3x + 2 = x2 + 2x + x + 2 = x(x + 2) + (x + 2) = (x + 1)(x + 2)
3) x2 - 7x + 12 = x2 - 3x - 4x + 12 = x(x - 3) - 4(x - 3) = (x - 3)(x - 4)
4) x2 + 7x + 12 = x2 + 3x + 4x + 12 = x(x + 3) + 4(x + 3) = (x + 3)(x + 4)
5) 16x - 5x2 - 3 = 15x - 5x2 + x - 3 = -5x(x - 3) + (x - 3) = (x - 3)(1 - 5x)
6) 6x2 + 7x - 3 = 6x2 - 2x + 9x - 3 = 2x(3x - 1) + 3(3x - 1) = (2x + 3)(3x - 1)
7) 3x2 - 3x - 6 = 3x2 - 6x + 3x - 6 = 3x(x - 2) + 3(x - 2) = (x - 2)(3x + 3) = 3(x - 2)(x + 1)
8) 3x2 + 3x - 6 = 3x2 - 3x + 6x - 6 = 3x(x - 1) + 6(x - 1) = (x - 1)(3x + 6) = 3(x - 1)(x + 2)
9) 6x2 - 13x + 6 = 6x2 - 9x - 4x + 6 = 3x(2x - 3) - 2(2x - 3) = (3x - 2)(2x - 3)
10) 6x2 + 15x + 6 = 6x2 + 12x + 3x + 6 = 6x(x + 2) + 3(x + 2) = (x + 2)(6x + 3) = 3(x + 2)(3x + 1)
11) 6x2 - 20x + 6 = 6x2 - 18x - 2x + 6 = 6x(x -3) - 2(x - 3) = (6x - 2)(x - 3) = 2(3x - 1)(x - 3)
12) 8x2 + 5x - 3 = 8x2 + 8x - 3x - 3 = 8x(x + 1) - 3(x + 1) = (x + 1)(8x - 3)
14 tháng 8 2021
Bài 1:
Để \(F\left(x\right)=G\left(x\right)\) thì \(3x^2-8x+4=3x+4\)
\(\Leftrightarrow3x^2-11x=0\)
\(\Leftrightarrow x\left(3x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{11}{3}\end{matrix}\right.\)
29 tháng 8 2023
a: x^3-7x-6
=x^3-x-6x-6
=x(x-1)(x+1)-6(x+1)
=(x+1)(x^2-x-6)
=(x-3)(x+2)(x+1)
b: =2x^3+x^2-2x^2-x+6x+3
=x^2(2x+1)-x(2x+1)+3(2x+1)
=(2x+1)(x^2-x+3)
c: =2x^3-3x^2-2x^2+3x+2x-3
=x^2(2x-3)-x(2x-3)+(2x-3)
=(2x-3)(x^2-x+1)
d: =2x^3+x^2+2x^2+x+2x+1
=(2x+1)(x^2+x+1)
e: =3x^3+x^2-3x^2-x+6x+2
=(3x+1)(x^2-x+2)
f: =27x^3-9x^2-18x^2+6x+12x-4
=(3x-1)(9x^2-6x+4)
29 tháng 8 2023
a) \(x^3-7x-6\)
\(=x^3-x-6x-6\)
\(=\left(x^3-x\right)-\left(6x+6\right)\)
\(=x\left(x^2-1\right)-6\left(x+1\right)\)
\(=x\left(x+1\right)\left(x-1\right)-6\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x-6\right)\)
b) \(2x^3-x^2+5x+3\)
\(=2x^3+x^2-2x^2-x+6x+3\)
\(=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)
\(=x^2\left(2x+1\right)-x\left(2x+1\right)+3\left(2x+1\right)\)
\(=\left(x^2-x+3\right)\left(2x+1\right)\)
c) \(2x^3-5x^2+5x+1\)
\(=2x^3-3x^2-2x^2+3x+2x-3\)
\(=\left(2x^3-3x^2\right)-\left(2x^2-3x\right)+\left(2x-3\right)\)
\(=x^2\left(2x-3\right)-x\left(2x-3\right)+\left(2x-3\right)\)
\(=\left(x^2-x+1\right)\left(2x-3\right)\)
d) \(2x^3+3x^2+3x+1\)
\(=2x^3+x^2+2x^2+x+2x+1\)
\(=\left(2x^3+x^2\right)+\left(2x^2+x\right)+\left(2x+1\right)\)
\(=x^2\left(2x+1\right)+x\left(2x+1\right)+\left(2x+1\right)\)
\(=\left(2x+1\right)\left(x^2+x+1\right)\)
e) \(3x^3-2x^2+5x+2\)
\(=3x^3+x^2-3x^2-x+6x+2\)
\(=\left(3x^3+x^2\right)-\left(3x^2+x\right)+\left(6x+2\right)\)
\(=x^2\left(3x+1\right)-x\left(3x+1\right)+2\left(3x+1\right)\)
\(=\left(3x-1\right)\left(x^2-x+2\right)\)
f) \(27x^3-27x^2+18x-4\)
\(=27x^3-9x^2-18x^2+6x+12x-4\)
\(=\left(27x^3-9x^2\right)-\left(18x^2-6x\right)+\left(12x-4\right)\)
\(=9x^2\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)\)
\(=\left(3x-1\right)\left(9x^2-6x+4\right)\)
HH
14 tháng 8 2017
My Nguyễn ơi,bạn truy cập vào đường link này để tìm câu hỏi tương tự của câu a/Bài 1 nhé
https://vn.answers.yahoo.com/question/index?qid=20110206184834AAokV5m&sort=N
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
4 tháng 9 2023
a,A(\(x\)) = 13\(x^4\) + 3\(x^2\) + 15\(x\) - 8\(x\) - 7 - 7\(x\) + 7\(x^2\) - 10\(x^4\)
A(\(x\)) = (13\(x^4\) - 10\(x^4\)) + (3\(x^2\) + 7\(x^2\)) + (15\(x\) - 8\(x\) - 7\(x\)) - 7
A(\(x\)) = 3\(x^4\) + 10\(x^2\) + 0 - 7
A(\(x\)) = 3\(x^4\) + 10\(x^2\) - 7
B(\(x\)) = -4\(x^4\) - 10\(x^2\) + 10 + 5\(x^4\) - 3\(x\) - 18 + 30 - 5\(x^2\)
B(\(x\)) = (-4\(x^4\) + 5\(x^4\)) - (10\(x^2\) + 5\(x^2\)) - 3\(x\) + (10 + 30 - 18)
B(\(x\)) = \(x^4\) - 15\(x^2\) - 3\(x\) + 22
b,C(\(x\)) = A(\(x\)) + B(\(x\)) = 3\(x^4\) + 10\(x^2\) - 7 + \(x^4\) - 15\(x^2\) - 3\(x\) + 22
C(\(x\)) = 4\(x^4\) - (15\(x^2\) - 10\(x^2\)) - 3\(x\) + 22
C(\(x\)) = 4\(x^4\) - 5\(x^2\) - 3\(x\) + 15
c, D(\(x\)) = B(\(x\)) - A(\(x\)) = \(x^4\) - 15\(x^2\) - 3\(x\) + 22 - 3\(x^4\) - 10\(x^2\) + 7
D(\(x\)) = (\(x^4\) - 3\(x^4\)) - (15\(x^2\) + 10\(x^2\)) + (22 + 7)
D(\(x\)) = - 2\(x^4\) - 25\(x^2\) + 29
d, Thay \(x\) = 1 vào C(\(x\)) ta có: C(1) = 4.14 - 5.12 -3.1 + 15 = 11 (xem lại đề bài em nhá)
