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a: \(f\left(x\right)+g\left(x\right)=x^3-x^2+5-2x^3+x^2+2x+1=2x+6\)
\(f\left(x\right)-g\left(x\right)=x^3-x^2+5+2x^3-x^2-2x-1\)
\(=3x^3-2x^2-2x+4\)
b: \(h\left(x\right)=2\cdot f\left(x\right)-g\left(x\right)\)
\(=2\left(x^3-x^2+5\right)+2x^3-x^2-2x-1\)
\(=2x^3-2x^2+10+2x^3-x^2-2x-1\)
\(=4x^3-3x^2-2x+9\)
Câu 1 :
\(\left(x-2\right)^2=x^2-4x+4\)
Câu 2:
\(2x^2\left(4x-5x^3\right)+10x^5-5x^3\)
\(=8x^3-10x^5+10x^5-5x^3\)
\(=3x^3\)
\(\left(x-2\right)\left(x^2-2x+4\right)+\left(x-4\right)\left(x-2\right)\)
\(=x^3-4x^2+8x-8+x^2-6x+8\)
\(=x^3-3x^2+2x\)
Còn lại tự làm nha dài lắm
T(x) = f(x) + g(x) = 5x2 - 2x + 3 (1)
H(x) = f(x) - g(X) = x2 - 2x + 5 (2)
Lấy (1) cộng (2) theo vế ta có
f(x) + g(x) + f(x) - g(x) = 5x2 - 2x + 3 + x2 - 2x + 5
=> 2.f(x) = 6x2 - 4x + 8
=> f(x) = 3x2 - 2x + 4
Thay f(x) vào (1) ta có
f(x) + g(x) = 5x2 - 2x + 3
=> (3x2 - 2x + 4) + g(x) = 5x2 - 2x + 3
=> g(x) = 5x2 - 2x + 3 - 3x2 + 2x - 4
=> g(x) = 2x2 - 1
Vậy f(x) = 3x2 - 2x + 4 ; g(x) = 2x2 - 1
c) thay x=1 vào đa thức f(x) ta có: f(1)=4.1^3-1^2+2.1-5
=4-2+2-5
=- 1
vậy 1 k phải là nghiệm của đa thức f(x)
MÌNH CHỈ LÀM ĐƯỢC C THÔI HOK TỐT
làm sai nha chỗ nào là 1 thì thay bằng -1 nha kq sẽ ra nha
a) Ta có:
\(f\left(x\right)+g\left(x\right)=\left(2x^3-x^2+5\right)+\left(x^2+2x-2x^3-1\right)\)
\(f\left(x\right)+g\left(x\right)=2x^3-x^2+5+x^2+2x-2x^3-1\)
\(f\left(x\right)+g\left(x\right)=2x-4\)
\(f\left(x\right)+g\left(x\right)=2\left(x-2\right)\)
Ta có:
\(f\left(x\right)-g\left(x\right)=\left(2x^3-x^2+5\right)-\left(x^2+2x-2x^3-1\right)\)
\(f\left(x\right)-g\left(x\right)=2x^3-x^2+5-x^2-2x+2x^3+1\)
\(f\left(x\right)-g\left(x\right)=4x^3-2x+6\)
b)
\(f\left(0\right)=2.0^3-0^2+5\)
\(f\left(0\right)=5\)
\(f\left(\dfrac{1}{2}\right)=2.\left(\dfrac{1}{2}\right)^3-\left(\dfrac{1}{2}\right)^2+5\)
\(f\left(\dfrac{1}{2}\right)=2.\dfrac{1}{8}-\dfrac{1}{4}+5\)
\(f\left(\dfrac{1}{2}\right)=\dfrac{1}{4}-\dfrac{1}{4}+5\)
\(f\left(\dfrac{1}{2}\right)=5\)
\(f\left(-5\right)=2.\left(-5\right)^3-\left(-5\right)^2+5\)
\(f\left(-5\right)=2.\left(-125\right)-25+5\)
\(f\left(-5\right)=-250-25+5\)
\(f\left(-5\right)=-270\)
c) Ta có:
\(f\left(x\right)+g\left(x\right)=0\)
\(\Leftrightarrow2\left(x-2\right)=0\)
\(\Rightarrow x-2=0\)
\(\Rightarrow x=2\)
Vậy nghiệm cùa f(x) + g(x) là 2
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)=\(5x^4+7x^3-6x^2+3x-7+4x^4-2x^3+5x^2-4x-5\)
=\(9x^4+5x^3-x^2-x-12\)
a )
\(f\left(x\right)+g\left(x\right)=x^5-4x^4-2x^2-7+-2x^5+6x^4-2x^2+6\)
\(\Rightarrow f\left(x\right)+g\left(x\right)=\left(x^5-2x^5\right)+\left(6x^4-4x^4\right)-\left(2x^2+2x^2\right)+\left(6-7\right)\)
\(\Rightarrow f\left(x\right)+g\left(x\right)=-x^5+2x^4-4x^2-1\)
\(f\left(x\right)-g\left(x\right)=x^5-4x^4-2x^2-7-\left(-2x^5+6x^4-2x^2+6\right)\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=x^5-4x^4-2x^2-7+2x^5-6x^4+2x^2-6\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=\left(x^5+2x^5\right)-\left(4x^4+6x^4\right)+\left(2x^2-2x^2\right)-\left(6+7\right)\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=3x^5-10x^4-13\)
Bài 1:
\(A=x^2y-y+xy^2-x=\left(x^2y+xy^2\right)-\left(x+y\right)\\ =xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\)
Voqis x=-1;y=3 ta có:
\(A=\left(-1+3\right)\left(-1\cdot3-1\right)=2\cdot\left(-4\right)=-8\)
b) \(B=x^2y^2+xy+x^3+y^3=\left(x^2y^2+x^3\right)+\left(xy+y^3\right)\\ =x^2\left(y^2+x\right)+y\left(x+y^2\right)=\left(x+y^2\right)\left(x^2+y\right)\)
Với x=-1;y=3 ta có:
\(B=\left(-1+3^2\right)\left(-1^2+3\right)=8\cdot2=16\)
c) \(C=2x+xy^2-x^2y-2y=\left(2x-2y\right)+\left(xy^2-x^2y\right)\\ =2\left(x-y\right)+xy\left(y-x\right)=\left(x-y\right)\left(2-xy\right)\)
Với x=-1;y=3 ta có:
\(C=\left(-1-3\right)\left(2-\left(-1\right)\cdot3\right)=-4\cdot5=-20\)
d) phân tích tt
a. \(f\left(x\right)+g\left(x\right)=\left(x^3-x^2+5\right)+\left(-2x^3+x^2+2x+1\right)\)
\(=x^3-x^2+5-2x^3+x^2+2x+1\)
\(=-x^3+2x+6\)
\(f\left(x\right)-g\left(x\right)=\left(x^3-x^2+5\right)-\left(-2x^3+x^2+2x+1\right)\)
\(=x^3-x^2+5+2x^3-x^2-2x-1\)
\(=3x^3-2x^2-2x+4\)
b. Ta co
\(2f\left(x\right)=2.\left(x^3-x^2+5\right)=2x^3-2x^2+10\)
\(2f\left(x\right)-g\left(x\right)=2x^3-2x^2+10-\left(-2x^3+x^2+2x+1\right)\)
\(=2x^3-2x^2+10+2x^3-x^2-2x-1\)
\(=4x^3-3x^2-2x+9\)
tick nha