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Thu gọn, sắp xếp đa thức theo lũy thừa giảm của biến:
* Ta có: f(x) = x5 – 3x2 + x3 – x2 – 2x + 5
= x5 – (3x2 + x2 ) + x3 - 2x + 5
= x5 – 4x2 + x3 – 2x + 5
= x5 + x3 – 4x2 – 2x + 5
Và g(x) = x2 – 3x + 1 + x2 – x4 + x5
= (x2 + x2 ) – 3x + 1 – x4 + x5
= 2x2 – 3x + 1 – x4 + x5
= x5 – x4 + 2x2 – 3x + 1
* f(x) + g(x):
a: \(F\left(x\right)=x^5-3x^2+x^3-x^2-2x+5\)
\(=x^5+x^3-4x^2-2x+5\)
\(G\left(x\right)=x^5-x^4+x^2-3x+x^2+1\)
\(=x^5-x^4+2x^2-3x+1\)
b: Ta có: \(H\left(x\right)=F\left(x\right)+G\left(x\right)\)
\(=x^5+x^3-4x^2-2x+5+x^5-x^4+2x^2-3x+1\)
\(=2x^5-x^4+x^3-2x^2-5x+6\)
Thu gọn, sắp xếp đa thức theo lũy thừa giảm của biến:
* Ta có: f(x) = x5 – 3x2 + x3 – x2 – 2x + 5 = x5 + x3 – 4x2 – 2x + 5
g(x) = x2 – 3x + 1 + x2 – x4 + x5 = x5 – x4 + 2x2 – 3x + 1
* f(x) + g(x):
Thu gọn, sắp xếp đa thức theo lũy thừa giảm của biến:
* Ta có: f(x) = x5 – 3x2 + x3 – x2 – 2x + 5 = x5 + x3 – 4x2 – 2x + 5
g(x) = x2 – 3x + 1 + x2 – x4 + x5 = x5 – x4 + 2x2 – 3x + 1
* f(x) + g(x): tự làm nha bạn
câu 4: b, đề bài là tính giá trị của A tại x =-1/2;y=-1
Tk
Bài 2
a) F(x)-G(x)+H(x)= \(x^3-2x^2+3x+1-\left(x^3+x-1\right)+\left(2x^2-1\right)\)
= \(x^3-2x^2+3x+1-x^3-x+1+2x^2-1\)
= \(x^3-x^3-2x^2+2x^2+3x-x+1+1-1\)
= 2x + 1
b) 2x + 1 = 0
2x = -1
x=\(\dfrac{-1}{2}\)
* Ta có:
f(x) = x5 – 3x2 + 7x4 – 9x3 + x2 - 1/4 x
= x5 – (3x2 – x2) + 7x4 – 9x3 -1/4.x
= x5 – 2x2 + 7x4 – 9x3 -1/4.x
= x5 + 7x4 – 9x3 – 2x2 - 1/4
g(x) = 5x4 – x5 + x2 – 2x3 + 3x2 - 1/4
= 5x4 –x5+ (x2 + 3x2) – 2x3 – 1/4
= 5x4 – x5 + 4x2 – 2x3 – 1/4
= -x5 + 5x4 – 2x3 + 4x2 - 1/4
* f(x) + g(x)
* f(x) - g(x)
Có \(f\left(x\right)=3x-2x+1\) và \(g\left(x\right)=2x2-3x+x-3\)
a) \(f\left(x\right)+g\left(x\right)=\left(3x-2x+1\right)+\left(2x2-3x+x-3\right)\)
\(f\left(x\right)+g\left(x\right)=\left(x+1\right)+\left(4x-3x+x-3\right)=\left(x+1\right)+\left(2x-3\right)\)
\(f\left(x\right)+g\left(x\right)=x+1+2x-3=\left(2x+x\right)\left(3-1\right)=3x-2\)
\(f\left(x\right)-g\left(x\right)=\left(3x-2x+1\right)-\left(2x2-3x+x-3\right)\)
\(f\left(x\right)-g\left(x\right)=\left(x+1\right)-\left(4x-3x+x-3\right)=\left(x+1\right)-\left(2x-3\right)\)
\(f\left(x\right)-g\left(x\right)=x+1-2x+3=\left(1+3\right)-\left(2x-x\right)=4-x\)
b) Có \(f\left(-1\right)=3\left(-1\right)-2\left(-1\right)+1=\left(-3\right)-\left(-2\right)+1=0\)
Và \(g\left(-1\right)=2.2\left(-1\right)-3\left(-1\right)+\left(-1\right)-3=\left(-4\right)-\left(-3\right)-1-3=-5\)
\(x=-1\Leftrightarrow\left[f\left(x\right)+g\left(x\right)\right]=0+\left(-5\right)=0-5=-5\)
Có \(f\left(-2\right)=3\left(-2\right)-2\left(-2\right)+1=\left(-6\right)-\left(-4\right)+1=-1\)
\(g\left(-2\right)=2.2\left(-2\right)-3\left(-2\right)+\left(-2\right)-3=\left(-8\right)-\left(-6\right)-2-3=-3\)
\(x=-2\Leftrightarrow\left[f\left(x\right)+g\left(x\right)\right]=\left(-1\right)+\left(-3\right)=-4\)
Cho f(x) = x3 - 2x + 1
g(x) = 2x2 - x3 + x - 3
a) f(x) + g(x) =3x -2
;f(x) - g(x) = 6x
b) Tính f(x) + g(x) tại x = - 1; x = - 2
\(#HaimeeOkk\)
\(a)\)
\(f ( x ) + g ( x ) = ( x ^3 − 2 x + 1 ) + ( 2 x ^2 − x ^3 + x − 3 ) \)
\(f ( x ) + g ( x ) = x ^3 − 2 x + 1 + 2 x ^2 − x ^3 + x − 3 \)
\(f ( x ) + g ( x ) = x ^3 − x ^3 + 2 x ^2 − 2 x + x + 1 − 3 \)
\(f ( x ) + g ( x ) = 2 x ^2 − x − 2\)
\(f ( x ) − g ( x ) = ( x ^3 − 2 x + 1 ) − ( 2 x ^2 − x ^3 + x − 3 ) \)
\(f ( x ) − g ( x ) =x^3- 2 x + 1 −2x^2+x^3-x+3\)
\(f ( x ) − g ( x ) = x ^3 + x ^3 − 2 x ^2 − 2 x − x + 1 + 3 \)
\(f ( x ) − g ( x ) = 2 x ^3 − 2 x ^2 − 3 x + 4\)
\(-----------------------------\)
\(b)\)
Thay \(x=-1\) vào \(f ( x ) + g ( x )\)
\(f ( x ) + g ( x ) = 2 x ^2 − x − 2\)
\(⇒ 2 ( − 1 ) ^2 − ( − 1 ) − 2 = 1\)
Thay \(x=-2\) vào \(f ( x ) + g ( x )\)
\(f ( x ) + g ( x ) = 2 x ^2 − x − 2\)
\(⇒ 2 ( − 2 ) ^2 − ( − 2 ) − 2 = 8\)
\(f\left(x\right)-g\left(x\right)=\left(x^5-3x^2+x^3-x^2-2x+5\right)-\left(x^2-3x+1+x^2-x^4+x^5\right)\)
\(f\left(x\right)-g\left(x\right)=x^5-3x^2+x^3-x^2-2x+5-x^2+3x-1-x^2+x^4-x^5\)
\(f\left(x\right)-g\left(x\right)=\left(x^5-x^5\right)+\left(-3x^2-x^2-x^2-x^2\right)+x^3+\left(-2x+3x\right)+\left(5-1\right)+x^4\)
\(f\left(x\right)-g\left(x\right)=-6x^2+x^3+x+4+x^4\)
\(f\left(x\right)-g\left(x\right)=x^4+x^3-6x^2+x+4\)
a) Ta có: f(x) = x100+x99+x98+...+x+1
=>2f(x) = x101+x100+x99+...+x+1
=>f(x) = 2f(x)-f(x)=(x101+x100+...+x+1)-(x100+x99+...+x+1)= x101-1
=>f(2) = 2101-1
=>f(-2) = (-2)101-1
b)câu còn lại tự giải :D
f(x) = x100+x99+x98+...+x+1
=>2f(x) = x101+x100+x99+...+x
=>f(x) = 2f(x)-f(x)=(x101+x100+...+x)-(x100+x99+...+x+1)= x101-1
=>f(2) = 2.101-1 = 201
=>f(-2) = (-2)101-1 = -203