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\(f\left(x\right)-g\left(x\right)=5x^2-2x+5-\left(5x^2-6x-\frac{1}{3}\right)\)
= \(5x^2-2x+5-5x^2+6x+\frac{1}{3}\)
=\(4x+\frac{16}{3}\)
1 )
a) f(x) + g(x) = (x2-5+x3-x ) + ( x+x4-4+x2)
= x2-5+x3-x + x+ x4-4 +x2
=( x2+x2) + (-5-4)+ x3+(-x+x)+x4
= 2x2 -9 + x3 + x4
= x4+x3+2x2-9
b) Có : g(x)-f(x)=h(x )
=> f(x) = g(x) - h(x)
Tiếp theo bn tự tính như phần a nhé
c ) Thay x=-1 , y=-1 vào đa thức rồi bn tự tính nhé ! dễ mà
Bài 1:
\(f(x)=ax^2+bx+c\Rightarrow \left\{\begin{matrix} f(-2)=a(-2)^2+b(-2)+c=4a-2b+c\\ f(3)=a.3^2+b.3+c=9a+3b+c\end{matrix}\right.\)
\(\Rightarrow f(-2)+f(3)=(4a-2b+c)+(9a+3b+c)\)
\(=13a+b+2c=0\)
\(\Rightarrow f(-2)=-f(3)\Rightarrow f(-2)f(3)=-f(3)^2\leq 0\) do \(f(3)^2\geq 0\)
Ta có đpcm.
Bài 2:
Thay $x=-3$ ta có:
\(f(-3)=a.(-3)+5=-2\)
\(\Rightarrow a=\frac{7}{3}\)
Vậy $a=\frac{7}{3}$
Bài 4:
Ta có: \(B=\frac{x^2+y^2+7}{x^2+y^2+2}=1+\frac{5}{x^2+y^2+2}\)
Vì \(x^2+y^2+2>0\) nên để \(\frac{5}{x^2+y^2+2}\) lớn nhất thì \(x^2+y^2+2\) nhỏ nhất.
Lại có:
\(\left\{\begin{matrix}x^2\ge0\\y^2\ge0\end{matrix}\right.\Rightarrow x^2+y^2\ge0\Rightarrow x^2+y^2+2\ge2\)
\(\Rightarrow\frac{5}{x^2+y^2+2}\le\frac{5}{2}\)
\(\Rightarrow1+\frac{5}{x^2+y^2+2}\le1+2,5\)
\(\Rightarrow B=\frac{x^2+y^2+7}{x^2+y^2+2}\le3,5\)
Vậy \(MAX_B=3,5\) khi \(x=y=0\)
5)Ta có 26y chẵn, 2000 chẵn \(\Rightarrow51x\)chẵn \(\Rightarrow x⋮2\)
Mà x nguyên tố \(\Rightarrow x=2\)
Thay x=2 vào ta có
51.2+26y=2000
\(\Rightarrow102+26y=2000\)
\(\Rightarrow26y=1898\)
\(\Rightarrow y=73\)
Vậy \(x=2,y=73\)
I . Trắc Nghiệm
1B . 2D . 3C . 5A
II . Tự luận
2,a,Ta có: A+(x\(^2\)y-2xy\(^2\)+5xy+1)=-2x\(^2\)y+xy\(^2\)-xy-1
\(\Leftrightarrow\) A=(-2x\(^2\)y+xy\(^2\)-xy-1) - (x\(^2\)y-2xy\(^2\)+5xy+1)
=-2x\(^2\)y+xy\(^2\)-xy-1 - x\(^2\)y+2xy\(^2\)-5xy-1
=(-2x\(^2\)y - x\(^2\)y) + (xy\(^2\)+ 2xy\(^2\)) + (-xy - 5xy ) + (-1 - 1)
= -3x\(^2\)y + 3xy\(^2\) - 6xy - 2
b, thay x=1,y=2 vào đa thức A
Ta có A= -3x\(^2\)y + 3xy\(^2\) - 6xy - 2
= -3 . 1\(^2\) . 2 + 3 .1 . 2\(^2\) - 6 . 1 . 2 -2
= -6 + 12 - 12 - 2
= -8
3,Sắp xếp
f(x) =9-x\(^5\)+4x-2x\(^3\)+x\(^2\)-7x\(^4\)
=9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x
g(x) = x\(^5\)-9+2x\(^2\)+7x\(^4\)+2x\(^3\)-3x
=-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x
b,f(x) + g(x)=(9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x) + (-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x)
=9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x
=(9-9)+(-x\(^5\)+x\(^5\))+(-7x\(^4\)+7x\(^4\))+(-2x\(^3\)+2x\(^3\))+(x\(^2\)+2x\(^2\))+(4x-3x)
= 3x\(^2\) + x
g(x)-f(x)=(-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x) - (9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x)
=-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x-9+x\(^5\)+7x\(^4\)+2x \(^3\)-x\(^2\)-4x
=(-9-9)+(x\(^5\)+x\(^5\))+(7x\(^4\)+7x\(^4\))+(2x\(^3\)+2x\(^3\))+(2x\(^2\)-x\(^2\))+(3x-4x)
= -18 + 2x\(^5\) + 14x\(^4\) + 4x\(^3\) + x\(^2\) - x
I . Trắc Nghiệm 1B . 2D . 3C . 5A II . Tự luận 2,a,Ta có: A+(x22y-2xy22+5xy+1)=-2x22y+xy22-xy-1 ⇔⇔ A=(-2x22y+xy22-xy-1) - (x22y-2xy22+5xy+1) =-2x22y+xy22-xy-1 - x22y+2xy22-5xy-1 =(-2x22y - x22y) + (xy22+ 2xy22) + (-xy - 5xy ) + (-1 - 1) = -3x22y + 3xy22 - 6xy - 2 b, thay x=1,y=2 vào đa thức A Ta có A= -3x22y + 3xy22 - 6xy - 2 = -3 . 122 . 2 + 3 .1 . 222 - 6 . 1 . 2 -2 = -6 + 12 - 12 - 2 = -8 3,Sắp xếp f(x) =9-x55+4x-2x33+x22-7x44 =9-x55-7x44-2x33+x22+4x g(x) = x55-9+2x22+7x44+2x33-3x =-9+x55+7x44+2x33+2x22-3x b,f(x) + g(x)=(9-x55-7x44-2x33+x22+4x) + (-9+x55+7x44+2x33+2x22-3x) =9-x55-7x44-2x33+x22+4x-9+x55+7x44+2x33+2x22-3x =(9-9)+(-x55+x55)+(-7x44+7x44)+(-2x33+2x33)+(x22+2x22)+(4x-3x) = 3x22 + x g(x)-f(x)=(-9+x55+7x44+2x33+2x22-3x) - (9-x55-7x44-2x33+x22+4x) =-9+x55+7x44+2x33+2x22-3x-9+x55+7x44+2x 33-x22-4x =(-9-9)+(x55+x55)+(7x44+7x44)+(2x33+2x33)+(2x22-x22)+(3x-4x) = -18 + 2x55 + 14x44 + 4x33 + x22 - x
a: \(h\left(x\right)=f\left(x\right)+g\left(x\right)=x^3-x^2+x-24\)
Bậc là 3
b: \(k\left(x\right)=f\left(x\right)-g\left(x\right)=7x^3-9x^2+11x+6\)
\(g\left(\dfrac{3}{2}\right)=-3\cdot\dfrac{27}{8}+4\cdot\dfrac{9}{4}-5\cdot\dfrac{3}{2}-15=-\dfrac{189}{8}\)
\(k\left(\dfrac{3}{2}\right)=7\cdot\dfrac{27}{8}-9\cdot\dfrac{9}{4}+11\cdot\dfrac{3}{2}+6=\dfrac{207}{8}\)
a: f(-1)=g(2)
nên \(-1-m-1+2m+m^2-1=12m+13m+m^2-3\)
\(\Leftrightarrow25m-3=m-3\)
=>m=0
b: \(s\left(x\right)=f\left(x\right)+g\left(x\right)=x^3+x^2\left(3m-m-1\right)+x\left(-2m+\dfrac{13}{2}m\right)+m^2-1+m^2-3\)
\(=x^3+\left(2m-1\right)x^2+\dfrac{9}{2}mx+2m^2-4\)
Vì m=1 nên \(s\left(x\right)=x^3+x^2+\dfrac{9}{2}x-2\)
Khi x=1 thì \(s=1+1+\dfrac{9}{2}-2=\dfrac{9}{2}\)
Khi x=-1 thì \(s=-1+1-\dfrac{9}{2}-2=-\dfrac{13}{2}\)