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Bài 1 :
\(M+N\)
\(=\left(2xy^2-3x+12\right)+\left(-xy^2-3\right)\)
\(=2xy^2-3x+12-xy^2-3\)
\(=\left(2xy^2-xy^2\right)-3x+\left(12-3\right)\)
\(=xy^2-3x+9\)
a)
\(P\left(x\right)=2x^3+x^2-3x-1+x^3-3x^2-5x+1\)
\(P\left(x\right)=\left(2x^3+x^3\right)+\left(x^2-3x^2\right)-\left(3x+5x\right)-\left(1-1\right)\)
\(P\left(x\right)=3x^3-2x^2-8x\)
tương tự làm nốt
b) Tìm nghiệm thì đặt bằng 0 rồi tính là OK
Học tốt~
\(h\left(x\right)+f\left(x\right)-g\left(x\right)=-2x^2-x+9\)
\(h\left(x\right)+\left(-5x^4+x^2-2x+6\right)-\left(-5x^4+x^3+3x^2-3\right)=-2x^2-x+9\)
\(h\left(x\right)-5x^4+x^2-2x+6+5x^4-x^3-3x^2-3=-2x^2-x+9\)
\(h\left(x\right)-\left(5x^4-5x^4\right)+\left(x^2-3x^2\right)-x^3-2x+\left(6-3\right)=-2x^2-x+9\)
\(h\left(x\right)-0-2x^2-x^3-2x+3=-2x^2-x+9\)
\(h\left(x\right)-x^3-2x^2-2x+3=-2x^2-x+9\)
\(h\left(x\right)+\left(-x^3-2x^2-2x+3\right)=-2x^2-x+9\)
\(h\left(x\right)=\left(-2x^2-x+9\right)-\left(-x^3-2x^2-2x+3\right)\)
\(h\left(x\right)=-2x^2-x+9+x^3+2x^2+2x-3\)
\(h\left(x\right)=\left(-2x^2+2x^2\right)-\left(x-2x\right)+\left(9-3\right)+x^3\)
\(h\left(x\right)=0+x+6+x^3\)
\(h\left(x\right)=x^3+x+6\)
d) Ta có : h(x) + f(x) - g(x) = -2x2 - x + 9
<=> h(x) = -2x2 - x + 9 - f(x) + g(x)
<=> h(x) = -2x2 - x + 9 - x2 + 2x + 5x4 - 6 + x3 - 5x4 + 3x2 - 3
<=> h(x) = x3 + x.
Vậy h(x) = x3 + x
a) \(A=-|x-2|\le0;\forall x\)
\(\Rightarrow-|x-2|+2019\le0+2019;\forall x\)
Hay \(A\le2019;\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow|x-2|=0\)
\(\Leftrightarrow x=2\)
Vậy \(A_{max}=2019\Leftrightarrow x=2\)
b) \(B=-2x^2+5x+3\)
\(=-2\left(x^2-\frac{5}{2}x-\frac{3}{2}\right)\)
\(=-2\left(x^2-2.x.\frac{5}{4}+\frac{25}{16}-\frac{25}{16}-\frac{3}{2}\right)\)
\(=-2\left(x-\frac{5}{4}\right)^2+\frac{49}{8}\)
Vì \(-2\left(x-\frac{5}{4}\right)^2\le0;\forall x\)
\(\Rightarrow-2\left(x-\frac{5}{4}\right)^2+\frac{49}{8}\le0+\frac{49}{8};\forall x\)
Hay \(B\le\frac{49}{8};\forall x\)
Dấu "="xảy ra \(\Leftrightarrow\left(x-\frac{5}{4}\right)^2=0\)
\(\Leftrightarrow x=\frac{5}{4}\)
Vậy \(B_{max}=\frac{49}{8}\Leftrightarrow x=\frac{5}{4}\)
c) \(-x^2-y^2+2x+8y+2028\)
\(=-\left(x^2+y^2-2x-8y-2028\right)\)
\(=-\left[\left(x^2-2x+1\right)+\left(y^2-8y+16\right)-2045\right]\)
\(=-\left(x-1\right)^2-\left(y-4\right)^2+2045\)
Vì \(\hept{\begin{cases}-\left(x-1\right)^2\le0;\forall x,y\\-\left(y-4\right)^2\le0;\forall x,y\end{cases}}\)
\(\Rightarrow-\left(x-1\right)^2-\left(y-4\right)^2\le0;\forall x,y\)
\(\Rightarrow-\left(x-1\right)^2-\left(y-4\right)^2+2045\le0+2045;\forall x,y\)
Hay \(C\le2045;\forall x,y\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}-\left(x-1\right)^2=0\\-\left(y-4\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=4\end{cases}}}\)
Vậy \(C_{max}=2045\Leftrightarrow\hept{\begin{cases}x=1\\y=4\end{cases}}\)
\(f\left(x\right)-g\left(x\right)=5x^2-2x+5-\left(5x^2-6x-\frac{1}{3}\right)\)
= \(5x^2-2x+5-5x^2+6x+\frac{1}{3}\)
=\(4x+\frac{16}{3}\)
a) \(f\left(x\right)-g\left(x\right)=\left[x\left(x^2-2x+7\right)-1\right]-\left[x\left(x^2-2x-1\right)-1\right]\)
\(f\left(x\right)-g\left(x\right)=x^3-2x^2+7x-1-x^3+2x^2+x+1\)
\(f\left(x\right)-g\left(x\right)=8x\)
\(f\left(x\right)+g\left(x\right)=x\left(x^2-2x+7\right)-1+x\left(x^2-2x-1\right)-1\)
\(f\left(x\right)+g\left(x\right)=x^3-2x^2+7x-1+x^3-2x^2-x-1\)
\(f\left(x\right)+g\left(x\right)=2x^3-4x^2+6x-2\)
b) 8x=0
=> x=0
=> Nghiệm đa thức f(x)-g(x)
c) Thay \(x=-\frac{3}{2}\)vào BT f(x)+g(x) ta được :
\(2.\left(-\frac{3}{2}\right)^3-4\left(-\frac{3}{2}\right)^2+6\left(-\frac{3}{2}\right)-2\)
\(=6,75+9-9-2\)
\(=4,75\)
#H
\(P\left(x\right)=3x^5+x^4-2x^2+2x-1\)
\(Q\left(x\right)=-3x^5+2x^2-2x+3\)
\(P\left(x\right)+Q\left(x\right)=3x^5+x^4-2x^2+2x-1-3x^5+2x^2-2x+3\)
\(=x^4+2\)
\(P\left(x\right)-Q\left(x\right)=3x^5+x^4-2x^2+2x-1+3x^5-2x^2+2x-3\)
\(=6x^5+x^4-4x^2+4x-4\)
Thu gọn + sắp xếp luôn
P(x) = 3x5 + x4 - 2x2 + 2x - 1
Q(x) = -3x5 + 2x2 - 2x + 3
P(x) + Q(x) = ( 3x5 + x4 - 2x2 + 2x - 1 ) + ( -3x5 + 2x2 - 2x + 3 )
= ( 3x5 - 3x5 ) + x4 + ( 2x2 -- 2x2 ) + ( 2x - 2x ) + ( 3 - 1 )
= x4 + 2
P(x) - Q(x) = ( 3x5 + x4 - 2x2 + 2x - 1 ) - ( -3x5 + 2x2 - 2x + 3 )
= 3x5 + x4 - 2x2 + 2x - 1 + 3x5 - 2x2 + 2x - 3
= ( 3x5 + 3x5 ) + x4 + ( -2x2 - 2x2 ) + ( 2x + 2x ) + ( -1 - 3 )
= 6x5 + x4 - 4x2 + 4x - 4
\(A=-\left(x^2-2x+1\right)-2\)
\(A=-\left(x-1\right)^2-2\)
Vì \(-\left(x-1\right)^2\le0;\forall x\)
\(\Rightarrow-\left(x-1\right)^2-2\le0-2;\forall x\)
Hay \(A\le-2;\forall x\)
Dấu "=" xảy ra\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x=1\)
Vậy MAX A=-2 \(\Leftrightarrow x=1\)
\(C=-2x^2+2xy-y^2+2x+4\)
\(C=-x^2+2xy-y^2-x^2+2x-1+5\)
\(C=-\left(x^2-2xy+y^2\right)-\left(x^2-2x+1\right)+5\)
\(C=-\left(x-y\right)^2-\left(x-1\right)^2+5\le5\)
Dấu = xảy ra khi :
\(\hept{\begin{cases}x-y=0\\x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=y\\x=1\end{cases}}\Leftrightarrow x=y=1\)
Vậy C max = 5 tại x = y = 1
a)Xét các TH:
\(\cdot f\left(1\right)=1+1^2+1^3+...+1^{2020}\)(có 2020 số)
\(=1+1+...+1\)(có 2020 số 1)
\(=1\cdot2020=2020\)
Làm câu b nx bn ơi !
Làm đc câu b thì mk k nha !~!~