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Để P(x)=Q(x) thì:\(3x^3+x^2-3x-1=-3x^3-x^2-x-15\)
Nếu \(3x^3+x^2-3x-1=-3x^3-x^2-x-15\)
=>\(\left(3x^3+x^2-3x-1\right)-\left(-3x^3-x^2-x-15\right)=0\)
=>\(3x^3+x^2-3x-1+3x^3+x^2+x+15=0\)
=>\(\left(3x^3+3x^3\right)+\left(x^2+x^2\right)+\left(-3x+x\right)+\left(-1+15\right)=0\)
=>\(6x^3+2x^2-2x+14=0\)
=>\(6x^3+2x^2-2x=-14\)
a) F(x) + Q(x) = ( x^2 + 3x^2 + 3x - 2 ) + ( - x^3 - x^2 - 5x + 2 )
= x^2 + 3x^2 + 3x - 2 - x^3 - x^2 - 5x + 2
= ( x^2 - x^2 +3x^2 ) + ( 3x - 5x ) + ( -2 + 2 )
= 3x^2 - 2x
b) F(x) - Q(x) = ( x^2 + 3x^2 + 3x - 2 ) - ( - x^3 - x^2 - 5x + 2 )
= x^2 + 3x^2 x+ 3x - 2 + x^3 + x^2 + 5x - 2
= ( x^2 + x^2 + 3x^2 ) + ( 3x + 5x ) + ( -2 - 2 )
= 5x^2 + 8x - 4
Bài 1 :
a) Ta thấy : \(\left(x^2-9\right)^2\ge0\)
\(\left|y-2\right|\ge0\)
\(\Leftrightarrow A=\left(x^2-9\right)^2+\left|y-2\right|-1\ge-1\)
Dấu " = " xảy ra :
\(\Leftrightarrow\hept{\begin{cases}x^2-9=0\\y-2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\in\left\{3;-3\right\}\\y=2\end{cases}}\)
Vậy \(Min_A=-1\Leftrightarrow\left(x;y\right)\in\left\{\left(3;2\right);\left(-3;2\right)\right\}\)
b) Ta thấy : \(B=x^2+4x-100\)
\(=\left(x+4\right)^2-104\ge-104\)
Dấu " = " xảy ra :
\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Vậy \(Min_B=-104\Leftrightarrow x=-4\)
c) Ta thấy : \(C=\frac{4-x}{x-3}\)
\(=\frac{3-x+1}{x-3}\)
\(=-1+\frac{1}{x-3}\)
Để C min \(\Leftrightarrow\frac{1}{x-3}\)min
\(\Leftrightarrow x-3\)max
\(\Leftrightarrow x\)max
Vậy để C min \(\Leftrightarrow\)\(x\)max
p/s : riêng câu c mình không tìm được C min :( Mong bạn nào giỏi tìm hộ mình
Bài 2 :
a) Ta thấy : \(x^2\ge0\)
\(\left|y+1\right|\ge0\)
\(\Leftrightarrow3x^2+5\left|y+1\right|-5\ge-5\)
\(\Leftrightarrow C=-3x^2-5\left|y+1\right|+5\le-5\)
Dấu " = " xảy ra :
\(\Leftrightarrow\hept{\begin{cases}x=0\\y+1=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\y=-1\end{cases}}\)
Vậy \(Max_A=-5\Leftrightarrow\left(x;y\right)=\left(0;-1\right)\)
b) Để B max
\(\Leftrightarrow\left(x+3\right)^2+2\)min
Ta thấy : \(\left(x+3\right)^2\ge0\)
\(\Leftrightarrow\left(x+3\right)^2+2\ge2\)
Dấu " = " xảy ra :
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)
Vậy \(Max_B=\frac{1}{2}\Leftrightarrow x=-3\)
c) Ta thấy : \(\left(x+1\right)^2\ge0\)
\(\Leftrightarrow x^2+2x+1\ge0\)
\(\Leftrightarrow-x^2-2x-1\le0\)
\(\Leftrightarrow C=-x^2-2x+7\le8\)
Dấu " = " xảy ra :
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy \(Max_C=8\Leftrightarrow x=-1\)
a) \(x\left(2x^2-3x+2\right)=2x^3-3x^2+2x\)
b) \(\left(2x^2\right).\left(x-x^2+3\right)=2x^3-2x^4+6x^2\)
\(=-2x^4+2x^3+6x^2\)
c) \(\left(x^2-5x-1\right).\left(-3x\right)^3=\left(x^2-5x-1\right).\left(-27x^3\right)\)
\(=-27x^5+135x^4+27x^3\)
d) \(\left(x^2-2x-1\right).\left(\frac{1}{2}x\right)^2=\left(x^2-2x-1\right).\frac{1}{4}x^2\)
\(=\frac{1}{4}x^4-\frac{1}{2}x^3-\frac{1}{4}x^2\)
a, x . ( 2x2 - 3x + 2) = x.2x2 - x3x + 2x
= 2x3 - 3x2 + 2x
b, (2x2) - (x - x2 +3) = 2x2 - x + x2 - 3
= 3x2 - x - 3
c, ( x2 - 5x - 1) . (-3x)3 = ( x2 - 5x - 1) . (-27).x3
= (-27).x3.x2 - (-27).x3.5x - (-27).x3
= -27x5 + 135x4 + 27x3
d, ( x2 - 2x - 1) . ( \(\frac{1}{2}\)x)2 = ( x2 - 2x - 1) . \(\frac{1}{4}\)x2
= \(\frac{1}{4}\)x2.x2 - \(\frac{1}{4}\)x2 .2x - \(\frac{1}{4}\)x2
= \(\frac{1}{4}\)x4 - \(\frac{1}{2}\)x3 - \(\frac{1}{4}\)x2
chia đa thức => phần dư=0
<=>A(x)=(x^2-3x+4).x^2-4(x^2-3x+4)+(a-3...
phân dư là (a-3).x+b+16=>a=3, b=-16