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\(f\left(x\right)=9-3x^3-2x^3+x^2+4x-6\)
\(g\left(x\right)=x^3-6x^3+2x^3+4x^2+7x-3x+3\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=9-3x^3-2x^3+x^2+4x-6-\left(x^3-6x^3+2x^3+4x^2+7x-3x+3\right)\)
Bạn tự phá dấu và trừ ra nhé, ghi ở đây dài lắm, kết quả bằng :
\(-2x^3-3x^2\)
Ta có:
\(f\left(x\right)=-5x^3+x^2+4x+3\)
\(g\left(x\right)=-3x^3+4x^2+4x+3\)
a) M + N = (6x2y + 8x + 7xy) + (5x2y + 7x + 3xy + 2)
= 6x2x + 8x + 7xy + 5x2y + 7x + 3xy + 2
= (6x2y + 5x2y) + (8x + 7x) + (7xy + 3xy) + 2
= 11x2y + 15x + 10xy + 2
M - N = (6x2y + 8x + 7xy) - (5x2y + 7x + 3xy + 2)
= 6x2y + 8x + 7xy - 5x2y - 7x - 3xy - 2
= (6x2y - 5x2y) + (8x - 7x) + (7xy - 3xy) - 2
= x2y + x + 7xy - 2
b) Sắp xếp : x4 + 2x3 + 3x2 - 5x
F(1) = 14 + 2.13 + 3.12 - 5.1
= 1 + 2 + 3 - 5
= 1
\(M+N\)
\(=\left(6x^2y+8x+7xy\right)+\left(5x^2y+7x+3xy+2\right)\)
\(=6x^2y+8x+7xy+5x^2y+7x+3xy+2\)
\(=\left(6x^2y+5x^2y\right)+\left(8x+7x\right)+\left(7xy+3xy\right)+2\)
\(=11x^2y+15x+10xy+2\)
a. M(x) = x2 -5x -2x3 + x4 + 1
= x4 - 2x3 + x2 - 5x + 1
N(x) = -5x3 -3 + 8x4 + x2
= 8x4 - 5x3 + x2 - 3
b. M(x) + N(x) = x4 - 2x3 + x2 - 5x + 1 + 8x4 - 5x3 + x2 - 3
= (x4 + 8x4) + (-2x3 - 5x3) + (x2 + x2 ) - 5x + (1 - 3)
= 9x4 - 7x3 + 2x2 - 5x - 2
M (x) - N (x) = x4 - 2x3 + x2 - 5x + 1 - ( 8x4 - 5x3 + x2 - 3)
= x4 - 2x3 + x2 - 5x + 1 - 8x4 + 5x3 - x2 + 3
= (x4 - 8x4 ) + ( -2x3 + 5x3 ) + (x2 - x2 ) - 5x + (1 + 3)
= -7x4 + 3x3 - 5x + 4
\(\left\{{}\begin{matrix}P\left(x\right)=x+x^2-x^3+2x^3+2=x^3+x^2+x+2\\Q\left(x\right)=1+3x-x^2-4x+x^3=x^3-x^2-x+1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}P\left(x\right)+Q\left(x\right)=2x^3+3\\P\left(x\right)-Q\left(x\right)=2x^2+2x+1\end{matrix}\right.\)
6-/2-x/=4
\(\Rightarrow\)/2-x/=6-4
\(\Rightarrow\)/2-x/=2
\(\Rightarrow\left\{\begin{matrix}2-x=2\\2-x=-2\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x=2-2\\x=2-\left(-2\right)\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(\Rightarrow x=\left\{0;4\right\}\)
a, Ta có: \(A=\left|x-1\right|+\left|x-2017\right|=\left|x-1\right|+\left|2017-x\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(A\ge\left|x-1+2017-x\right|=\left|-2016\right|=2016\)
Dấu " = " khi \(\left\{{}\begin{matrix}x-1\ge0\\2017-x\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge1\\x\le2017\end{matrix}\right.\Rightarrow1\le x\le2017\)
Vậy \(MIN_A=2016\) khi \(1\le x\le2017\)
b, Ta có: \(\left\{{}\begin{matrix}\left(x-5\right)^2\ge0\\\left|x-5\right|\ge0\end{matrix}\right.\Rightarrow\left(x-5\right)^2+\left|x-5\right|\ge0\)
\(\Rightarrow B=\left(x-5\right)^2+\left|x-5\right|+2014\ge2014\)
Dấu " = " khi \(\left\{{}\begin{matrix}\left(x-5\right)^2=0\\\left|x-5\right|=0\end{matrix}\right.\Rightarrow x=5\)
Vậy \(MIN_B=2014\) khi x = 5
b may cho chú là chung nghiệm là x=5 nếu (x-6)^2+|x-5| thì sao? cần phải nhớ (x-6)^2=|x-6|^2 sau đó áp dụng |a|+|b|>=|a+b|
a) \(M\left(x\right)+N\left(x\right)=\left(x^4+5x^3-x^2+x-0,5\right)+\left(3x^4-5x^2-x-2,5\right)\)
\(=\left(x^4+3x^4\right)+5x^3-\left(x^2+5x^2\right)+\left(x-x\right)-\left(0,5+2,5\right)\)
\(=4x^4+5x^3-6x^2-3\)
b) \(M\left(x\right)-N\left(x\right)=\left(x^4+5x^3-x^2+x-0,5\right)-\left(3x^4-5x^2-x-2,5\right)\)
\(=x^4+5x^3-x^2+x-0,5-3x^4+5x^2+x+2,5\)
\(=\left(x^4-3x^4\right)+5x^3-\left(x^2-5x^2\right)+\left(x+x\right)-\left(0,5-2,5\right)\)
\(=-2x^4+5x^3+4x^2+2x+2\)
a)\(M\left(x\right)=x^4-2x^3+x^2-5x+1\)
\(N\left(x\right)=8x^4-5x^3+x^2-3\)
b)\(M\left(x\right)+N\left(x\right)=9x^4-7x^3+2x^2-5x-2\)
\(M\left(x\right)-N\left(x\right)=x^4-2x^3+x^2-5x+1-8x^4+5x^3-x^2+3\)
\(M\left(x\right)-N\left(x\right)=-7x^3+3x^3-5x+4\)