Tính \(f\left(x\right)+g\left(x\right)-h\left(x\right)\) biết :
\(f\left(x\right)=x^5-4x^3+x^2-2x+1\)
\(g\left(x\right)=x^5-2x^4+x^2-5x+3\)
\(h\left(x\right)=x^4-3x^2+2x-5\)
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Những câu hỏi liên quan
28 tháng 3 2018
Ta có: \(F\left(x\right)+G\left(x\right)-H\left(x\right)=0\)
\(\Leftrightarrow4x^2+3x-2+3x^2-2x+5-5x^2+2x-3=0\\ \Leftrightarrow2x^2+3x=0\\ \Rightarrow x\left(2x+3\right)=0\\ \Rightarrow x=0;x=\dfrac{-3}{2}\)
Vậy tìm được x thỏa mãn là: \(x=0;x=\dfrac{-3}{2}\)
15 tháng 5 2017
a) Thu gọn, sắp xếp các đa thức theo lũy thừa tăng của biến
= -9 - 2x2 + 3x3 - 6x5 - 3x7
b) Tính -9 - 2x2 + 3x3 - 6x5 - 3x7 ) + (-12 + 3x3 + x4 + x5 - x6 - 6x7 - 5x8 ) - (2x - 3x2 + 4x3 +4x5 -4x6 - 10x7)
= - 9 - 2x2 + 3x3 - 6x5 - 3x7 -12 + 3x3 + x4 + x5 - x6 - 6x7 - 5x8 - 2x + 3x2 - 4x3 - 4x5 + 4x6 + 10x7
= -21 - 2x + x2 + 2x3 + x4 - 9x5 + 3x6 + x7 - 5x8
27 tháng 12 2019
\(f\left(x\right)+h\left(x\right)-g\left(x\right)\)
\(=\left(5x^4+3x^2+x-1\right)+\left(-x^4+3x^3-2x^2-x+2\right)\)
\(-\left(2x^4-x^3+x^2+2x+1\right)\)
\(=\left(5x^4-x^4-2x^4\right)+\left(3x^3+x^3\right)+\left(3x^2-2x^2-x^2\right)\)
\(+\left(x-x-2x\right)+\left(-1+2-1\right)\)
\(=2x^4+4x^3-2x\)
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 3 2020
1.
\(f\left(x\right)=\frac{x-7}{\left(x-4\right)\left(4x-3\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định tại \(x=\left\{\frac{3}{4};4\right\}\)
\(f\left(x\right)=0\Rightarrow x=7\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}\frac{3}{4}< x< 4\\x>7\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{3}{4}\\4< x< 7\end{matrix}\right.\)
2.
\(f\left(x\right)=\frac{11x+3}{-\left(x-\frac{5}{2}\right)^2-\frac{3}{4}}\)
Vậy:
\(f\left(x\right)=0\Rightarrow x=-\frac{3}{11}\)
\(f\left(x\right)>0\Rightarrow x< -\frac{3}{11}\)
\(f\left(x\right)< 0\Rightarrow x>-\frac{3}{11}\)
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 3 2020
3.
\(f\left(x\right)=\frac{3x-2}{\left(x-1\right)\left(x^2-2x-2\right)}\)
Vậy:
\(f\left(x\right)\) ko xác định khi \(x=\left\{1;1\pm\sqrt{3}\right\}\)
\(f\left(x\right)=0\Rightarrow x=\frac{2}{3}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< 1-\sqrt{3}\\\frac{2}{3}< x< 1\\x>1+\sqrt{3}\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}1-\sqrt{3}< x< \frac{2}{3}\\1< x< 1+\sqrt{3}\end{matrix}\right.\)
4.
\(f\left(x\right)=\frac{\left(x-2\right)\left(x+6\right)}{\sqrt{6}\left(x+\frac{\sqrt{6}}{4}\right)^2+\frac{8\sqrt{2}-3\sqrt{6}}{8}}\)
Vậy:
\(f\left(x\right)=0\Rightarrow x=\left\{-6;2\right\}\)
\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< -6\\x>2\end{matrix}\right.\)
\(f\left(x\right)< 0\Rightarrow-6< x< 2\)
7 tháng 4 2019
a) \(f\left(x\right)=5x^3-7x^2+x+7+4x^5\)
\(f\left(-1\right)=5.\left(-1\right)^3-7.\left(-1\right)^2+\left(-1\right)+7+4.\left(-1\right)^5\)
\(f\left(-1\right)=\left(-5\right)-7+\left(-1\right)+7+\left(-4\right)\)
\(f\left(-1\right)=-10\)
\(\Rightarrow f\left(x\right)=-10\)
\(g\left(x\right)=4x^5-3x^3-7x^2+2x+5\)
\(g\left(0\right)=4.0^5-3.0^3-7.0^2+2.0+5\)
\(g\left(0\right)=5\)
\(\Rightarrow g\left(x\right)=0\)
\(h\left(x\right)=x^2-4x-5\)
\(h\left(-\frac{1}{2}\right)=\left(-\frac{1}{2}\right)^2-4.\left(-\frac{1}{2}\right)-5\)
\(h\left(-\frac{1}{2}\right)=\frac{1}{4}-\left(-2\right)-5\)
\(h\left(-\frac{1}{2}\right)=-\frac{11}{4}\)
\(\Rightarrow h\left(x\right)=-\frac{11}{4}\)
7 tháng 4 2019
\(f\left(-1\right)=5\left(-1\right)^3-7\left(-1\right)^2+\left(-1\right)+7+4\left(-1\right)^5\)
\(f\left(-1\right)=-5-7-1+7-4\)
\(f\left(-1\right)=-10\)
\(g\left(0\right)=4.0^5-3.0^3-7.0^2+2.0+5\)
\(g\left(0\right)=0-0-0+0+5\)
\(g\left(0\right)=5\)
\(h\left(-\frac{1}{2}\right)=\left(-\frac{1}{2}\right)^2-4\left(-\frac{1}{2}\right)-5\)
\(h\left(-\frac{1}{2}\right)=\frac{1}{4}-\left(-2\right)-5\)
\(h\left(-\frac{1}{2}\right)=\frac{1}{4}+2-5\)
\(h\left(-\frac{1}{2}\right)=-\frac{11}{4}\)
26 tháng 12 2021
a) \(\Rightarrow\dfrac{1}{3}x\left(x-2\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow\left(x+5\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
c) \(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
e) \(\Rightarrow\left(x+2\right)\left(x+2-x+2\right)=0\Rightarrow\left(x+2\right).4=0\Rightarrow x=-2\)
f) \(\Rightarrow x\left(2x-3\right)+2\left(2x-3\right)=0\Rightarrow\left(2x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)
g) \(\Rightarrow2\left(3x-2\right)^2-\left(3x-2\right)\left(3x+2\right)=0\Rightarrow\left(3x-2\right)\left(3x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)
h) \(\Rightarrow x\left(x+1\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)
i) \(\Rightarrow4x\left(x+1\right)+5\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(4x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{5}{4}\end{matrix}\right.\)
f(x) + g(x) - h(x) = (x5 - 4x3 + x2 - 2x + 1) + (x5 - 2x4 + x2 - 5x + 3) - (x4 - 3x2 + 2x - 5)
= x5 - 4x3 + x2 - 2x + 1 + x5 - 2x4 + x2 - 5x + 3 - x4 + 3x2 - 2x + 5
= (x5 + x5) - (2x4 + x4) - 4x3 + ( x2 + x2 + 3x2) - (2x + 5x + 2x) + (1 + 3 + 5)
= 2x5 - 3x4 - 4x3 + 5x2 - 9x + 9
f(x)=
f(x) + g(x) - h(x) = (x5 - 4x3 + x2 - 2x + 1) + (x5 - 2x4 + x2 - 5x + 3) - (x4 - 3x2 + 2x - 5)
= x5 - 4x3 + x2 - 2x + 1 + x5 - 2x4 + x2 - 5x + 3 - x4 + 3x2 - 2x + 5
= (x5 + x5) - (2x4 + x4) - 4x3 + ( x2 + x2 + 3x2) - (2x + 5x + 2x) + (1 + 3 + 5)
= 2x5 - 3x4 - 4x3 + 5x2 - 9x + 9