19x......>194
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N
0
8 tháng 5 2018
Ta có: \(C=\frac{3x^2-7x^2-12+45}{3x^3-19x^2+33x-9}\) ĐKXĐ: x khác 3, 1/3
\(=\frac{\left(x-3\right)^2\left(2x+5\right)}{\left(x-3\right)^2\left(3x-1\right)}\)
\(=\frac{2x+5}{3x-1}\)
Để C>0, ta có:
-5/2<x<1/3 (thỏa mãn ĐKXĐ)
ND
0
6 tháng 9 2017
997+37997+3749+194
=994+3+37997+3743+6+194
=994+(3+37997)+3743+(6+194)
=994+38000+3743+200
=38994+3943
=38994+1006+2937
=(38994+1006)+2937
=40000+2937
=42937
PH
1
2 tháng 5 2019
Ta có:\(x=18\Rightarrow\hept{\begin{cases}x+1=19\\-x-1=-19\end{cases}}\)
Thay vào BT D ta được:
\(D=x^{12}+\left(-x-1\right)x^{11}+\left(x+1\right)x^{10}+\left(-x-1\right)x^9+...+\left(x+1\right)x^2+\left(-x-1\right)x+1\)
\(=x^{12}-x^{12}-x^{11}+x^{11}+x^{10}-x^{10}-x^9+...+x^3+x^2-x^2-x+1\)
\(=1-x\)
DT
19 tháng 11 2021
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19 tháng 11 2021
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19 tháng 4 2018
\(B=\dfrac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}\)
\(=\dfrac{2x^3+5x^2-12x^2-30x+18x+45}{3x^3-x^2-18x^2+6x+27x-9}\)
\(=\dfrac{\left(2x^3+5x^2\right)-\left(12x^2+30x\right)+\left(18x+45\right)}{\left(3x^3-x^2\right)-\left(18x^2-6x\right)+\left(27x-9\right)}\)
\(=\dfrac{x^2\left(2x+5\right)-6x\left(2x+5\right)+9\left(2x+5\right)}{x^2\left(3x-1\right)-6x\left(3x-1\right)+9\left(3x-1\right)}\)
\(=\dfrac{\left(2x+5\right)\left(x^2-6x+9\right)}{\left(3x-1\right)\left(x^2-6x+9\right)}\)
\(=\dfrac{\left(2x+5\right)\left(x-3\right)^2}{\left(3x-1\right)\left(x-3\right)^2}\)
ĐKXĐ : \(\left\{{}\begin{matrix}3x-1\ne0\\x-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{1}{3}\\x\ne3\end{matrix}\right.\)
\(a,B=\dfrac{\left(2x+5\right)\left(x-3\right)^2}{\left(3x-1\right)\left(x-3\right)^2}=\dfrac{2x+5}{3x-1}\)
b,Để \(B>0\)
\(\Leftrightarrow\dfrac{2x+5}{3x-1}>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x+5>0\\3x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x+5< 0\\3x-1< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-\dfrac{5}{2}\\x>\dfrac{1}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x< -\dfrac{5}{2}\\x< \dfrac{1}{3}\end{matrix}\right.\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x>\dfrac{1}{3}\\x< -\dfrac{5}{2}\end{matrix}\right.\) thì B > 0
19 tháng 4 2018
a) ĐKXĐ:\(x\ne\dfrac{1}{3};x\ne3\)
\(B=\dfrac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}\)
\(B=\dfrac{\left(2x^3-12x^2+18x\right)+\left(5x^2-30x+45\right)}{\left(3x^3-18x^2+27x\right)-\left(x^2-6x+9\right)}\)
\(B=\dfrac{2x\left(x^2-6x+9\right)+5\left(x^2-6x+9\right)}{3x\left(x^2-6x+9\right)-\left(x^2-6x+9\right)}\)
\(B=\dfrac{\left(2x+5\right)\left(x^2-6x+9\right)}{\left(3x-1\right)\left(x^2-6x+9\right)}\)
\(B=\dfrac{2x+5}{3x-1}\)
b) Để \(B>0\Leftrightarrow\dfrac{2x+5}{3x-1}>0\Leftrightarrow2x+5\)và \(3x-1\) cùng dấu
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x+5>0\\3x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x+5< 0\\3x-1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{-5}{2}\\x>\dfrac{1}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{-5}{2}\\x< \dfrac{1}{3}\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1}{3}\\x< -\dfrac{5}{2}\end{matrix}\right.\)
19 \(\times\)... . > 194
... > \(\dfrac{194}{19}\)
.
Dễ , 19 × 11 > 194