b bài 19
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DT
30 tháng 12 2021
18a
\(\dfrac{x^2-x}{xy}+\dfrac{1-4x}{xy}=\dfrac{x^2-x+1-4x}{xy}=\dfrac{x^2-5x+1}{xy}\)
b,
\(\dfrac{5xy^2-x^2y}{3xy}+\dfrac{4xy^2+x^2y}{3xy}=\dfrac{5xy^2-x^2y+4xy^2+x^2y}{3xy}\\ =\dfrac{9xy^2}{3xy}=3y\)
c,
\(\dfrac{2x^2-xy}{x-y}+\dfrac{xy+y^2}{y-x}+\dfrac{2y^2-x^2}{x-y}\\ =\dfrac{2x^2-xy-xy-y^2+2y^2-x^2}{x-y}=\dfrac{x^2-2xy+y^2}{x-y}\\ =\dfrac{\left(x-y\right)\left(x-y\right)}{x-y}=x-y\)
XO
29 tháng 11 2019
Ta có : \(A=\frac{19^{30}+15}{19^{31}+15}\)
\(\Rightarrow19A=\frac{19^{31}+285}{19^{31}+15}=\frac{19^{31}+15+270}{19^{31}+15}=1+\frac{270}{19^{31}+15}\)
Lại có \(B=\frac{19^{31}+15}{19^{32}+15}\)
\(\Rightarrow19B=\frac{19^{32}+285}{19^{32}+15}=\frac{19^{32}+15+270}{19^{32}+15}=1+\frac{270}{19^{32}+15}\)
Vì \(\frac{270}{19^{32}+15}< \frac{270}{19^{31}+15}\Rightarrow1+\frac{270}{19^{32}+5}< 1+\frac{270}{19^{31}+15}\Rightarrow19B< 19A\Rightarrow B< A\)
NH
bài 19 nha
1
VX
1
6 tháng 1 2016
a) Cách 1:
15.12-3.5.10
=180-15.10
180-150
=30
Cách 2:
15.12-3.5.10
=15.12-15.10
=15.(12-10)
=15.2
=30
b)45-9.(13+5)
=45-9.18
=45-162
=-117
Cách 2:
45-9.(13+5)
=45-9.13+9.5
=45-117+45
=45-162
=-117
c)Cách 1
29.(19-13)-19.(19-13)
=29.3-19.6
=87-114
=-27
Cách 2
29.(19-13)-19.(19-13)
=29.19-29.13-19.19-19.13
=551-377-377-3
21 tháng 3 2020
\(\text{a)A=2021+{750-[2021-(+50)}\)
\(A=2021+750-2021+50\)
\(A=\left(2021-2021\right)+\left(750+50\right)\)
\(A=0+800\)
\(A=800\)
\(\text{b)B=-215.[19+(-1236)]+215.(19-236)}\)
\(B=-215.19+215.1236+215.19-215.236\)
\(B=19.\left(-215+215\right)+215.\left(1236-236\right)\)
\(B=19.0+215.1000\)
\(B=0+215000\)
\(B=215000\)
học tốt
VT
1
bài 19 :
a) \(\left(x+1\right)\left(x-2\right)< 0\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x+1< 0\\x-2>0\end{matrix}\right.\\\left[{}\begin{matrix}x+1>0\\x-2< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\\\left[{}\begin{matrix}x>-1\\x< 2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\in\varnothing\\-1< x< 2\end{matrix}\right.\) vậy \(-1< x< 2\)
b) \(\left(x-2\right)\left(x+\dfrac{2}{3}\right)>0\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x-2>0\\x+\dfrac{2}{3}>0\end{matrix}\right.\\\left[{}\begin{matrix}x-2< 0\\x+\dfrac{2}{3}< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>2\\x>\dfrac{-2}{3}\end{matrix}\right.\\\left[{}\begin{matrix}x< 2\\x< \dfrac{-2}{3}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>2\\x< \dfrac{-2}{3}\end{matrix}\right.\) vậy \(x>2\) hoặc \(x< \dfrac{-2}{3}\)