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b: \(\dfrac{2x+3}{3-x}\le0\)
\(\Leftrightarrow\dfrac{2x+3}{x-3}\ge0\)
=>x>3 hoặc x<=-3/2
c: \(\dfrac{x+5}{x+3}>1\)
\(\Leftrightarrow\dfrac{x+5-x-3}{x+3}>0\)
=>2/(x+3)>0
=>x+3>0
hay x>-3
Bài 1:
|\(x\)| = 1 ⇒ \(x\) \(\in\) {-\(\dfrac{1}{3}\); \(\dfrac{1}{3}\)}
A(-1) = 2(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)) + 5
A(-1) = \(\dfrac{2}{9}\) + 1 + 5
A (-1) = \(\dfrac{56}{9}\)
A(1) = 2.(\(\dfrac{1}{3}\) )2- \(\dfrac{1}{3}\).3 + 5
A(1) = \(\dfrac{2}{9}\) - 1 + 5
A(1) = \(\dfrac{38}{9}\)
|y| = 1 ⇒ y \(\in\) {-1; 1}
⇒ (\(x;y\)) = (-\(\dfrac{1}{3}\); -1); (-\(\dfrac{1}{3}\); 1); (\(\dfrac{1}{3};-1\)); (\(\dfrac{1}{3};1\))
B(-\(\dfrac{1}{3}\);-1) = 2.(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)).(-1) + (-1)2
B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) - 1 + 1
B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\)
B(-\(\dfrac{1}{3}\); 1) = 2.(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)).1 + 12
B(-\(\dfrac{1}{3};1\)) = \(\dfrac{2}{9}\) + 1 + 1
B(-\(\dfrac{1}{3}\); 1) = \(\dfrac{20}{9}\)
B(\(\dfrac{1}{3};-1\)) = 2.(\(\dfrac{1}{3}\))2 - 3.(\(\dfrac{1}{3}\)).(-1) + (-1)2
B(\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) + 1 + 1
B(\(\dfrac{1}{3}\); -1) = \(\dfrac{20}{9}\)
B(\(\dfrac{1}{3}\); 1) = 2.(\(\dfrac{1}{3}\))2 - 3.(\(\dfrac{1}{3}\)).1 + (1)2
B(\(\dfrac{1}{3}\); 1) = \(\dfrac{2}{9}\) - 1 + 1
B(\(\dfrac{1}{3}\);1) = \(\dfrac{2}{9}\)
Tk mình đi mọi người mình bị âm nè!
Ai tk mình mình tk lại cho
Bài 1:
a: \(\left(2x-1\right)^4=16\)
=>2x-1=2 hoặc 2x-1=-2
=>2x=3 hoặc 2x=-1
=>x=3/2 hoặc x=-1/2
b: \(\left(2x-y+7\right)^{2012}+\left|x-3\right|^{2013}< =0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-y+7=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2x+7=y=2\cdot3+7=13\end{matrix}\right.\)
c: \(10800=2^4\cdot3^3\cdot5^2\)
mà \(2^{x+2}\cdot3^{x+1}\cdot5^x=10800\)
nên \(\left\{{}\begin{matrix}x+2=4\\x+1=3\\x=2\end{matrix}\right.\Leftrightarrow x=2\)
1) a)\(A=\dfrac{1-2x}{x+3}=\dfrac{-2x+1}{x+3}\)
\(A\in Z\Rightarrow-2x+1⋮x+3\)
\(\Rightarrow-2x-6+7⋮x+3\)
\(\Rightarrow-2\left(x+3\right)+7⋮x+3\)
\(\Rightarrow7⋮x+3\)
\(\Rightarrow x+3\inƯ\left(7\right)\)
\(Ư\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(\Rightarrow\left[{}\begin{matrix}x+3=1\\x+3=-1\\x+3=7\\x+3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-4\\x=4\\x=-10\end{matrix}\right.\)
b)\(A=\dfrac{x+3}{x-2}=\dfrac{x-2+5}{x-2}=\dfrac{x-2}{x-2}+\dfrac{5}{x-2}=1+\dfrac{5}{x-2}\)
\(\Rightarrow5⋮x-2\Rightarrow x-2\inƯ\left(5\right)\)
\(Ư\left(5\right)=\left\{\pm1;\pm5\right\}\)
\(\Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\\x-2=5\\x-2=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\\x=7\\x=-3\end{matrix}\right.\)
2)
\(25-y^2=8\left(x-2009\right)^2\)
\(\left\{{}\begin{matrix}8\left(x-2009\right)^2\ge0\\8\left(x-2009\right)^2⋮8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}25-y^2\ge0\\25-y^2⋮8\end{matrix}\right.\)
Vậy \(0\le y^2\le25\)
\(\Leftrightarrow0\le y^2\le5^2\)
Vì \(y\in Z\) nên: \(y\in\left\{0;\pm1;\pm2;\pm3;\pm4;\pm5\right\}\)
\(y\in\left\{0;1;4;9;16;25\right\}\)
mà chỉ có :
\(25-25=0⋮8\Rightarrow y^2=25\Leftrightarrow y=\pm5\)
\(\Leftrightarrow8\left(x-2009\right)^2=0\Leftrightarrow x=2009\)
Vậy \(\left\{{}\begin{matrix}y=5\\x=2009\end{matrix}\right.\) và \(\left\{{}\begin{matrix}y=-5\\x=2009\end{matrix}\right.\)
thanks bạn nhìu, theo dõi nhau nha để có thể giúp nhau.
Nice to meet you
a, Với x = 1 thì \(A=\frac{3x+2}{x-3}=\frac{3\cdot1+2}{1-3}=\frac{5}{-2}=\frac{-5}{2}\)
Với x = 2 thì \(A=\frac{3x+2}{x-3}=\frac{3\cdot2+2}{2-3}=\frac{8}{-1}=-\frac{8}{1}=-8\)
Với x =\(\frac{5}{2}\)thì : \(A=\frac{3x+2}{x-3}=\frac{3\cdot\frac{5}{2}+2}{\frac{5}{2}-3}=\frac{\frac{15}{2}+2}{\frac{5}{2}-3}=\frac{\frac{19}{2}}{-\frac{1}{2}}=\frac{19}{2}\cdot(-2)=\frac{19}{1}\cdot(-1)=-19\)
b, Ta có : \(\frac{3x+2}{x-3}=\frac{3x-9+11}{x-3}=\frac{3(x-3)+11}{x-3}=3+\frac{11}{x-3}\)
\(\Leftrightarrow11⋮x-3\Leftrightarrow x-3\inƯ(11)=\left\{\pm1;\pm11\right\}\)
Lập bảng :
| x - 3 | 1 | -1 | 11 | -11 |
| x | 4 | 2 | 14 | -8 |
c,Để suy nghĩ đã
Làm tiếp :v
c, \(B=\frac{x^2+3x-7}{x+3}=\frac{x(x+3)-7}{x+3}=x-\frac{7}{x+3}\)
\(\Rightarrow7⋮x+3\Leftrightarrow x+3\inƯ(7)=\left\{\pm1;\pm7\right\}\)
Lập bảng :
| x + 3 | 1 | -1 | 7 | -7 |
| x | -2 | -4 | 4 | -10 |
d, Tương tự
Bài 2:
a: Để B=1 thì \(2x^2+1=4\)
\(\Leftrightarrow x^2=\dfrac{3}{2}\)
hay \(x=\pm\dfrac{\sqrt{6}}{2}\)
b: Để B là số nguyên thì \(2x^2+1\inƯ\left(4\right)\)
\(\Leftrightarrow2x^2+1\in\left\{1;2;4\right\}\)
hay \(x\in\left\{0;\dfrac{\sqrt{2}}{2};-\dfrac{\sqrt{2}}{2};-\dfrac{\sqrt{6}}{2};\dfrac{\sqrt{6}}{2}\right\}\)