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

1.

a, Để \(\dfrac{x+1}{x^2-2}\) có nghĩa \(\Leftrightarrow x^2-2\ne0\Leftrightarrow x^2\ne2\Leftrightarrow\left\{{}\begin{matrix}x\ne\sqrt{2}\\x\ne-\sqrt{2}\end{matrix}\right.\)

b, Để \(\dfrac{x-1}{x^2+1}\)có nghĩa \(\Leftrightarrow x^2+1\ne0\Leftrightarrow x^2\ne-1\)

Vì \(x^2\ge0\forall x\in R\).

Vậy biểu thức trên luôn luôn có nghĩa.

c, Để \(\dfrac{ax+by+c}{xy-3y}cónghĩa\Leftrightarrow xy-3y=y\left(x-3\right)\ne0\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\y\ne3\end{matrix}\right.\).