tìm điều kiện bài toán:
a) \(y=\dfrac{1}{x}-\dfrac{\sqrt{2x-1}}{x^2-3x+2}\)
b) \(y=\dfrac{1}{x^2-1}-\sqrt{7-2x}\)
c) \(y=\dfrac{2}{x}+\dfrac{3}{4-2x+x^2}\)
d) \(y=\sqrt{25-x^2}-2\sqrt{x}+3\)
Tìm điều kiện để các biểu thức sau xác định
a)\(\sqrt{x+1}-\dfrac{1}{2}\)
b)\(2.\sqrt{1-2x}-\dfrac{\sqrt{3}-1}{4}\)
c)\(\sqrt{x+1}-\sqrt{x-2}\)
d)\(\sqrt{2-3x}-\sqrt{1-2x}\)
e)\(2.\sqrt{\sqrt{3}-2x}+\dfrac{1}{x-1}\)
f)\(\dfrac{1}{2}.\sqrt{x-\dfrac{\sqrt{3}}{2}}-\dfrac{1}{\sqrt{x}-1}\)
g)\(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+2}\)
h)\(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x^2+2}}\)
a, \(x+1\ge0\Leftrightarrow x\ge-1\)
b, \(1-2x\ge0\Leftrightarrow x\le\dfrac{1}{2}\)
c, \(\left\{{}\begin{matrix}x+1\ge0\\x-2\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\ge2\end{matrix}\right.\Leftrightarrow x\ge2\)
d, \(\left\{{}\begin{matrix}2-3x\ge0\\1-2x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{2}{3}\\x\le\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x\le\dfrac{1}{2}\)
e, \(\left\{{}\begin{matrix}\sqrt{3}-2x\ge0\\x-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{\sqrt{3}}{2}\\x\ne1\end{matrix}\right.\Leftrightarrow x\le\dfrac{\sqrt{3}}{2}\)
f, \(\left\{{}\begin{matrix}x-\dfrac{\sqrt{3}}{2}\ge0\\x\ge0\\\sqrt{x}-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{\sqrt{3}}{2}\\x\ge0\\x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{\sqrt{3}}{2}\\x\ne1\end{matrix}\right.\)
g, \(\left\{{}\begin{matrix}\sqrt{x}-1\ne0\\\sqrt{x}+2\ne0\\x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
1. Tìm hàm số xác định của các hàm số sau.
a) \(y=\dfrac{x}{x^2-3x+2}\)
b)\(y=\dfrac{x-1}{2x^2-5x+2}\)
c)\(y=\dfrac{x-1}{x^3+1}\)
d) \(y=\dfrac{1}{x^4+2x^2-3}\)
e) \(y=\sqrt{x+3-2\sqrt{x+2}}\)
a)x khác 1;2 b)x khác 2;1/2 c)x khác -1 d)x khác 1 e x>/=-2
Bài 1: Tìm x; y ϵ \(ℤ\)
a) 2x - y\(\sqrt{6}\) = 5 + (x + 1)\(\sqrt{6}\)
b) 5x + y - (2x -1)\(\sqrt{7}\) = y\(\sqrt{7}\) + 2
Bài 2: So sánh M và N
M = \(\dfrac{\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{6}{4}+\dfrac{6}{5}+\dfrac{6}{7}-\dfrac{6}{11}}\)
N = \(\dfrac{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}{\dfrac{6}{2}+\dfrac{6}{5}-\dfrac{6}{7}-\dfrac{6}{11}}\)
Bài 3: Chứng minh:
\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)
Bài 3 :
\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}\)
\(\dfrac{1}{2!}=\dfrac{1}{2.1}=1-\dfrac{1}{2}< 1\)
\(\dfrac{1}{3!}=\dfrac{1}{3.2.1}=1-\dfrac{1}{2}-\dfrac{1}{3}< 1\)
\(\dfrac{1}{4!}=\dfrac{1}{4.3.2.1}< \dfrac{1}{3!}< \dfrac{1}{2!}< 1\)
.....
\(\)\(\dfrac{1}{2023!}=\dfrac{1}{2023.2022....2.1}< \dfrac{1}{2022!}< ...< \dfrac{1}{2!}< 1\)
\(\Rightarrow\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)
Tìm điều kiện có nghĩa:
1) \(\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
2) \(\sqrt{\dfrac{2}{x^2+2x+2}}\)
3) \(\sqrt{\dfrac{-3}{x^2-4x+5}}\)
1) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\y\ge0\\\sqrt{x}+\sqrt{y}\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x>0\\y>0\end{matrix}\right.\)
2) ĐKXĐ: \(x^2+2x+2>0\Leftrightarrow\left(x^2+2x+1\right)+1>0\Leftrightarrow\left(x+1\right)^2+1>0\left(đúng\forall x\right)\)
3) ĐKXĐ: \(x^2-4x+5< 0\Leftrightarrow\left(x^2-4x+4\right)+1< 0\Leftrightarrow\left(x-2\right)^2+1< 0\left(VLý.do.\left(x-2\right)^2+1\ge1>0\right)\)
Vậy biểu thức không xác định với mọi x
Đkien
a) \(\left\{{}\begin{matrix}x\ge0;y\ge0\\\sqrt[]{x}+\sqrt{y}\ne0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge0,y>0\\x>0,y\ge0\end{matrix}\right.\)
b) \(\dfrac{2}{x^2+2x+2}\ge0\Leftrightarrow x^2+2x+2>0\)
\(\Leftrightarrow x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}>0\forall x\)
=> PT luôn xác định
c) \(-\dfrac{3}{x^2-4x+5}\ge0\Leftrightarrow x^2-4x+5< 0\)
\(\)=> vô nghiệm
Vậy căn thức k xác định
Tìm tập xác định của hàm số :
a. y=\(\dfrac{1}{x^2-2x}+\sqrt{x^2-1}\)
b.y=\(\sqrt{x+1}+\sqrt{5-3x}\)
c.y=\(\sqrt{5x+3}+\dfrac{2x}{\sqrt{3-x}}\)
d.y=\(\dfrac{3x}{\sqrt{4-x^2}}+\sqrt{1+x}\)
e.y=\(\dfrac{5-2x}{(2-3x)\sqrt{1-6x}}\)
a: ĐKXĐ: x^2-2x<>0 và x^2-1>0
=>(x>1 và x<>2) hoặc x<-1
b: ĐKXĐ: x+1>0 và 5-3x>0
=>x>-1 và 3x<5
=>-1<x<5/3
c: DKXĐ: 5x+3>=0 và 3-x>0
=>x>=-3/5 và x<3
=>-3/5<=x<3
d: ĐKXĐ: 4-x^2>0 và 1+x>=0
=>x^2<4 và x>=-1
=>-2<x<2 và x>=-1
=>-1<=x<2
e: ĐKXĐ: 2-3x<>0 và 1-6x>0
=>x<>2/3 và x<1/6
=>x<1/6
1) \(\dfrac{x-3x^2}{2}+\sqrt{2x^4-x^3+7x^2-3x+3}=2\)
2) \(1+\sqrt{\dfrac{x-2}{1-x}}=\dfrac{2x^2-2x+1}{x^2-2x+2}\)
3) \(x+y+z+\dfrac{3}{x-1}+\dfrac{3}{y-1}+\dfrac{3}{z-1}=2\left(\sqrt{x+2}+\sqrt{y+2}+\sqrt{z+2}\right)\) với x ,y ,z > 1
4) \(\sqrt[3]{x+6}+x^2=7-\sqrt{x-1}\)
5) \(x^4-2x^3+x-\sqrt{2\left(x^2-x\right)}=0\)
Tính đạo hàm:
a) y= \(\dfrac{x^3+2\sqrt{x-1}}{x-1}\)
b) y= \(\dfrac{4x^3+2x-3}{\sqrt{x^2+2}}\)
c) y= \(|x^3+x+1|\)
d) y= \(\sqrt{7-6x^4+x^3}\)
e) y= \(\dfrac{x^5+1}{2-\sqrt{x^2+3}}\)
a/ \(y'=\dfrac{\left(x^3+2\sqrt{x-1}\right)'\left(x-1\right)-\left(x-1\right)'\left(x^3+2\sqrt{x-1}\right)}{\left(x-1\right)^2}\)
\(y'=\dfrac{\left(2x^2+\dfrac{1}{\sqrt{x-1}}\right)\left(x-1\right)-x^3-2\sqrt{x-1}}{\left(x-1\right)^2}=\dfrac{x^3-2x^2-\sqrt{x-1}}{\left(x-1\right)^2}\)
b/ \(y'=\dfrac{\left(4x^3+2x-3\right)'\left(\sqrt{x^2+2}\right)-\left(\sqrt{x^2+2}\right)'\left(4x^3+2x-3\right)}{x^2+2}\)
\(y'=\dfrac{\left(12x^2+2\right)\sqrt{x^2+2}-\dfrac{x}{\sqrt{x^2+2}}\left(4x^3+2x-3\right)}{x^2+2}\) (ban tu rut gon nhe)
c/ \(y'=\dfrac{\left(x^3+x+1\right)'\left(x^3+x+1\right)}{\left|x^3+x+1\right|}=\dfrac{\left(3x^2+1\right)\left(x^3+x+1\right)}{\left|x^3+x+1\right|}\)
d/ \(y'=\dfrac{3x^2-24x^3}{2\sqrt{x^3-6x^4+7}}\)
e/ \(y'=\dfrac{\left(x^5+1\right)'\left(2-\sqrt{x^2+3}\right)-\left(x^5+1\right)\left(2-\sqrt{x^2+3}\right)'}{\left(2-\sqrt{x^2+3}\right)^2}\)
\(y'=\dfrac{5x^4\left(2-\sqrt{x^2+3}\right)+\left(x^5+1\right)\dfrac{x}{\sqrt{x^2+3}}}{\left(2-\sqrt{x^2+3}\right)^2}\)
Tìm điều kiện xác định
\(A=\sqrt{x^2-5x+6}\)
\(B=\dfrac{x}{\sqrt{7x^2-8}}\)
\(C=\sqrt{-9x^2+6x-1}-\dfrac{1}{\sqrt{x^2+x+2}}\)
\(D=\sqrt{3-x^2}-\sqrt{\dfrac{2021}{3x+2}}\)
\(E=\sqrt{\dfrac{3x^2}{2x+1}-1}\)
\(F=\sqrt{25x^2-10x+1}+\dfrac{1}{1-5x}\)
a: ĐKXĐ: \(\left[{}\begin{matrix}x\ge3\\x\le2\end{matrix}\right.\)
b: ĐKXĐ: \(\left[{}\begin{matrix}x>\dfrac{2\sqrt{14}}{7}\\x< -\dfrac{2\sqrt{14}}{7}\end{matrix}\right.\)
c: ĐKXĐ: \(x=\dfrac{1}{3}\)
d: ĐKXĐ: \(-\dfrac{2}{3}< x\le\sqrt{3}\)
Tìm tập xác định
a) y=\(\dfrac{x-1}{\left(2x^2-5x+2\right)\left(x^3+1\right)}\)
b)y=\(\dfrac{3x\left(x^2-1\right)}{\left(x^2+2x+2\right)\left(x+5\right)}\)
c)y=\(\dfrac{x-1}{x^4-1}\)
d)\(\dfrac{1}{x^4+2x^2-3}\)
e)y=\(\dfrac{x+2}{x^3+2x^2-3x-6}\)
g) y=\(\sqrt{4-x}+\sqrt{5x+1}\)
h)y=\(\dfrac{1+x}{\left(x^2+2x-8\right)\sqrt{x-1}}\)
i)y=\(\dfrac{\sqrt{5-2x}}{\left(2x^2-5x+2\right)\sqrt{x-1}}\)
a: ĐKXĐ: \(\left(2x^2-5x+2\right)\left(x^3+1\right)< >0\)
=>(2x-1)(x-2)(x+1)<>0
hay \(x\notin\left\{\dfrac{1}{2};2;-1\right\}\)
b: ĐKXĐ: x+5<>0
=>x<>-5
c: ĐKXĐ: x4-1<>0
hay \(x\notin\left\{1;-1\right\}\)
d: ĐKXĐ: \(x^4+2x^2-3< >0\)
=>\(x\notin\left\{1;-1\right\}\)