Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) TXĐ: \(D=R\).
b) \(TXD=D=R\backslash\left\{4\right\}\)
c) Đkxđ: \(\left\{{}\begin{matrix}4x+1\ge0\\-2x+1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{-1}{4}\\x\le\dfrac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\dfrac{-1}{4}\le x\le\dfrac{1}{2}\).
TXĐ: D = \(\left[\dfrac{-1}{4};\dfrac{1}{2}\right]\)
a) Đkxđ: \(\left\{{}\begin{matrix}x+9\ge0\\x^2+8x-20\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-9\\\left\{{}\begin{matrix}x\ne2\\x\ne-10\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-9\\x\ne2\end{matrix}\right.\)
Txđ: D = [ - 9; 2) \(\cup\) \(\left(2;+\infty\right)\)
b) Đkxđ: \(\left\{{}\begin{matrix}2x+1\ne0\\x-3\ne0\end{matrix}\right.\Leftrightarrow\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{-1}{2}\\x\ne3\end{matrix}\right.\)
Txđ: \(D=R\backslash\left\{\dfrac{-1}{2};3\right\}\)
c) \(x^2+2x-5\ne0\Leftrightarrow\left\{{}\begin{matrix}x\ne-1+\sqrt{6}\\x\ne-1-\sqrt{6}\end{matrix}\right.\)
Txđ: \(D=R\backslash\left\{-1+\sqrt{6};-1-\sqrt{6}\right\}\)
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: \(x^4+2x^2-3< >0\)
=>\(x\notin\left\{1;-1\right\}\)
5,\(hpt\Leftrightarrow\left\{{}\begin{matrix}x\left(x+y\right)\left(x+2\right)=0\\2\sqrt{x^2-2y-1}+\sqrt[3]{y^3-14}=x-2\end{matrix}\right.\)
Thay từng TH rồi làm nha bạn
3,\(hpt\Leftrightarrow\left\{{}\begin{matrix}x-y=\frac{1}{x}-\frac{1}{y}=\frac{y-x}{xy}\\2y=x^3+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(1+\frac{1}{xy}\right)=0\\2y=x^3+1\end{matrix}\right.\)
thay nhá
Bài 1:ĐKXĐ: \(2x\ge y;4\ge5x;2x-y+9\ge0\)\(\Rightarrow2x\ge y;x\le\frac{4}{5}\Rightarrow y\le\frac{8}{5}\)
PT(1) \(\Leftrightarrow\left(x-y-1\right)\left(2x-y+3\right)=0\)
+) Với y = x - 1 thay vào pt (2):
\(\frac{2}{3+\sqrt{x+1}}+\frac{2}{3+\sqrt{4-5x}}=\frac{9}{x+10}\) (ĐK: \(-1\le x\le\frac{4}{5}\))
Anh quy đồng lên đê, chắc cần vài con trâu đó:))
+) Với y = 2x + 3...
1: ĐKXĐ: \(\left|x^2-4\right|+\left|x+2\right|< >0\)
\(\Leftrightarrow x\ne-2\)
2: ĐKXĐ: \(\left|x-2\right|-\left|x+1\right|< >0\)
\(\Leftrightarrow\left|x-2\right|< >\left|x+1\right|\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2< >x+1\\x-2< >-x-1\end{matrix}\right.\Leftrightarrow2x< >1\Leftrightarrow x< >\dfrac{1}{2}\)
3: ĐKXĐ: \(\left\{{}\begin{matrix}2x+11>=0\\\left\{{}\begin{matrix}3x-2< >4\\3x-2< >-4\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{11}{2}\\x\notin\left\{2;-\dfrac{2}{3}\right\}\end{matrix}\right.\)
a. R / \(\left\{-2\right\}\)
b. R / \(\left\{4;-1\right\}\)
c. R ( mẫu luôn > 0 )
d. \(\left(2;+\infty\right)\)
e. \(\left(-\infty;\dfrac{5}{6}\right)\)
f. \(\left(2;+\infty\right)\)
g. \(\left(1;3\right)\)
h. \(\left(5;+\infty\right)\)
i. \(\left(1;+\infty\right)\)
k. \(\left(-\infty;2\right)\)
l. R/\(\left\{\pm3\right\}\)
m. \(\left(-2;+\infty\right)/\left\{3\right\}\)
a: ĐKXĐ: x-1>0 và x+2<>0
=>x>1
b: DKXĐ: (x-2)căn x-1<>0
=>x-1>0 và x-2<>0
=>x>1 và x<>2
c: ĐKXĐ: 2x-1>=0 và 3-x>0
=>x>=1/2 và x<3
d: ĐKXĐ: x-1>0 và x-2<>0
=>x>1 và x<>2
e: ĐKXĐ: x3+1>=0
=>x>=-1
1) \(y=\dfrac{2x^2+1}{x^3-5x+4}\)
ĐK \(x^3-5x+4\ne0\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne\dfrac{\sqrt{17}-1}{2}\\x\ne\dfrac{-\sqrt{17}-1}{2}\end{matrix}\right.\)
TXĐ \(D=R\backslash\left\{1;\dfrac{\sqrt{17}-1}{2};\dfrac{-\sqrt{17}-1}{2}\right\}\)
2) \(y=\dfrac{\sqrt{x-2}}{\left(x-3\right)^3-1}\)
ĐK \(\left\{{}\begin{matrix}x-2\ge0\\x-3\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ne4\end{matrix}\right.\)
TXĐ \(D=[2;+\infty)\backslash\left\{4\right\}\)
3) \(y=\sqrt{x-2}-\dfrac{2}{\sqrt[3]{x-1}}\)
ĐK\(\left\{{}\begin{matrix}x+2\ge0\\x-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\x\ne1\end{matrix}\right.\)
TXĐ \(D=[-2;+\infty)\backslash\left\{1\right\}\)
4) \(y=\dfrac{x^2+2}{\sqrt{\left(x+3\right)^2}}=\dfrac{x^2+2}{\left|x-3\right|}\)
ĐK \(x-3\ne0\Leftrightarrow x\ne3\)
TXĐ \(D=R\backslash\left\{3\right\}\)
5) \(y=\dfrac{\sqrt{x^2-2}}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
ĐK \(\left\{{}\begin{matrix}x^2-2\ge0\\x>0\\\sqrt{x}-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in(-\infty;-\sqrt{2}]\cap[\sqrt{2};+\infty)\\x>0\\x\ne9\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\ge\sqrt{2}\\x\ne9\end{matrix}\right.\)
TXĐ \(D=[\sqrt{2};+\infty)\backslash\left\{9\right\}\)
6) \(y=\sqrt{1-\sqrt{1+x}}\)
ĐK \(\left\{{}\begin{matrix}x+1\ge0\\1-\sqrt{1+x}\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\1\ge\sqrt{1+x}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\1\ge1+x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\le0\end{matrix}\right.\)
TXĐ \(D=\left[0;-1\right]\)
Lời giải:
\(y=\frac{\sqrt{2x-5}}{|x|-3}\)
ĐK: \(\left\{\begin{matrix} 2x-5\geq 0\\ |x|-3\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{5}{2}\\ x\neq \pm 3\end{matrix}\right.\)
\(\Rightarrow x\geq \frac{5}{2}; x\neq 3\)
Vậy TXĐ là \(x\in [\frac{5}{2}; +\infty)\setminus \left\{3\right\}\)
------------
\(y=\frac{|x|}{\sqrt{x-2}}+\frac{5x^2}{-x^2+6x-5}\)
ĐK: \(\left\{\begin{matrix} x-2>0\\ -x^2+6x-5\neq 0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x>2\\ (5-x)(x-1)\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x>2\\ x\neq 1; x\neq 5\end{matrix}\right.\)
Vậy TXĐ: \(x\in (2;+\infty)\setminus \left\{1;5\right\}\)
-----------
\(y=\frac{2x}{\sqrt{x+1}}+\frac{3x}{x^2+1}\)
ĐK: \(\left\{\begin{matrix} x+1>0\\ x^2+1\neq 0\end{matrix}\right.\Leftrightarrow x>-1\)
Vậy TXĐ: \(x\in (-1;+\infty)\)
5. \(y=\dfrac{-3x}{x+2}\)
xác định khi: \(x+2\ne0\Leftrightarrow x\ne-2\)
vậy D= (\(-\infty;+\infty\))\{-2}
6. \(y=\sqrt{-2x-3}\)
xác định khi: \(-2x-3\ge0\Leftrightarrow x\le\dfrac{-3}{2}\)
vậy D= (\(-\infty;\dfrac{-3}{2}\)]
7. \(y=\dfrac{3-x}{\sqrt{x-4}}\)
xác định khi: x-4 >0 <=> x>4
vậy D= (\(4;+\infty\))
8. \(y=\dfrac{2x-5}{\left(3-x\right)\sqrt{5-x}}\)
xác định khi: \(\left\{{}\begin{matrix}3-x\ne0\\5-x>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x< 5\end{matrix}\right.\)
vậy D= (\(-\infty;5\))\ {3}
9.\(y=\sqrt{2x+1}+\sqrt{4-3x}\)
xác định khi: \(\left\{{}\begin{matrix}2x+1\ge0\\4-3x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{-1}{2}\\x\le\dfrac{4}{3}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{-1}{2}\le x\le\dfrac{4}{3}\)
vậy D= [\(\dfrac{-1}{2};\dfrac{4}{3}\)]
1. \(y=\dfrac{3x-2}{x^2-4x+3}\)
xác định khi : \(x^2-4x+3\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne1\end{matrix}\right.\)
vậy tập xác định là: D = \(\left(-\infty;+\infty\right)\backslash\left\{3;1\right\}\)
2.\(y=2\sqrt{5-4x}\)
xác định khi \(5-4x\ge0\Leftrightarrow x\le\dfrac{5}{4}\)
vậy D= (\(-\infty;\dfrac{5}{4}\)]
3. \(y=\dfrac{2}{\sqrt{x+3}}+\sqrt{5-2x}\)
xác định khi: \(\left\{{}\begin{matrix}x+3>0\\5-2x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-3\\x\le\dfrac{5}{2}\end{matrix}\right.\)
\(\Leftrightarrow-3< x\le\dfrac{5}{2}\)
vậy D= (\(-3;\dfrac{5}{2}\)]
4.\(\sqrt{9-x}+\dfrac{1}{\sqrt{x+2}-2}\)
xác định khi: \(\left\{{}\begin{matrix}9-x\ge0\\x+2\ge0\\x\ne2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le9\\x\ge-2\\x\ne2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2\le x\le9\\x\ne2\end{matrix}\right.\)
Vậy D= [\(-2;9\)]\{2}