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a) Để biểu thức xác định thì \(3x^2+2\ne0\forall x\in R\)
vậy với mọi x thì biểu thức trên luôn xác định.
b) Để .......
\(\left\{{}\begin{matrix}2x+5\ge0\\x-1>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-\frac{5}{2}\\x>1\end{matrix}\right.\)
vậy biểu thức trên xác định khi x>1.
c) Để ..........
\(\left\{{}\begin{matrix}x+1\ge0\\x^2-2x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\\left\{{}\begin{matrix}x\ne0\\x\ne2\end{matrix}\right.\end{matrix}\right.\)
Vậy để biểu thức xđ khi \(x\in[-1;+\infty)\backslash\left\{0;2\right\}\)
d) Để ........
\(\left\{{}\begin{matrix}2x+3\ge0\\5-x\ge\\2-\sqrt{5-x}\ne0\end{matrix}\right.0\Leftrightarrow\left\{{}\begin{matrix}x\ge-\frac{3}{2}\\x\le5\\x\ne1\end{matrix}\right.\)
Vậy để btxđ khi \(x\in\left[-\frac{3}{2};5\right]\backslash\left\{1\right\}\)
e) Để ......
\(\left\{{}\begin{matrix}x+2\ge0\\3-2x\ge0\\\left|x\right|-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\x\le\\\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\end{matrix}\right.\frac{3}{2}\)
Vậy để btxđ khi ....
a/ ĐKXĐ: ...
\(\Leftrightarrow3\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)-7\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow a^2=x+\frac{1}{4x}+1\)
\(\Rightarrow x+\frac{1}{4x}=a^2-1\)
Pt trở thành:
\(3a=2\left(a^2-1\right)-7\)
\(\Leftrightarrow2a^2-3a-9=9\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=3\)
\(\Leftrightarrow2x-6\sqrt{x}+1=0\)
\(\Rightarrow\sqrt{x}=\frac{3+\sqrt{7}}{2}\Rightarrow x=\frac{8+3\sqrt{7}}{2}\)
b/ ĐKXĐ:
\(\Leftrightarrow5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow x+\frac{1}{4x}=a^2-1\)
\(\Rightarrow5a=2\left(a^2-1\right)+4\Leftrightarrow2a^2-5a+2=0\)
\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{2\sqrt{x}}=2\\\sqrt{x}+\frac{1}{2\sqrt{x}}=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-4\sqrt{x}+1=0\\2x-\sqrt{x}+1=0\left(vn\right)\end{matrix}\right.\)
c/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)
\(\Leftrightarrow\frac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)
\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\frac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)
\(\Leftrightarrow2x^2-8x+5=0\)
d/ ĐKXĐ: ...
\(\Leftrightarrow x+1-\frac{15}{6}\sqrt{x}+\sqrt{x^2-4x+1}-\frac{1}{2}\sqrt{x}=0\)
\(\Leftrightarrow\frac{x^2-\frac{17}{4}x+1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{x^2-\frac{17}{4}x+1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}=0\)
\(\Leftrightarrow\left(x^2-\frac{17}{4}x+1\right)\left(\frac{1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}\right)=0\)
\(\Leftrightarrow x^2-\frac{17}{4}x+1=0\)
\(\Leftrightarrow4x^2-17x+4=0\)
a) y xác định \(\Leftrightarrow2x^2-5x+2\ne0\Leftrightarrow\left(x-2\right)\left(2x-1\right)\ne0\Leftrightarrow\left\{{}\begin{matrix}x-2\ne0\\2x-1\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ne\frac{1}{2}\end{matrix}\right.\). Vậy tập xác định D = R / { 2; 1/2}
b) y xác định \(\Leftrightarrow\left\{{}\begin{matrix}x-1\ne0\\2x+4\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ge-2\end{matrix}\right.\).
Vậy tập xác định D = \([-2;+\infty)/1\)
y xác định \(\Leftrightarrow x^2-3x+m-1\ne0\forall x\in R\)
suy ra phương trình x2 - 3x + m - 1 = 0 vô nghiệm
\(\Rightarrow\Delta=9-4\left(m-1\right)< 0\Leftrightarrow9-4m+4< 0\Leftrightarrow m>\frac{13}{4}\)
\(\Rightarrow m\in\left(\frac{13}{4};+\infty\right)\)
a)
ĐK: $x-2\geq 0\Leftrightarrow x\geq 2$
TXĐ: $[2;+\infty)$
b)
ĐK: $4x-3\geq 0\Leftrightarrow x\geq \frac{3}{4}$
TXĐ: $[\frac{3}{4};+\infty)$
c) ĐK: \(x+2>0\Leftrightarrow x>-2\)
TXĐ: $(-2;+\infty)$
d)
ĐK: $3-x>0\Leftrightarrow x< 3$
TXĐ: $(-\infty; 3)$
e)
$4-3x>0\Leftrightarrow x< \frac{4}{3}$
TXĐ: $(-\infty; \frac{4}{3})$
f)
ĐK:\(\left\{\begin{matrix} x^2+2\geq 0\\ x\geq 0\end{matrix}\right.\Leftrightarrow x\geq 0\)
TXĐ: $[0;+\infty)$
g) ĐK: \(\left\{\begin{matrix} x^2-2x+1\geq 0\\ 2-3x\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} (x-1)^2\geq 0\\ x\leq\frac{2}{3}\end{matrix}\right.\Leftrightarrow x\leq \frac{2}{3}\)
TXĐ: $(-\infty; \frac{2}{3}]$
h)
ĐK: \(\left\{\begin{matrix} 2+x\geq 0\\ x-2\geq 0\end{matrix}\right.\Leftrightarrow x\geq 2\)
TXĐ: $[2;+\infty)$
i)
ĐK: \(\left\{\begin{matrix} 2+x\geq 0\\ 2-x\geq 0\end{matrix}\right.\Leftrightarrow 2\geq x\geq -2\)
TXĐ: $[-2;2]$
8.
ĐKXĐ: \(x\ge\frac{2}{3}\)
\(\Leftrightarrow\frac{9\left(x+3\right)}{\sqrt{4x+1}+\sqrt{3x-2}}=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\left(l\right)\\\frac{9}{\sqrt{4x+1}+\sqrt{3x-2}}=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{4x+1}+\sqrt{3x-2}=9\)
\(\Leftrightarrow\sqrt{4x+1}-5+\sqrt{3x-2}-4=0\)
\(\Leftrightarrow\frac{4\left(x-6\right)}{\sqrt{4x+1}+5}+\frac{3\left(x-6\right)}{\sqrt{3x-2}+4}=0\)
\(\Leftrightarrow\left(x-6\right)\left(\frac{4}{\sqrt{4x+1}+5}+\frac{3}{\sqrt{3x-2}+4}\right)=0\)
\(\Leftrightarrow x=6\)
6.
ĐKXD: ...
\(\Leftrightarrow2\left(x^2-6x+9\right)+\left(x+5-4\sqrt{x+1}\right)=0\)
\(\Leftrightarrow2\left(x-3\right)^2+\frac{\left(x-3\right)^2}{x+5+4\sqrt{x+1}}=0\)
\(\Leftrightarrow\left(x-3\right)^2\left(2+\frac{1}{x+5+4\sqrt{x+1}}\right)=0\)
\(\Leftrightarrow x=3\)
7.
\(\sqrt{x-\frac{1}{x}}-\sqrt{2x-\frac{5}{x}}+\frac{4}{x}-x=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x-\frac{1}{x}}=a\ge0\\\sqrt{2x-\frac{5}{x}}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=\frac{4}{x}-x\)
\(\Rightarrow a-b+a^2-b^2=0\)
\(\Leftrightarrow\left(a-b\right)\left(a+b+1\right)=0\)
\(\Leftrightarrow a=b\Leftrightarrow x-\frac{1}{x}=2x-\frac{5}{x}\)
\(\Leftrightarrow x=\frac{4}{x}\Rightarrow x=\pm2\)
Thế nghiệm lại pt ban đầu để thử (hoặc là bạn tìm ĐKXĐ từ đầu)