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Tìm miền xác định phải không
a)
\(1-\sqrt{2x-x^2}\)
a xác định \(\Leftrightarrow2x-x^2\ge0\)
\(0\le x\le2\)
b)
\(\sqrt{-4x^2+4x-1}\)
b xác định
\(\Leftrightarrow-4x^2+4x-1\ge0\)
\(-\left(4x^2-4x+1\right)\ge0\)
\(4x^2-4x+1\le0\)
\(\left(2x-1\right)^2\le0\)
2x - 1 = 0
x = 1/2
c)
\(\frac{x}{\sqrt{5x^2-3}}\)
c xác định
\(\Leftrightarrow5x^2-3>0\)
\(5x^2>3\)
\(x^2>\frac{3}{5}\)
\(\orbr{\begin{cases}x< -\frac{\sqrt{15}}{5}\\x>\frac{\sqrt{15}}{5}\end{cases}}\)
d)
d xác định
\(\Leftrightarrow\sqrt{x-\sqrt{2x-1}}>0\)
\(x-\sqrt{2x-1}>0\)
\(x>\sqrt{2x-1}\)
\(\hept{\begin{cases}2x-1\ge0\\x^2>2x-1\end{cases}}\)
\(\hept{\begin{cases}x\ge\frac{1}{2}\\x^2-2x+1>0\end{cases}}\)
\(\hept{\begin{cases}x\ge\frac{1}{2}\\\left(x-1\right)^2>0\end{cases}}\)
\(\hept{\begin{cases}x\ge\frac{1}{2}\\x-1\ne0\end{cases}}\)
\(\hept{\begin{cases}x\ge\frac{1}{2}\\x\ne1\end{cases}}\)
e)
e xác định
\(\Leftrightarrow\frac{-2x^2}{3x+2}\ge0\)
\(3x+2< 0\) ( vì \(-2x^2\le0\forall x\) )
\(x< -\frac{2}{3}\)
f)
f xác định
\(\Leftrightarrow x^2+x-2>0\)
\(\orbr{\begin{cases}x< -2\\x>1\end{cases}}\)
+) Ta có: \(4\sqrt{3x}+\sqrt{12x}=\sqrt{27x}+6\) \(\left(ĐK:x\ge0\right)\)
\(\Leftrightarrow4\sqrt{3x}+2\sqrt{3x}=3\sqrt{3x}+6\)
\(\Leftrightarrow3\sqrt{3x}=6\)
\(\Leftrightarrow\sqrt{3x}=2\)
\(\Leftrightarrow3x=4\)
\(\Leftrightarrow x=\frac{4}{3}\left(TM\right)\)
Vậy \(S=\left\{\frac{4}{3}\right\}\)
+) Ta có:\(\sqrt{x^2-1}-4\sqrt{x-1}=0\) \(\left(ĐK:x\ge1\right)\)
\(\Leftrightarrow\sqrt{x-1}.\sqrt{x+1}-4\sqrt{x-1}=0\)
\(\Leftrightarrow\sqrt{x-1}.\left(\sqrt{x+1}-4\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-1}=0\\\sqrt{x+1}-4=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-1=0\\\sqrt{x+1}=4\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-1=0\\x+1=16\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=1\left(TM\right)\\x=15\left(TM\right)\end{cases}}\)
Vậy \(S=\left\{1,15\right\}\)
+) Ta có: \(\frac{\sqrt{x}-2}{2\sqrt{x}}< \frac{1}{4}\) \(\left(ĐK:x\ge0\right)\)
\(\Leftrightarrow\frac{\sqrt{x}-2}{2\sqrt{x}}-\frac{1}{4}< 0\)
\(\Leftrightarrow\frac{2.\left(\sqrt{x}-2\right)-\sqrt{x}}{4\sqrt{x}}< 0\)
\(\Leftrightarrow\frac{2\sqrt{x}-4-\sqrt{x}}{4\sqrt{x}}< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-4}{4\sqrt{x}}< 0\)
Để \(\frac{\sqrt{x}-4}{4\sqrt{x}}< 0\)mà \(4\sqrt{x}\ge0\forall x\)
\(\Rightarrow\)\(\sqrt{x}-4< 0\)
\(\Leftrightarrow\)\(\sqrt{x}< 4\)
\(\Leftrightarrow\)\(x< 16\)
Kết hợp ĐKXĐ \(\Rightarrow\)\(0\le x< 16\)
Vậy \(S=\left\{\forall x\inℝ/0\le x< 16\right\}\)
\(4\sqrt{3x}+\sqrt{12x}=\sqrt{27x}+6\) (Đk: x \(\ge\)0)
<=> \(4\sqrt{3x}+2\sqrt{3x}-3\sqrt{3x}=6\)
<=> \(3\sqrt{3x}=6\)
<=> \(\sqrt{3x}=2\)
<=> \(3x=4\)
<=> \(x=\frac{4}{3}\)
\(\sqrt{x^2-1}-4\sqrt{x-1}=0\) (đk: x \(\ge\)1)
<=> \(\sqrt{x-1}.\sqrt{x+1}-4\sqrt{x-1}=0\)
<=> \(\sqrt{x-1}\left(\sqrt{x+1}-4\right)=0\)
<=> \(\orbr{\begin{cases}\sqrt{x-1}=0\\\sqrt{x+1}-4=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x-1=0\\x+1=16\end{cases}}\)
<=> \(\orbr{\begin{cases}x=1\\x=15\end{cases}}\)(tm)
\(\frac{\sqrt{x}-2}{2\sqrt{x}}< \frac{1}{4}\) (Đk: x > 0)
<=> \(\frac{\sqrt{x}-2}{2\sqrt{x}}-\frac{1}{4}< 0\)
<=>\(\frac{2\sqrt{x}-4-\sqrt{x}}{4\sqrt{x}}< 0\)
<=> \(\frac{\sqrt{x}-4}{4\sqrt{x}}< 0\)
Do \(4\sqrt{x}>0\) => \(\sqrt{x}-4< 0\)
<=> \(\sqrt{x}< 4\) <=> \(x< 16\)
Kết hợp với đk => S = {x|0 < x < 16}
a, ĐKXĐ : \(\left[{}\begin{matrix}x\ge0\\ y>0\end{matrix}\right.\) hoặc \(\left[{}\begin{matrix}x>0\\y\ge0\end{matrix}\right.\)
Ta có :\(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
= \(\frac{\sqrt{x^2}\sqrt{x}+\sqrt{y^2}\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=\frac{\sqrt{x^3}+\sqrt{y^3}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
= \(\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-\left(x-2\sqrt{xy}+y\right)\)
= \(\left(x-\sqrt{xy}+y\right)-\left(x-2\sqrt{xy}+y\right)\)
= \(x-\sqrt{xy}+y-x+2\sqrt{xy}-y\)
= \(\sqrt{xy}\)
\(\sqrt{\frac{\sqrt{a}-1}{\sqrt{b}+1}}:\sqrt{\frac{\sqrt{b}-1}{\sqrt{a}+1}}\) \(=\sqrt{\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{b}+1\right)\left(\sqrt{b}-1\right)}}\)\(=\sqrt{\frac{a^2-1}{b^2-1}}\) (*)
Thay a=7,25 và b= 3,25 vào (*) ta có:
\(\sqrt{\frac{7,25^2-1}{3,25^2-1}}\) \(=\frac{5\sqrt{33}}{4}:\frac{3\sqrt{17}}{4}=\frac{5\sqrt{33}}{3\sqrt{17}}=\frac{5\sqrt{561}}{51}\)
Bình phương 2 vế lên là được