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\(A=\frac{3}{4}.4.x^2\left(8-x^2\right)\le\frac{3}{4}\left(x^2+8-x^2\right)^2=48\)
\(A_{max}=48\) khi \(x^2=8-x^2\Rightarrow x=\pm2\)
\(B=\frac{1}{2}\left(2x-1\right)\left(6-2x\right)\le\frac{1}{8}\left(2x-1+6-2x\right)^2=\frac{25}{8}\)
\(B_{max}=\frac{25}{8}\) khi \(2x-1=6-2x\Rightarrow x=\frac{7}{4}\)
\(C=\frac{1}{\sqrt{3}}.\sqrt{3}x\left(3-\sqrt{3}x\right)\le\frac{1}{4\sqrt{3}}\left(\sqrt{3}x+3-\sqrt{3}x\right)^2=\frac{3\sqrt{3}}{4}\)
\(C_{max}=\frac{3\sqrt{3}}{4}\) khi \(\sqrt{3}x=3-\sqrt{3}x=\frac{\sqrt{3}}{2}\)
\(D=\frac{1}{20}.20x\left(32-20x\right)\le\frac{1}{80}\left(20x+32-20x\right)^2=\frac{64}{5}\)
\(D_{max}=\frac{64}{5}\) khi \(20x=32-20x\Rightarrow x=\frac{4}{5}\)
\(E=\frac{4}{5}\left(5x-5\right)\left(8-5x\right)\le\frac{1}{5}\left(5x-5+8-5x\right)=\frac{9}{5}\)
\(E_{max}=\frac{9}{5}\) khi \(5x-5=8-5x\Leftrightarrow x=\frac{13}{10}\)
A = \(\frac{3x}{2}+\frac{2}{x-1}=3.\frac{x-1}{2}+\frac{2}{x-1}+\frac{3}{2}\)\(\ge2\sqrt{3}+\frac{3}{2}\)
\(\Rightarrow\)min A = \(2\sqrt{3}+\frac{3}{2}\Leftrightarrow x=\frac{2}{\sqrt{3}}+1\)(thỏa mãn)
B = \(x+\frac{3}{3x-1}=\frac{1}{3}\left(3x-1+\frac{9}{3x-1}+1\right)\)\(\ge\frac{1}{3}\left(2\sqrt{9}+1\right)=\frac{7}{3}\)
\(\Rightarrow\)min B = \(\frac{7}{3}\Leftrightarrow x=\frac{4}{3}\)
\(A\) \(=\) \(3x^2\left(8-x^2\right)\le3\frac{\left(x^2+8-x^2\right)^2}{4}=48\)
\(\Rightarrow\) maxA = 48 \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)(thỏa mãn)
\(B=\) \(4x\left(8-5x\right)\)\(=\frac{4}{5}.5x\left(8-5x\right)\le\frac{4}{5}.\frac{\left(5x+8-5x\right)^2}{4}=\frac{64}{5}\)
\(\Rightarrow\)max B = \(\frac{64}{5}\Leftrightarrow x=\frac{4}{5}\)(thỏa mãn)
a. \(\sqrt{x+8}=x+2\)
đk x ≥ -2
⇔ \(\left(\sqrt{x+8}\right)^2\) = (x + 2 )2
⇔ x + 8 = x2 + 4x + 4
⇔ x2 + 3x - 4 = 0
⇔ (x - 1)(x + 4) = 0
⇔\(\left[{}\begin{matrix}x=1\\x=-4\left(L\right)\end{matrix}\right.\)
S = \(\left\{1\right\}\)
c: \(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{7}{3}\\9x^2-42x+49-5x-3=0\end{matrix}\right.\)
=>x>=7/3 và 9x^2-47x+46=0
=>\(x=\dfrac{47+\sqrt{553}}{18}\)
d: \(\left\{{}\begin{matrix}x>=-\dfrac{1}{3}\\3x^2-2x-1=9x^2+6x+1\end{matrix}\right.\)
=>x>=-1/3 và -6x^2-8x-2=0
=>x=-1/3
e: =>3x-5=16
=>3x=21
=>x=7
g: =>x<=3 và x^2+x+1=x^2-6x+9
=>x=8/7
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}
a) \(\sqrt{1+x}-\sqrt{8-x}+\sqrt{\left(1+x\right)\left(8-x\right)}=3\)
đặt t \(=\sqrt{1+x}-\sqrt{8-x}\)
\(\Leftrightarrow t^2=1+x-2\sqrt{\left(1+x\right)\left(8-x\right)}+8-x\)
\(\Leftrightarrow\sqrt{\left(1+x\right)\left(8-x\right)}=\dfrac{9-t^2}{2}\)
pt \(\Rightarrow t+\dfrac{9-t^2}{2}=3\)
\(\Leftrightarrow t^2-2t-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-1\\t=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{1+x}-\sqrt{8-x}=-1\\\sqrt{1+x}-\sqrt{8+x}=3\end{matrix}\right.\)
suy ra tìm đc x
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: ...
\(\Leftrightarrow x+1-\frac{15}{6}\sqrt{x}+\sqrt{x^2-4x+1}-\frac{1}{2}\sqrt{x}=0\)
\(\Leftrightarrow\frac{x^2-\frac{17}{4}x+1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{x^2-\frac{17}{4}x+1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}=0\)
\(\Leftrightarrow\left(x^2-\frac{17}{4}x+1\right)\left(\frac{1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}\right)=0\)
\(\Leftrightarrow x^2-\frac{17}{4}x+1=0\)
\(\Leftrightarrow4x^2-17x+4=0\)
8) ĐKXĐ: $-2\leq x\leq 1$
PT $\Leftrightarrow (2x+4)-4\sqrt{2x+4}+4+[(1-x)-2\sqrt{1-x}+1]=0$
$\Leftrightarrow (\sqrt{2x+4}-2)^2+(\sqrt{1-x}-1)^2=0$
Dễ thấy: $(\sqrt{2x+4}-2)^2; (\sqrt{1-x}-1)^2\geq 0$ với mọi $x\in [-2;1]$ nên để tổng của chúng bằng $0$ thì:
$(\sqrt{2x+4}-2)^2=(\sqrt{1-x}-1)^2=0$
$\Leftrightarrow \sqrt{2x+4}=2; \sqrt{1-x}-1=0$
$\Leftrightarrow x=0$ (thỏa mãn)
Vậy.....
7)
ĐKXĐ: $x\geq -1$
PT $\Leftrightarrow x^2+[(x+1)-2\sqrt{x+1}+1]=0$
$\Leftrightarrow x^2+(\sqrt{x+1}-1)^2=0$
Ta thấy:
$x^2\geq 0; (\sqrt{x+1}-1)^2\geq 0$ với mọi $x\geq -1$
Do đó để tổng của chúng bằng $0$ thì $x^2=(\sqrt{x+1}-1)^2=0$
$\Leftrightarrow x=0$ (thỏa mãn)
Vậy.......
Lời giải:
Áp dụng BĐT Bunhiacopxky:
\(y^2=(3x+\sqrt{8-x^2})^2\leq [x^2+(8-x^2)](3^2+1^2)\)
\(\Leftrightarrow y^2\leq 80\Rightarrow y\leq 4\sqrt{5}\)
Đáp án C