Tìm GTNN:
A= \(x+2\sqrt{x}+3\)
B= \(\left|2x+1\right|+3\)
C= \(\left(\sqrt{x}+10\right)^2\)
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27 tháng 8 2021
a) \(\sqrt{2x-1}=3\left(đk:x\ge\dfrac{1}{2}\right)\)
\(\Leftrightarrow2x-1=9\Leftrightarrow2x=10\Leftrightarrow x=5\)(thỏa đk)
b) \(\sqrt{1-3x}=\dfrac{1}{2}\left(đk:x\le\dfrac{1}{3}\right)\)
\(\Leftrightarrow1-3x=\dfrac{1}{4}\Leftrightarrow3x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{4}\)(thỏa đk)
c) \(\sqrt{\left(x-1\right)^2}=\dfrac{1}{2}\)
\(\Leftrightarrow\left|x-1\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{2}\\x-1=-\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
d) \(\sqrt{\left(1+2x\right)^2}=\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow\left|1+2x\right|=\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}1+2x=\dfrac{\sqrt{3}}{2}\\1+2x=-\dfrac{\sqrt{3}}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2+\sqrt{3}}{4}\\x=-\dfrac{2+\sqrt{3}}{4}\end{matrix}\right.\)
e) \(\sqrt{\left(1-2x\right)^2}=\left|x-1\right|\)
\(\Leftrightarrow\left|1-2x\right|=\left|x-1\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}1-2x=x-1\\1-2x=1-x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=0\end{matrix}\right.\)
25 tháng 8 2021
a)√x−1=2(x≥1)
\(x-1=4
\)
x=5
b)
\(\sqrt{3-x}=4\) (x≤3)
\(\left(\sqrt{3-x}\right)^2=4^2\)
x-3=16
x=19
26 tháng 8 2021
a: Ta có: \(\sqrt{x-1}=2\)
\(\Leftrightarrow x-1=4\)
hay x=5
b: Ta có: \(\sqrt{3-x}=4\)
\(\Leftrightarrow3-x=16\)
hay x=-13
c: Ta có: \(2\cdot\sqrt{3-2x}=\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{3-2x}=\dfrac{1}{4}\)
\(\Leftrightarrow-2x+3=\dfrac{1}{16}\)
\(\Leftrightarrow-2x=-\dfrac{47}{16}\)
hay \(x=\dfrac{47}{32}\)
d: Ta có: \(4-\sqrt{x-1}=\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{7}{2}\)
\(\Leftrightarrow x-1=\dfrac{49}{4}\)
hay \(x=\dfrac{53}{4}\)
e: Ta có: \(\sqrt{x-1}-3=1\)
\(\Leftrightarrow\sqrt{x-1}=4\)
\(\Leftrightarrow x-1=16\)
hay x=17
f:Ta có: \(\dfrac{1}{2}-2\cdot\sqrt{x+2}=\dfrac{1}{4}\)
\(\Leftrightarrow2\cdot\sqrt{x+2}=\dfrac{1}{4}\)
\(\Leftrightarrow\sqrt{x+2}=\dfrac{1}{8}\)
\(\Leftrightarrow x+2=\dfrac{1}{64}\)
hay \(x=-\dfrac{127}{64}\)
Y
21 tháng 7 2023
\(a,\sqrt{72x}\) xác định \(\Leftrightarrow72x\ge0\Leftrightarrow x\ge0\)
\(b,\dfrac{2x+3}{\sqrt{x^2-4}}\) xác định \(\Leftrightarrow x^2-4>0\Leftrightarrow\left(x-2\right)\left(x+2\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-2>0\\x+2>0\end{matrix}\right.\\\left[{}\begin{matrix}x-2< 0\\x+2< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x>2\\x>-2\end{matrix}\right.\\\left[{}\begin{matrix}x< 2\\x< -2\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -2\end{matrix}\right.\)
\(c,\sqrt{\left(2x+1\right)\left(x+2\right)}\) xác định \(\Leftrightarrow\left(2x+1\right)\left(x+2\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}2x+1\ge0\\x+2\ge0\end{matrix}\right.\\\left[{}\begin{matrix}2x+1\le0\\x+2\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-\dfrac{1}{2}\\x\ge-2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-\dfrac{1}{2}\\x\le-2\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ge-\dfrac{1}{2}\\x\le-2\end{matrix}\right.\)
\(d,3-\sqrt{16x^2-1}\) xác định \(\Leftrightarrow16x^2-1\ge0\Leftrightarrow\left(4x-1\right)\left(4x+1\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}4x-1\ge0\\4x+1\ge0\end{matrix}\right.\\\left[{}\begin{matrix}4x-1\le0\\4x+1\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge\dfrac{1}{4}\\x\ge-\dfrac{1}{4}\end{matrix}\right.\\\left[{}\begin{matrix}x\le\dfrac{1}{4}\\x\le-\dfrac{1}{4}\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{1}{4}\\x\le-\dfrac{1}{4}\end{matrix}\right.\)
\(e,\sqrt{\dfrac{3+x}{4-x}}\) xác định \(\Leftrightarrow\left[{}\begin{matrix}3+x\ge0\\4-x>0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ge-3\\x>4\end{matrix}\right.\) \(\Leftrightarrow x>4\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 11 2019
a/ ĐKXĐ: \(-\frac{3}{2}\le x\le4\)
\(\sqrt{2x+3}+\sqrt{4-x}=6x-3\left(x+7-2\sqrt{\left(2x+3\right)\left(4-x\right)}\right)-10\)
\(\Leftrightarrow\sqrt{2x+3}+\sqrt{4-x}=3\left(x+7+2\sqrt{\left(2x+3\right)\left(4-x\right)}\right)-52\)
Đặt \(\sqrt{2x+3}+\sqrt{4-x}=a>0\Rightarrow a^2=x+7+2\sqrt{\left(2x+3\right)\left(4-x\right)}\)
Phương trình trở thành:
\(a=3a^2-52\Leftrightarrow3a^2-a-52=0\Rightarrow\left[{}\begin{matrix}a=-4\left(l\right)\\a=\frac{13}{3}\end{matrix}\right.\)
\(\sqrt{2x+3}+\sqrt{4-x}=\frac{13}{3}\)
Phương trình này vô nghiệm nên ko muốn giải tiếp, bạn bình phương lên và chuyển vế thôi :(
b/ ĐKXĐ: \(-\frac{1}{4}\le x\le1\)
Đặt \(\sqrt{4x+1}+2\sqrt{1-x}=a>0\Rightarrow a^2=5+4\sqrt{-4x^2+3x+1}\)
\(\Rightarrow\sqrt{-4x^2+3x+1}=\frac{a^2-5}{4}\)
Pt trở thành:
\(a+10\left(\frac{a^2-5}{4}\right)=13\)
\(\Leftrightarrow5a^2+2a-51=0\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{17}{5}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{-4x^2+3x+1}=\frac{a^2-5}{4}=1\)
\(\Leftrightarrow-4x^2+3x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{3}{4}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 11 2019
c/ \(\Leftrightarrow x^2\left(x^2+2\right)=12-x\sqrt{2x^2+4}\)
\(\Leftrightarrow x^2\left(2x^2+4\right)=24-2x\sqrt{2x^2+4}\)
Đặt \(x\sqrt{2x^2+4}=a\) ta được:
\(a^2=24-2a\Leftrightarrow a^2+2a-24=0\Leftrightarrow\left[{}\begin{matrix}a=4\\a=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\sqrt{2x^2+4}=4\left(x>0\right)\\x\sqrt{2x^2+4}=-6\left(x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2\left(2x^2+4\right)=16\\x^2\left(2x^2+4\right)=36\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^4+2x^2-8=0\\x^4+2x^2-18=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2=2\\x^2=-4\left(l\right)\\x^2=\sqrt{19}-1\\x^2=-\sqrt{19}-1\left(l\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}< 0\left(l\right)\\x=-\sqrt{\sqrt{19}-1}\\x=\sqrt{\sqrt{19}-1}>0\left(l\right)\end{matrix}\right.\)
\(A=x+2\sqrt{x}+3=\left(\sqrt{x}+1\right)^2+2\ge3\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)
\(B=\left|2x+1\right|+3\ge3\)
Dấu "=" xảy ra \(\Leftrightarrow\left|2x+1\right|=0\)
\(\Leftrightarrow x=\frac{-1}{2}\)
\(C=\left(\sqrt{x}+10\right)^2\ge100\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)