help Pain " ad duyệt dùm "
ở phần học bài của OLM " toán 9 căn bậc 2 " phút " 9,27
có bài tìm min " OLM làm như sau "
\(A=\sqrt{12x+1}-3\)
do \(\sqrt{12x+1}\ge0\)\(\Rightarrow\sqrt{12x+1}-3\ge-3\)
giải thích hộ pain tại sao \(\sqrt{12x+1}\ge0\)
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H9
HT.Phong (9A5)
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24 tháng 7 2023
9) Sửa: \(2\sqrt{8\sqrt{3}}-2\sqrt{5\text{ }\sqrt{3}}-3\sqrt{20\sqrt{3}}\)
\(=2\sqrt{2^2\cdot2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{2^2\cdot5\sqrt{3}}\)
\(=2\cdot2\sqrt{2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\cdot2\sqrt{5\sqrt{3}}\)
\(=4\sqrt{2\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}\)
\(=4\sqrt{2\sqrt{3}}-8\sqrt{5\sqrt{3}}\)
10) \(\sqrt{12x}-\sqrt{48x}-3\sqrt{3x}+27\)
\(=\sqrt{2^2\cdot3x}-\sqrt{4^2\cdot3x}-3\sqrt{3x}+27\)
\(=2\sqrt{3x}-4\sqrt{3x}-3\sqrt{3x}+27\)
\(=-5\sqrt{3x}++27\)
11) \(\sqrt{18x}-5\sqrt{8x}+7\sqrt{18x}+28\)
\(=\sqrt{3^2\cdot2x}-5\sqrt{2^2\cdot2x}+7\sqrt{3^2\cdot2x}+28\)
\(=3\sqrt{2x}-5\cdot2\sqrt{2x}+7\cdot3\sqrt{2x}+28\)
\(=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+28\)
\(=14\sqrt{2x}+28\)
12) \(\sqrt{45a}-\sqrt{20a}+4\sqrt{45a}+\sqrt{a}\)
\(=\sqrt{3^2\cdot5a}-\sqrt{2^2\cdot5a}+4\sqrt{3^2\cdot5a}+\sqrt{a}\)
\(=3\sqrt{5a}-2\sqrt{5a}+4\cdot3\sqrt{5a}+\sqrt{a}\)
\(=3\sqrt{5a}-2\sqrt{5a}+12\sqrt{5a}+\sqrt{a}\)
\(=13\sqrt{5a}+\sqrt{a}\)
30 tháng 7 2021
1)\(\sqrt{4x^2+12x+9}=2-x\)
\(\Leftrightarrow\sqrt{\left(2x+3\right)^2}=2-x\)
\(\Leftrightarrow\left|2x+3\right|=2-x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=2-x\\2x+3=x-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-5\end{matrix}\right.\)
\(\)
18 tháng 11 2021
a, ĐKXĐ: \(x\ge-\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{3}{2}.2\sqrt{1+3x}-\dfrac{5}{3}.3\sqrt{1+3x}-\dfrac{1}{4}.4\sqrt{1+3x}=1\\ \Leftrightarrow3\sqrt{1+3x}-5\sqrt{1+3x}-\sqrt{1+3x}=1\\ \Leftrightarrow-3\sqrt{1+3x}=1\\ \Leftrightarrow\sqrt{1+3x}=-\dfrac{1}{3}\left(vô.lí\right)\)
b, \(\Leftrightarrow\sqrt{\left(x-\dfrac{1}{2}\right)^2}=3\\ \Leftrightarrow\left|x-\dfrac{1}{2}\right|=3\\ \Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=3\\x-\dfrac{1}{2}=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
18 tháng 11 2021
a) ĐKXĐ: \(x\ge-\dfrac{1}{3}\)
\(pt\Leftrightarrow3\sqrt{3x+1}-5\sqrt{3x+1}-\sqrt{3x+1}=1\)
\(\Leftrightarrow-3\sqrt{3x+1}=1\Leftrightarrow\sqrt{3x+1}=-\dfrac{1}{3}\left(VLý\right)\)
Vậy \(S=\varnothing\)
b) \(pt\Leftrightarrow\sqrt{\left(x-\dfrac{1}{2}\right)^2}=3\Leftrightarrow\left|x-\dfrac{1}{2}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=3\\x-\dfrac{1}{2}=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
26 tháng 12 2021
a: \(\Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
15 tháng 8 2017
Bài 2:Áp dụng BĐT AM-GM ta có:
\(\frac{1}{x}+\frac{1}{y}\ge2\sqrt{\frac{1}{xy}}\)
\(\frac{1}{y}+\frac{1}{z}\ge2\sqrt{\frac{1}{yz}}\)
\(\frac{1}{x}+\frac{1}{z}\ge2\sqrt{\frac{1}{xz}}\)
CỘng theo vế 3 BĐT trên có:
\(2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge2\left(\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\right)\)
Khi x=y=z
15 tháng 8 2017
Ta có: \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}\)
\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}\)
\(\frac{1}{\sqrt{3}}>\frac{1}{\sqrt{100}}\)
\(..........................\)
\(\frac{1}{\sqrt{99}}>\frac{1}{\sqrt{100}}\)
\(\frac{1}{\sqrt{100}}=\frac{1}{\sqrt{100}}\)
Cộng theo vế ta có:
\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}>\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}=\frac{100}{10}=10\)
25 tháng 11 2023
2: ĐKXĐ: x>=0
\(\sqrt{3x}-2\sqrt{12x}+\dfrac{1}{3}\cdot\sqrt{27x}=-4\)
=>\(\sqrt{3x}-2\cdot2\sqrt{3x}+\dfrac{1}{3}\cdot3\sqrt{3x}=-4\)
=>\(\sqrt{3x}-4\sqrt{3x}+\sqrt{3x}=-4\)
=>\(-2\sqrt{3x}=-4\)
=>\(\sqrt{3x}=2\)
=>3x=4
=>\(x=\dfrac{4}{3}\left(nhận\right)\)
3:
ĐKXĐ: x>=0
\(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18}=0\)
=>\(3\sqrt{2x}+5\cdot2\sqrt{2x}-20-3\sqrt{2}=0\)
=>\(13\sqrt{2x}=20+3\sqrt{2}\)
=>\(\sqrt{2x}=\dfrac{20+3\sqrt{2}}{13}\)
=>\(2x=\dfrac{418+120\sqrt{2}}{169}\)
=>\(x=\dfrac{209+60\sqrt{2}}{169}\left(nhận\right)\)
4: ĐKXĐ: x>=-1
\(\sqrt{16x+16}-\sqrt{9x+9}=1\)
=>\(4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>\(\sqrt{x+1}=1\)
=>x+1=1
=>x=0(nhận)
5: ĐKXĐ: x<=1/3
\(\sqrt{4\left(1-3x\right)}+\sqrt{9\left(1-3x\right)}=10\)
=>\(2\sqrt{1-3x}+3\sqrt{1-3x}=10\)
=>\(5\sqrt{1-3x}=10\)
=>\(\sqrt{1-3x}=2\)
=>1-3x=4
=>3x=1-4=-3
=>x=-3/3=-1(nhận)
6: ĐKXĐ: x>=3
\(\dfrac{2}{3}\sqrt{x-3}+\dfrac{1}{6}\sqrt{x-3}-\sqrt{x-3}=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}\cdot\left(\dfrac{2}{3}+\dfrac{1}{6}-1\right)=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}\cdot\dfrac{-1}{6}=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}=\dfrac{2}{3}:\dfrac{1}{6}=\dfrac{2}{3}\cdot6=\dfrac{12}{3}=4\)
=>x-3=16
=>x=19(nhận)