\sqrt{9x^2}=\left|12\right|
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5 tháng 7 2021
a) Ta có: \(\sqrt{25x+75}+3\sqrt{x-2}=2\sqrt{x-2}+\sqrt{9x-18}\)
\(\Leftrightarrow5\sqrt{x+3}+3\sqrt{x-2}=2\sqrt{x-2}+3\sqrt{x-2}\)
\(\Leftrightarrow\sqrt{25x+75}=\sqrt{4x-8}\)
\(\Leftrightarrow25x-4x=-8-75\)
\(\Leftrightarrow21x=-83\)
hay \(x=-\dfrac{83}{21}\)
b) Ta có: \(\sqrt{\left(2x-1\right)^2}=4\)
\(\Leftrightarrow\left|2x-1\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
c) Ta có: \(\sqrt{\left(2x+1\right)^2}=3x-5\)
\(\Leftrightarrow\left|2x+1\right|=3x-5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=3x-5\left(x\ge-\dfrac{1}{2}\right)\\2x+1=5-3x\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3x=-5-1\\2x+3x=5-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(nhận\right)\\x=\dfrac{4}{5}\left(loại\right)\end{matrix}\right.\)
d) Ta có: \(\sqrt{4x-12}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)
\(\Leftrightarrow2\sqrt{x-3}-2\sqrt{x-2}=3\sqrt{x-2}+8\)
\(\Leftrightarrow2\sqrt{x-3}-5\sqrt{x-2}=8\)
\(\Leftrightarrow4\left(x-3\right)+25\left(x-2\right)-20\sqrt{x^2-5x+6}=8\)
\(\Leftrightarrow4x-12+25x-50-8=20\sqrt{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow20\sqrt{\left(x-2\right)\left(x-3\right)}=29x-70\)
\(\Leftrightarrow x^2-5x+6=\dfrac{\left(29x-70\right)^2}{400}\)
\(\Leftrightarrow x^2-5x+6=\dfrac{841}{400}x^2-\dfrac{203}{20}x+\dfrac{49}{4}\)
\(\Leftrightarrow\dfrac{-441}{400}x^2+\dfrac{103}{20}x-\dfrac{25}{4}=0\)
\(\Delta=\left(\dfrac{103}{20}\right)^2-4\cdot\dfrac{-441}{400}\cdot\dfrac{-25}{4}=-\dfrac{26}{25}\)(Vô lý)
vậy: Phương trình vô nghiệm
NV
Nguyễn Việt Lâm
Giáo viên
31 tháng 1 2019
Do \(12=\sqrt{144}>\sqrt{135}\) nên \(x>0\)
Đặt \(a=\sqrt[3]{\dfrac{12+\sqrt{135}}{3}}+\sqrt[3]{\dfrac{12-\sqrt{135}}{3}}\) \(\Rightarrow x=\dfrac{1}{3}\left(a+1\right)\)
\(a^3=8+3\left(\sqrt[3]{\dfrac{12+\sqrt{135}}{3}}+\sqrt[3]{\dfrac{12-\sqrt{135}}{3}}\right)=8+3a\)
Ta có: \(x=\dfrac{1}{3}\left(a+1\right)\Rightarrow3x=a+1\Rightarrow9x^2=a^2+2a+1\)
Lại có: \(x^3=\dfrac{1}{27}\left(a+1\right)^3\Leftrightarrow9x^3=\dfrac{1}{3}\left(a^3+3a^2+3a+1\right)\)
\(\Leftrightarrow9x^3=\dfrac{1}{3}\left(8+3a+3a^2+3a+1\right)=a^2+2a+3\)
\(\Rightarrow M=\left(a^2+2a+3-a^2-2a-1-3\right)^2=\left(-1\right)^2=1\)
ND
3
NH
9 tháng 7 2023
a)
\(\sqrt{9x^2}=12\\ < =>\left(\sqrt{9x^2}\right)^2=12^2\\ < =>9x^2=144\\ < =>x^2=16\\ < =>\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
b)
\(\sqrt{25x^2}=\left|-50\right|\\ < =>\sqrt{25x^2}=50\left(vì-50< 0\right)\)
\(< =>\left(\sqrt{25x^2}\right)^2=50^2\\ =>25x^2=2500\\ < =>x^2=100\\ < =>\left[{}\begin{matrix}x=10\\x=-10\end{matrix}\right.\)
8 tháng 1 2016
Đặt \(a=\sqrt[3]{\frac{12+\sqrt{135}}{3}}+\sqrt[3]{\frac{12-\sqrt{135}}{3}}\)
\(\Rightarrow a^3=\left(\sqrt[3]{4+\sqrt{15}}+\sqrt[3]{4-\sqrt{15}}\right)^3\)
Có (a+b)^3=a^3+b^3+3ab(a+b)
\(\Rightarrow a^3=4+\sqrt{15}+4-\sqrt{15}+3\sqrt[3]{\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)}a\)
\(\Rightarrow a^3=8+3a\Rightarrow a^3-3a-8=0\)-> khó