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Đặt \(\sqrt{7+3x}=a;\sqrt{13-3x}=b\)
=>a+b+5ab=46
=>(a+b)^2=46-5ab
=>a^2+b^2+2ab=2116-460ab+25a^2b^2
=>25a^2b^2-460ab+2116=7+3x+13-3x+2ab
=>25a^2b^2-462ab+2096=0
=>\(\left[{}\begin{matrix}ab=\dfrac{262}{25}\\ab=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(7+3x\right)\cdot\left(13-3x\right)=109.8304\\\left(7+3x\right)\left(13-3x\right)=64\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}91-21x+39x-9x^2=109.8304\\91-21x+39x-9x^2=64\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-9x^2+18x-18.8304=0\\-9x^2+18x+27=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
\(\left(\sqrt{7+3x}-4\right)+\left(\sqrt{13-3x}-2\right)+5.\left(\sqrt{\left(7+3x\right)\left(13-3x\right)}-8\right)=0\)
=) \(\frac{7+3x-16}{\sqrt{7+3x}+4}+\frac{13-3x-4}{\sqrt{13-3x}+2}+5.\left(\sqrt{91+18x-9x^2}-8\right)=0\)
=) \(\frac{3\left(x-3\right)}{\sqrt{7+3x}+4}+\frac{3\left(3-x\right)}{\sqrt{13-3x}+2}+\frac{5\left(27+18x-9x^2\right)}{\sqrt{91+18x-9x^2}+8}=0\)
=) \(\frac{3\left(x-3\right)}{\sqrt{7+3x}+4}-\frac{3\left(x-3\right)}{\sqrt{13-3x}+2}-\frac{45\left(x+1\right)\left(x-3\right)}{\sqrt{91+18x-9x^2}+8}=0\)
=) đến đây chắc là tự làm đc rồi
a) \(x^2+8=3\sqrt{x^3+8}\)
\(\left(x^2+8\right)^2=\left(3\sqrt{x^2+8}\right)^2\)
\(x^4+16x^2+64=9x^2+72\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
a: =>\(\sqrt{3x-5}+2=x+1\)
\(\Leftrightarrow\sqrt{3x-5}=x-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=1\\x^2-2x+1-3x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
b: \(\Leftrightarrow x-15\sqrt{x}+56=x+11\)
=>-15 căn x=-45
=>x=9
c: =>căn 3x+1=3x-1
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{3}\\9x^2-6x+1-3x-1=0\end{matrix}\right.\Leftrightarrow x=1\)
d: =>(3x+7)/(x+3)=16
=>16x+48=3x+7
=>13x=-41
=>x=-41/13
b) đặt \(\sqrt{3x+1}=a\)(\(a\ge0\))
\(PT\Leftrightarrow\dfrac{a^2-1}{\sqrt{a^2+9}}+1=a\)
\(\Leftrightarrow\left(a-1\right)\left(1-\dfrac{a+1}{\sqrt{a^2+9}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a+1=\sqrt{a^2+9}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a=4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)(tm)
c) bunyalovsky:
\(VT^2\le2\left(7-x+x-5\right)=4\)
\(\Leftrightarrow VT\le2\)
\(VF=\left(x-6\right)^2+2\ge2\)
Dấu = xảy ra khi x=6