Giá trị của x thỏa mãn 5\(\sqrt{x}\)-17 = 108 ( x > 0 )
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Đáp án: x=-1 Giải thích các bước giải:
x(x-5)(x+5)-(x+2)( x 2 -2x+4 )=17
⇔x( x 2 +5x-5x-25)-( x 3 -2 x 2 +4x+2 x 2 -4x+8)=17
⇔ x 3 -25x - x 3 -8-17=0
⇔ x 3 -25x - x 3 -25=0
⇔-25x-25=0
⇔-25x=25
⇒x=-1
\(\Leftrightarrow x\left(x^2-25\right)-x^3-8=17\\ \Leftrightarrow x^3-25x-x^3=25\\ \Leftrightarrow-25x=25\Leftrightarrow x=-1\)
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Ta có: \(\left|x-4\right|-x+17=0\)
\(\Leftrightarrow\left|x-4\right|=x-17\)
\(\Leftrightarrow x-4=17-x\left(x< 4\right)\)
\(\Leftrightarrow2x=21\)
hay \(x=\dfrac{21}{2}\left(loại\right)\)
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\(P=\dfrac{x}{\sqrt{x+y-x}}+\dfrac{y}{\sqrt{x+y-y}}=\dfrac{x}{\sqrt{y}}+\dfrac{y}{\sqrt{x}}\)
\(=\dfrac{x^2}{x\sqrt{y}}+\dfrac{y^2}{y\sqrt{x}}\ge\dfrac{\left(x+y\right)^2}{x\sqrt{y}+y\sqrt{x}}=\dfrac{\left(x+y\right)^2}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}\)
\(\ge\dfrac{\left(x+y\right)^2}{\dfrac{x+y}{2}.\left(1.\sqrt{x}+1.\sqrt{y}\right)}\ge\dfrac{\left(x+y\right)^2}{\dfrac{x+y}{2}.\sqrt{\left(1^2+1^2\right)\left(x+y\right)}}=\dfrac{1}{\dfrac{1}{2}\sqrt{2}}=\sqrt{2}\)
"=" khi x = y = 1/2
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Đặt \(t=\sqrt{x-3}\)\(\left(t\ge0\right)\)
\(\sqrt{8+t}+\sqrt{5-t}=5\)
\(\Leftrightarrow\left(\sqrt{8+t}+\sqrt{5-t}\right)^2=25\)
\(\Leftrightarrow8+t+5-t+2\sqrt{\left(8+t\right)\left(5-t\right)}=25\)
\(\Leftrightarrow2\sqrt{\left(8+t\right)\left(5-t\right)}=12\)
\(\Leftrightarrow\sqrt{\left(8+t\right)\left(5-t\right)}=6\)
\(\Leftrightarrow\left(8+t\right)\left(5-t\right)=36\)
\(\Leftrightarrow t^2+3t-4=0\)
\(\Leftrightarrow\orbr{\begin{cases}t=1\\t=-4\left(l\right)\end{cases}}\)
thay t=1 = căn (x-3) => x=4
điều kiện x-3 \(\ge0;5-\sqrt{x-3}\ge0\)(1)
đặt \(\sqrt{8+\sqrt{x-3}}=a\left(a\ge\sqrt{8}\right);\sqrt{5-\sqrt{x-3}}=b\left(b\ge0\right)\)
\(\hept{\begin{cases}a+b=5\\a^2+b^2=13\end{cases}< =>\hept{\begin{cases}a=5-b\\\left(5-b\right)^2+b^2=13\end{cases}< =>}}\)\(\hept{\begin{cases}a=5-b\\2b^2-10b+12=0\end{cases}< =>\hept{\begin{cases}a=3\\b=2\end{cases};\hept{\begin{cases}a=2\\b=3\end{cases}}}}\)
chỉ có a=3 là thoảm= mãn a \(\ge\sqrt{8}\)
\(\hept{\begin{cases}a=3\\b=2\end{cases}< =>\hept{\begin{cases}8+\sqrt{x-3}=9\\5-\sqrt{x-3}=4\end{cases}< =>x=4}}\)(thỏa mãn (1))
vậy x=4
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\(\Leftrightarrow0\le\sqrt{3-2x}\le\sqrt{5}\\ \Leftrightarrow0\le3-2x\le5\\ \Leftrightarrow-1\le x\le\dfrac{3}{2}\)
\(5\sqrt{x}-17=108\)
\(\Rightarrow5\sqrt{x}=17+108=125\)
\(\Rightarrow\sqrt{x}=25\)
\(\Rightarrow x=25^2=625\)