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\(a,\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x^3=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow4x-3-x-5=30-3x\\ \Leftrightarrow4x-x+3x=30+5+3\\ \Leftrightarrow6x=38\\ \Leftrightarrow x=\dfrac{19}{3}\)
\(E=-3x^2-x+6=-3\left(x^2+\dfrac{x}{3}\right)+6=-3\left(x^2+2x.\dfrac{1}{6}\right)+6=-3\left(x^2+2x.\dfrac{1}{6}+\dfrac{1}{36}\right)+6+\dfrac{1}{12}\le-3.0+6+\dfrac{1}{12}=6\dfrac{1}{12}\)
cái này ko tìm dc Min nha bạn (với x dương thì x càng lớn E càng nhỏ)
3(x+3)-x(x+3)=0
(x+3)(3-x) =0
x+3 =0 hoặc 3-x=0 =>x={-3;3}
\(2x\left(x-3\right)=x^2-3x\)
\(\Rightarrow2x\left(x-3\right)=x\left(x-3\right)\)
\(\Rightarrow2x=x\)
\(\Rightarrow x=0\)
`C=-2x^2+x+1`
`C=-2(x^2-x/2)+1`
`C=-2(x^2-2*x*1/4+1/16)+1+1/8`
`C=-2(x-1/4)^2+9/8<=9/8`
Dấu "=" `<=>x=1/4.`
Ta có: \(C=-2x^2+x+1\)
\(=-2\left(x^2-2\cdot x\cdot\dfrac{1}{4}+\dfrac{1}{16}-\dfrac{9}{16}\right)\)
\(=-2\left(x-\dfrac{1}{4}\right)^2+\dfrac{9}{8}\le\dfrac{9}{8}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{4}\)
Đk x>=0
A=\(\frac{2\sqrt{x}}{\sqrt{x}+3}\)=\(\frac{2\sqrt{x}+6-6}{\sqrt{x}+3}\)=\(\frac{2\left(\sqrt{x}+3\right)-6}{\sqrt{x}+3}\)=\(2-\frac{6}{\sqrt{x}+3}\)
Để A nguyên thì \(\frac{6}{\sqrt{x}+3}\)nguyên
=> 6\(⋮\)\(\sqrt{x}+3\)=>\(\sqrt{x}+3\in\left\{1;2;3;6\right\}\)=>\(\sqrt{x}\in\left\{0;3\right\}\)vì \(\sqrt{x}\ge0\)
vậy x\(\in\left\{0;9\right\}\)
\(ĐK:x\ge0\)
\(A=\frac{2\sqrt{x}}{\sqrt{x}+3}=\frac{2\sqrt{x}+6-6}{\sqrt{x}+3}=\frac{2\left(\sqrt{x}+3\right)-6}{\sqrt{x}+3}=2-\frac{6}{\sqrt{x}+3}\)
Để A nguyên thì \(\frac{6}{\sqrt{x}+3}\inℤ\Leftrightarrow\sqrt{x}+3\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
lập bảng xét nốt nhé:)
\(\dfrac{3}{x-2}=\dfrac{-2}{x-4}\left(dk:x\ne2;x\ne4\right)\)
\(\Rightarrow3\cdot\left(x-4\right)=-2\cdot\left(x-2\right)\)
\(\Rightarrow3x-12=-2x+4\)
\(\Rightarrow3x+2x=4+12\)
\(\Rightarrow5x=16\)
\(\Rightarrow x=\dfrac{16}{5}\left(tm\right)\)
\(ĐK:x\ne2;x\ne4\\ Có:\dfrac{3}{x-2}=\dfrac{-2}{x-4}\\ \Leftrightarrow3\left(x-4\right)=-2\left(x-2\right)\\ \Leftrightarrow3x-12=-2x+4\\ \Leftrightarrow3x+2x=4+12\\ \Leftrightarrow5x=16\\ \Leftrightarrow x=\dfrac{16}{5}\left(TM\right)\\ Vậy:x=\dfrac{16}{5}\)
\(\frac{x+3}{\sqrt{x}}=\sqrt{x}+\frac{3}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\frac{3}{\sqrt{x}}}=2\sqrt{3}\)
dấu "=" xảy ra khi và chỉ khi \(\sqrt{x}=\frac{3}{\sqrt{x}}< =>x=3\)
\(MIN=2\sqrt{3}\)