`[x-4]/[x+4]-x/[x-4]=[3x-14]/[x^2-16]`         `ĐK: x \ne +-4`

`<=>[(x-4)^2-x(x+4)]/[(x-4)(x+4)]=[3x-14]/[(x-4)(x+4)]`

   `=>x^2-8x+16-x^2-4x=3x-14`

`<=>3x+8x+4x=16+14`

`<=>15x=30`

`<=>x=2` (t/m)

Vậy `S={2}`

`(x - 4)/(x + 4) - x/(x - 4) = (3x - 14)/(x^2 - 16)`

`=>` `x = 2`

ĐK: \(x\ge0\)

Dễ thấy \(1-\sqrt{2\left(x^2-x+1\right)}\le1-\sqrt{2}< 0\)

Khi đó bất phương trình tương đương:

\(x-\sqrt{x}\le1-\sqrt{2\left(x^2-x+1\right)}\)

\(\Leftrightarrow\sqrt{x}-\dfrac{1}{\sqrt{x}}-1+\sqrt{2\left(x+\dfrac{1}{x}-1\right)}\le0\)

\(\Leftrightarrow\sqrt{x}-\dfrac{1}{\sqrt{x}}-1+\sqrt{2\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)^2+2}\le0\)

\(\Leftrightarrow t-1+\sqrt{2t^2+2}\le0\)

They can play volleyball and so can we

They can play volleyball and we can too

Mr Tam won't come to the party tonight and neither will his wife

Mr Tam won't come to the party tonight and his wife won't either

My father didn't drink beer and neither did my uncle/ my uncle did either
She is learning English and so am I/ I am too
I'm not a doctor and neither are they / they aren't either

\(\Leftrightarrow2cos4x\left(cos2x-sin2x\right)=0\)

\(\Leftrightarrow cos4x=0\) (do \(cos4x=cos^22x-sin^22x\) đã bao hàm \(cos2x-sin2x\))

\(\Rightarrow4x=\dfrac{\pi}{2}+k\pi\)

\(\Rightarrow x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\)

3x(2-x)-5=1-(3x²+2)

<=>6x-3x²-5=-3x²-2

<=>6x=3

<=>x=1/2

ĐKXĐ: \(x\ge1\)

Đặt \(\left\{{}\begin{matrix}\sqrt[]{x-1}=a\ge0\\\sqrt[3]{2-x}=b\end{matrix}\right.\) \(\Rightarrow a^2+b^3=1\)

Ta được hệ: 

\(\left\{{}\begin{matrix}a+b=1\\a^2+b^3=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=1-a\\a^2+b^3=1\end{matrix}\right.\)

\(\Rightarrow a^2+\left(1-a\right)^3=1\)

\(\Leftrightarrow a^3-4a^2+3a=0\)

\(\Leftrightarrow a\left(a-1\right)\left(a-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a=0\\a=1\\a=3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt[]{x-1}=0\\\sqrt[]{x-1}=1\\\sqrt[]{x-1}=3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=10\end{matrix}\right.\)

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