b/ \(\sqrt{12-\dfrac{12}{x^2}}+\sqrt{x^2-\dfrac{12}{x^2}}=x^2\)

\(\Leftrightarrow x-\sqrt{12-\dfrac{12}{x^2}}=\sqrt{x^2-\dfrac{12}{x^2}}\)

Bình phương 2 vế rút gọn

\(\Leftrightarrow x^4-x^2-4\sqrt{3\left(x^4-x^2\right)}+12=0\)

Đặt \(\sqrt{x^4-x^2}=a\)

\(\Rightarrow a^2-4\sqrt{3}a+12=0\)

\(\Leftrightarrow a=2\sqrt{3}\)

\(\Leftrightarrow x^4-x^2=12\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Câu a xem lại đề đúng không b. Do nghiệm xấu lắm

a/ \(\Leftrightarrow2\left(x^2+4x+4\right)-2x-\frac{7}{2}\ge0\)

\(\Leftrightarrow2x^2+6x+\frac{9}{2}\ge0\)

\(\Leftrightarrow2\left(x+\frac{3}{2}\right)^2\ge0\) (luôn đúng)

Vậy nghiệm của BPT là \(D=R\)

b/ \(\Leftrightarrow\frac{x^2+x-1}{1-x}+x>0\)

\(\Leftrightarrow\frac{x^2+x-1+x-x^2}{1-x}>0\)

\(\Leftrightarrow\frac{2x-1}{1-x}>0\Rightarrow\frac{1}{2}< x< 1\)

c/ \(x^2+x+12=\left(x+\frac{1}{2}\right)^2+\frac{47}{4}>0\) \(\forall x\) nên BPT tương đương:

\(x^2+x+12>x^2+x+12\)

\(\Leftrightarrow0>0\)

Vậy BPT vô nghiệm

Dự đoán điểm rơi: x=3 ; y =4;z =2

ÁP dụng AM-Gm ta có:

\(\dfrac{8}{xyz}+\dfrac{x}{9}+\dfrac{y}{12}+\dfrac{z}{6}\ge4\sqrt[4]{\dfrac{8}{9.12.6}}=\dfrac{4}{3}\)

\(\dfrac{2}{xy}+\dfrac{x}{18}+\dfrac{y}{24}\ge3\sqrt[3]{\dfrac{2}{18.24}}=\dfrac{1}{2}\)

\(\dfrac{2}{yz}+\dfrac{y}{16}+\dfrac{z}{8}\ge3\sqrt[3]{\dfrac{2}{16.8}}=\dfrac{3}{4}\)

\(\dfrac{2}{xz}+\dfrac{z}{6}+\dfrac{x}{9}\ge3\sqrt[3]{\dfrac{2}{6.9}}=1\)

\(\dfrac{13}{18}x+\dfrac{13}{24}y\ge2\sqrt{\dfrac{169}{18.24}xy}\ge\dfrac{13}{3}\)

\(\dfrac{13}{24}z+\dfrac{13}{48}y\ge2\sqrt{\dfrac{169}{24.48}.yz}\ge\dfrac{13}{6}\)

Cộng tất cả theo vế ,ta thu được Đpcm.

a/ ĐKXĐ: ...

\(\Leftrightarrow x+8+\sqrt{x+8}-\left(x+8\right)=\sqrt{x}+\sqrt{x+3}\)

\(\Leftrightarrow\sqrt{x+8}=\sqrt{x}+\sqrt{x+3}\)

\(\Leftrightarrow x+8=2x+3+2\sqrt{x^2+3x}\)

\(\Leftrightarrow5-x=2\sqrt{x^2+3x}\) (\(x\le5\))

\(\Leftrightarrow x^2-10x+25=4\left(x^2+3x\right)\)

\(\Leftrightarrow...\)

b/ ĐKXĐ: \(2\le x\le5\)

\(\Leftrightarrow2\left(x-2\right)+\sqrt{2\left(x-2\right)}\left(\sqrt{5-x}-\sqrt{3x-3}\right)=0\)

\(\Leftrightarrow\sqrt{2\left(x-2\right)}\left(\sqrt{2x-4}+\sqrt{5-x}-\sqrt{3x-3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\sqrt{2x-4}+\sqrt{5-x}=\sqrt{3x-3}\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x+1+2\sqrt{\left(2x-4\right)\left(5-x\right)}=3x-3\)

\(\Leftrightarrow\sqrt{\left(2x-4\right)\left(5-x\right)}=x-2\)

\(\Leftrightarrow\left(2x-4\right)\left(5-x\right)=\left(x-2\right)^2\)

\(\Leftrightarrow...\)

c/ ĐKXĐ: \(x\le12\)

\(\Leftrightarrow\sqrt[3]{24+x}\sqrt{12-x}-6\sqrt{12-x}+12-x=0\)

\(\Leftrightarrow\sqrt{12-x}\left(\sqrt[3]{24+x}-6+\sqrt{12-x}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=12\\\sqrt[3]{24+x}+\sqrt{12-x}=6\left(1\right)\end{matrix}\right.\)

Xét (1):

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{24+x}=a\\\sqrt{12-x}=b\ge0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=6\\a^3+b^2=36\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=6-a\\a^3+b^2=36\end{matrix}\right.\)

\(\Leftrightarrow a^3+\left(6-a\right)^2=36\)

\(\Leftrightarrow a^3+a^2-12a=0\)

\(\Leftrightarrow a\left(a^2+a-12\right)=0\Rightarrow\left[{}\begin{matrix}a=0\\a=3\\a=-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt[3]{24+x}=0\\\sqrt[3]{24+x}=3\\\sqrt[3]{24+x}=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}24+x=0\\24+x=27\\24+x=-64\end{matrix}\right.\)

uỵttyutyutu

\(\Leftrightarrow-x^2+2x+3+4\sqrt{-x^2+2x+3}\le m\)

Đặt \(\sqrt{-x^2+2x+3}=\sqrt{4-\left(x-1\right)^2}=t\Rightarrow0\le t\le2\)

BPT trở thành:

\(f\left(t\right)=t^2+4t\le m\)

Để BPT nghiệm đúng với mọi \(t\in\left[0;2\right]\)

\(\Leftrightarrow m\ge\max\limits_{\left[0;2\right]}f\left(t\right)=12\)

\(\Rightarrow m\ge12\)