\(P=\sqrt{2x+\sqrt{4x-1}}+\sqrt{2x-\sqrt{4x-1}}\) với \(\dfrac{1}{4}< x< \dfrac{1}{2}\)

\(\Leftrightarrow\sqrt{2}P=\sqrt{4x+2\sqrt{4x-1}}+\sqrt{4x-2\sqrt{4x-1}}\)

\(=\sqrt{\left(\sqrt{4x-1}\right)^2+2\sqrt{4x-1}+1}+\sqrt{\left(\sqrt{4x-1}\right)^2-2\sqrt{4x-1}+1}\)

\(=\sqrt{4x-1}+1+\left|\sqrt{4x-1}-1\right|\)

Do \(\dfrac{1}{4}< x< \dfrac{1}{2}\Leftrightarrow0< \sqrt{4x-1}< 1\)

\(\Rightarrow P=\dfrac{1}{\sqrt{2}}\left(\sqrt{4x-1}+1+1-\sqrt{4x-1}\right)=\sqrt{2}\)

Vậy \(P=\sqrt{2}\).

\(1,4x\left(1-x\right)-8=1-\left(4x^2+3\right)\\ \Leftrightarrow4x-4x^2-8=1-4x^2-3\\ \Leftrightarrow4x-4x^2-8-1+4x^2+3=0\\ \Leftrightarrow4x-6=0\\ \Leftrightarrow x=\dfrac{3}{2}\)

\(2,\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\\ \Leftrightarrow\left(2-3x\right)\left(x+11\right)-\left(2-3x\right)\left(5x-2\right)=0\\ \Leftrightarrow\left(2-3x\right)\left(x+11-5x+2\right)=0\\ \Leftrightarrow\left(2-3x\right)\left(-4x+13\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{13}{4}\end{matrix}\right.\)

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: 

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

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

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

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

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

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

3x(2-x)-5=1-(3x²+2)

<=>6x-3x²-5=-3x²-2

<=>6x=3

<=>x=1/2

= -2³/3 + 2²/2 + 2.2 - [-(-1)³/3 + (-1)²/2 + 2.(-1)]

= -8/3 + 2 + 4 - 1/3 - 1/2 + 2

= 8 - 3 - 1/2

= 9/2

\(\int\limits^2_{-1}\left(-x^2+x+2\right)dx=\left(-\dfrac{x^3}{3}+\dfrac{x^2}{2}+2x\right)|^2_{-1}=\dfrac{9}{2}\)

\(\dfrac{x}{\dfrac{4}{2}}=\dfrac{4}{\dfrac{x}{2}}\)

\(\Rightarrow\dfrac{x}{2}=\dfrac{8}{x}\)

\(\Rightarrow x^2=8\cdot2\)

\(\Rightarrow x^2=16\)

\(\Rightarrow x^2=4^2\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

4x+2x=60

nên 6x=60

hay x=10

4x + 2x = 60

=> (4 + 2).x = 60

=> 6.x          = 60

=>    x          = 60 : 6

=>    x          = 10

Vậy x = 10

Bo thi:>

+ đk x > 0 , x khác 1