\(ĐK:x\ge0\\ P=\dfrac{x}{\sqrt{x}+1}+\sqrt{x}\cdot\dfrac{\sqrt{x}+2+\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}+\dfrac{2}{\sqrt{x}+2}\\ P=\dfrac{x}{\sqrt{x}+1}+\dfrac{2x+3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}+\dfrac{2}{\sqrt{x}+2}\\ P=\dfrac{x\sqrt{x}+2x+2x+3\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}\\ P=\dfrac{x\sqrt{x}+4x+5\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}\\ P=\dfrac{x\sqrt{x}+2x+2x+4\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}\\ P=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}=\sqrt{x}+1\)

\(a,\Leftrightarrow5-4\sqrt{x}=0\left(\text{do }\sqrt{x}+9\ge9>0\right)\\ \Leftrightarrow\sqrt{x}=\dfrac{5}{4}\Leftrightarrow x=\dfrac{25}{16}\\ \text{Vậy }S=\left\{\dfrac{25}{16}\right\}\\ b,\Leftrightarrow x-6\sqrt{x}+\sqrt{x}-6=0\\ \Leftrightarrow\left(\sqrt{x}-6\right)\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow\sqrt{x}-6=0\left(\text{do }\sqrt{x}+1\ge1>0\right)\\ \Leftrightarrow\sqrt{x}=6\Leftrightarrow x=36\\ \text{Vậy }S=\left\{36\right\}\)

\(a,ĐK:x\ge0;x\ne4\\ b,A=\dfrac{x+3\sqrt{x}+2-x+4\sqrt{x}-4-12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ A=\dfrac{7\sqrt{x}-14}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{7\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\\ A=\dfrac{7}{\sqrt{x}+2}\)

Xem lại $b)$

\(x^2+2x+1-y^2=\left(x+1\right)^2-y^2=\left(x-y+1\right)\left(x+y+1\right)\\ 25x^2+1-9y^2-10x=\left(5x-1\right)^2-9y^2=\left(5x-3y-1\right)\left(5x+3y-1\right)\\ x^2-1+4y-4y^2=x^2-\left(2y-1\right)^2=\left(x-2y+1\right)\left(x+2y-1\right)\\ 9x^2+15x+5y+6xy+y^2\\ =\left(9x^2+6xy+y^2\right)+5\left(3x+y\right)\\ =\left(3x+y\right)^2+5\left(3x+y\right)\\ =\left(3x+y\right)\left(3x+y+5\right)\\ 9-x^2+6xy-9y^2=9-\left(x-3y\right)^2=\left(3-x+3y\right)\left(3+x-3y\right)\\ x^2+4x-2xy-4y+y^2\\ =\left(x^2-2xy+y^2\right)+\left(4x-4y\right)\\ =\left(x-y\right)^2+4\left(x-y\right)\\ =\left(x-y\right)\left(x-y+4\right)\\ 7y-4xy+14x-y^2-4x^2\\ =\left(-4x^2-4xy-y^2\right)+\left(7y+14x\right)\\ =7\left(2x+y\right)-\left(2x+y\right)^2\\ =\left(2x+y\right)\left(7-2x-y\right)\\ 3a-3b-a^2+2ab-b^2\\ =3\left(a-b\right)-\left(a-b\right)^2=\left(a-b\right)\left(3-a+b\right)\)

\(ĐK:\left\{{}\begin{matrix}\dfrac{x-1}{x+2}\ge0\\x+2\ne0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1\ge0\\x+2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1\le0\\x+2< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge1\\x< -2\end{matrix}\right.\)

\(d,\\ =\left(x^2+2xy+y^2\right)-25\\ =\left(x+y\right)^2-25\\ =\left(x+y+5\right)\left(x+y-5\right)\\ e,\\ =\left(a+b\right)^2-c\left(a+b\right)\\ =\left(a+b\right)\left(a+b-c\right)\\ f,\\ =\left(x^2-4y^2\right)-\left(2x+4y\right)\\ =\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\\ =\left(x+2y\right)\left(x-2y-2\right)\)

\(\left|2x-\dfrac{1}{2}\right|\ge0\\ \to-\left|2x-\dfrac{1}{2}\right|\le0\\ \to A=-\left|2x-\dfrac{1}{2}\right|+5\le5\\ "="\Leftrightarrow2x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{4}\)

\(\text{Viet: }\left\{{}\begin{matrix}x_1+x_2=4\\x_1x_2=-7\end{matrix}\right.\\ T=\dfrac{x_1^2+x_2^2}{x_1x_2}-2\\ T=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2}{-7}-2\\ T=\dfrac{4^2-2\left(-7\right)}{-7}-2=-\dfrac{44}{7}\)

\(\begin{array}{l}
XH_4\to X(IV)\\
YO\to Y(II)\\
CTTQ:X_m^{IV}Y_n^{II}\\
QTHT:m.IV=n.II\\
\to \frac{m}{n}=\frac{1}{2}\\
\to CTHH:XY_2
\end{array}\)

\(\begin{array}{l}
a)\\
NaOH+HCl\to NaCl+H_2O\\
n_{NaOH}=\frac{16}{40}=0,4(mol)\\
\to n_{NaCl}=n_{NaOH}=0,4(mol)\\
\to m_{NaCl}=0,4.58,5=23,4(g)\\
b)\\
m_{dd\,spu}=16+1,1=17,1(g)\\
C\%_{NaCl}=\frac{23,4}{17,1}.100\%>100\%
\end{array}\)

Xem lại \(m_{dd\,HCl}\)