\(b,B=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-8}{x-5\sqrt{x}+6}\left(x\ge0;x\ne4;x\ne9\right)\\ B=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)+\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ B=\dfrac{x-4+\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\)

\(c,B< A\Leftrightarrow\dfrac{\sqrt{x}-4}{\sqrt{x}-2}< \dfrac{\sqrt{x}+1}{\sqrt{x}-2}\Leftrightarrow\dfrac{\sqrt{x}-4}{\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}< 0\\ \Leftrightarrow\dfrac{-5}{\sqrt{x}-2}< 0\Leftrightarrow\sqrt{x}-2>0\left(-5< 0\right)\\ \Leftrightarrow x>4\\ d,P=\dfrac{B}{A}=\dfrac{\sqrt{x}-4}{\sqrt{x}-2}:\dfrac{\sqrt{x}+1}{\sqrt{x}-2}=\dfrac{\sqrt{x}-4}{\sqrt{x}+1}=1-\dfrac{5}{\sqrt{x}+1}\in Z\\ \Leftrightarrow5⋮\sqrt{x}+1\Leftrightarrow\sqrt{x}+1\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{-6;-2;0;4\right\}\\ \Leftrightarrow x\in\left\{0;16\right\}\left(\sqrt{x}\ge0\right)\)

\(e,P=1-\dfrac{5}{\sqrt{x}+1}\)

Ta có \(\sqrt{x}+1\ge1,\forall x\Leftrightarrow\dfrac{5}{\sqrt{x}+1}\ge5\Leftrightarrow1-\dfrac{5}{\sqrt{x}+1}\le-4\)

\(P_{max}=-4\Leftrightarrow x=0\)

Bài 1:

3: ĐKXĐ: x>=1

\(x-\sqrt{x+3+4\sqrt{x-1}}=1\)

=>\(x-\sqrt{x-1+2\cdot\sqrt{x-1}\cdot2+4}=1\)

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

=>\(x-\left|\sqrt{x-1}+2\right|=1\)

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

=>\(x-\sqrt{x-1}-2-1=0\)

=>\(x-1-\sqrt{x-1}-2=0\)

=>\(\left(\sqrt{x-1}\right)^2-2\sqrt{x-1}+\sqrt{x-1}-2=0\)

=>\(\left(\sqrt{x-1}-2\right)\left(\sqrt{x-1}+1\right)=0\)

=>\(\sqrt{x-1}-2=0\)

=>\(\sqrt{x-1}=2\)

=>x-1=4

=>x=5(nhận)

\(\left\{{}\begin{matrix}3x+1< x-7\\1-2x>x+1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x< -8\\3x< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< -4\\x< 0\end{matrix}\right.\) \(\Rightarrow x< -4\)

Vậy nghiệm của hệ là \(S=\left(-\infty;-4\right)\)

43.a)  \(m_{HCl\left(bđ\right)}=200.10,95\%=21,9\left(g\right)\)

=> \(n_{HCl\left(bđ\right)}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)

b) HCl phản ứng với NaOH là HCl dư

 \(HCl+NaOH\rightarrow NaCl+H_2O\)

\(n_{HCl\left(dư\right)}=n_{NaOH}=0,05.2=0,1\left(mol\right)\)

=> \(n_{HCl\left(pứ\right)}=n_{HCl\left(bđ\right)}-n_{HCl\left(dư\right)}=0,6-0,1=0,5\left(mol\right)\)

c) \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\)

\(n_{CaCO_3}=\dfrac{1}{2}n_{HCl\left(pứ\right)}=0,25\left(mol\right)\)

=> \(m_{CaCO_3}=0,25.100=25\left(g\right)\)

d) \(n_{CO_2}=\dfrac{1}{2}n_{HCl\left(pứ\right)}=0,25\left(mol\right)\)

=> \(V_{CO_2}=0,25.22,4=5,6\left(l\right)\)

e) \(m_{ddsaupu}=25+200-0,25.44=214\left(g\right)\)

Dung dịch A gồm CaCl2 và HCl dư

\(n_{CaCl_2}=\dfrac{1}{2}n_{HCl\left(pứ\right)}=0,25\left(mol\right)\)

\(C\%_{CaCl_2}=\dfrac{0,25.111}{214}.100=12,97\%\)

\(C\%_{HCl\left(dư\right)}=\dfrac{0,1.36,5}{214}.100=1,71\%\)

e:

\(E=\left(\dfrac{\sqrt{15}-\sqrt{20}}{2-\sqrt{3}}+\dfrac{\sqrt{21}-\sqrt{7}}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)

\(=\left(-\dfrac{\sqrt{5}\left(2-\sqrt{3}\right)}{2-\sqrt{3}}-\dfrac{\sqrt{7}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right)\cdot\dfrac{\sqrt{7}-\sqrt{5}}{1}\)

\(=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)

=-2

f: \(F=\sqrt{3}+1+2-\sqrt{3}=3\)

a: Thay \(x=\dfrac{1}{4}\) vào A, ta được:

\(A=\left(\dfrac{1}{2}+1\right):\left(\dfrac{1}{2}-2\right)=\dfrac{3}{2}:\dfrac{-5}{2}=\dfrac{-3}{5}\)

Bài II:

a: Thay \(x=\dfrac{1}{4}\) vào A, ta được:

\(A=\left(\dfrac{1}{2}+1\right):\left(\dfrac{1}{2}-2\right)=\dfrac{3}{2}:\dfrac{-5}{2}=\dfrac{3}{2}\cdot\dfrac{-2}{5}=\dfrac{-3}{5}\)

b: Ta có: \(B=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-8}{x-5\sqrt{x}+6}\)

\(=\dfrac{x-4+\sqrt{x}-8}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{\sqrt{x}+4}{\sqrt{x}-2}\)

a, Th1 : \(m-1=0\Rightarrow m=1\)

\(\Rightarrow-x+3=0\\ \Rightarrow x=3\)

Th2 : \(m\ne1\)

\(\Delta=\left(-1\right)^2-4.\left(m-1\right).3\\ =1-12m+12\\=13-12m \)

phương trình có nghiệm \(\Delta\ge0\)

\(\Rightarrow13-12m\ge0\\ \Rightarrow m\le\dfrac{13}{12}\)

b, Áp dụng hệ thức vi ét : \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{1}{m-1}\\x_1x_1=\dfrac{3}{m-1}\end{matrix}\right.\)

Tổng bình phương hai nghiệm bằng 12 \(\Rightarrow x^2_1+x^2_2=12\)

\(\left(x_1+x_2\right)^2-2x_1x_2=12\\ \Leftrightarrow\left(\dfrac{1}{m-1}\right)^2-2.\left(\dfrac{3}{m-1}\right)=12\\ \Leftrightarrow\dfrac{1}{\left(m-1\right)^2}-\dfrac{6}{m-1}=12\\ \Leftrightarrow1-6\left(m-1\right)=12\left(m-1\right)^2\\ \Leftrightarrow1-6m+6=12\left(m^2-2m+1\right)\\ \Leftrightarrow7-6m-12m^2+24m-12=0\\ \Leftrightarrow-12m^2+18m-5=0\\ \Leftrightarrow\left[{}\begin{matrix}m=\dfrac{9-\sqrt{21}}{12}\\m=\dfrac{9+\sqrt{21}}{12}\end{matrix}\right.\Rightarrow m=\dfrac{9+\sqrt{21}}{12}\)