Câu 7: B

Câu 8: C

Câu 10: A

7.

ĐKXĐ: \(3\le x\le7\)

\(x-5+\sqrt{7-x}-\sqrt{x-3}=0\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x-3}=a\ge0\\\sqrt{7-x}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=2\left(x-5\right)\)

Pt trở thành:

\(\dfrac{a^2-b^2}{2}+b-a=0\)

\(\Leftrightarrow\left(a-b\right)\left(a+b-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a+b=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=\sqrt{7-x}\\\sqrt{x-3}+\sqrt{7-x}=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=7-x\\4+2\sqrt{\left(x-3\right)\left(7-x\right)}=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=5\\x=7\end{matrix}\right.\)

8.

ĐKXĐ: \(-2\le x\le2\)

Đặt \(\sqrt{x+2}+\sqrt{2-x}=t\Rightarrow2\le t\le2\sqrt{2}\)

\(t^2=4+2\sqrt{4-x^2}\Rightarrow2\sqrt{-x^2+4}=t^2-4\)

Pt trở thành:

\(t+t^2-4+2m+3=0\)

\(\Leftrightarrow t^2+t+1=-2m\)

Xét hàm \(f\left(t\right)=t^2+t+1\) trên \(\left[2;2\sqrt{2}\right]\)

\(a=1>0;-\dfrac{b}{2a}=-\dfrac{1}{2}< 2\Rightarrow f\left(t\right)\) đồng biến trên đoạn đã cho

\(\Rightarrow f\left(2\right)\le f\left(t\right)\le f\left(2\sqrt{2}\right)\Rightarrow7\le f\left(t\right)\le9+2\sqrt{2}\)

\(\Rightarrow7\le-2m\le9+2\sqrt{2}\Rightarrow-\dfrac{9+2\sqrt{2}}{2}\le m\le-\dfrac{7}{2}\)

\(\Rightarrow m=\left\{-5;-4\right\}\)

\(x^2-6x+1>\left(2x-3\right)\sqrt{x^2+1}\)

\(\Leftrightarrow\left(x^2+1-9\right)-3\left(2x-3\right)-\left(2x-3\right)\sqrt{x^2+1}>0\)

\(\Leftrightarrow\left(\sqrt{x^2+1}-3\right)\left(\sqrt{x^2+1}+3\right)-\left(2x-3\right)\left(\sqrt{x^2+1}+3\right)>0\)

\(\Leftrightarrow\left(\sqrt{x^2+1}+3\right)\left(\sqrt{x^2+1}-3-\left(2x-3\right)\right)>0\)

\(\Leftrightarrow\sqrt{x^2+1}-2x>0\) (do \(\sqrt{x^2+1}+3>0\) với mọi x)

\(\Leftrightarrow\sqrt{x^2+1}>2x\)

\(\Leftrightarrow\left[{}\begin{matrix}x\le0\\\left\{{}\begin{matrix}x>0\\x^2+1>4x^2\end{matrix}\right.\end{matrix}\right.\) 

\(\Leftrightarrow\left[{}\begin{matrix}x\le0\\\left\{{}\begin{matrix}x>0\\-\dfrac{\sqrt{3}}{3}< x< \dfrac{\sqrt{3}}{3}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow x< \dfrac{\sqrt{3}}{3}\)

Pt bậc 2 có 2 nghiệm trái dấu khi \(ac< 0\)

\(\Leftrightarrow\left(m^2-4\right)m< 0\)

\(\Leftrightarrow m\in\left(-\infty;-2\right)\cup\left(0;2\right)\)

a.Theo định luật bảo toàn khối lượng, ta có:

\(m_{Cl_2}=36,2-7,8=28,4g\)

\(n_{Cl_2}=\dfrac{m}{M}=\dfrac{28,4}{71}=0,4mol\)

\(2Al+3Cl_2\rightarrow\left(t^o\right)2AlCl_3\)

  2         3                    2  ( mol )

\(2Mg+Cl_2\rightarrow\left(t^o\right)2MgCl_2\)

  2          1                  2    ( mol )

Gọi \(n_{Al}=x\)

       \(n_{Mg}=y\)

\(\Rightarrow m_{Al}=27x\)

\(\Rightarrow m_{Mg}=24y\)

Ta có:

\(\left\{{}\begin{matrix}m_{hh}=27x+24y=7,8\\\dfrac{3}{2}x+\dfrac{1}{2}y=0,4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{19}{75}\\y=\dfrac{1}{25}\end{matrix}\right.\)

\(\Rightarrow m_{Al}=27.\dfrac{19}{75}=6,84g\)

\(\Rightarrow m_{Mg}=24.\dfrac{1}{25}=0,96g\)

\(\%m_{Al}=\dfrac{6,84.100}{7.8}=87,7\%\)

\(\%m_{Mg}=100\%-87,7\%=12,3\%\)

ĐKXĐ: \(x\ge4\)

\(\sqrt{x+11}\ge\sqrt{x-4}+\sqrt{2x-1}\)

\(\Leftrightarrow x+11\ge3x-5+2\sqrt{\left(x-4\right)\left(2x-1\right)}\)

\(\Leftrightarrow8-x\ge\sqrt{2x^2-9x+4}\)

\(\Leftrightarrow\left\{{}\begin{matrix}8-x\ge0\\\left(8-x\right)^2\ge2x^2-9x+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le8\\x^2+7x-60\le0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le8\\-12\le x\le5\end{matrix}\right.\) \(\Rightarrow-12\le x\le5\)

Kết hợp ĐKXD ta được nghiệm của BPT là: \(4\le x\le5\)

\(a,Ca+2HCl\rightarrow CaCl_2+H_2\left(1\right)\\ CaO+2HCl\rightarrow CaCl_2+H_2O\left(2\right)\\ Đặt:n_{Ca}=a\left(mol\right);n_{CaO}=b\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}111a+111b=22,2\\22,4a=2,24\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\\ \Rightarrow\%m_{CaO}=\dfrac{0,1.56}{0,1.56+0,1.40}.100\approx52,83\%\\ \Rightarrow\%m_{Ca}=47,17\%\\ b,n_{HCl}=2.\left(a+b\right)=0,4\left(mol\right)\\ \Rightarrow m_{ddHCl}=\dfrac{0,4.36,5.100}{36,5}=40\left(g\right)\)

Bài 30:

\(Đặt:n_{KBr}=a\left(mol\right);n_{NaCl}=b\left(mol\right)\left(a,b>0\right)\\ KBr+AgNO_3\rightarrow AgBr\downarrow+KNO_3\\ NaCl+AgNO_3\rightarrow AgCl\downarrow+NaNO_3\\ \Rightarrow\left\{{}\begin{matrix}119a+58,5b=23,6\\188a+143,5b=47,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\\ a,m_{KBr}=0,1.119=11,9\left(g\right)\\ m_{NaCl}=0,2.58,5=11,7\left(g\right)\\ b,n_{AgNO_3}=a+b=0,3\left(mol\right)\\ \Rightarrow V_{ddAgNO_3}=\dfrac{0,3}{0,5}=0,6\left(l\right)=600\left(ml\right)\)