Là do trọng lực của trái đất

`@`\(85-\left(14+3x\right):5=81\)

\(\Rightarrow\left(14+3x\right):5=4\)

\(\Rightarrow14+3x=20\)

`=>3x=6`

`=>x=2`

`@`\(\left(x-6\right)^2=16\)

`=>(x-6)^2=4^2`

\(\Rightarrow\left[{}\begin{matrix}x-6=4\\x-6=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=10\\x=2\end{matrix}\right.\)

`@`\(3^{x+2}-3^x=72\)

\(\Rightarrow3^x.9-3^x=72\)

\(\Rightarrow3^x\left(9-1\right)=72\)

\(\Rightarrow3^x.8=72\)

\(\Rightarrow3^x=9\)

\(\Rightarrow x=2\)

Gọi quãng đường là `x`\(\left(km\right)\) ; \(\left(x>0\right)\)

Thời gian đi là: \(\dfrac{x}{42}\left(h\right)\)

Thời gian về là: \(\dfrac{x}{42+6}=\dfrac{x}{48}\left(h\right)\)

Theo đề bài ta có pt:

\(\dfrac{x}{42}+\dfrac{x}{48}=5\)

\(\Leftrightarrow x\left(\dfrac{1}{42}+\dfrac{1}{48}\right)=5\)

\(\Leftrightarrow x=112\left(tm\right)\)

Vậy quãng đường AB dài 112 km

`a.`\(4\dfrac{1}{2}+\dfrac{11}{8}\times1\dfrac{2}{11}\)

\(=\dfrac{9}{2}+\dfrac{11}{8}\times\dfrac{13}{11}\)

\(=\dfrac{9}{2}+\dfrac{13}{8}\)

\(=\dfrac{49}{8}\)

`b.`\(\left(\dfrac{9}{4}-1\dfrac{1}{5}\right):\dfrac{9}{5}\)

\(=\left(\dfrac{9}{4}-\dfrac{6}{5}\right):\dfrac{9}{5}\)

\(=\dfrac{21}{20}:\dfrac{9}{5}\)

\(=\dfrac{7}{12}\)

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

\(A=\dfrac{\sqrt{\dfrac{1}{4}}+1}{\sqrt{\dfrac{1}{4}}-3}=-\dfrac{3}{5}\)

`b.`\(B=\dfrac{x-3}{x-9}+\dfrac{1}{\sqrt{x}+3}-\dfrac{2}{3-\sqrt{x}}\)

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

\(B=\dfrac{x-3+\sqrt{x}-3+2\sqrt{x}+6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(B=\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(B=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(B=\dfrac{\sqrt{x}}{\sqrt{x}-3}\)

`c.`\(P=B:A\)

\(P=\dfrac{\sqrt{x}}{\sqrt{x}-3}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

\(P=\dfrac{\sqrt{x}}{\sqrt{x}-3}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(P=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)

\(P=\dfrac{5}{2}\)

\(\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}+1}=\dfrac{5}{2}\)

\(\Leftrightarrow2\sqrt{x}=5\sqrt{x}+5\)

\(\Leftrightarrow-3\sqrt{x}=5\) ( vô lý )

Vậy không có giá trị `x` thỏa mãn

`d.`\(P=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)

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

Để P nguyên thì \(\dfrac{1}{\sqrt{x}+1}\in Z\Rightarrow\sqrt{x}+1\inƯ\left(1\right)=\left\{\pm1\right\}\)

`@`\(\sqrt{x}+1=1\Rightarrow\sqrt{x}=0\Rightarrow x=0\left(tm\right)\)

`@`\(\sqrt{x}+1=-1\) ( vô lý )

Vậy \(x=0\) thì P nguyên

`a.` \(\dfrac{6x+5}{2}-\dfrac{10x+3}{4}\ge2x+\dfrac{2x+1}{2}\)

\(\Leftrightarrow\dfrac{12x+10-10x-3}{4}\ge\dfrac{4x+2x+1}{2}\)

\(\Leftrightarrow\dfrac{2x+7}{4}\ge\dfrac{6x+1}{2}\)

\(\Leftrightarrow\dfrac{2x+7}{4}\ge\dfrac{12x+2}{4}\)

\(\Leftrightarrow2x+7\ge12x+2\)

\(\Leftrightarrow-10x\ge-5\)

\(\Leftrightarrow x\le\dfrac{1}{2}\)

Vậy \(S=\left\{x|x\le\dfrac{1}{2}\right\}\)

`b.`

Ta có:

\(x+y=3\)

\(\Leftrightarrow\dfrac{x}{2}+\dfrac{x}{2}+y=3\)

Áp dụng BĐT AM-GM, ta có:

\(\dfrac{x}{2}+\dfrac{x}{2}+y\ge3\sqrt[3]{\dfrac{x.x.y}{2.2}}\)

\(\Leftrightarrow\dfrac{x}{2}+\dfrac{x}{2}+y\ge3\sqrt[3]{\dfrac{x^2y}{4}}\)

\(\Leftrightarrow3\ge3\sqrt[3]{\dfrac{x^2y}{4}}\)

\(\Leftrightarrow1\ge\sqrt[3]{\dfrac{x^2y}{4}}\)

\(\Leftrightarrow1\ge\dfrac{x^2y}{4}\)

\(\Leftrightarrow4\ge x^2y\) hay \(x^2y\le4\left(đfcm\right)\)

`a.`\(x=\sqrt[3]{-125}+\sqrt[3]{216}\)

\(\Leftrightarrow x=-5+6\)

`<=>x=1` thế vào A, ta được:

\(A=\dfrac{\sqrt{1}-2}{\sqrt{1}+2}=-\dfrac{1}{3}\)

`b.`\(A>\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+2}>\dfrac{1}{3}\)

\(\Leftrightarrow3\sqrt{x}-6>\sqrt{x}+2\)

\(\Leftrightarrow2\sqrt{x}>8\)

\(\Leftrightarrow\sqrt{x}>4\)

\(\Leftrightarrow x>16\)

`c.` \(B=\dfrac{\sqrt{x}+3}{\sqrt{x}+2}+\dfrac{10}{2-\sqrt{x}}+\dfrac{4}{x-4}\)

\(B=\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)-10\left(\sqrt{x}+2\right)+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(B=\dfrac{x+\sqrt{x}-6-10\sqrt{x}-20+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(B=\dfrac{x-9\sqrt{x}-22}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(B=\dfrac{\left(\sqrt{x}-11\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(B=\dfrac{\sqrt{x}-11}{\sqrt{x}-2}\) ( sửa đề )

\(B=\dfrac{\sqrt{x}-2-9}{\sqrt{x}-2}=1-\dfrac{9}{\sqrt{x}-2}\)

Để B nguyên thì \(\dfrac{9}{\sqrt{x}-2}\in Z\Rightarrow\sqrt{x}-2\inƯ\left(9\right)=\left\{\pm1;\pm3;\pm9\right\}\)

Mà \(\sqrt{x}\ge0\Rightarrow\sqrt{x}-2\ge-2\)

`@`\(\sqrt{x}-2=-1\Rightarrow\sqrt{x}=1\Rightarrow x=1\left(tm\right)\)

`@`\(\sqrt{x}-2=1\Rightarrow\sqrt{x}=3\Rightarrow x=9\left(tm\right)\)

`@`\(\sqrt{x}-2=3\Rightarrow\sqrt{x}=5\Rightarrow x=25\left(tm\right)\)

`@`\(\sqrt{x}-2=9\Rightarrow\sqrt{x}=11\Rightarrow x=121\left(tm\right)\)

Vậy \(x\in\left\{1;9;25;121\right\}\) thì B nguyên

`d.`\(P=A.B\)

\(P=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}.\dfrac{\sqrt{x}-11}{\sqrt{x}-2}\)

\(P=\dfrac{\sqrt{x}-11}{\sqrt{x}+2}\)

\(P=\dfrac{\sqrt{x}+2-13}{\sqrt{x}+2}=1-\dfrac{13}{\sqrt{x}+2}\)

Để P nguyên thì \(\dfrac{13}{\sqrt{x}+2}\in Z\Rightarrow\sqrt{x}+2\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)

Mà \(\sqrt{x}+2\ge2\)

`@`\(\sqrt{x}+2=13\Rightarrow\sqrt{x}=11\Rightarrow x=121\left(tm\right)\)

Vậy \(x=121\) thì P nguyên

`@`\(A=2\sqrt{8}-\dfrac{2}{3}\sqrt{18}+\sqrt{50}\)

\(A=4\sqrt{2}-2\sqrt{2}+5\sqrt{2}\)

\(A=7\sqrt{2}\)

`@`\(B=\dfrac{3+\sqrt{3}}{\sqrt{3}}-\dfrac{2}{\sqrt{3}-1}\)

\(B=\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}}-\dfrac{2}{\sqrt{3}-1}\)

\(B=\left(\sqrt{3}+1\right)-\dfrac{2}{\sqrt{3}-1}\)

\(B=\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)-2}{\sqrt{3}-1}\)

\(B=\dfrac{3-1-2}{\sqrt{3}-1}=\dfrac{0}{\sqrt{3}-1}=0\)

\(n_{HCl}=\dfrac{200.14,6\%}{36,5}=0,8\left(mol\right)\)

\(CuO+2HCl\rightarrow CuCl_2+H_2O\)

  0,4         0,8                           ( mol )

\(m_{CuO}=0,4.80=32\left(g\right)\)

\(n_{CO_2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\)

\(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2\uparrow+H_2O\) 

  0,02             0,04         0,02     0,02                ( mol )

\(C_{M_{HCl}}=\dfrac{0,04}{0,02}=2\left(M\right)\)

\(m_{Na_2CO_3}=0,02.106=2,12\left(g\right)\)

\(m_{NaCl\left(hh\right)}=5-2,12=2,88\left(g\right)\)

\(m_{muối}=2,88+\left(0,02.58,5\right)=4,05\left(g\right)\)

`c.\(\left\{{}\begin{matrix}\%m_{Na_2CO_3}=\dfrac{2,12}{5}.100=42,4\%\\\%m_{NaCl}=100\%-42,4\%=57,6\%\end{matrix}\right.\)