a)\(n_{Fe_2O_3}=0,2\left(mol\right)\)

PT:\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

     \(0,2\)          \(1,2\)          \(0,4\)

\(\Rightarrow n_{FeCl_3}=0,4\left(mol\right)\)

\(\Rightarrow m_{FeCl_3}=65\left(g\right)\)

b) \(n_{HCl}=\dfrac{218.30\%}{35,5+1}=\dfrac{654}{365}\left(mol\right)\)

Từ PT \(\Rightarrow\)\(n_{HClpư}=1,2\left(mol\right)\)

\(\Rightarrow n_{HCldư}=\dfrac{654}{365}-1,2=\dfrac{216}{365}\left(mol\right)\)

\(\Rightarrow m_{HCldư}=21,6\left(g\right)\)

\(m_{dd}=32+218=250\left(g\right)\)

\(C\%_{FeCl_3}=\dfrac{65}{250}.100\%=26\left(\%\right)\)

\(C\%_{HCldu}=\dfrac{21,6}{250}.100\%=8,64\%\)

\(\Delta=m^2-4\left(m-4\right)=\left(m^2-4m+4\right)+12=\left(m-2\right)^2+12>0;\forall m\)

Suy ra pt luôn có hai nghiệm pb với mọi m

Theo viet có:\(\left\{{}\begin{matrix}x_1+x_2=m\\x_1.x_2=m-4\end{matrix}\right.\)

\(\left(5x_1-1\right)\left(5x_2-1\right)< 0\)

\(\Leftrightarrow25x_1x_2-5\left(x_1+x_2\right)+1< 0\)

\(\Leftrightarrow25\left(m-4\right)-5m+1< 0\)

\(\Leftrightarrow m< \dfrac{99}{20}\)

Vậy...

a)\(P=x^3+6x^2+12x+8+x^3-6x^2+12x-8-2x^3-24x=0\)

Vậy g/t P không phụ thuộc vào biến.

b)\(Q=x^3-3x^2+3x-1-\left(x^3+3x^2+3x+1\right)+6\left(x^2-1\right)=-6x^2-2+6x^2-6=-8\)

Vậy g/t Q không phụ thuộc vào biến.

a) Pt \(\Leftrightarrow\sqrt[3]{x+7}=4\)

\(\Leftrightarrow x+7=64\)\(\Leftrightarrow x=57\)

Vậy...

b) Pt \(\Leftrightarrow\sqrt[3]{x}+1+\sqrt[3]{x+1}+\sqrt[3]{x+2}-1=0\)

\(\Leftrightarrow\dfrac{x+1}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}+\sqrt[3]{x+1}+\dfrac{x+2-1}{\sqrt[3]{\left(x+2\right)^2}+\sqrt[3]{x+2}+1}=0\)

\(\Leftrightarrow\sqrt[3]{x+1}\left(\dfrac{\sqrt[3]{\left(x+1\right)^2}}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}+1+\dfrac{\sqrt[3]{\left(x+1\right)^2}}{\sqrt[3]{\left(x+2\right)^2}+\sqrt[3]{x+2}+1}\right)=0\)

\(\Leftrightarrow\sqrt[3]{x+1}=0\) (vì \(\dfrac{\sqrt[3]{\left(x+1\right)^2}}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}+1+\dfrac{\sqrt[3]{\left(x+1\right)^2}}{\sqrt[3]{\left(x+2\right)^2}+\sqrt[3]{x+2}+1}>0\) với mọi x)

\(\Leftrightarrow x=-1\)

Vậy...

c)\(\sqrt[3]{2+x}+\sqrt[3]{5-x}=1\)

\(\Leftrightarrow2+x+3\sqrt[3]{\left(2+x\right)\left(5-x\right)}\left(\sqrt[3]{2+x}+\sqrt[3]{5-x}\right)+5-x=1\)

\(\Leftrightarrow7+3\sqrt[3]{\left(2+x\right)\left(5-x\right)}.1=1\)

\(\Leftrightarrow\sqrt[3]{\left(2+x\right)\left(5-x\right)}=-2\)

\(\Leftrightarrow10+3x-x^2=-8\)

\(\Leftrightarrow18+3x-x^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-3\end{matrix}\right.\)

Vậy ...

À không, ý a \(\Leftrightarrow2x+1< 0\Leftrightarrow x< -\dfrac{1}{2}\)

a)ĐK:\(-\dfrac{5}{2x+1}\ge0\) và \(2x+1\ne0\)

\(\Leftrightarrow2x+1>0\) \(\Leftrightarrow x>-\dfrac{1}{2}\)

Vậy \(x< -\dfrac{1}{2}\) thì căn thức có nghĩa

b)\(\sqrt[3]{64}+\sqrt[3]{-27}-\sqrt[3]{-4}.\sqrt[3]{2}=\sqrt[3]{4^3}+\sqrt[3]{-3^3}-\sqrt[3]{-8}\)

\(=4+\left(-3\right)-\left(-2\right)\)

\(=3\)

a)Pt \(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\dfrac{1}{3}+\dfrac{1}{2}\)

\(\Leftrightarrow\left|2x-1\right|=\dfrac{5}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\dfrac{5}{6}\\2x-1=-\dfrac{5}{6}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{12}\\x=\dfrac{1}{12}\end{matrix}\right.\)

Vậy...

b)Đk:\(x\ge3\)

Pt \(\Leftrightarrow\sqrt{x-3}\left(x-4\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\x-4=0\\x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=4\left(tm\right)\\x=2\left(ktm\right)\end{matrix}\right.\)

Vậy...

c)Đk:\(x\ge1\)

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

\(\Leftrightarrow\sqrt{x-1}=13-x\)

\(\Leftrightarrow\left\{{}\begin{matrix}13-x\ge0\\x-1=x^2-26x+169\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}13\ge x\\x^2-27x+170=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}13\ge x\\x^2-17x-10x+170=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}13\ge x\\\left(x-17\right)\left(x-10\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}13\ge x\\\left[{}\begin{matrix}x=17\\x=10\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow x=10\) (tm)

Vậy...

a)Pt \(\Leftrightarrow\left|2x-1\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

Vậy...

b)Đk:\(x\ge-1\)

Pt\(\Leftrightarrow2\sqrt{x+1}-3\sqrt{x+1}-2\sqrt{x+1}=5\)

\(\Leftrightarrow-3\sqrt{x+1}=5\) (vô nghiệm)

Vậy...

\(P=1:\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{x-1}\right)\) (Đk:\(x\ge0;x\ne1\))

\(=1:\left[\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\)

\(=1:\left[\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right]\)

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

\(=1:\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=1:\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

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

b) Áp dụng AM-GM có:

\(\sqrt{x}+\dfrac{1}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\dfrac{1}{\sqrt{x}}}=2\)

Dấu "=" xảy ra khi \(\sqrt{x}=\dfrac{1}{\sqrt{x}}\Leftrightarrow x=1\left(ktm\right)\)

\(\Rightarrow\)Dấu "=" không xảy ra

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

hay P>3

Vậy...

c)\(C=\sqrt{1-2x+x^2}+\sqrt{x^2-6x+9}\)

\(=\sqrt{\left(1-x\right)^2}+\sqrt{\left(x-3\right)^2}\)\(=\left|1-x\right|+\left|x-3\right|\ge\left|1-x+x-3\right|=2\)

Dấu "=" xảy ra khi \(\left(1-x\right)\left(x-3\right)\ge0\Leftrightarrow1\le x\le3\)

Vậy \(C_{min}=2\)