\(\left(x-7\right)^4=64\\ \left(x-7\right)^4=\left(2\sqrt{2}\right)^4\\ x-7=2\sqrt{2}\\ x=2\sqrt{2}+7\)

Theo bảo toàn nguyên tố H có:

\(n_H\) trong \(n_{HCl}\) \(=\) với \(n_{HCl}\) 

\(n_H\) trong \(n_{H_2O}\) \(=2n_{H_2O}\)

Dể hiểu hơn là \(\left\{{}\begin{matrix}n_{H\left(HCl\right)}=n_{HCl}\left(1\right)\\n_{H\left(H_2O\right)}=2n_{H_2O}\left(2\right)\end{matrix}\right.\)

Từ (1) và (2) có \(n_{HCl\left(thamgia.pứ\right)}=2n_{H_2O\left(sản.phẩm\right)}\)

a Đề sai: )

b

\(a^3-a^2x-ay+xy\\ =a^2\left(a-x\right)-y\left(a-x\right)\\ =\left(a-x\right)\left(a^2-y\right)\)

c

\(4x^2-y^2+4x+1\\ =\left(2x\right)^2+2.2x.1+1-y^2\\ =\left(2x+1\right)^2-y^2\\ =\left(2x+1-y\right)\left(2x+1+y\right)\)

d

\(x^4+2x^3+x^2\\ =x^4+x^3+x^3+x^2\\ =x^3\left(x+1\right)+x^2\left(x+1\right)\\ =\left(x^3+x^2\right)\left(x+1\right)\)

e

\(5x^2-10xy+5y^2-5z^2\\ =5\left(x^2-2xy+y^2-z^2\right)\\ =5\left[\left(x-y\right)^2-z^2\right]\\ =5\left(x-y-z\right)\left(x-y+z\right)\)

\(n_{O\left(oxit\right)}=\dfrac{17,1-11,5}{16}=0,35\left(mol\right)\)

\(\Rightarrow n_{HCl}=2n_{H_2O}=2n_{O\left(oxit\right)}=2.0,35=0,7\left(mol\right)\)

\(\Rightarrow V=V_{HCl}=\dfrac{0,7}{2}=0,35=350\left(ml\right)\)

a

\(xy+3x-7y-21\\ =\left(xy+3x\right)-\left(7y+21\right)\\ =x\left(y+3\right)-7\left(y+3\right)\\ =\left(y+3\right)\left(x-7\right)\)

b

\(2xy-15-6x+5y\\ =\left(2xy-6x\right)-\left(15-5y\right)\\ =2x\left(y-3\right)-5\left(3-y\right)\\ =2x\left(y-3\right)+5\left(y-3\right)\\ =\left(y-3\right)\left(2x+5\right)\)

c Đề phải là \(\left(2x^2y+2xy^2-x-y\right)\) mới phân tích được: )

\(=2xy\left(x+y\right)-\left(x+y\right)\\ =\left(x+y\right)\left(2xy-1\right)\)

d

\(7x^3y-3xyz-21x^2+9z\\ =\left(7x^3y-21x^2\right)-\left(3xyz-9z\right)\\ =7x^2\left(xy-3\right)-3z\left(xy-3\right)\\ =\left(xy-3\right)\left(7x^2-3z\right)\)

e

\(4x^2-2x-y^2-y\\ =\left(2x\right)^2-y^2-\left(2x+y\right)\\ =\left(2x-y\right)\left(2x+y\right)-\left(2x+y\right)\\ =\left(2x+y\right)\left(2x-y-1\right)\)

f

\(9x^2-25y^2-6x+10y\\ =\left(3x\right)^2-\left(5y\right)^2-\left(6x-10y\right)\\ =\left(3x-5y\right)\left(3x+5y\right)-2\left(3x-5y\right)\\ =\left(3x-5y\right)\left(3x+5y-2\right)\)

a

\(\sqrt{6}x+\sqrt{6}=\sqrt{54}+\sqrt{24}\\ \Leftrightarrow\sqrt{6}\left(x+1\right)=\sqrt{3^2.6}+\sqrt{2^2.6}\\ \Leftrightarrow\sqrt{6}\left(x+1\right)=3\sqrt{6}+2\sqrt{6}\\ \Leftrightarrow\sqrt{6}\left(x+1\right)=5\sqrt{6}\\ \Leftrightarrow x+1=5\\ \Leftrightarrow x=4\)

b

\(\dfrac{x^2}{10}-\sqrt{1,21}=0\\ \Leftrightarrow\dfrac{x^2}{10}-\sqrt{\left(1,1\right)^2}=0\\ \Leftrightarrow\dfrac{x^2}{10}-\sqrt{\left(\dfrac{11}{10}\right)^2}=0\\ \Leftrightarrow\dfrac{x^2}{10}-\dfrac{11}{10}=0\\ \Leftrightarrow x^2-11=0\\ \Leftrightarrow x^2=11\\ \Leftrightarrow x=\pm\sqrt{11}\)

c

ĐK: \(x\ne-1,x\ge-\dfrac{3}{4}\)

\(\sqrt{\dfrac{4x+3}{x+1}}=3\\ \Leftrightarrow\dfrac{4x+3}{x+1}=3^2=9\\ \Leftrightarrow9x+9-4x-3=0\\ \Leftrightarrow5x+6=0\\ \Leftrightarrow x=-\dfrac{6}{5}\left(nhận\right)\)

d

ĐK: \(x\ge\dfrac{3}{2}\)

\(\dfrac{\sqrt{2x-3}}{\sqrt{x-1}}=2\\ \Leftrightarrow\sqrt{\dfrac{2x-3}{x-1}}=2\\ \Leftrightarrow\dfrac{2x-3}{x-1}=2^2=4\\ \Leftrightarrow4x-4-2x+3=0\\ \Leftrightarrow2x-1=0\\ \Leftrightarrow x=\dfrac{1}{2}\left(loại\right)\)

Vậy PT vô nghiệm

e

ĐK: \(x\ge0\)

Đặt \(t=\sqrt{x}\left(t\ge0\right)\)

Khi đó PT trở thành:

\(t^2-3t-5=0\\ \Leftrightarrow\left[{}\begin{matrix}t=\dfrac{\sqrt{29}+3}{2}\left(nhận\right)\\t=\dfrac{-\sqrt{29}+3}{2}\left(loại\right)\end{matrix}\right.\)

Với \(t=\dfrac{\sqrt{29}+3}{2}\Rightarrow x=\left(\dfrac{\sqrt{29}+3}{2}\right)^2=\dfrac{19+3\sqrt{29}}{2}\) (nhận)

\(x^2+6x-7=0\\ \Leftrightarrow x^2+7x-x-7=0\\ \Leftrightarrow\left(x^2+7x\right)-\left(x+7\right)=0\\ \Leftrightarrow x\left(x+7\right)-\left(x+7\right)=0\\ \Leftrightarrow\left(x+7\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-7\\x=1\end{matrix}\right.\)

\(\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\dfrac{6+2\sqrt{6}}{\sqrt{3}+\sqrt{2}}\\ =\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{4}\right)}{\sqrt{5}-\sqrt{4}}+\dfrac{\sqrt{2}.\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}\\ =\sqrt{3}+\sqrt{12}\\ =\sqrt{3}+\sqrt{2^2.3}\\ =\sqrt{3}+2\sqrt{3}\\ =3\sqrt{3}\)

a

\(x^2\left(2x+15\right)+4\left(2x+15\right)=0\\ \Leftrightarrow\left(2x+15\right)\left(x^2+4\right)=0\\ \Leftrightarrow2x+15=0\left(x^2+4>0\forall x\right)\\ \Leftrightarrow2x=-15\\ \Leftrightarrow x=-\dfrac{15}{2}\)

b

\(5x\left(x-2\right)-3\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\5x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0+2=2\\x=\dfrac{0+3}{5}=\dfrac{3}{5}\end{matrix}\right.\)

c

\(2\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow2\left(x+3\right)-\left(x^2+3x\right)=0\\ \Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\2-x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0-3=-3\\x=2-0=2\end{matrix}\right.\)

\(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\\ =\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}\\ =\sqrt{5}+\dfrac{\sqrt{5}}{2}\\ =\dfrac{2\sqrt{5}}{2}+\dfrac{\sqrt{5}}{2}\\ =\dfrac{2\sqrt{5}+\sqrt{5}}{2}\\ =\dfrac{3\sqrt{5}}{2}\)