\(a,=6x\left(2x+y-1\right)\\ b,=\left(2x+1\right)^2-16y^2=\left(2x+1-4y\right)\left(2x+1+4y\right)\\ c,=2x^2-2x-5x+5=\left(x-1\right)\left(2x-5\right)\)

\(=\dfrac{7x+6-3x-6}{2x\left(x+7\right)}=\dfrac{4x}{2x\left(x+7\right)}=\dfrac{2}{x+7}\)

\(n_{HCl}=\dfrac{10,95}{36,5}=0,3\\ (n_{CaO};n_{ZnO})=(x;y)\\ \Rightarrow 56x+81y=12,1(1)\\ CaO+2HCl\to CaCl_2+H_2O\\ ZnO+2HCl\to ZnCl_2+H_2O\\ \Rightarrow 2x+2y=0,3(2)\\ (1)(2)\Rightarrow \begin{cases} x=0,002(mol)\\ y=0,148(mol) \end{cases}\Rightarrow \begin{cases} \%_{CaO}=\dfrac{0,002.56}{12,1}.100\%=0,93\%\\ \%_{ZnO}=100\%-0,93\%=99,07\% \end{cases}\)

Câu b ko hiểu đề :v

hỏi lại thầy bạn coi :v

\(\Rightarrow x-\left(-x+x+3\right)-\left(x+3-x+2\right)=0\\ \Rightarrow x-3-5=0\\ \Rightarrow x=8\)

lmj có sao mà vote :v

Gọi số cây 7A,7B,7C lần lượt là \(a,b,c\in \mathbb{N^*},cây\)

Ta có \(\left\{{}\begin{matrix}\dfrac{8}{9}a=b\\\dfrac{17}{16}b=c\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{9}{8}\\\dfrac{b}{c}=\dfrac{16}{17}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{9}=\dfrac{b}{8}\\\dfrac{b}{16}=\dfrac{c}{17}\end{matrix}\right.\Rightarrow\dfrac{a}{18}=\dfrac{b}{16}=\dfrac{c}{17}\)

\(\Rightarrow\dfrac{a}{18}=\dfrac{b}{16}=\dfrac{c}{17}=\dfrac{a+b+c}{18+16+17}=\dfrac{1020}{51}=20\\ \Rightarrow\left\{{}\begin{matrix}a=360\\b=320\\c=340\end{matrix}\right.\)

Vậy ...

\(a,=2x\left(x^2-25\right)=2x\left(x-5\right)\left(x+5\right)\\ b,=\left(x-3\right)^2-4y^2=\left(x-2y-3\right)\left(x+2y-3\right)\\ c,=3x\left(x-2\right)-4\left(x-2\right)=\left(3x-4\right)\left(x-2\right)\\ d,=x^2-2x-5x+10=\left(x-2\right)\left(x-5\right)\\ e,=x\left(x-3\right)+y\left(x-3\right)=\left(x-3\right)\left(x+y\right)\)

\(a,ĐK:x,y\ne2\)

Đặt \(\left\{{}\begin{matrix}x-2=a\\y-2=b\end{matrix}\right.\)

\(HPT\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{a}+\dfrac{3}{b}=5\\\dfrac{3}{a}+\dfrac{2}{b}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{a}+\dfrac{9}{b}=15\\\dfrac{6}{a}+\dfrac{4}{b}=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{a}+\dfrac{3}{b}=5\\\dfrac{5}{b}=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{a}+3=5\\b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\Leftrightarrow x=y=3\left(tm\right)\)

\(b,ĐK:x\ge3;y\ge1\)

Sửa: \(\sqrt{x-3}-\sqrt{y-1}=4\)

Đặt \(\left\{{}\begin{matrix}a=\sqrt{x-3}\ge0\\b=\sqrt{y-1}\ge0\end{matrix}\right.\)

\(HPT\Leftrightarrow\left\{{}\begin{matrix}a-2b=2\\a-b=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a-b=4\\-b=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=6\\b=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-3=36\\y-1=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=39\\y=5\end{matrix}\right.\)