\(\dfrac{0,8\left(\dfrac{4}{5}.1,25\right)}{0,64-\dfrac{1}{25}}+\dfrac{\left(10,8-\dfrac{2}{25}\right):\dfrac{4}{7}}{\left(6\dfrac{5}{9}-3\dfrac{1}{4}.2\dfrac{2}{17}\right)}\)

\(=\dfrac{0,8}{0,6}+\dfrac{\dfrac{268}{25}.\dfrac{7}{4}}{\left(\dfrac{59}{9}-\dfrac{13}{4}.\dfrac{36}{17}\right)}\)

\(=\dfrac{4}{3}+\dfrac{\dfrac{469}{25}}{\dfrac{-50}{153}}\)

\(=\dfrac{4}{3}-57,4056\)

\(=-56,07\)

\(M=\dfrac{x+12}{x-4}+\dfrac{1}{\sqrt{x}+2}-\dfrac{4}{\sqrt{x}-2}\)

\(M=\dfrac{x+12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{1}{\sqrt{x}+2}-\dfrac{4}{\sqrt{x}-2}\)

\(M=x+12+\sqrt{x}-2-4\left(\sqrt{x}+2\right)\)

\(M=x+12+\sqrt{x}-2-4\sqrt{x}-8\)

\(M=x-3\sqrt{x}+2\)

\(M=x-\sqrt{x}-2\sqrt{x}+2\)

\(M=\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)\)

\(M=\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\)

Ei mà bài 21 tui lấy số mol đẩy sang chia ra cũng ra M = 56

21. đề sai nha cụ

23. \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

\(n_{CuCl_2}=\dfrac{67,5.60\%}{135}=0,3\left(mol\right)\)

Pt: \(Zn+CuCl_2\rightarrow ZnCl_2+Cu\)

0,2 mol 0,3mol \(\rightarrow\) 0,2mol \(\rightarrow\)0,2mol

Lập tỉ số: \(n_{Zn}:n_{CuCl_2}=0,2< 0,3\)

Zn hết, CuCl₂ dư

\(m_{Cu}=0,2.64=12,8\left(g\right)\)

\(\Sigma_{m\left(spu\right)}=13+67,5-12,8=67,7\left(g\right)\)

\(C\%_{CuCl_2\left(dư\right)}=\dfrac{\left(0,3-0,2\right).135}{67,6}.100=19,94\%\)

\(C\%_{ZnCl_2}=\dfrac{0,2.136}{67,7}.100=40,18\%\)

Í :v làm lại

20. Gọi cthc : Fe_xO_y

Pt: \(Fe_xO_y+2yHCl\rightarrow xFeCl_{\dfrac{2y}{x}}+yH_2O\)

56x + 16y 56x + 71y

14,4g 25,4g

\(\Rightarrow\dfrac{56x+16y}{14,4}=\dfrac{56x+71y}{25,4}\)

\(\Rightarrow x=y\)

Vậy cthc: FeO

1. Gọi cthc: Fe₂O_x

Pt: \(Fe_2O_x+2xHCl\rightarrow2FeCl_x+xH_2O\)

112 +16x 112 + 71x

14,4 g 25,4g

\(\Rightarrow\dfrac{112+16x}{14,4}=\dfrac{112+71x}{25,4}\)

\(\Rightarrow x=2\)

Vậy cthc: FeO

\(n_{HCl}=0,2.0,2=0,04\left(mol\right)\)

Pt: \(NaOH+HCl\rightarrow NaCl+H_2O\)

0,04mol <---0,04mol

\(V_{NaOH}=\dfrac{0,04}{0,1}=0,4\left(l\right)=400\left(ml\right)\)

b) Pt: \(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)

0,02mol<--- 0,04mol

\(m_{dd_{Ca\left(OH\right)_2}}=\dfrac{0,02.74.100}{6}=24,67\left(g\right)\)

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

Thế x = -1 ta có \(\left(-1-2\right)^2=9\)

2. \(a=\left(x-1\right)^2+\left(x+1\right)\left(3-x\right)\)

\(\Leftrightarrow a=x^2-2x+1+3x-x^2+3-x\)

\(\Leftrightarrow a=4\)

Vậy biểu thức a không phụ thuộc vào giá trị biến

3. a) \(2x-\left(x-3\right)^2=5-x\)

\(\Leftrightarrow2x-x^2+6x-9=5-x\)

\(\Leftrightarrow-x^2+9x-14=0\)

\(\Leftrightarrow-x^2+2x+7x-14=0\)

\(\Leftrightarrow-x\left(x-2\right)+7\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(-x+7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\-x+7=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=7\end{matrix}\right.\)

Vậy..

b) \(\left(5x-2x\right)^2-\left(x+2\right)^2=0\)

\(\Leftrightarrow\left(5x-2x+x+2\right)\left(5x-2x-x-2\right)=0\)

\(\Leftrightarrow\left(4x+2\right)\left(2x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}4x+2=0\\2x-2=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=1\end{matrix}\right.\)

Vậy...

\(n_{SO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Pt: \(2Al+6H_2SO_4\left(đ\right)\underrightarrow{t^o}Al_2\left(SO_4\right)_3+3SO_2+6H_2O\)

0,1mol \(\leftarrow\) ------------------------------------- 0,15mol

\(m_{Al}=0,1.27=2,7\left(g\right)\)