a) \(4x^2-8x=0\)

\(\Rightarrow4x\left(x-2\right)=0\)

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

Vậy...

b) \(\left(x+5\right)-3x\left(x+5\right)=0\)

\(\Rightarrow\left(x+5\right)\left(1-3x\right)=0\)

\(\left[{}\begin{matrix}x+5=0\\1-3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy..

d) \(3x\left(x-1\right)+\left(1-x\right)^2=0\)

\(\Rightarrow3x\left(x-1\right)-\left(x-1\right)^2=0\)

\(\Rightarrow\left(x-1\right)\left(3x-x+1\right)=0\)

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

Vậy..

Đổi: 15 p=\(\frac{1}{4}\) h

1 giờ máy bay bay đc số km là:   240:\(\frac{1}{4}\)=240 x 4=960(km)

Đ/s:960 km.

\(n_{Mg}=\dfrac{8,4}{24}=0,35\left(mol\right)\)

Pt: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

0,35mol\(\rightarrow\)0,7mol\(\rightarrow\)0,35mol\(\rightarrow\)0,35mol

\(C\%_{HCl}=\dfrac{0,7.36,5}{146}.100=17,5\%\)

\(m_{dd\left(spu\right)}=m_{Mg}+m_{HCl}-m_{H_2}\)

= 8,4 + 146 - 0,35.2 = 153,7 (g)

\(C\%_{MgCl_2}=\dfrac{0,35.95}{153,7}.100=21,63\%\)

a) \(\Delta vgCHD\) có: BH = BC (gt)
KH = KD (gt)

\(\Rightarrow BK\) là đường trung bình \(\Delta vgCHD\)

\(\Rightarrow BK=\dfrac{CD}{2}\)

\(\Rightarrow BK=\dfrac{24}{2}=12\left(cm\right)\)

\(\Delta vgCHD\) có HD² = HC² + CD²

\(\Rightarrow HD=\sqrt{HC^2+CD^2}\)

\(\Rightarrow HD=\sqrt{18^2+24^2}\)

\(\Rightarrow HD=30\left(cm\right)\)

\(\Delta vgCHD\) có HK = DK (gt)

\(\Rightarrow CK=\dfrac{HD}{2}=\dfrac{30}{2}=15\left(cm\right)\)

\(m_{H_2SO_4}=20\%.100=20\left(g\right)\)

\(n_{H_2SO_4}=\dfrac{20}{98}=0,2\left(mol\right)\)

\(n_{CuO}=\dfrac{1,6}{80}=0,02\left(mol\right)\)

Pt: \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

0,02 mol 0,2mol -> 0,02mol

Lập tỉ số: \(n_{CuO}\) : \(n_{H_2SO_4}\)= 0,02 < 0,2

\(\Rightarrow\) CuO hết, H₂SO₄ dư

\(n_{H_2SO_4\left(dư\right)}=0,2-0,02=0,18\left(mol\right)\)

\(m_{H_2SO_4\left(dư\right)}=0,18.98=17,64\left(g\right)\)

\(m_{dd\left(spu\right)}=1,6+100=101,6\left(g\right)\)

\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{17,64}{101,6}.100=17,36\%\)

\(C\%_{CuSO_4}=\dfrac{0,02.160.100}{101,6}=3,15\%\)

=(3992+5988)+(7984+1996)

=9980+9980

=19960

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