Xét ΔvgHAB và ΔvgKAC

có \(\widehat{A}\) chung

⇒ΔvgHAB ~ ΔvgKAC

⇒\(\dfrac{AH}{AK}=\dfrac{AB}{AC}\)

\(\Rightarrow AH.AC=AK.AB\)

https://hoc24.vn/hoi-dap/question/280059.html Câu tương tự.

là 142015  =10452016

\(m_{Fe_2O_3}=60\%.20=12\left(g\right)\)

\(n_{Fe_2O_3}=\dfrac{12}{160}=0,075\left(mol\right)\)

\(m_{CuO}=40\%.20=8\left(g\right)\)

\(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)

Pt: \(Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\)

0,075mol\(\rightarrow\)0,225mol\(\rightarrow\) 0,15mol

Pt: CuO + H₂ \(\rightarrow\) Cu + H₂O

0,1mol\(\rightarrow\)0,1mol\(\rightarrow\)0,1mol

a) \(m_{Fe}=0,15.56=8,4\left(g\right)\)

\(m_{Cu}=0,1.64=6,4\left(g\right)\)

b) \(n_{H_2\left(pư\right)}=0,225+0,1=0,325\left(mol\right)\)

\(\frac{9}{10}\)

f) \(4x^2-12x+9=0\)

\(\Rightarrow\left(2x-3\right)^2=0\)

\(\Rightarrow2x-3=0\)

\(\Rightarrow x=\dfrac{3}{2}\)

Vậy..

g) \(3x^2+7x+2=0\)

\(\Rightarrow3x^2+6x+x+2=0\)

\(\Rightarrow3x\left(x+2\right)+\left(x+2\right)=0\)

\(\Rightarrow\left(x+2\right)\left(3x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+2=0\\3x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-1}{3}\end{matrix}\right.\)

Vậy..

h) \(x^2-4x+1=0\)

\(\Rightarrow x^2-4x+4-3=0\)

\(\Rightarrow\left(x-2\right)^2-3=0\)

\(\Rightarrow\left(x-2-\sqrt{3}\right)\left(x-2+\sqrt{3}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2-\sqrt{3}=0\\x-2+\sqrt{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2+\sqrt{3}\\x=2-\sqrt{3}\end{matrix}\right.\)

Vậy..

j) \(3x^2+4x-4=0\)

\(\Rightarrow3x^2+6x-2x-4=0\)

\(\Rightarrow3x\left(x+2\right)-2\left(x+2\right)=0\)

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

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

Vậy..

Tình bạn chân chính là viên ngọc quý :v

\(2x-2x^2+13\)

\(\Rightarrow-2\left(x^2-x-\dfrac{13}{2}\right)\)

\(\Rightarrow-2\left(x^2-2x\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{27}{4}\right)\)

\(\Rightarrow-2\left(x-\dfrac{1}{2}\right)^2+\dfrac{27}{2}\)

\(\Rightarrow\dfrac{27}{2}-2\left(x-\dfrac{1}{2}\right)^2\le\dfrac{27}{2}\)

B/thức_max\(=\dfrac{27}{2}\) khi \(x-\dfrac{1}{2}=0\)

\(x=\dfrac{1}{2}\)

Vậy...

b) \(2x\left(x-4\right)^2-\left(x-4\right)=0\)

\(\Rightarrow\)\(\left(x-4\right)\left(2x^2-8x-1\right)=0\)

( Vì \(2x^2-8x-1>0\) )

\(\Rightarrow x-4=0\)

\(\Rightarrow x=4\)

Vậy..

( câu này hơi không chắc :v )

e) \(5x\left(x-9\right)^2-\left(9-x\right)^3=0\)

\(\Rightarrow\)\(5x\left(x-9\right)^2+\left(x-9\right)^3=0\)

\(\Rightarrow\left(x-9\right)^2\left(5x+x-9\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-9\right)^2=0\\6x-9=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=9\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy..