THAM KHẢO

Gọi x là v.tốc dự định của xe(x>0, km/h)

Nửa quãng đường xe đi là: 120:2=60(km)

=> Vận tốc đi nửa quãng đường là: 120x(h)120x(h)

Vì nửa qquangx đường sau xe đi với thời gian là: 60x+60x+10=120x−0.560x+60x+10=120x−0.5

Gải được x=40(tmđk)

Vậy v.tốc dự định là 40km/h

 \(a)P=\left(\dfrac{x^2+2}{x^3-1}+\dfrac{x+1}{x^2+x+1}+\dfrac{1}{1-x}\right).\left(\dfrac{x^2}{x+1}+1\right).\left(x\ne1;x\ne-1\right).\\ P=\dfrac{x^2+2+x^2-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}.\dfrac{x^2+x+1}{x+1}.\\ P=\dfrac{x^2-x}{x-1}.\dfrac{1}{x+1}.\\ P=\dfrac{x\left(x-1\right)}{x-1}.\dfrac{1}{x+1}.\\ P=x.\dfrac{1}{x+1}.\\ P=\dfrac{x}{x+1}.\)

\(P=\dfrac{1}{4}.\Rightarrow\dfrac{x}{x+1}=\dfrac{1}{4}.\\ \Leftrightarrow4x-x-1=0.\\ \Leftrightarrow3x-1=0.\\ \Leftrightarrow x=\dfrac{1}{3}\left(TM\right).\)

Bài 2:

a: Ta có: \(5x\left(x-1\right)+10x-10=0\)

\(\Leftrightarrow\left(x-1\right)\left(5x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

b: Ta có: \(\left(x+2\right)\left(x+3\right)-2x=6\)

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

\(\Leftrightarrow x\left(x+3\right)=0\)

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

c: Ta có: \(\left(x-1\right)\left(x-2\right)-2=0\)

\(\Leftrightarrow x\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

\(1,\widehat{D}=360-\widehat{A}-\widehat{B}-\widehat{C}=360-120-50-90=100\)

\(2,\widehat{D}+\widehat{C}=360-\widehat{A}-\widehat{B}=360-50-110=200\\ \Rightarrow4\widehat{D}=200\Rightarrow\widehat{D}=50\Rightarrow\widehat{C}=50\cdot3=150\)

\(3,\\ \widehat{D}+\widehat{C}=360-\widehat{A}-\widehat{B}=160\)

Áp dụng tính chất dãy tỉ số bằng nhau

\(\dfrac{\widehat{D}}{5}=\dfrac{\widehat{C}}{3}=\dfrac{\widehat{D}+\widehat{C}}{5+3}=\dfrac{160}{8}=20\\ \Rightarrow\left\{{}\begin{matrix}\widehat{D}=100\\\widehat{C}=60\end{matrix}\right.\)