1) a) \(3x\left(x-\dfrac{2}{3}\right)=0\Leftrightarrow\left\{{}\begin{matrix}3x=0\\x-\dfrac{2}{3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
vậy \(x=0;x=\dfrac{3}{2}\)
b) \(7\left(x-1\right)+2x\left(1-x\right)=0\Leftrightarrow7x-7+2x-2x^2=0\)
\(\Leftrightarrow\) \(-2x^2+9x-7=0\)
\(\Delta=9^2-4.\left(-2\right)\left(-7\right)=81-56=25>0\)
\(\Rightarrow\) phương trình có 2 nghiệm phân biệt
\(x_1=\dfrac{-9+5}{-4}=1\)
\(x_2=\dfrac{-9-5}{-4}=\dfrac{7}{2}\)
vậy \(x=1;x=\dfrac{7}{2}\)
B1:
\(a,3x\left(x-\dfrac{2}{3}\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}3x=0\\x-\dfrac{2}{3}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
B4:
\(A=3x+8xy+3y\)
\(=3\left(x+y\right)+8xy\)
Mà \(x+y=\dfrac{4}{3}\) và \(xy=-2\)
\(\Rightarrow3.\dfrac{4}{3}+8.\left(-2\right)=4-16=-12\)
MK nghĩ là đề câu B sai ,theo mk :
\(B=x^5-5x^4+5x^3+5x^2+x+2017\)
Ta có : 5 = 4+ 1 = x+ 1
Thay 5 = x+1 vào B ,có :
= \(x^5-\left[\left(x+1\right)x^4\right]+\left(x+1\right)x^3+\left(x+1\right)x^2+x+2017\)
= \(x^5-\left(x^5+x^4\right)+x^4+x^3-x^3-x^2+x^2+x+2017\)
= \(x^5-x^5-x^4+x^4+x+2017\)
= x + 2017
= 4 + 2017
= 2021
