Question

[nhok hanahmoon]

giúp mk với các bạn @Nguyễn Thanh Hằng

Answer 1

[Nguyễn Thị Huyền Trang]

Bài 11:

+)\(A=10x^2+20xy+10y-90=10.\left(x^2+2xy+y\right)-10.9\)

\(=10.\left(x+y\right)^2-10.9=10.\left[\left(x+y\right)^2-9\right]\)

+)\(B=x^3y-3x^2y-4xy+12y=\left(x^3y-3x^2y\right)-\left(4xy-12y\right)\)

\(=x^2y.\left(x-3\right)-4y.\left(x-3\right)=\left(x^2y-4y\right)\left(x-3\right)\)

\(=y.\left(x^2-4\right)\left(x-3\right)=y.\left(x+2\right)\left(x-2\right)\left(x-3\right)\)

+) \(C=125x^3-10x^2+2x-1=\left(125x^3-1\right)-\left(10x^2-2x\right)\)

\(=\left(5x-1\right)\left(25x^2+5x+1\right)-2x\left(5x-1\right)\)

\(=\left(5x-1\right)\left(25x^2+3x+1\right)\)

Bài 12:

1) \(x^3-7x^2-9x+63=0\Rightarrow\left(x^3-7x^2\right)-\left(9x-63\right)=0\)

\(\Rightarrow x^2.\left(x-7\right)-9.\left(x-7\right)=0\Rightarrow\left(x-3\right)\left(x+3\right)\left(x-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x+3=0\\x-7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=7\end{matrix}\right.\)

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

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

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

Answer 2

[Nguyễn Thanh Hằng]

anh nghĩ sao khi tag e vào bài này! Lớp 7 chưa học sao làm lớp 8 giờ! Để e nhờ ng` khác.

Nguyễn Huy Tú Tuấn Anh Phan Nguyễn văn tài Nguyễn Thị Huyền Trang