Question

1/phân tích thành phân tử

a. x^6 - 64

b.  x^3+3x^2 + 3x+1- y^3

c. x^3 - 27+ x( x-3)

d, y^6 - 625

2/tìm x biết 

a.  25x^2 -1=0

b.  4 (x-1)^2 -9 =0 

c.  1/4 -9(x-1)^2 =0 

d.  1/16 - ( 2x + 3/4) ^2 =0

1/4 nghĩa là 1 phần 4 á nhé,  nhờ giúp đỡ 

Answer 1

2/

a/ \(25x^2-1=0\)

<=> \(\left(5x\right)^2-1=0\)

<=> \(\left(5x-1\right)\left(5x+1\right)=0\)

<=> \(\orbr{\begin{cases}5x-1=0\\5x+1=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=\frac{1}{5}\\x=-\frac{1}{5}\end{cases}}\)

b/ \(4\left(x-1\right)^2-9=0\)

<=> \(\left[2\left(x-1\right)\right]^2-3^2=0\)

<=> \(\left(2x-2\right)^2-3^2=0\)

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

<=> \(\left(2x-5\right)\left(2x+1\right)=0\)

<=> \(\orbr{\begin{cases}2x-5=0\\2x+1=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{1}{2}\end{cases}}\)

c/ \(\frac{1}{4}-9\left(x+1\right)^2=0\)

<=> \(\left(\frac{1}{2}\right)^2-\left[3\left(x-1\right)\right]^2=0\)

<=> \(\left(\frac{1}{2}\right)^2-\left(3x-3\right)^2=0\)

<=> \(\left(\frac{1}{2}-3x+3\right)\left(\frac{1}{2}+3x-3\right)=0\)

<=> \(\left(\frac{7}{2}-3x\right)\left(-\frac{5}{2}+3x\right)=0\)

<=> \(\orbr{\begin{cases}\frac{7}{2}-3x=0\\-\frac{5}{2}+3x=0\end{cases}}\)<=> \(\orbr{\begin{cases}3x=\frac{7}{2}\\3x=\frac{5}{2}\end{cases}}\)

<=> \(\orbr{\begin{cases}x=\frac{7}{6}\\x=\frac{5}{6}\end{cases}}\)

d/ \(\frac{1}{16}-\left(2x+\frac{3}{4}\right)^2=0\)

<=> \(\left(\frac{1}{4}\right)^2-\left(2x+\frac{3}{4}\right)^2=0\)

<=> \(\left(\frac{1}{4}-2x-\frac{3}{4}\right)\left(\frac{1}{4}+2x+\frac{3}{4}\right)=0\)

<=> \(\left(-\frac{1}{2}-2x\right)\left(1+2x\right)=0\)

<=> \(2\left(-\frac{1}{4}-x\right)\left(1+2x\right)=0\)

<=> \(\orbr{\begin{cases}-\frac{1}{4}-x=0\\1+2x=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{1}{2}\end{cases}}\)

Answer 2

Còn Bài 1 nữa, nhờ các bạn giúp với