Question

Đề: a/ (3x-x)^2 (3x+1) (3x+1)=29

       b/(4x-1)+(9-4x) (3+4x)=-8

Answer 1

a, \(\left(3x-x\right)^2\left(3x+1\right)\left(3x+1\right)=29\)

<=> \(4x^2\left(3x+1\right)^2=29\)

<=> \(4x^2;\left(3x+1\right)^2\inƯ\left(29\right)=\left\{\pm1;\pm29\right\}\)

4x^2	1	-1	29	-29
(3x+1)^2	29	-29	1	-1
x	1/2	ktm	\(\sqrt{\frac{29}{4}}\)	ktm
x	\(\frac{\sqrt{29}-1}{3}\)	ktm	0	ktm

b, Tương tự 

Answer 2

b) ( 4x - 1 ) + ( 9 - 4x )( 3 + 4x ) = -8

<=> ( 4x - 1 ) + ( 27 + 24x - 16x² ) = -8

<=> 4x - 1 + 27 + 24x - 16x² = -8

<=> -16x² + 28x + 26 = -8

<=> -16x² + 28x + 26 + 8 = 0

<=> -16x² + 28x + 34 = 0

<=> -2( 8x² - 14x - 17 ) = 0

=> 8x² - 14x - 17 = 0

\(\Delta'=b'^2-ac=\left(\frac{b}{2}\right)^2-ac=\left(\frac{-14}{2}\right)^2-\left(-17\right)\cdot8=185\)

\(\Delta'>0\)nên phương trình đã cho có hai nghiệm phân biệt : 

\(x_1=\frac{-b'+\sqrt{\Delta'}}{a}=\frac{-\left(-7\right)+\sqrt{185}}{8}=\frac{7+\sqrt{185}}{8}\)

\(x_2=\frac{-b'-\sqrt{\Delta'}}{a}=\frac{-\left(-7\right)-\sqrt{185}}{8}=\frac{7-\sqrt{185}}{8}\)

Lớp 7 mà nghiệm xấu nhỉ ?