Question

Giải phương trình

a) (3x-1)²+(5-x)(3x-1)=0

b) (x+3)(2-3x)=x²+6x+9

c) (x-1) ( x² +5x+2)( x³ -1)=0

d) x² +( x+3)(10-2x)=9

Answer 1

c,

<=> \(\left[\begin{matrix}x-1=0\\x^2+5x+2=0\\x^3-1=0\end{matrix}\right.\)

+/ x - 1 = 0 <=> x = 1

+/x² + 5x + 2 =0 <=> (x + \(\frac{5}{2}\))² - \(\frac{17}{4}\)= 0 <=> (x + \(\frac{5}{2}\))² = \(\frac{17}{4}\)<=> x + \(\frac{5}{2}\)= \(\pm\)\(\sqrt{\frac{17}{4}}\)

<=> x = \(\pm\)\(\sqrt{\frac{17}{4}}\) - \(\frac{5}{2}\)

+/ x³ - 1 = 0 <=.> ( x - 1 )(x² + x + 1 ) = 0

<=> x = 1

Vậy phương trình có Nghiệm là x = 1 và x = \(\pm\)\(\sqrt{\frac{17}{4}}\) - \(\frac{5}{2}\)

Answer 2

d,

x² + (x + 3)(10 -2x ) = 9

<=> x² + 10x - 2x² + 30 - 6x -9 = 0

<=> x² + 4x + 21 = 0

<=> 7x - x² + 21 -3x = 0

<=> (x +3)(7-x) =0

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

Vậy pt có nghiệm là x = -3 và x = 7

Answer 3

a/

(3x-1)²+(5-x)(3x-1)=0

\(\Leftrightarrow\)(3x-1)(3x-1+5-x)=0

​\(\Leftrightarrow\)​(3x-1)(2x+4)=0

\(\Leftrightarrow\)\(\left[\begin{matrix}3x-1=0\\2x+4=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[\begin{matrix}x=\frac{1}{3}\\x=-2\end{matrix}\right.\)

vậy x=1/3, x=-2

b/(x+3)(2-3x)=x²+6x+9

\(\Leftrightarrow\)(x+3)(2-3x)=(x+3)² \(\Leftrightarrow\)(x+3)(2-3x)-(x+3)² =0 \(\Leftrightarrow\)(x+3)(2-3x-x-3) =0 \(\Leftrightarrow\)(x+3)(-4x-1)=0 \(\Leftrightarrow\)\(\left[\begin{matrix}x+3=0\\-4x-1=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[\begin{matrix}x=-3\\x=-\frac{1}{4}\end{matrix}\right.\) vậy x=-3,x=-1/4

Answer 4

a) (3x-1)²+(5-x)(3x-1)=0

\(\Leftrightarrow\left(3x-1\right)\left(3x-1+5-x\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(2x+4\right)=0\)

\(\Leftrightarrow\left[\begin{matrix}3x-1=0\\2x+4=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}3x=1\\2x=-4\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=\frac{1}{3}\\x=-2\end{matrix}\right.\)

Vậy: \(x=-2;\frac{1}{3}\)

b) (x+3)(2-3x)=x²+6x+9

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

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

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

\(\Leftrightarrow\left(x+3\right)\left(-4x-1\right)=0\)

\(\Leftrightarrow\left[\begin{matrix}x+3=0\\-4x-1=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-3\\x=-\frac{1}{4}\end{matrix}\right.\)

Vậy: \(x=-3;-\frac{1}{4}\)