Question

Tìm x

a) 4x² – 25 – (2x–5)(2x + 7) =0

b)2x³ + 3x² + 2x +3 = 0

Answer 1

a) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left[\left(2x\right)^2-5^2\right]-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x+5\right)\left(2x-5\right)-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(-2\right)=0\)

\(\Leftrightarrow10-4x=0\)

\(\Leftrightarrow4x=10\)

\(\Leftrightarrow x=\dfrac{10}{4}=\dfrac{5}{2}=2,5\)

Vậy: \(x=2,5\)

b) \(2x^3+3x^2+2x+3=0\)

\(\Leftrightarrow2x^3+2x+3x^2+3=0\)

\(\Leftrightarrow2x\left(x^2+1\right)+3\left(x^2+1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\2x+3=0\end{matrix}\right.\)\(\Leftrightarrow2x=-3\)\(\Leftrightarrow x=-\dfrac{3}{2}\)

Vậy: \(x=-\dfrac{3}{2}\)

_Chúc bạn học tốt_

Answer 2

a) 4x²-25-(2x-5)(2x+7)=0

<=> (4x²-25)-(2x-5)(2x+7)=0

<=> [(2x)²-5²]-(2x-5)(2x+7)=0

<=> (2x-5).(2x+5)-(2x-5)(2x+7)=0

<=> (2x-5).[(2x+5)-(2x+7)]=0

<=> (2x-5).(2x+5-2x-7)=0

<=> (2x-5).(-2)=0

=> 2x-5=0

<=> 2x=5

<=> x=5/2

Vậy x=5/2

b) 2x³+3x²+2x+3=0

<=> (2x³+2x)+(3x²+3)=0

<=> 2x(x²+1)+3(x²+1)=0

<=> (x²+1).(2x+3)=0

x²+1=0 x²= -1(vô lí)

<=> <=>

2x+3=0 x= -3/2

Vậy x= -3/2

Answer 3

a, 4x\(^2\)-25-(2x-5)(2x+7)=0

=>(2x-5)(2x+5)-(2x-5)(2x+7)=0

=>(2x-5)(2x+5-2x-7)=0

=>(2x-5)(-2)=0

=>2x-5=0

=>2x=5

=>x=\(\dfrac{5}{2}\)

b,2x\(^3\)+3x\(^2\)+2x+3=0

=>2x(x\(^2\)+1)+3(x\(^2\)+1)=0

=>(x\(^2\)+1)(2x+3)=0

=>\(\left[{}\begin{matrix}x^2-1=0\\2x+3=0\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=\pm1\\x=\dfrac{-3}{2}\end{matrix}\right.\)