Question

Tìm x,biết:
a)5x.(x+1)-5.(x+1).(x-2)=0
b)(4x+1).(x-2)-(2x-3)²=4

Answer 1

a)5(x+1)(x-x-2)=0

=>5(x+1).-2=0

=>5(x+1)=0

=>x+1=0

=>x=-1

Answer 2

a)5x.(x+1)-5.(x+1).(x-2)=0

⇒5x(x+1)-(5x-10)(x+1)=0

⇒(x+1)(5x-5x+10)=0

⇒10(x+1)=0

⇒x+1=0⇒x=-1

 

Answer 3

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

\(\Leftrightarrow5\left(x+1\right)\left(x-x+2\right)=0\)

\(\Leftrightarrow10\left(x+1\right)=0\)

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

b) \(\left(4x+1\right)\left(x-2\right)-\left(2x-3\right)^2=4\)

\(\Leftrightarrow4x^2-7x-2-4x^2+12x-9=4\)

\(\Leftrightarrow5x=15\Leftrightarrow x=3\)

Answer 4

\(a,\Leftrightarrow5x^2+5x-5x^2+5x+10=0\\ \Leftrightarrow10x=-10\\ \Leftrightarrow x=1\\ b,\Leftrightarrow4x^2-7x-2-4x^2+12x-9=4\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\)

Answer 5

b)4x^2-8x+x-2-4x^2+12x-9=4

=>5x-11=4

=>5x=15

=>x=3

Answer 6

b)(4x+1).(x-2)-(2x-3)²=4

⇒4x²-8x+x-2-(4x²-12x+9)=4

⇒4x²-8x+x-2-4x²+12x-9=4

⇒5x-15=0⇒x=3

Answer 7

a)5(x+1)(x-x-2)=0

=>5(x+1).-2=0

=>5(x+1)=0

=>x+1=0

=>x=-1

b) (4x+1)(x−2)−(2x−3)2=4(4x+1)(x−2)−(2x−3)2=4

⇔4x2−7x−2−4x2+12x−9=4⇔4x2−7x−2−4x2+12x−9=4

⇔5x=15⇔x=3

Answer 8

a: Ta có: \(5x\left(x+1\right)-5\left(x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow5\left(x+2\right)\left(x-2+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

b: Ta có: \(\left(4x+1\right)\left(x-2\right)-\left(2x-3\right)^2=4\)

\(\Leftrightarrow4x^2-8x+x-2-4x^2+12x-9=4\)

\(\Leftrightarrow x=3\)