Question

Đề: Tìm x

a)2x.(3x+5)-x.(6x-1)=33 k)5(x+3)-2x(x+3)=0

b)x(3x-1)+12x-4=0 i)5x(x-2)-(2-x)=0

c)5x(2x+1)-12x-6=0 m)x(x-1)-2(1-x)=0

d)x³-5x²+4x-20=0

e)2x³-5x²+2x-5=0

g)(x-2)³-x(x+1).(x-1)+6²=5

Answer 1

a)2x.(3x+5)-x.(6x-1)=33

=>\(6x^2+10x-6x^2+x=33\)

=>11x=33

=>x=3

Answer 2

b)x(3x-1)+12x-4=0

=>x(3x-1)+4(3x-1)=0

=>(x-4)(3x-1)=0

=>x-4=0 hoặc 3x-1=0

+)x-4=0 +)3x-1=0

=>x=4 =>x=\(\frac{1}{3}\)

Answer 3

c)5x(2x+1)-12x-6=0

=>10x\(^2\)+5x-12x-6=0

=>10x\(^2\)-7x-6=0

=>(10x\(^2\)+5x)-(12x+6)=0

=>5x(2x+1)-6(2x+1)=0

=>(5x-6)(2x+1)=0

=>\(\left[{}\begin{matrix}5x-6=0\\2x+1=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\frac{6}{5}\\x=\frac{-1}{2}\end{matrix}\right.\)

Answer 4

d)x\(^3\)-5x\(^2\)+4x-20=0

=>x\(^2\)(x-5)+4(x-5)=0

=>(x\(^2\)+4)(x-5)=0

=>\(\left[{}\begin{matrix}\text{x^2+4=0}\\x-5=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x^2=-4\left(loại\right)\\x=5\end{matrix}\right.\)