Question

[Trần Anh Thư]

Giải hộ em với ạ

Answer 1

[Nguyễn Lê Phước Thịnh]

Bài 3:

a: \(\left(2x+1\right)\left(x^2+2\right)=0\)

mà \(x^2+2\ge2>0\forall x\)

nên 2x+1=0

=>2x=-1

=>\(x=-\frac12\)

b: \(\left(x^2+4\right)\left(7x-3\right)=0\)

mà \(x^2+4\ge4>0\forall x\)

nên 7x-3=0

=>7x=3

=>\(x=\frac37\)

c: \(\left(x^2+x+1\right)\left(6-2x\right)=0\)

mà \(x^2+x+1=x^2+x+\frac14+\frac34=\left(x+\frac12\right)^2+\frac34\ge\frac34>0\forall x\)

nên 6-2x=0

=>2x=6

=>x=3

d: \(\left(8x-4\right)\left(x^2+2x+2\right)=0\)

mà \(x^2+2x+2=x^2+2x+1+1=\left(x+1\right)^2+1\ge1>0\forall x\)

nên 8x-4=0

=>8x=4

=>\(x=\frac48=\frac12\)

Bài 4:

a: \(\left(x-2\right)\left(3x+5\right)=\left(2x-4\right)\left(x+1\right)\)

=>(x-2)(3x+5)=(x-2)(2x+2)

=>(x-2)(3x+5-2x-2)=0

=>(x-2)(x+3)=0

=>\(\left[\begin{array}{l}x-2=0\\ x+3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=-3\end{array}\right.\)

b: \(\left(2x+5\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)

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

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

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

=>3x(x-4)=0

=>x(x-4)=0

=>\(\left[\begin{array}{l}x=0\\ x-4=0\end{array}\right.=>\left[\begin{array}{l}x=0\\ x=4\end{array}\right.\)

c: \(9x^2-1=\left(3x+1\right)\left(2x-3\right)\)

=>(3x+1)(3x-1)=(3x+1)(2x-3)

=>(3x+1)(3x-1)-(3x+1)(2x-3)=0

=>(3x+1)(3x-1-2x+3)=0

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

=>\(\left[\begin{array}{l}3x+1=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac13\\ x=-2\end{array}\right.\)

d: \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)

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

=>\(\left(3x+1\right)\left(6x+2-x+2\right)=0\)

=>(3x+1)(5x+4)=0

=>\(\left[\begin{array}{l}3x+1=0\\ 5x+4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac13\\ x=-\frac45\end{array}\right.\)

e: \(27x^2\left(x+3\right)-12\left(x^2+3x\right)=0\)

=>\(27x^2\left(x+3\right)-12x\left(x+3\right)=0\)

=>3x(x+3)(9x-4)=0

=>x(x+3)(9x-4)=0

=>\(\left[\begin{array}{l}x=0\\ x+3=0\\ 9x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-3\\ x=\frac49\end{array}\right.\)

f: \(16x^2-8x+1=4\left(x+3\right)\left(4x-1\right)\)

=>\(\left(4x-1\right)^2=\left(4x+12\right)\left(4x-1\right)\)

=>(4x+12)(4x-1)-\(\left(4x-1\right)^2=0\)

=>(4x-1)(4x+12-4x+1)=0

=>13(4x-1)=0

=>4x-1=0

=>4x=1

=>\(x=\frac14\)