https://www.mathway.com/vi/popular-problems/Algebra/242673

https://www.mathway.com/vi/Algebra

a) \(x^2+2x+7=0\)

\(\Leftrightarrow\left(x+1\right)^2=-6\) ( vô lý )

Vậy pt vô nghiệm

b) \(x^3-x^2-21x+45=0\)

\(\Leftrightarrow x^3-3x^2+2x^2-6x-15x+45=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2+2x-15\right)=0\)

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

\(\Leftrightarrow\left(x-3\right)^2\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+5=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)

Vậy pt có tập nghiệm \(S=\left\{3,-5\right\}\)

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

\(Th1:x-3=0\Leftrightarrow x=3\)

\(Th2:\left[\left(2x+1\right)-\left(1-2x\right)\right]=0\)

\(\Leftrightarrow2x+1-1+2x=0\)

\(\Leftrightarrow2x+2x=0-1+1\)

\(\Leftrightarrow4x=0\Leftrightarrow x=0\)

\(Th3:x+3=0\Leftrightarrow x=-3\)

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

\(\Leftrightarrow2x^2-5x-3+2x^2+5x-3=0\)

\(\Leftrightarrow4x^2-6=0\Leftrightarrow x^2=\frac{3}{2}\)

\(\Leftrightarrow x=\pm\sqrt{\frac{3}{2}}\)

Từ biểu thức, ta suy ra:

(x+2)(3-4x)=(x+2)²

<=> (x+2)(3-4x)-(x+2)²=0

<=>(x+2)(3-4x-x-2)=0

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

<=>x+2=0 hoặc 1-5x=0

<=>x=-2 hoặc x=1/5

Vậy phương trình có tập nghiệm S={-2;1/5}

(x + 2)(3 - 4x) = x² + 4x + 4

<=> 3x - 4x² + 6 - 8x = x² + 4x + 4

<=> -5x - 4x² + 6 = x² + 4x + 4

<=> 5x + 4x² - 6 + x² + 4x + 4 = 0

<=> 9x + 5x² - 2 = 0

<=> 5x² + 10x - x - 2 = 0

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

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

<=> x + 2 = 0 hoặc 5x - 1 = 0

<=> x = -2 hoặc x = 1/5