*)\(b^2+c^2=a^2\)

\(\Leftrightarrow b^2=a^2-c^2\)

\(\Leftrightarrow b=\sqrt{a^2-c^2}\)

Ta có: \(\sqrt{a^2-c^2}>c\Leftrightarrow a^2-c^2>c^2\)

\(\Leftrightarrow a^2>2c^2\)(luôn đúng)

=> c<b

*) \(a^2=b^2+c^2\Leftrightarrow\hept{\begin{cases}c=3\\b=4\\a=5\end{cases}\Leftrightarrow c=b+1}\)

1) x²-1=0 <=> x²=1 <=> \(\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

2) x²+1=0

<=> x²=-1

Mà x² >=0 với mọi x; -1<0

=> không có x thỏa mãn

3) \(x^3-x^2-21x+45=0\)

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

<=> x²(x-3) +2x(x-3)-15(x-3)=0

<=> (x-3)(x² +2x-15)=0

<=> (x-3)(x²+5x-3x-15)=0

<=> (x-3)[x(x+5)-3(x+5)]=0

<=> (x-3)²(x+5)=0

\(\Leftrightarrow\orbr{\begin{cases}\left(x-3\right)^2\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+5=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\x=-5\end{cases}}}\)

\(3x\left(x+5\right)-\left(x+2\right)^2=2x^2+7\)

\(\Leftrightarrow3x^2+15x-x^2-4x-4=2x^2+7\)

\(\Leftrightarrow3x^2-2x^2-x^2+15x-4x=7+4\)

\(\Leftrightarrow11x=11\)

\(\Leftrightarrow x=1\)

 https://www.youtube.com/channel/UCT23clmdY5azigRNMRDxGfw

đăng kí hộ

\(\frac{3x+2}{x-1}+\frac{2x-4}{x+2}=5\)

<=> \(\frac{3x+2}{x-1}+\frac{2\left(x-2\right)}{x+2}=5\)

<=> (3x + 2)(x + 2) + 2(x - 2)(x - 1) = 5(x - 1)(x + 2)

<=> 3x² + 6x + 2x + 4 + 2x² - 2x - 4x + 4 = 5x² + 10x - 5x - 10

<=> 5x² + 2x + 8 = 5x² + 5x - 10

<=> 5x² + 2x + 8 - 5x² = 5x - 10

<=> 2x + 8 = 5x - 10

<=> 2x + 8 - 5x = -10

<=> -3x + 8 = -10

<=> -3x = -10 - 8

<=> -3x = -18

<=> x = 6

sai òi 

lạc đề kìa

\(3\left(x-1\right)+4\left(2x-1\right)=-8\)

\(\Leftrightarrow3x-3+8x-4=-8\)

\(\Leftrightarrow3x+8x=-8+4+3\)

\(\Leftrightarrow11x=-1\)

\(\Leftrightarrow x=-\frac{1}{11}\)

3(x - 1) + 4(2x - 1) = -8

<=> 3x - 3 + 8x - 4 = -8

<=> 11x = -8 + 4 +3 

<=> 11x = -1

<=> x = \(\frac{-1}{11}\)