Question

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x^2 - 5x + 4 = 0

 

Answer 1

x²-x-4x+4=0

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

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

\(\orbr{\begin{cases}x-1=0\\x-4=0\end{cases}}\)

\(\orbr{\begin{cases}x=1\\x=4\end{cases}}\)

Answer 2

x² - 5x + 4 = 0

=> x² - x - 4x + 4 = 0

=> (x² - x) - (4x - 4) = 0

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

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

=> x - 4 = 0 hoặc x - 1 = 0

=> x = 4 hoặc x = 1

vậy_

Answer 3

\(x^2-5x+4=0\)\(=>x^2-2x.\frac{5}{2}+\left(\frac{5}{2}\right)^2-\frac{9}{4}=0\)\(=>\left(x-\frac{5}{2}\right)^2-\frac{9}{4}=0\)\(=>\left(x-\frac{5}{2}\right)^2=\frac{9}{4}>0\)\(=>\orbr{\begin{cases}x-\frac{5}{2}=\frac{9}{4}\\x-\frac{5}{2}=\frac{-9}{4}\end{cases}}\)\(=>\orbr{\begin{cases}x=\frac{19}{4}\\x=\frac{1}{4}\end{cases}}\)

Answer 4

Vậy \(x=\frac{19}{4}\) và \(x=\frac{1}{4}\)

Answer 5

k cho mình nha

Answer 6

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

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

\(x\left(x-4\right)-\left(x-4\right)=0\)

\(\left(x-4\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=1\end{cases}}}\)