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\(x^2+6x+9=0\\ \Leftrightarrow x^2+2.3.x+3^2=0\\ \Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x+3=0\\ \Leftrightarrow x=-3\)
Vậy \(x=-3\)
a) x2 - 4x - 5 = 0
=> x2 - 5x + x - 5 = 0
=> x(x - 5) + (x - 5) = 0
=> (x + 1)(x - 5) = 0
=> \(\orbr{\begin{cases}x+1=0\\x-5=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-1\\x=5\end{cases}}\)
b) 4x2 + 7x - 11 = 0
=> 4x2 + 11x - 4x - 11 = 0
=> x(4x + 11) - (4x + 11) = 0
=> (x - 1)(4x + 11) = 0
=> \(\orbr{\begin{cases}x-1=0\\4x+11=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=1\\x=-\frac{11}{4}\end{cases}}\)
c) -7x2 + 6x + 1 = 0
=> -7x2 + 7x - x + 1 = 0
=> -7x(x - 1) - (x - 1) = 0
=> (-7x - 1)(x - 1) = 0
=> \(\orbr{\begin{cases}-7x-1=0\\x-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}-7x=1\\x=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{1}{7}\\x=1\end{cases}}\)
d) -10x2 + 7x + 3 = 0
=> -10x2 + 10x - 3x + 3 = 0
=> -10x(x - 1) - 3(x - 1) = 0
=> (-10x - 3)(x - 1) = 0
=> \(\orbr{\begin{cases}-10x-3=0\\x-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}-10x=3\\x=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{3}{10}\\x=1\end{cases}}\)
\(x\left(3x-5\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\3x-5=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=\frac{5}{3}\end{cases}}}\)
Vậy \(x\in\left\{0;\frac{5}{3}\right\}\)
a) \(x\left(3x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\3x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{5}{3}\end{cases}}}\)
b) \(3x^2-27=0\)
\(\Leftrightarrow3x^2=27\)
\(\Leftrightarrow x^2=9\)
\(\Leftrightarrow x=\pm3\)
c) \(\left(x-5\right)^2=x-5\)
\(\Leftrightarrow x^2-10x+25-x+5=0\)
\(\Leftrightarrow x^2-11x+30=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=6\\x=5\end{cases}}}\)
d) \(2\left(x+7\right)-x^2-7x=0\)
\(\Leftrightarrow2x+14-x^2-7x=0\)
\(\Leftrightarrow-x^2-5x+14=0\)
\(\Leftrightarrow\left(x-7\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=2\end{cases}}}\)
e)\(7x\left(x-3\right)+2.3x=0\)
\(\Leftrightarrow7x^2-21x+6x=0\)
\(\Leftrightarrow7x^2-15x=0\)
\(\Leftrightarrow x\left(7x-15\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\7x-15=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{15}{7}\end{cases}}}\)
#H
\(\Leftrightarrow\left(7x+1\right)\left(x-5\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{7}\\x=5\end{cases}}\)
ta có : 7x(x-5)-(5-x) = 0
=> 7x(x-5)+(x-5) = 0
=>7x(x^2-25)=0
=>x=-5
=>x=5
(3x+2).(x+1)=3x.(5+x)
\(\Rightarrow\)\(3x^2+3x+2x+2=15x+3x^2\)
\(\Rightarrow3x^2+5x+2=15x+3x^2\)
\(\Rightarrow5x-15x+2=3x^2-3x^2\)
\(\Rightarrow-10x+2=0\)
\(-10x=-2\)
\(x=\frac{1}{5}\)
a.
\(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
b.
\(6x^2-7x+2=0\)
\(\Leftrightarrow6x^2-3x-4x+2=0\)
\(\Leftrightarrow3x\left(2x-1\right)-2\left(2x-1\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)
x2 + 7x + \(\frac{49}{4}\) - \(\frac{23}{4}\) = 0
<=> x2 + 7x + \(\frac{49}{4}\) = \(\frac{23}{4}\)
<=> (x + \(\frac{7}{2}\))2 = \(\frac{23}{4}\)
Tự gải tiếp
\(x^2+7x+5=0\)
\(x^2+2.x.\frac{7}{2}+\left(\frac{7}{2}\right)^2-\frac{29}{4}=0\)
\(\left(x+\frac{7}{2}\right)^2-\left(\frac{\sqrt{29}}{2}\right)^2=0\)
\(\left(x+\frac{7}{2}-\frac{\sqrt{29}}{2}\right)\left(x+\frac{7}{2}+\frac{\sqrt{29}}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{7}{2}-\frac{\sqrt{29}}{2}=0\\x+\frac{7}{2}+\frac{\sqrt{29}}{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{29}-7}{2}\\x=-\frac{\sqrt{29}+7}{2}\end{cases}}}\)
Vậy \(\Rightarrow\orbr{\orbr{\begin{cases}x=\frac{\sqrt{29}-7}{2}\\x=-\frac{\sqrt{29}+7}{2}\end{cases}}}\)