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\(\left(x^2+3x+2\right)\left(x^2+11x+30\right)-60=0\)
\(\Rightarrow\left[\left(x+1\right)\left(x+2\right)\right].\left[\left(x+5\right)\left(x+6\right)\right]-60=0\)
\(\Rightarrow\left[\left(x+1\right)\left(x+6\right)\right].\left[\left(x+2\right)\left(x+5\right)\right]-60=0\)
\(\Rightarrow\left(x^2+7x+6\right)\left(x^2+7x+10\right)-60=0\left(1\right)\)
Đặt \(x^2+7x+6=a\Rightarrow x^2+7x+10=a+4\)
Thay vào (1), ta có:
\(a\left(a+4\right)-60=0\)
\(\Rightarrow a^2+4a-60=0\)
\(\Rightarrow a^2+10a-6a-60=0\)
\(\Rightarrow a\left(a+10\right)-6\left(a+10\right)=0\)
\(\Rightarrow\left(a-6\right)\left(a+10\right)=0\)
\(\Rightarrow\orbr{\begin{cases}a=6\\a=-10\end{cases}}\)
- Nếu \(x^2+7x+6=6\)
\(\Rightarrow x^2+7x=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=-7\end{cases}}\)
- Nếu \(x^2+7x+6=-10\)
\(\Rightarrow x^2+7x+16=0\)
Mà \(x^2+7x+16=x^2+2.x.\frac{7}{2}+\frac{49}{4}+\frac{15}{4}=\left(x+\frac{7}{2}\right)^2+\frac{15}{4}>0\forall x\)
Vậy \(x=0,x=-7\)
Học tốt.
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\(1\) . \(x\left(x-2\right)-x+2=0\)
\(\Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\Rightarrow x=2\\x-1=0\Rightarrow x=1\end{matrix}\right.\)
2 . \(5x\left(x-3\right)-x+3=0\)
\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\Rightarrow x=3\\5x-1=0\Rightarrow x=\dfrac{1}{5}\end{matrix}\right.\)
3 . \(3x\left(x-5\right)-\left(x-1\right)\left(2+3x\right)=30\)
\(\Leftrightarrow3x^2-15x-2x-3x^2+2+3x=30\)
\(\Leftrightarrow-14x+2=30\)
\(\Leftrightarrow-14x=28\)
\(\Rightarrow x=-2\)
4 . \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow x^2+3x+2x+6-x^2-5x+2x+10=0\)
\(\Leftrightarrow2x+16=0\)
\(\Rightarrow x=-8\)
Chúc bạn học tốt nhoaaaaaaaaaaaaaaaaaaaaaaaaa :D
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\(a,\Leftrightarrow x^2+6x+9-x^2+3x+10=1\\ \Leftrightarrow9x=-18\Leftrightarrow x=-2\\ b,\Leftrightarrow4x^2-4x+1-4x^2+17x+15=3\\ \Leftrightarrow13x=-13\Leftrightarrow x=-1\\ c,\Leftrightarrow3x\left(x-2\right)+4\left(x-2\right)=0\\ \Leftrightarrow\left(3x+4\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=2\end{matrix}\right.\\ d,\Leftrightarrow2x\left(3x+5\right)-6\left(3x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{3}\end{matrix}\right.\)
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x3- 6x3 -x + 30 = 0
x3 + 2x2- 8x2- 16x + 15x + 30 = 0
x2 ( x + 2 ) - 8x ( x + 2 ) + 15 ( x + 2 ) = 0
( x + 2 )( x2 - 8x + 15 ) = 0
x + 2 = 0 hoặc x2 - 8x + 15 = 0
x = - 2 hoặc ( x - 4 )2 - 1 = 0
x = - 2 hoặc ( x - 4 - 1 ) ( x - 4 + 1 ) = 0
x = - 2 hoặcx = 5 hoặc x = 3
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d) \(4x^2-9-x\left(2x-3\right)=0\)
\(\Leftrightarrow4x^2-9-2x^2+3x=0\)
\(\Leftrightarrow2x^2+3x-9=0\)
\(\Delta=3^2-4.2.\left(-9\right)=9+72=81\)
Vậy pt có 2 nghiệm phân biệt
\(x_1=\frac{-3+\sqrt{81}}{4}=\frac{-3}{2}\);\(x_1=\frac{-3-\sqrt{81}}{4}=-3\)
e) \(x^3+5x^2+9x=-45\)
\(\Leftrightarrow x^3+5x^2+9x+45=0\)
\(\Leftrightarrow x^2\left(x+5\right)+9\left(x+5\right)=0\)
\(\Leftrightarrow\left(x^2+9\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+9=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm3i\\x=-5\end{cases}}\)
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a, \(x^2-x-30=0\)
\(\Leftrightarrow x^2-6x+5x-30=0\)
\(\Leftrightarrow x\left(x-6\right)+5\left(x-6\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)
Vậy x = -5 hoặc x = 6
\(x^2-x-30=0\)
\(x^2+5x-6x-30=0\)
\(x\left(x+5\right)-6\left(x+5\right)\)
\(\left(x-6\right)\left(x+5\right)=0\)
\(\Rightarrow x-6=0\Rightarrow x=6\)
\(\Rightarrow x+5=0\Rightarrow x=-5\)
=> -(x-5)(x+6)=0
=> (x-5)(x+6) = 0
=> x-5=0 hoặc x+6=0
=> x = 5 hoặc x = - 6