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\(a,\Leftrightarrow2x^2+10x-2x^2=12\Leftrightarrow x=\dfrac{12}{10}=\dfrac{6}{5}\\ b,\Leftrightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\\ \Leftrightarrow\left(1-2x\right)\left(9-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\\ d,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ e,\Leftrightarrow4x^2-4x+1-4x^2+196=0\\ \Leftrightarrow-4x=-197\Leftrightarrow x=\dfrac{197}{4}\)
\(f,\Leftrightarrow x^2+8x+16-x^2+1=16\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\\ g,Sửa:\left(3x+1\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(3x+1-x-1\right)\left(3x+1+x+1\right)=0\\ \Leftrightarrow2x\left(4x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\\ h,\Leftrightarrow x^2+8x-x-8=0\\ \Leftrightarrow\left(x+8\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-8\end{matrix}\right.\\ i,\Leftrightarrow2x^2-13x+15=0\\ \Leftrightarrow2x^2+2x-15x-15=0\\ \Leftrightarrow\left(x+1\right)\left(2x-15\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{15}{2}\end{matrix}\right.\)
Bài 2:
a: =>2x^2-4x+1=x^2+x+5
=>x^2-5x-4=0
=>\(x=\dfrac{5\pm\sqrt{41}}{2}\)
b: =>11x^2-14x-12=3x^2+4x-7
=>8x^2-18x-5=0
=>x=5/2 hoặc x=-1/4
Bài 1:
a: \(\Leftrightarrow x^2-5x+6< =0\)
=>(x-2)(x-3)<=0
=>2<=x<=3
b: \(\Leftrightarrow\left(x-6\right)^2< =0\)
=>x=6
c: \(\Leftrightarrow x^2-2x+1>=0\)
\(\Leftrightarrow\left(x-1\right)^2>=0\)
hay \(x\in R\)
Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5
b) 5x(x-2000)-x+2000=0
\(\Rightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\\ \Rightarrow\left(x-2000\right)\left(5x-1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-2000=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0+2000\\5x=0+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\5x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\x=\dfrac{1}{5}\end{matrix}\right.\)
Bài 2:
a) \(3x^2-7x-10=\left(x+1\right)\left(3x-10\right)\)
b) \(x^2+6x+9-4y^2=\left(x+3\right)^2-\left(2y\right)^2=\left(x+3-2y\right)\left(x+3+2y\right)\)
c) \(x^2-2xy+y^2-5x+5y=\left(x-y\right)^2-5\left(x-y\right)=\left(x-y\right)\left(x-y-5\right)\)
d) \(4x^2-y^2-6x+3y=\left(2x-y\right)\left(2x+y\right)-3\left(2x-y\right)=\left(2x-y\right)\left(2x+y-3\right)\)
e) \(1-2a+2bc+a^2-b^2-c^2=\left(a-1\right)^2-\left(b-c\right)^2=\left(a-1-b+c\right)\left(a-1+b-c\right)\)
f) \(x^3-3x^2-4x+12=\left(x+2\right)\left(x-3\right)\left(x-2\right)\)
g) \(x^4+64=\left(x^2+8\right)^2-16x^2=\left(x^2+8-4x\right)\left(x^2+6+4x\right)\)h) \(x^4-5x^2+4=\left(x+2\right)\left(x+1\right)\left(x-1\right)\left(x-2\right)\)
i) \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+16=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+16=\left(x^2+8x+7\right)^2+8\left(x^2+8x+7\right)+16=\left(x^2+8x+11\right)^2\)
a: \(3x^2-7x-10\)
\(=3x^2+3x-10x-10\)
\(=\left(x+1\right)\left(3x-10\right)\)
b: \(x^2+6x+9-4y^2\)
\(=\left(x+3\right)^2-4y^2\)
\(=\left(x+3-2y\right)\left(x+3+2y\right)\)
c: \(x^2-2xy+y^2-5x+5y\)
\(=\left(x-y\right)^2-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-5\right)\)
Bài 1
a/ \(x\left(x^2+1\right)+2\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+1\right)=0\Rightarrow x=-2\)
b/
\(\Leftrightarrow x^3-6x^2+9x+5x^2-30x+45=0\)
\(\Leftrightarrow x\left(x-3\right)^2+5\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-3\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)
1.
c/ \(\Leftrightarrow x^3+2x^2+2x+x^2+2x+2=0\)
\(\Leftrightarrow x\left(x^2+2x+2\right)+x^2+2x+2=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+2x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x^2+2x+2=0\left(vn\right)\end{matrix}\right.\)
d/
\(\Leftrightarrow x^4+x^3-2x^2-x^3-x^2+2x+4x^2+4x-8=0\)
\(\Leftrightarrow x^2\left(x^2+x-2\right)-x\left(x^2+x-2\right)+4\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x^2-x+4\right)\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+4=0\left(vn\right)\\x^2+x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
a: \(\Leftrightarrow\left(-x+3\right)\left(x+6\right)=18\)
\(\Leftrightarrow-x^2-6x+3x+18-18=0\)
\(\Leftrightarrow-x\left(x+3\right)=0\)
=>x=0 hoặc x=-3
b: \(\Leftrightarrow x\left(3x^2+6x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x^2+6x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+2x-\dfrac{4}{3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x+1\right)^2=\dfrac{7}{3}\end{matrix}\right.\Leftrightarrow x\in\left\{0;\dfrac{\sqrt{21}}{3}-1;\dfrac{-\sqrt{21}}{3}-1\right\}\)
c: =>x(3x-5)=0
=>x=0 hoặc x=5/3
d: =>(x-2)(x+2)=0
=>x=2 hoặc x=-2
Bài 1.
Bài 1 :
a) \(x^2+7x+12=0\)
\(\Leftrightarrow x^2+2.x.\frac{7}{2}+\left(\frac{7}{2}\right)^2-\frac{1}{4}=0\)
\(\Leftrightarrow\left(x+\frac{7}{2}\right)^2=\frac{1}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{7}{2}=\frac{1}{2}\\x+\frac{7}{2}=-\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x=-4\end{cases}}\)
Vậy : \(x\in\left\{-3,-4\right\}\)