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a, \(\left(x^2-5x+7\right)^2-\left(2x-5\right)^2=0\)
\(\Leftrightarrow\left(x^2-5x+7-2x+5\right)\left(x^2-5x+7+2x-5\right)=0\)
\(\Leftrightarrow\left(x^2-7x+12\right)\left(x^2-3x+2\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-3\right)\left(x-1\right)\left(x-2\right)=0\Leftrightarrow x=1;x=2;x=3;x=4\)
Vậy tập nghiệm phương trình là S = { 1 ; 2 ; 3 ; 4 }
b, \(\left|2x-1\right|=5\Leftrightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là S = { -2 ; 3 }
c, \(\left|2x-1\right|=\left|x+5\right|\Leftrightarrow\left(2x-1\right)^2=\left(x+5\right)^2\)
\(\Leftrightarrow\left(2x-1\right)^2-\left(x+5\right)^2=0\Leftrightarrow\left(2x-1-x-5\right)\left(2x-1+x+5\right)=0\Leftrightarrow x=6;x=-\dfrac{4}{3}\)
Vậy tập nghiệm của phương trình là S = { -4/3 ; 6 }
d, \(\left|3x+1\right|=x-2\)
TH1 : \(3x+1=x-2\Leftrightarrow2x=-3\Leftrightarrow x=-\dfrac{3}{2}\)
TH2 : \(3x+1=-x+2\Leftrightarrow4x=1\Leftrightarrow x=\dfrac{1}{4}\)
Vậy tập nghiệm của phương trình là S = { -3/2 ; 1/4 }
các ý còn lại tương tự
a) Ta có: \(\left(x^2-5x+7\right)^2-\left(2x-5\right)^2=0\)
\(\Leftrightarrow\left(x^2-5x+7-2x+5\right)\left(x^2-5x+7+2x-5\right)=0\)
\(\Leftrightarrow\left(x^2-7x+12\right)\left(x^2-3x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\\x=1\\x=2\end{matrix}\right.\)
Vậy: S={3;4;1;2}
a)
\(\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)\)
\(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x+1\right)\left(x-1\right)\)
\(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[9x^2-4-\left[\left(3x+2\right)\left(x-1\right)\right]\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left[9x^2-4-\left(3x^2-3x+2x-2\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-3x^2+3x-2x+2\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(6x^2+x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\6x^2+x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(2x-1\right)\left(3x+2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-2}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{1;\dfrac{-2}{3};\dfrac{1}{2}\right\}\)
b)
\(\left(x-1\right)^2-1+x^2=\left(1-x\right)\left(x+3\right)\)
\(\Leftrightarrow x^2-2x+1-1+x^2=x+3-x^2-3x\)
\(\Leftrightarrow2x^2-2x=x^2-2x+3\)
\(\Leftrightarrow3x^2=3\)
\(\Leftrightarrow x^2=1\)
\(\Leftrightarrow x=\left(\pm1\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy \(x\in\left\{1;-1\right\}\)
a) \(\left(3x-2\right)\left(4x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{2}{3};-\dfrac{5}{4}\right\}\)
b) \(\left(2,3x-6,9\right)\left(0,1x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-20\end{matrix}\right.\)
c) \(\left(4x+2\right)\left(x^2+1\right)=0\)
Vì \(x^2+1\ge1>0\forall x\)
\(\Rightarrow4x+2=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)
d) \(\left(2x+7\right)\left(x-5\right)\left(5x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+7=0\\x-5=0\\5x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\\x=-\dfrac{1}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{7}{2};5;-\dfrac{1}{5}\right\}\)
e) \(\left(x-1\right)\left(2x+7\right)\left(x^2+2\right)=0\)
Vì \(x^2+2\ge2>0\forall x\)
\(\Rightarrow\left(x-1\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)
f) \(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)
\(\Leftrightarrow\left(3x+2\right)\left(x-1\right)\left(x+1\right)-\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[\left(3x+2\right)\left(x+1\right)\right].\left(x-1-3x+2\right)=0\)
\(\Leftrightarrow\left(3x^2+5x+2\right)\left(-2x+1\right)=0\)
\(\Leftrightarrow\left(3x^2+3x+2x+2\right)\left(-2x+1\right)=0\)
\(\Leftrightarrow\left[3x\left(x+1\right)+2\left(x+1\right)\right]\left(-2x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x+2\right)\left(-2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x+2=0\\-2x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{2}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{-1;-\dfrac{2}{3};\dfrac{1}{2}\right\}\)
a) Ta có: \(\left(x^2-2x\right)^2-6x^2+12x+9=0\)
\(\Leftrightarrow\left(x^2-2x\right)^2-6\left(x^2-2x\right)+9=0\)
\(\Leftrightarrow\left(x^2-2x-3\right)^2=0\)
\(\Leftrightarrow x^2-2x-3=0\)
\(\Leftrightarrow x^2-3x+x-3=0\)
\(\Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Vậy: S={3;-1}
b) Ta có: \(\left(x^2+x+1\right)\left(x^2+x+2\right)=12\)
\(\Leftrightarrow\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12=0\)
\(\Leftrightarrow\left(x^2+x\right)^2+5\left(x^2+x\right)-2\left(x^2+x\right)-10=0\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x+5\right)-2\left(x^2+x+5\right)=0\)
\(\Leftrightarrow\left(x^2+x+5\right)\left(x^2+x-2\right)=0\)
\(\Leftrightarrow x^2+x-2=0\)(Vì \(x^2+x+5>0\forall x\))
\(\Leftrightarrow x^2+2x-x-2=0\)
\(\Leftrightarrow x\left(x+2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
Vậy: S={-2;1}
2 ý a và b anh CTV nãy đã làm rồi nha, còn câu c này thì làm dài dòng+không chắc :VVV
c)\(\left(2x^2-3x+1\right)\left(2x^2+5x+1\right)-9x^2=0\)
\(\Leftrightarrow\left(2x^2-3x+1\right)\left(2x^2-3x+1+8x\right)-9x^2=0\)
\(\Leftrightarrow\left(2x^2-3x+1\right)^2+8x\left(2x^2-3x+1\right)+16x^2-25x^2=0\)
\(\Leftrightarrow\left(2x^2-3x+1+4x\right)^2-25x^2=0\)
\(\Leftrightarrow\left(2x^2+x+1\right)^2-25x^2=0\)
\(\Leftrightarrow\left(2x^2+x+1-5x\right)\left(2x^2+x+1+5x\right)=0\)
\(\Leftrightarrow\left(2x^2-4x+1\right)\left(2x^2+6x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(2x^2-4x+1\right)=0\\\left(2x^2+6x+1\right)=0\end{matrix}\right.\)
Rồi đến đây tự giải nhé, không phân tích được thì bấm máy tính là ra nha:vv
( 9 x 2 – 4 ) ( x + 1 ) = ( 3 x + 2 ) ( x 2 - 1 )
⇔ (3x – 2)(3x + 2)(x + 1) - (3x + 2)(x - 1)(x + 1) = 0
⇔(3x+ 2)(x + 1)(3x – 2 – x + 1) = 0
⇔ (3x + 2)(x + 1)(2x – 1) = 0
a) 8x-3=0
⇔8x=3
⇔x=\(\dfrac{3}{8}\)
Vậy...
b) -5x+7=-3x-9
⇔-5x+3x=-9-7
⇔-2x=-16
⇔x=8
Vậy...
e)
\(\dfrac{1}{x-2}+4=\dfrac{x+3}{x-2}\)
⇔\(\dfrac{1}{x-2}-\dfrac{x+3}{x-2}=4\)
⇔\(\dfrac{-x-2}{x-2}=4\)
⇔\(x+2=4x-8\)
⇔\(-3x=-10\)
⇔\(x=\dfrac{10}{3}\)
\(a,x-5\left(x-2\right)=6x\\ \Leftrightarrow x-5x+10-6x=0\\ \Leftrightarrow-10x+10=0\\ \Leftrightarrow x=1\\ b,2^3+3x^2-32x=48\\ \Leftrightarrow3x^2-32x+8=48\\ \Leftrightarrow3x^2-32x-40=0\)
Nghiệm xấu lắm bn
\(c,\left(3x+1\right)\left(x-3\right)^2=\left(3x+1\right)\left(2x-5\right)^2\\ \Leftrightarrow c,\left(3x+1\right)\left[\left(2x-5\right)^2-\left(x-3\right)^2\right]\\ \Leftrightarrow\left(3x+1\right)\left(2x-5-x+3\right)\left(2x-5+x-3\right)=0\\ \Leftrightarrow\left(3x+1\right)\left(x-2\right)\left(3x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=2\\x=\dfrac{8}{3}\end{matrix}\right.\)
\(d,9x^2-1=\left(3x+1\right)\left(4x+1\right)\\ \Leftrightarrow\left(3x+1\right)\left(4x+1\right)-\left(3x-1\right)\left(3x+1\right)=0\\ \Leftrightarrow\left(3x+1\right)\left(4x+1-3x+1\right)=0\\ \Leftrightarrow\left(3x+1\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-2\end{matrix}\right.\)
\(b,2x^3+3x^2-32x-48=0\\ \Leftrightarrow\left(2x^3-8x^2\right)+\left(11x^2-44x\right)+\left(12x-48\right)=0\\ \Leftrightarrow2x^2\left(x-4\right)+11x\left(x-4\right)+12\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(2x^2+11x+12\right)=0\\ \Leftrightarrow\left(x-4\right)\left[\left(2x^2+8x\right)+\left(3x+12\right)\right]=0\\ \Leftrightarrow\left(x-4\right)\left[2x\left(x+4\right)+3\left(x+4\right)\right]=0\\ \Leftrightarrow\left(x-4\right)\left(2x+3\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{3}{2}\\x=-4\end{matrix}\right.\)
a: =>3x=-12
hay x=-4
b: =>2x=8
hay x=4
c: =>-2x=2
hay x=-1
a. \(2x+x+12=0\\ \Leftrightarrow2x+x=0-12\\ \Leftrightarrow3x=-12\\ \Leftrightarrow x=-4\)
Vậy S = { -4 }
b. \(x-5=3-x\\ \Leftrightarrow x+x=3+5\\ \Leftrightarrow2x=8\\ \Leftrightarrow x=4\)
Vậy S = { 4 }
c. \(7-3x=9-x\\ \Leftrightarrow-3x+x=9-7\\ \Leftrightarrow-2x=2\\ \Leftrightarrow x=-1\)
Vậy S = { -1 }
a: \(x\left(x-1\right)+2x^2-2=0\)
=>\(x\left(x-1\right)+2\left(x-1\right)\left(x+1\right)=0\)
=>\(\left(x-1\right)\left(x+2x+2\right)=0\)
=>(x-1)(3x+2)=0
=>\(\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{3}\end{matrix}\right.\)
b: \(9x^2-1=\left(3x+1\right)\left(2x-3\right)\)
=>\(\left(3x+1\right)\left(3x-1\right)-\left(3x+1\right)\left(2x-3\right)=0\)
=>\(\left(3x+1\right)\left(3x-1-2x+3\right)=0\)
=>(3x+1)(x+2)=0
=>\(\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-2\end{matrix}\right.\)
a: x(x−1)+2x2−2=0
=>x(x−1)+2(x−1)(x+1)=0
=>(x−1)(x+2x+2)=0
=>(x-1)(3x+2)=0
=>⎡⎣x=1x=−23
b: 9x2−1=(3x+1)(2x−3)9
=>(3x+1)(3x−1)−(3x+1)(2x−3)=0
=>(3x+1)(3x−1−2x+3)=0
=>(3x+1)(x+2)=0
=>⎡⎣x=−13x=−2