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`a,(x+3)(x^2+2021)=0`
`x^2+2021>=2021>0`
`=>x+3=0`
`=>x=-3`
`2,x(x-3)+3(x-3)=0`
`=>(x-3)(x+3)=0`
`=>x=+-3`
`b,x^2-9+(x+3)(3-2x)=0`
`=>(x-3)(x+3)+(x+3)(3-2x)=0`
`=>(x+3)(-x)=0`
`=>` $\left[ \begin{array}{l}x=0\\x=-3\end{array} \right.$
`d,3x^2+3x=0`
`=>3x(x+1)=0`
`=>` $\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.$
`e,x^2-4x+4=4`
`=>x^2-4x=0`
`=>x(x-4)=0`
`=>` $\left[ \begin{array}{l}x=0\\x=4\end{array} \right.$
1) a) \(\left(x+3\right).\left(x^2+2021\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2021=0\end{matrix}\right.\\\left[{}\begin{matrix}x=-3\left(nhận\right)\\x^2=-2021\left(loại\right)\end{matrix}\right. \)
=> S={-3}
b: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)
\(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=0\)
\(\Leftrightarrow\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24=0\)
\(\Leftrightarrow x^2+7x+6=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)
\(a,4\left(x-3\right)^2-\left(2x-1\right)^2\ge12\)
\(\Leftrightarrow4x^2-24x+36-4x^2-4x+1\ge12\)
\(\Leftrightarrow-28x+37\ge12\)
\(\Leftrightarrow-28x\ge12-37\)
\(\Leftrightarrow-28x\ge-25\)
\(\Leftrightarrow x\le\dfrac{25}{28}\)
Vậy \(S=\left\{x\left|x\le\dfrac{25}{28}\right|\right\}\)
b, \(\left(x-4\right)\left(x+4\right)\ge\left(x+3\right)^2+5\)
\(\Leftrightarrow x^2-16\ge x^2+6x+9+5\)
\(\Leftrightarrow x^2-x^2-6x\ge9+5+16\)
\(\Leftrightarrow-6x\ge30\)
\(\Leftrightarrow x\le-5\)
Vậy \(S=\left\{x\left|x\le-5\right|\right\}\)
\(c,\left(3x-1\right)^2-9\left(x+2\right)\left(x-2\right)< 5x\)
\(\Leftrightarrow9x^2-6x-1-9x^2+36< 5x\)
\(\Leftrightarrow9x^2-9x^2-6x-5x+36+1< 0\)
\(\Leftrightarrow-11x+37< 0\)
\(\Leftrightarrow-11x< -37\)
\(\Leftrightarrow x>\dfrac{37}{11}\)
vậy \(S=\left\{x\left|x>\dfrac{37}{11}\right|\right\}\)
a: =>|x-7|=3-2x
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(-2x+3\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(2x-3-x+7\right)\left(2x-3+x-7\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(x+4\right)\left(3x-10\right)=0\end{matrix}\right.\Leftrightarrow x=-4\)
b: =>|2x-3|=4x+9
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(4x+9-2x+3\right)\left(4x+9+2x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(2x+12\right)\left(6x+6\right)=0\end{matrix}\right.\Leftrightarrow x=-1\)
c: =>3x+5=2-5x hoặc 3x+5=5x-2
=>8x=-3 hoặc -2x=-7
=>x=-3/8 hoặc x=7/2
a) \(\left(3x+2\right)^2-\left(3x-2\right)^2=5x+8\)
\(\Rightarrow\left(3x+2+3x-2\right)\left(3x+2-3x+2\right)=5x+8\)
\(\Rightarrow4.6x=5x+8\Rightarrow24x=5x+8\)
\(\Rightarrow19x=8\Rightarrow x=\frac{8}{19}\)
b) \(3\left(x-2\right)^2+9\left(x-1\right)=3\left(x^2+x-3\right)\)
\(\Rightarrow3\left(x^2-4x+4\right)+9x-9=3x^2+3x-9\)
\(\Rightarrow3x^2-12x+12+9x-9=3x^2+3x-9\)
\(\Rightarrow-12x+12+9x-9=3x-9\)
\(\Rightarrow-3x+3=3x-9\)
\(\Rightarrow6x=12\Rightarrow x=2\)
Ta có : \(x\left(x+1\right)\left(x-1\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2-x+2x-2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)
\(\Leftrightarrow\left(x^2+x-1+1\right)\left(x^2+x-1-1\right)=24\)
\(\Leftrightarrow\left(x^2+x-1\right)^2-1=24\)
\(\Leftrightarrow\left(x^2+x-1\right)^2=25\)
<=> 2 trường hợp sảy ra là bằng 5 hoặc -5 nhé
a: =>2x-4+3+3y=-2 và 3x-6-2-2y=-3
=>2x+3y=-2+4-3=2-3=-1 và 3x-2y=-3+6+2=5
=>x=1; y=-1
b: =>x^2-x+xy-y=x^2+x-xy-y+2xy
=>-x-y=x-y và y^2+y-yx-x=y^2-2y+xy-2x-2xy
=>x=0 và y-x=-2y-2x
=>x=0 và y=0
Bài 4 :
24 phút = \(\dfrac{24}{60} = \dfrac{2}{5}\) giờ
Gọi thời gian dự định đi từ A đến B là x(giờ) ; x > 0
Suy ra quãng đường AB là 36x(km)
Khi vận tốc sau khi giảm là 36 -6 = 30(km/h)
Vì giảm vận tốc nên thời gian đi hết AB là x + \(\dfrac{2}{5}\)(giờ)
Ta có phương trình:
\(36x = 30(x + \dfrac{2}{5})\\ \Leftrightarrow x = 2\)
Vậy quãng đường AB dài 36.2 = 72(km)
Nhân 2 vế với 2 rồi chuyển vế và rút gọn
Bạn Tên Là Long
a/ \(\Leftrightarrow2x^3+9x^2-27=0\)
\(\Leftrightarrow2x^3+12x^2+18x-3x^2-18x-27=0\)
\(\Leftrightarrow2x\left(x^2+6x+9\right)-3\left(x^2+6x+9\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x+3\right)^2=0\)
\(\Leftrightarrow...\)
b/ \(\Leftrightarrow x^3-3x^2+3x-1+x^3+x^3+3x^2+3x+1=x^3+6x^2+12x+8\)
\(\Leftrightarrow x^3-3x^2-3x-4=0\)
\(\Leftrightarrow\left(x-4\right)\left(x^2+x+1\right)=0\)
c/ \(x\left(x+1\right)\left(x-1\right)\left(x+2\right)-24=0\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)-24=0\)
Đặt \(x^2+x=t\)
\(t\left(t-2\right)-24=0\Leftrightarrow t^2-2t-24=0\Rightarrow\left[{}\begin{matrix}t=6\\t=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2+x=6\\x^2+x=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+x-6=0\\x^2+x+4=0\end{matrix}\right.\)
d/ \(\Leftrightarrow\left(x-7\right)\left(x-2\right)\left(x-4\right)\left(x-5\right)-72=0\)
\(\Leftrightarrow\left(x^2-9x+14\right)\left(x^2-9x+20\right)-72=0\)
Đặt \(x^2-9x+14=0\)
\(t\left(t+6\right)-72=0\Leftrightarrow t^2+6t-72=0\Rightarrow\left[{}\begin{matrix}t=6\\t=-12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2-9x+14=6\\x^2-9x+14=-12\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-9x+8=0\\x^2-9x+26=0\end{matrix}\right.\)