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a,\(\left(x-4-5\right)\left(x-4+5\right)=0\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\Leftrightarrow x=9;x=-1\)
b, \(\left(x-3-x-1\right)\left(x-3+x+1\right)=0\Leftrightarrow2x-2=0\Leftrightarrow x=1\)
c, \(\left(x^2-4\right)\left(2x-3\right)-\left(x^2-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(2x-3-x+1\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-2\right)=0\Leftrightarrow x=-2;x=2\)
d, \(\left(3x-7\right)^2-\left(2x+2\right)^2=0\Leftrightarrow\left(3x-7-2x-2\right)\left(3x-7+2x+2\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left(5x-5\right)=0\Leftrightarrow x=1;x=9\)
Bài 3:
b: \(\Leftrightarrow x^2\left(x+1\right)^2=0\)
hay \(x\in\left\{0;-1\right\}\)
c: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=0\)
=>x-1=0
hay x=1
d: \(\Leftrightarrow6x^2-3x-4x+2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(3x-2\right)=0\)
hay \(x\in\left\{\dfrac{1}{2};\dfrac{2}{3}\right\}\)
a: =>x-2=0 hoặc x+3=0
=>x=2 hoặc x=-3
b:=>x-7=0 hoặc x+2=0
=>x=7 hoặc x=-2
c: =>4x+2=0 hoặc 3x-4=0
=>x=4/3 hoặc x=-1/2
d: =>2x+1=0 hoặc x-3=0
=>x=3 hoặc x=-1/2
a)
`(x-2)(x+3)=0`
`<=> x-2=0` hoặc `x+3=0`
`<=>x=2` hoặc `x=-3`
b)
`(x-7)(2+x)=0`
`<=>x-7=0` hoặc `2+x=0`
`<=>x=7` hoặc `x=-2`
c)
`(4x+2)(3x-4)=0`
`<=>4x+2=0` hoặc `3x-4=0`
`<=>x=-1/2` hoặc `x=4/3`
d)
`(2x+1)(x-3)=0`
`<=>2x+1=0` hoặc `x-3=0`
`<=>x=-1/2` hoặc `x=3`
e)
`(0,1x-3)(x+0,5)=0`
`<=>0,1x-3=0` hoặc `x+0,5=0`
`<=>x=30` hoặc `x=-0,5`
f)
`(0,2x-0,4)(0,1x+0,7)=0`
`<=>0,2x-0,4=0` hoặc `0,1x+0,7=0`
`<=>x=2` hoặc `x=-7`
2:
a: =>x-1=0 hoặc 3x+1=0
=>x=1 hoặc x=-1/3
b: =>x-5=0 hoặc 7-x=0
=>x=5 hoặc x=7
c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)
d: =>x=0 hoặc x^2-1=0
=>\(x\in\left\{0;1;-1\right\}\)
(x-1)(2x^2-8)=0
\(\Leftrightarrow\left(x-1\right)\left(2x^2-8\right)=0\\ \left(2x^3-8x-2x^2+8\right)=0\)
\(\Leftrightarrow2x\left(x-1\right)-8\left(x-1\right)=0\)
\(\Leftrightarrow x=1;x=\dfrac{8}{2}\)
3x^2-8x+5=0
áp dụng công thức bậc 2 ta có:
\(x=\dfrac{-\left(-8\right)\pm\sqrt{\left(-8\right)^2-4.3.5}}{2.3}\)
\(\Rightarrow x=\dfrac{5}{3};x=1\)
(7x-1).2x-7x+1=0
\(\Leftrightarrow\left(7x-1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow x=\dfrac{1}{7};x=\dfrac{1}{2}\)
a: =>x(x+3)=0
=>x=0 hoặc x=-3
b: =>x(1-2x)=0
=>x=0 hoặc x=1/2
c: =>(x-7)(2x+3-x)=0
=>(x-7)(x+3)=0
=>x=7 hoặc x=-3
d: =>(x-2)(3x-1-x-3)=0
=>(x-2)(2x-4)=0
=>x=2
a)
`x^2 +3x=0`
`<=>x(x+3)=0`
\(< =>\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
b)
`x-2x^2 =0`
`<=>x(1-2x)=0`
\(< =>\left[{}\begin{matrix}x=0\\1-2x=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)
c)
`(x-7)(2x+3)=x(x-7)`
`<=>(x-7)(2x+3)-x(x-7)=0`
`<=>(x-7)(2x+3-x)=0`
`<=>(x-7)(x+3)=0`
\(< =>\left[{}\begin{matrix}x-7=0\\x+3=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
d)
`(x-2)(x+3)=(x-2)(3x-1)`
`<=>(x-2)(x+3)-(x-2)(3x-1)=0`
`<=>(x-2)(x+3-3x+1)=0`
`<=>(x-2)(-2x+4)=0`
\(< =>\left[{}\begin{matrix}x-2=0\\-2x+4=0\end{matrix}\right.\\ < =>x=2\)
\(a,\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
\(b,\left(x-2\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
\(c,\left(x+3\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
\(d,\left(x+\dfrac{1}{2}\right)\left(4x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\4x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\4\left(x+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
\(e,\left(x-4\right)\left(5x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
\(f,\left(2x-1\right)\left(3x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-2\end{matrix}\right.\)
`a,(x-1)(x+2)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
`b,(x -2)(x -5)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
`c,(x +3)(x -5)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
`d,(x + 1/2)(4x + 4)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\4x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\4x=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
`e,(x -4)(5x -10)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5x-10=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\5x=10\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
`f,(2x -1)(3x +6)=0`
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\3x=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-2\end{matrix}\right.\)
`g,(2,3x -6,9)(0,1x -2)=0`
\(\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2,3x=6,9\\0,1x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=20\end{matrix}\right.\)
a, 3x - 7 = 0
<=> 3x = 7
<=> x = 7/3
b, 8 - 5x = 0
<=> -5x = -8
<=> x = 8/5
c, 3x - 2 = 5x + 8
<=> -2x = 10
<=> x = -5
e) Ta có: \(\left(5x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-1\\x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=3\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{1}{5};3\right\}\)
Bài 1: Tìm x
a) (x-5) (x-3)+ 2(x-5)=0
b) (x-2)(x^2+2x+4)-(x+2)(x^2-2x+4)=2(x+2)
giúp e với ạ, e cảm ơn
a) (x - 5)(x - 3) + 2(x - 5) = 0
(x - 5)(x - 3 + 2) = 0
(x - 5)(x - 1) = 0
x - 5 = 0 hoặc x - 1 = 0
*) x - 5 = 0
x = 5
*) x - 1 = 0
x = 1
Vậy x = 1; x = 5
b) (x - 2)(x² + 2x + 4) - (x + 2)(x² - 2x + 4) = 2(x + 2)
x³ - 8 - x³ - 8 = 2x + 4
2x = -8 - 8 - 4
2x = -20
x = -20 : 2
x = -10
a)
\(\left(x-5\right)\left(x-3\right)+2\left(x-5\right)=0\)
\(\left(x-5\right)\left(x-3+2\right)=0\)
\(\left(x-5\right)\left(x-1\right)=0\)
\(x-5=0\) hoặc \(x-1=0\)
+) \(x-5=0\\ \Rightarrow x=5\)
+) \(x-1=0\\ \Rightarrow x=1\)
Vậy \(x=1\) hoặc \(x=5\)
b) \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x+2\right)\left(x^2-2x+4\right)=2\left(x+2\right)\)
\(x^3-8-x^3-8=2x+4\)
\(2x=-8-8-4\)
\(2x=-20\)
\(x=-20:2\)
\(x=-10\)
Vậy \(x=-10\)
Trả lời:
\(a,3x\left(x^2+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x=0\\x-2=0\end{cases}}\)( vì x2 + 1 > 0 )
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Vậy S = { 0; 2 }
\(b,\left(x-5\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left[\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right]=0\)
\(\Leftrightarrow x-5=0\)( vì \(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\))
\(\Leftrightarrow x=5\)
Vậy S = { 5 }
\(c,\left(x-1\right)\left(x+2\right)+\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow x^2+x-2+x^2-2x-3=0\)
\(\Leftrightarrow2x^2-x-5=0\)
\(\Leftrightarrow2\left(x^2-\frac{1}{2}x-\frac{5}{2}\right)=0\)
\(\Leftrightarrow x^2-\frac{1}{2}x-\frac{5}{2}=0\)
\(\Leftrightarrow\left(x^2-2.x.\frac{1}{4}+\frac{1}{16}\right)-\frac{41}{16}=0\)
\(\Leftrightarrow\left(x-\frac{1}{4}\right)^2=\frac{41}{16}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{4}=\frac{\sqrt{41}}{4}\\x-\frac{1}{4}=-\frac{\sqrt{41}}{4}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1+\sqrt{41}}{4}\\x=\frac{1-\sqrt{41}}{4}\end{cases}}\)
Vậy \(S=\left\{\frac{1+\sqrt{41}}{4};\frac{1-\sqrt{41}}{4}\right\}\)
\(a,3x\left(x^3+x-2x^2-2\right)=0\)
\(3x\left[x^2\left(x-2\right)+\left(x-2\right)\right]=0\)\
\(\left(3x\right)\left(x-2\right)\left(x^2+1\right)=0\)
\(x^2+1=0\left(ktm\right)\)
\(\orbr{\begin{cases}3x=0\\x-2\end{cases}\orbr{\begin{cases}x=0\\x=2\end{cases}\left(TM\right)}}\)
\(b,x^3-5x^2+x^2-5x+x-5=0\)
\(x^3-4x^2-4x-5=0\)
vì phương trình này lẻ lên bạn bấm máy nha
\(c,x^2-x+2x-2+x^2+x-3x-3=0\)
\(2x^2-x-5=0\)
pt này cũng lẻ nha lên bạn tham khảo cách này
\(a=2,b=-1,c=-5\)
\(\Delta=-1^2-\left(4.2.-5\right)=41>0\)
pt có 2 n0 pb
\(\sqrt{\Delta}=\sqrt{41}\)
\(x_1=\frac{\sqrt{41}+1}{2.2}=\frac{\sqrt{41}+1}{4}\)
\(x_2=\frac{-\sqrt{41}+1}{2.2}=\frac{-\sqrt{41}+1}{4}\)