tìm x biết: a.x( x+5) =0
b. (2x-6)(-4x-8)=0
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a) 3x(4x-3)-2x(5-6x)=0
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow24x^2-19x=0\)
\(\Leftrightarrow x\left(24x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)
Vậy x=0 hoặc x=\(\dfrac{19}{24}\)
a) \(\text{5x(x-2)+(2-x)=0}\)
\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(\text{x(2x-5)-10x+25=0}\)
\(\Rightarrow x\left(2x-5\right)-5\left(2x-5\right)=0\\ \Rightarrow\left(x-5\right)\left(2x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\2x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=2,5\end{matrix}\right.\)
c) \(\dfrac{25}{16}-4x^2+4x-1=0\)
\(\Rightarrow\dfrac{9}{16}-4x^2+4x=0\)
\(\Rightarrow-4x^2+4x+\dfrac{9}{16}=0\)
\(\Rightarrow-4x^2-\dfrac{1}{2}x+\dfrac{9}{2}x+\dfrac{9}{16}=0\)
\(\Rightarrow\left(-4x^2-\dfrac{1}{2}x\right)+\left(\dfrac{9}{2}x+\dfrac{9}{16}\right)=0\)
\(\Rightarrow-\dfrac{1}{2}x\left(8x+1\right)+\dfrac{9}{16}\left(8x+1\right)=0\)
\(\Rightarrow\left(-\dfrac{1}{2}x+\dfrac{9}{16}\right)\left(8x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-\dfrac{1}{2}x+\dfrac{9}{16}=0\\8x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=\dfrac{-1}{8}\end{matrix}\right.\)
a: \(x^2-3x+2=0\)
=>(x-2)(x-1)=0
=>x=2 hoặc x=1
b: 2x+6>0
=>2x>-6
hay x>-3
c: \(x^2+4x+5=0\)
\(\text{Δ}=4^2-4\cdot1\cdot5< 0\)
Do đó: Phương trình vô nghiệm
\(a,\) \(x^2-3x+2=0\) có \(2\) nghiệm \(x=1;x=3\)
\(\Rightarrow D=\left\{1;3\right\}\)
\(b,\) \(2x+6>0\) \(\Leftrightarrow x>-3\)
\(\Rightarrow\) \(D=\left\{-3;+\infty\right\}\)
\(c,\) \(x^2+4x+5=0\Leftrightarrow\left(x+2\right)^2+1=0\) \(\Rightarrow ptvn\)
\(\Rightarrow\) \(D=\varnothing\)
\(\Rightarrow\left(x+3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x+3=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
\(2\left(x+3\right)+x\left(3+x\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
a) x2 - 5x - 6 = 0
=> x2 - 2x - 3x - 6 = 0
=> (x2 - 2x) + (-3x - 6) = 0
=> x(x - 2) - 3 (x - 2) = 0
=> (x - 2) (x - 3) = 0
=> x - 2 = 0 => x = 2
x - 3 = 0 => x = 3
còn lại tương tự nhé!! 46566578768698945635655675656788787868789789879789098089364556546
a,(2x-5^2)-4x(x-3)=0
=> 2x-25-4x2+12x=0
=>-4x2+14x-25=0
đề bài ý a sai nha
b, 6x2-7x=0
=>x(6x-7)=0
=>x=0 và 6x-7=0
=>x=0 và x=7/6
vậy x=0 và x=7/6
a) \(3\left(x-1\right)^2\cdot3x\left(x-5\right)=0\)
\(\Rightarrow9x\left(x-1\right)^2\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=5\end{matrix}\right.\)
b) \(\left(x+3\right)^2-5x-15=0\)
\(\Rightarrow\left(x+3\right)^2-5\left(x+3\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x+3-5\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
c) \(2x^5-4x^3+2x=0\)
\(\Rightarrow2x\left(x^4-2x^2+1\right)=0\)
\(\Rightarrow2x\left[\left(x^2\right)^2-2\cdot x^2\cdot1+1^2\right]=0\)
\(\Rightarrow2x\left(x^2-1\right)^2=0\)
\(\Rightarrow2x\left(x-1\right)^2\left(x+1\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
\(\text{#}Toru\)
a) \(\Leftrightarrow x^2-4x-x^2+6x-9=0\\ \Leftrightarrow2x=9\\ \Leftrightarrow x=4,5\)
b) \(\Leftrightarrow x^2-3x-10=0\\ \Leftrightarrow\left(x^2+2x\right)-\left(5x+10\right)=0\\ \Leftrightarrow x\left(x+2\right)-5\left(x+2\right)=0\\ \left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
c) \(\Leftrightarrow\left(2x-3-7\right)\left(2x-3+7\right)=0\\ \Leftrightarrow\left(2x-10\right)\left(2x+4\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
d) \(\Leftrightarrow\left(2x+7\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\end{matrix}\right.\)
`a,x^2 +4x-5=0`
`<=> x^2-x+5x-5=0`
`<=> x(x-1)+5(x-1)=0`
`<=>(x-1)(x+5)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
`b, x^2 -x-12=0`
`<=> x^2 +3x-4x-12=0`
`<=>(x^2+3x)-(4x+12)=0`
`<=>x(x+3)-4(x+3)=0`
`<=>(x+3)(x-4)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)
`c, (2x-7)^2 - 6(2x-7)(x-3)=0`
`<=>(2x-7)(2x-7 -6x+18)=0`
`<=>(2x-7) ( -4x+11)=0`
\(\Leftrightarrow\left[{}\begin{matrix}2x-7=0\\-4x+11=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=7\\-4x=-11\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=\dfrac{11}{4}\end{matrix}\right.\)
a: =>(x+5)(x-1)=0
=>x=1 hoặc x=-5
b: =>(x-4)(x+3)=0
=>x=4 hoặc x=-3
c: =>(2x-7)(2x-7-6x+18)=0
=>(2x-7)(-4x+11)=0
=>x=11/4 hoặc x=7/2
làm ăn vầy là chết tui rồi
\(x.\left(x+5\right)=0\)
\(\Leftrightarrow\)\(x=0\Rightarrow x=0\)
\(\Leftrightarrow\)\(x+5=0\Rightarrow x=-5\)
Vậy \(x\in\left\{0;-5\right\}\)
b) \(\left(2x-6\right)\left(-4x-8\right)\)
\(\Leftrightarrow\)\(2x-6=0\Rightarrow x=\frac{6}{2}=3\)
\(\Leftrightarrow\)\(-4x-8=0\Rightarrow x=-2\)
Vậy \(x\in\left\{3;-2\right\}\)
a) x . ( x + 5 ) = 0
\(\Rightarrow\orbr{\begin{cases}x=0\\x+5=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
b) ( 2x - 6 ) . ( -4x - 8 ) = 0
\(\Rightarrow\orbr{\begin{cases}2x-6=0\\-4x-8=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=6\\-4x=8\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)