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NN
2
H
10 tháng 7 2023
`a,x^2+2x+1=9`
`<=>x^2+2.x.1+1^2=9`
`<=>(x+1)^2=3^2`
`<=>(x+1)^2=+-3`
\(\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
`b, x^2-4x-21=0`
`<=>x^2+3x-7x-21=0`
`<=>x(x+3) - 7(x+3)=0`
`<=>(x+3)(x-7)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)
`c,x^2+10x-24=0`
`<=>x^2+12x-2x-24=0`
`<=>x(x+12)-2(x+12)=0`
`<=>(x+12)(x-2)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+12=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-12\\x=2\end{matrix}\right.\)
10 tháng 7 2023
a: =>(x+1)^2=9
=>(x+1+3)(x+1-3)=0
=>(x+4)(x-2)=0
=>x=2 hoặc x=-4
b: =>x^2-7x+3x-21=0
=>(x-7)(x+3)=0
=>x=7;x=-3
c: =>x^2+12x-2x-24=0
=>(x+12)(x-2)=0
=>x=2 hoặc x=-12
6 tháng 7 2018
\(1.6x\left(x-10\right)-2x+20=0\)
⇔\(6x\left(x-10\right)-2\left(x-10\right)=0\)
⇔ \(2\left(x-10\right)\left(3x-1\right)=0\)
⇔ x = 10 hoặc x = \(\dfrac{1}{3}\)
KL....
\(2.3x^2\left(x-3\right)+3\left(3-x\right)=0\)
⇔ \(3\left(x-3\right)\left(x^2-1\right)=0\)
⇔ \(x=+-1\) hoặc \(x=3\)
KL....
\(3.x^2-8x+16=2\left(x-4\right)\)
⇔ \(\left(x-4\right)^2-2\left(x-4\right)=0\)
⇔ \(\left(x-4\right)\left(x-6\right)=0\)
⇔ \(x=4\) hoặc \(x=6\)
KL.....
\(4.x^2-16+7x\left(x+4\right)=0\)
\(\text{⇔}4\left(x+4\right)\left(2x-1\right)=0\)
⇔ \(x=-4hoacx=\dfrac{1}{2}\)
KL.....
\(5.x^2-13x-14=0\)
⇔ \(x^2+x-14x-14=0\)
\(\text{⇔}\left(x+1\right)\left(x-14\right)=0\)
\(\text{⇔}x=14hoacx=-1\)
KL......
Còn lại tương tự ( dài quá ~ )
S
13 tháng 10 2019
\(do:x=9\Rightarrow x+1=10\Rightarrow A=x^{16}-\left(x+1\right)x^{15}+\left(x+1\right)x^{14}-....+\left(x+1\right)=x^{16}-x^{16}-x^{15}+x^{15}+x^{14}-x^{14}-x^{13}+x^{13}+.....-x+x+1=1\)
\(-x^2+3x-4=-x^2+3x-2,25-1,75=-\left(x-\frac{3}{2}\right)^2-1,75< 0\left(đpcm\right)\)
L
24 tháng 8
a)
\(\left(x+2\right)^2-9=0\)
\(\Rightarrow\left(x+2\right)^2=9=3^2\)
\(\Rightarrow x+2=\pm3\)
\(\Rightarrow x=-5;1\)
b)
\(25x^2-10x+1=0\)
\(\left(5x\right)^2-2\cdot5x+1^2=0\)
\(\Rightarrow\left(5x+1\right)^2=0\)
\(\Rightarrow5x+1=0\)
\(\Rightarrow5x=-1;x=\dfrac{-1}{5}\)
c)
\(x^2+14x+49=0\)
\(\Rightarrow x^2+2\cdot7x+7^2=0\)
\(\Rightarrow\left(x+7\right)^2=0;x+7=0\)
\(\Rightarrow x=-7\)
d)
\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(4x^2-4x+1+x^2+6x+9-5x^2+5\cdot49=0\)
\(\Rightarrow5x^2-5x^2-4x+6x+10+245=0\)
\(\Rightarrow2x+255=0\)
\(\Rightarrow2x=-255\)
\(\Rightarrow x=\dfrac{-255}{2}\)
26 tháng 9 2018
1)x^2 - 10x + 25 = 0
X^2-5X-5X+25=0
X(X-5)-5(X-5)=0
(X-5).(X-5)=0
(X-5)^2 =0
X-5=0
X=5
vậy x=5
22 tháng 10 2020
a) \(15y\left(4y-9\right)-3\left(4y-9\right)=0\)
\(\Leftrightarrow\left(15y-3\right)\left(4y-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}15y-3=0\\4y-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=\frac{1}{5}\\y=\frac{9}{4}\end{matrix}\right.\)
Vây \(y\in\left\{\frac{1}{5};\frac{9}{4}\right\}\)
b) \(8\left(25z+7\right)-27z\left(25z+7\right)=0\)
\(\Leftrightarrow\left(8-27z\right)\left(25z+7\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}z=\frac{8}{27}\\z=-\frac{7}{25}\end{matrix}\right.\)
Vậy \(z\in\left\{\frac{8}{27};-\frac{7}{25}\right\}\)
c) \(13y\left(y-8\right)-2y+16=0\)
\(\Leftrightarrow13y\left(y-8\right)-2\left(y-8\right)=0\)
\(\Leftrightarrow\left(13y-2\right)\left(y-8\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}y=\frac{2}{13}\\y=8\end{matrix}\right.\)
Vậy \(y\in\left\{\frac{2}{13};8\right\}\)
d) \(-10y\left(y+2\right)-y-2=0\)
\(\Leftrightarrow\left(-10y-1\right)\left(y+2\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}y=-2\\y=-\frac{1}{10}\end{matrix}\right.\)
Vậy \(y\in\left\{-2;-\frac{1}{10}\right\}\)
e) \(x\left(x+19\right)^2-\left(x+19\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+19\right)^2=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-19\end{matrix}\right.\)
Vậy \(x\in\left\{1;-19\right\}\)
17 tháng 11 2021
\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)
\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)
17 tháng 11 2021
\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)
\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)
LV
1
26 tháng 2 2022
a: =>5x-5+17x=1-12x-4
=>22x-5=-12x-3
=>34x=2
hay x=1/17
b: =>\(\left(x-3\right)^2-4x\left(x-3\right)=0\)
=>(x-3)(-3x-3)=0
=>x=3 hoặc x=-1
c: =>(x-4)(x-6)=0
=>x=4 hoặc x=6