3x.(x-18)=0
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6x(3x + 5) - 2x(3x - 2) + (17 - x)(x - 1) + x(x - 18) = 0
=> (18x2 - 6x2 - x2 + x2) + (30x + 4x - 16x - 18x) - 17 = 0
=> 12x2 - 17 = 0
=> 12x2 = 17
=> x2 = 17/12
=> \(\orbr{\begin{cases}x=\sqrt{\frac{17}{12}}\\x=-\sqrt{\frac{17}{12}}\end{cases}}\)
a) \(x^2-x+x=4\)
\(x^2=4\)
\(x=\pm2\)
b) \(3x\left(x-5\right)-2\left(x-5\right)=0\)
\(\left(x-5\right)\left(3x-2\right)=0\)
\(\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\end{matrix}\right.\)
c) Ta có: \(a+b+c=5-3-2=0\)
\(\left[{}\begin{matrix}x=1\\x=\dfrac{c}{a}=\dfrac{-2}{5}\end{matrix}\right.\)
d) Đặt \(x^2=t\left(t\ge0\right)\) . Lúc đó phương trình trở thành :
\(t^2-11t+18=0\)
\(\left[{}\begin{matrix}t=9\left(tmđk\right)\\t=2\left(tmđk\right)\end{matrix}\right.\)
\(t=9\rightarrow x^2=9\rightarrow x=\pm3\)
\(t=2\rightarrow x^2=2\rightarrow x=\pm\sqrt{2}\)
a: =>x+5>0 và x-2<0
=>-5<x<2
=>x thuộc {-4;-3;...;1}
b: =>(x-5)(x+5)>0
=>x>5 hoặc x<-5
=>x thuộc Z\{-5;-4;-3;...;3;4;5}
c: =>(x+6)(x-7)>0
=>x>7 hoặc x<-6
a:Ta có: \(x\left(x-1\right)+x=4\)
\(\Leftrightarrow x^2-x+x=4\)
\(\Leftrightarrow x^2=4\)
hay \(x\in\left\{2;-2\right\}\)
b: Ta có: \(3x\left(x-5\right)-2x+10=0\)
\(\Leftrightarrow\left(x-5\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\end{matrix}\right.\)
c: Ta có: \(5x^2-3x-2=0\)
\(\Leftrightarrow5x^2-5x+2x-2=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{5}\end{matrix}\right.\)
d: Ta có: \(x^4-11x^2+18=0\)
\(\Leftrightarrow x^4-9x^2-2x^2+18=0\)
\(\Leftrightarrow x^2\left(x^2-9\right)-2\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)
a) x(x-1)+x=4
⇔x2=4⇔\(x=\pm2\)
b)3x(x-5)-2x+10=0
⇔3x(x-5)-2(x-5)=0
⇔(x-5)(3x-1)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{3}\end{matrix}\right.\)
c)5x2-3x-2=0
⇔ 5x(x-1)+2(x-1)=0
⇔ (x-1)(5x+2)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{5}\end{matrix}\right.\)
d)x4-11x2+18=0
⇔ x2(x2-2)-9(x2-2)=0
⇔ (x2-2)(x2-9)=0
\(\Leftrightarrow\left[{}\begin{matrix}x^2=2\\x^2=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm\sqrt{2}\\x=\pm3\end{matrix}\right.\)
x2+3x-18=0
x2-3x+6x-18=0
x(x-3)+6.(x-3)=0
(x-3)(x+6)=0
=> x-3=0 hoặc x+6=0
•x-3=0=> x=3
•x+6=0=>x=-6
Vậy x thuộc {3;-6}
5x(1 - 2x) - 3x(x + 18) = 0
=> 5x - 10x2 - 3x2 - 54x = 0
=> -13x2 - 49x = 0
=> x.(-13x - 49) = 0
=> x = 0
hoặc -13x - 49 = 0 => -13x = 49 => x = -49/13
Vậy x = 0, x = -49/13
\(5x\left(1-2x\right)-3x\left(x+18\right)=0\)
\(\Rightarrow x\left(5\left(1-2x\right)-3\left(x+18\right)\right)=0\)
\(\Rightarrow x\left(5-10x-3x+54\right)=0\)
\(\Rightarrow x\left(59-13x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\59-13x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\13x=59\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=\frac{59}{13}\end{cases}}}\)
5x(1-2x)-3x(x+18)=0
=> \(5x-10x^2-3x^2-54x=0\)
=> \(-49x-13x^2=0\)
=> \(x\left(-49-13x\right)=0\)
Suy ra
- \(x=0\)
- \(-49-13x=0\)=> \(-13x=49\)=> \(x=\frac{-49}{13}\)
Vậy \(x=\frac{-49}{13}; x=0\)
`3x(x-18)=0`
`TH1:3x=0`
`=>x=0:3`
`=>x=0`
`TH2:x-18=0`
`=>x=18`
Vậy: `x∈{0;18}`
3x(x -18) = 0
x = 0 hoặc x - 18 = 0 ⇒ x = 18
Vậy x ϵ {0; 18)