a: \(\Leftrightarrow x^2\left(9x^2-4\right)=0\)

\(\Leftrightarrow x^2\left(3x-2\right)\left(3x+2\right)=0\)

hay \(x\in\left\{0;\dfrac{2}{3};-\dfrac{2}{3}\right\}\)

b: \(\Leftrightarrow2x^4-4x^2+3x^2-6=0\)

\(\Leftrightarrow x^2-2=0\)

hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)

d: \(\Leftrightarrow x^4-9x^2+6x^2-54=0\)

\(\Leftrightarrow x^2-9=0\)

=>x=3 hoặc x=-3

a) 5 (x-2)-3x=0

=>5x-10-3x=0

=>2x-10=0

=>x=5

b) => x^2=49 =>x=+-7

c)3x^2 +7x-18=0

=> ( vô ngiệm)

d)3x (x^2-16)=0

=>3x=0 hoặc x^2-16=0

=>x=0 hoặc x=+-4

K mk nha bn, 

1) 5(x-2)-3x=0

=> 5x-10-3x=0

=> 2x-10=0

=> 2x=10

=>x=5

2) x²-49=0

=> x²=49

=> x=+-7

3) 3x²+7x-18=0

=> \(x\in\varnothing\)

4) 3x(x²-16)=0

=> 3x(x-4)(x+4)=0

=>\(\hept{\begin{cases}x=0\\x-4=0\\x+4=0\end{cases}}\)=>\(\hept{\begin{cases}x=0\\x=4\\x=-4\end{cases}}\)

vậy x=0 hoặc x=+-4

Ta có: \(6x\left(3x+5\right)-2x\left(3x-2\right)+\left(17-x\right)\left(x-1\right)+x\left(x-18\right)=0\)

\(\Leftrightarrow18x^2+30x-6x^2+4x+17x-17-x^2+x+x^2-18x=0\)

\(\Leftrightarrow12x^2-34x-17=0\)

\(\Leftrightarrow12\left(x^2-\frac{34}{12}x-\frac{17}{12}\right)=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\frac{17}{12}+\frac{289}{144}-\frac{493}{144}=0\)

\(\Leftrightarrow\left(x-\frac{17}{12}\right)^2=\frac{493}{144}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{17}{12}=\frac{\sqrt{493}}{12}\\x-\frac{17}{12}=-\frac{\sqrt{493}}{12}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{17+\sqrt{493}}{12}\\x=\frac{17-\sqrt{493}}{12}\end{matrix}\right.\)

Vậy: \(S=\left\{\frac{17+\sqrt{493}}{12};\frac{17-\sqrt{493}}{12}\right\}\)

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

=> \(\orbr{\begin{cases}x=\sqrt{\frac{17}{12}}\\x=-\sqrt{\frac{17}{12}}\end{cases}}\)

\(6x\left(3x+5\right)-2x\left(3x-2\right)+\left(17-x\right)\left(x-1\right)+x\left(x-18\right)=0\)

\(\Leftrightarrow9x^2+30x-6x^2+4x+17x-17-x^2+x+x^2-18x=0\)

\(\Leftrightarrow3x^2+34x-17=0\) ( vô nghiệm ) 

6x( 3x + 5 ) - 2x( 3x - 2 ) + ( 17 - x )( x - 1 ) + x( x - 18 ) = 0

<=> 18x² + 30x - 6x² + 4x - x² + 18x - 17 + x² - 18x = 0

<=> 12x² + 34x - 17 = 0

\(\Delta'=b'^2-ac=\left(\frac{b}{2}\right)^2-ac=\left(\frac{34}{2}\right)^2-12\cdot\left(-17\right)=289+204=493\)( không muốn xài Delta nữa đâu nhưng ... :)) )

\(\Delta'>0\)nên phương trình đã cho có hai nghiệm phân biệt

\(\hept{\begin{cases}x_1=\frac{-b'+\sqrt{\Delta'}}{a}=\frac{-17+\sqrt{493}}{12}\\x_2=\frac{-b'-\sqrt{\Delta'}}{a}=\frac{-17-\sqrt{493}}{12}\end{cases}}\)

Vậy ...

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 :

\(t^2-11t+18=0\)

\(\left[{}\begin{matrix}t=9\left(tmđk\right)\\t=2\left(tmđk\right)\end{matrix}\right.\)

\(t=9\rightarrow x^2=9\rightarrow x=\pm3\)

\(t=2\rightarrow x^2=2\rightarrow x=\pm\sqrt{2}\)

3x.(x^2 -2) - (3x^3 -18) =0 

suy ra x=0

1) Ta có: \(\left(x-2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

Vậy \(x=2\) hoặc \(x=-1\)

2) Ta có: \(\left(3-x\right)x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3-x=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)

Vậy \(x=3\) hoặc \(x=0\)

3) Ta có: \(2x-17=-\left(3x-18\right)\)

\(\Leftrightarrow2x-17=18-3x\)

\(\Leftrightarrow2x+3x=18+17\)

\(\Leftrightarrow5x=35\Leftrightarrow x=\dfrac{35}{5}=7\)

Vậy \(x=7\)

(1)X1=2 và X2=-2

(2)X1=3 và X2=0

(3) X=7

a)5x ( 1 - 2x ) - 3x ( x + 18 ) = 0

5x - 10x^2 - 3x^2 - 54x =0

-13x^2 - 49x =0

(-13x - 49)x =0

\(\Rightarrow\orbr{\begin{cases}-13x-49=0\\x=0\end{cases}\Rightarrow\orbr{\begin{cases}-13x=49\\x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{49}{13}\\x=0\end{cases}}}\)

Vậy x= -49/13 hoặc x=0

b)5x - 10x^2 - 3x^2 - 54 = 0

(câu b giống câu a)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2=2\\x^2=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm\sqrt{2}\\x=\pm3\end{matrix}\right.\)