ĐKXĐ: x<>1

Đặt a=x; \(b=\frac{x}{x-1}\)

\(a+b=x+\frac{x}{x-1}=\frac{x^2-x+x}{x-1}=\frac{x^2}{x-1}\)

\(ab=x\cdot\frac{x}{x-1}=\frac{x^2}{x-1}\)

TA có: \(x^3+\frac{x^3}{\left(x-1\right)^3}+\frac{3x^2}{x-1}=2\)

=>\(a^3+b^3+3ab=2\)

=>\(\left(a+b\right)^3-3ab\left(a+b\right)+3ab=2\)

=>\(\left(\frac{x^2}{x-1}\right)^3-3\cdot\frac{x^2}{x-1}\cdot\frac{x^2}{x-1}+3\cdot\frac{x^2}{x-1}=2\)

=>\(\left(\frac{x^2}{x-1}\right)^3-3\cdot\left(\frac{x^2}{x-1}\right)^2+3\cdot\frac{x^2}{x-1}-1=1\)

=>\(\left(\frac{x^2}{x-1}-1\right)^3=1\)

=>\(\frac{x^2}{x-1}-1=1\)

=>\(\frac{x^2}{x-1}=2\)

=>\(x^2=2x-2\)

=>\(x^2-2x+2=0\)

=>\(\left(x-1\right)^2+1=0\) (vô lý)

=>x∈∅

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

<=> 3x^2-x-3x^2+15x-2x+10=0

<=>12x+10=0

<=>12x=-10

<=>x=-5/6

Link : Hoc24

a: \(\left(3x-2\right)\left(2x-1\right)-\left(6x^2-3x\right)=0\)

\(\Leftrightarrow6x^2-3x-4x+2-6x^2+3x=0\)

\(\Leftrightarrow-4x=-2\)

hay \(x=\dfrac{1}{2}\)

b: \(x^3-\left(x+1\right)\left(x^2-x+1\right)=x\)

\(\Leftrightarrow x=x^3-x^3-1\)

hay x=-1

Đk:\(x\ge1\)

Pt \(\Leftrightarrow\sqrt{x-1}=\sqrt{3x-2}+\sqrt{5x-1}\)

\(\Leftrightarrow x-1=8x-3+2\sqrt{15x^2-13x+2}\)

\(\Leftrightarrow2-7x=2\sqrt{15x^2-13x+2}\) (1)

Với \(x\ge1\Rightarrow\)\(\left\{{}\begin{matrix}2-7x\le2-7.1=-5< 0\\2\sqrt{15x^2-13x+2}=4>0\end{matrix}\right.\)

Từ (1) => Dấu "=" không xảy ra

Vậy pt vô nghiệm.

(2x^2-3x+1)(2x^2+5x+1)=9x^2

<=> (2x^2+5x+1- 8x)(2x^2 +5x+1)=9x^2

<=> (2x^2+5x+1)^2 -8x(2x^2+5x+1)=9x^2

<=>  (2x^2+5x+1)^2 -2*(4x)*(2x^2+5x+1)=9x^2

<=>  (2x^2+5x+1)^2 -2*(4x)*(2x^2+5x+1)+(4x)^2=9x^2+16x^2

<=> (2x^2+5x+1 - 4x)^2=25x^2

<=> (2x^2+x+1)^2=25x^2

<=> (2x^2+x+1)^2 - 25x^2 =0

<=>(2x^2+x+1-5x)(2x^2+x+1+5x)=0

<=>(2x^2-4x+1)(2x^2+6x+1)=0

<=> (2x^2-4x+1)=0 => 2( x^2 - 2x + 1/2)=0

                                <=> x^2-2x +1/2 =0

                                <=> (x^2-2x+1) -1/2 =0

                                <=> (x-1)^2 =1/2     =>  x-1 =căn(1/2)  => x=căn(1/2)+1

                                                              => x-1=-(căn(1/2)) => x=- (căn(1/2)) +1

Hoặc  2x^2 +6x +1=0 

         <=> x^2 + 3x +1/2 =0                

         <=> (x^2 + 2*(1.5)x + (1.5)^2) -(1.5)^2+1/2 =0

         <=> (x+1.5)^2 - 7/4 =0

         <=> (x+1.5)^2 = 7/4    =>        x+1.5 = căn(7/4) => x=căn(7/4) -1.5

                                           =>      x+1.5 =- căn(7/4) => x=-căn(7/4) -1.5

nhớ thanks bạn (+_+)

( 3x-1) ( x²+ 9) = (3x-1) (7x-10)

⇒( 3x-1) ( x²+ 9) - (3x-1) (7x-10) = 0

⇒( 3x-1) (( x²+ 9)-(7x-10)) = 0

⇒( 3x-1)(x²+9-7x+10)=0

⇒( 3x-1)(x²-7x+19)=0

⇒\(\left[{}\begin{matrix}3x-1=0\\x^2-7x+19=0\end{matrix}\right.\)

3x-1=0

⇒x=\(\dfrac{1}{3}\)

x²-7x+19=0

⇒ \(x^2-\dfrac{7}{2}x-\dfrac{7}{2}x+\left(\dfrac{7}{2}\right)^2+\dfrac{27}{4}=0\)

⇒ \(\left(x-\dfrac{7}{2}\right)^2+\dfrac{27}{4}=0\)

vì \(\left(x-\dfrac{7}{2}\right)^2\ge0\); \(\dfrac{27}{4}>0\)

⇒ \(\left(x-\dfrac{7}{2}\right)^2+\dfrac{27}{4}>0\)

⇒ x vô nghiệm

Vậy x= \(\dfrac{1}{3}\)

\(\left(3x-1\right)\left(x^2+9\right)=\left(3x-1\right)\left(7x-10\right)\\ \Leftrightarrow\left(3x-1\right)\left(x^2+9\right)-\left(3x-1\right)\left(7x-10\right)\\ \Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\\ \Leftrightarrow\left(3x-1\right)\left(x^2-4x-3x+12\right)=0\\ \Leftrightarrow\left(3x-1\right)\left[x\left(x-4\right)-3\left(x-4\right)\right]=0\\ \Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}3x-1=0\\x-3=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\x=3\\x=4\end{matrix}\right.\)

bạn ơi x^2-1 bạn nhé 

`@`` \text {Ans}`

`\downarrow`

Sửa lại đúng lớp đi c nhé.