(3x-1) (x² +2) = (3x-1)(7x-10)

=> (3x-1) (x²+2)-(3x-1)(7x-10)=0

=>(3x-1)(x²+2-7x+10)=0

=>(3x-1)(x²-7x+12)=0

=>(3x-1)(x-3)(x-4)=0

=>3x-1=0 => x= 1/3

    x-3=0 =>  x=3

    x-4=0 =>  x=4

vậy pt có tập nghiệm S={ 1/3; 3; 4}

mk lam xong roi ban moi giai

vg còn bài nào ko ak

Câu 1: ĐKXĐ: ...

\(\Leftrightarrow4x\left(3x-1\right)+x-1=4x\sqrt{3x+1}\)

\(\Leftrightarrow12x^2-3x-1-4x\sqrt{3x+1}=0\)

\(\Leftrightarrow16x^2-\left(4x^2+4x\sqrt{3x+1}+3x+1\right)=0\)

\(\Leftrightarrow16x^2-\left(2x+\sqrt{3x+1}\right)^2=0\)

\(\Leftrightarrow\left(2x-\sqrt{3x+1}\right)\left(6x+\sqrt{3x+1}\right)=0\)

\(\Leftrightarrow...\)

Câu 2:

\(\Leftrightarrow\left\{{}\begin{matrix}x\left(x^2-4\right)=y^3+2y\\x^2-4=-3y^2\end{matrix}\right.\)

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

\(\Leftrightarrow y\left(y^2+3xy+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=0\Rightarrow...\\y^2+3xy+2=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow3xy=-y^2-2\Rightarrow x=\frac{-y^2-2}{3y}\)

\(\Rightarrow\left(\frac{y^2+2}{3y}\right)^2-1=3\left(1-y^2\right)\)

\(\Leftrightarrow\left(\frac{y^2-3y+2}{3y}\right)\left(\frac{y^2+3y+2}{3y}\right)=3\left(1-y^2\right)\)

\(\Leftrightarrow\frac{\left(y-1\right)\left(y-2\right)\left(y+1\right)\left(y+2\right)}{9y^2}=3\left(1-y^2\right)\)

\(\Leftrightarrow\frac{\left(y^2-1\right)\left(y^2-4\right)}{9y^2}=3\left(1-y^2\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}y^2-1=0\\\frac{y^2-4}{9y^2}=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y^2-1=0\\28y^2=4\end{matrix}\right.\)

\(3x-1+\frac{x-1}{4x}=\sqrt{3x+1}\)

\(\Leftrightarrow\frac{4x\left(3x-1\right)+x-1}{4x}=\sqrt{3x+1}\)

\(\Leftrightarrow\frac{12x^2-4x+x-1}{4x}=\sqrt{3x+1}\)

\(\Leftrightarrow\frac{12x^2-3x-1}{4x}=\sqrt{3x+1}\)

\(\Leftrightarrow\frac{\left(12x^2-3x-1\right)^2}{16x^2}=3x+1\)

\(\Leftrightarrow\left(12x^2-3x-1\right)^2=16x^2\left(3x+1\right)\)

\(\Leftrightarrow144x^4-120x^3-31x^2+6x+1=0\)

\(\Leftrightarrow144x^4-144x^3+24x^3-24x^2-7x^2+7x-x+1=0\)

\(\Leftrightarrow144x^3\left(x-1\right)+24x^2\left(x-1\right)+7x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(144x^3+24x^2+7x-1\right)=0\)

Tìm được mỗi nghiệm thôi à :v

a, ĐKXĐ:...

\(\sqrt{5x+10}=8-x\\ \Leftrightarrow5x+10=64-16x+x^2\\ \Leftrightarrow x^2-21x+54=0\)

.....

b, ĐKXĐ:...

\(\sqrt{4x^2+x-12}=3x-5\\ \Leftrightarrow4x^2+x-12=9x^2-30x+25\\ \Leftrightarrow5x^2-31x+37=0\)

.....

ta có pt

<=> \(2\left(2x+1\right)\sqrt{x+8}=4x^2+4x+1+x+8-x^2+2x-1\)

\(\Leftrightarrow2\left(2x+1\right)\sqrt{x+8}=\left(2x+1\right)^2+x+8-\left(x-1\right)^2\)

\(\Leftrightarrow\left(2x+1\right)^2-2\left(2x+1\right)\sqrt{x+8}+x+8-\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(2x+1-\sqrt{x+8}\right)^2-\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(2x+1-\sqrt{x+8}+x-1\right)\left(2x+1-\sqrt{x+8}-x+1\right)=0\)

\(\Leftrightarrow\left(3x-\sqrt{x+8}\right)\left(x+2-\sqrt{x+8}\right)=0\)

đến đây thì dễ rồi nhé

\(x^4-2x^3+3x^2-2x+1=0\)

Chia cả hai vé cho \(x^2\)

\(\Leftrightarrow x^2-2x+3-\dfrac{2}{x}+\dfrac{1}{x^2}\)

\(\Leftrightarrow x^2+2+\dfrac{1}{x^2}-2\left(x+\dfrac{1}{x}\right)+1=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)^2-2\left(x+\dfrac{1}{x}\right)+1=0\)

Đặt x+1/x = a, ta có:

\(a^2-2a+1=0\)

\(\Leftrightarrow\left(a-1\right)^2=0\)

\(\Leftrightarrow a=1\)

\(\Leftrightarrow x+\dfrac{1}{x}=1\)

\(\Leftrightarrow x^2+1=x\)

\(\Leftrightarrow x^2-x+1=0\)

\(\Leftrightarrow x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\)

Do \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+3>0\)

Do đó phương trình vô nghiệm