Phương trình tích - Hỏi đáp

2c/

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

\(\Leftrightarrow\left(x^2-x\right)^2+3\left(x-\frac{1}{2}\right)^2+\frac{5}{4}=0\)

Phương trình vô nghiệm

d/

Nhận thấy \(x=1\) không phải là nghiệm, nhân 2 vế với \(x-1\)

\(\Leftrightarrow\left(x-1\right)\left(x^6+x^5+x^4+x^3+x^2+x+1\right)=0\)

\(\Leftrightarrow x^7-1=0\Rightarrow x=1\) (ktm)

Vậy pt vô nghiệm

2.

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=0\left(vn\right)\\x^2-x+1=0\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}=0\left(vn\right)\end{matrix}\right.\)

Vậy pt vô nghiệm

b.

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

\(\Leftrightarrow x\left(x^3+1\right)+x^3+1+x^2=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^3+1\right)+x^2=0\)

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

Mà \(\left\{{}\begin{matrix}\left(x+1\right)^2\left(x^2-x+1\right)\ge0\\x^2\ge0\end{matrix}\right.\)

Nên dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}x+1=0\\x=0\end{matrix}\right.\) ko tồn tại x thỏa mãn

đặt (x-2)^2 =t=> t>=0< => x^2 -4x =t-4

<=>[t -4)^2 +2t =43

đặt t-4 =y => y>=-4

<=> y^2 +2y =35

đặt y+1 =z => z>=-3

<=> z^2=36 => z=6 => y=5 =>t=9 =>|x-2| =3 => x=(-1;5)

Đặt \(x-3=t\)

\(\Rightarrow\left(t-1\right)^4+\left(t+1\right)^4-82=0\)

\(\Leftrightarrow2t^4+12t^2-80=0\)

\(\Rightarrow\left[{}\begin{matrix}t^2=4\\t^2=-10\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}t=2\\t=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)

Lời giải:

Lần sau bạn chú ý ghi đầy đủ đề bài.

Đặt $x^2+3=a$ thì PT trở thành:

$(a-3x)(a-2x)=2x^2$

$\Leftrightarrow a^2-5ax+4x^2=0$

$\Leftrightarrow (a^2-ax)-(4ax-4x^2)=0$

$\Leftrightarrow a(a-x)-4x(a-x)=0$

$\Leftrightarrow (a-x)(a-4x)=0$

Nếu $a-x=0$

$\Leftrightarrow x^2-x+3=0$

$\Leftrightarrow (x-\frac{1}{2})^2=\frac{-11}{4}< 0$ (vô lý- loại)

Nếu $a-4x=0$

$\Leftrightarrow x^2-4x+3=0$

$\Leftrightarrow (x-1)(x-3)=0$

$\Rightarrow x=1$ hoặc $x=3$

Vậy.......

Lời giải:
Đặt $x^2-2x+3=a$ thì PT trở thành:

$(a-x)a=2x^2$

$\Leftrightarrow a^2-ax-2x^2=0$

$\Leftrightarrow a^2-2ax+ax-2x^2=0$

$\Leftrightarrow a(a-2x)+x(a-2x)=0$

$\Leftrightarrow (a+x)(a-2x)=0$

Xét 2 TH:

TH1: $a+x=0$

$\Leftrightarrow x^2-x+3=0\Leftrightarrow (x-\frac{1}{2})^2=\frac{-11}{4}< 0$ (vô lý- loại)

TH2: $a-2x=0$

$\Leftrightarrow x^2-4x+3=0$

$\Leftrightarrow (x-1)(x-3)=0\Rightarrow x=1$ hoặc $x=3$

Vậy......

5.

PT \(\Leftrightarrow (x-1)(x-2)(2x-3)(2x-5)=30\)

\(\Leftrightarrow [(x-1)(2x-5)][(x-2)(2x-3)]=30\)

\(\Leftrightarrow (2x^2-7x+5)(2x^2-7x+6)=30\)

Đặt \(2x^2-7x+5=a\) thì:

PT \(\Leftrightarrow a(a+1)=30\)

\(\Leftrightarrow a^2+a-30=0\)

\(\Leftrightarrow (a-5)(a+6)=0\Rightarrow \left[\begin{matrix} a-5=0\\ a+6=0\end{matrix}\right.\)

Nếu \(a-5=0\Leftrightarrow 2x^2-7x=0\Leftrightarrow x(2x-7)=0\)

\(\Rightarrow \left[\begin{matrix} x=0\\ x=\frac{7}{2}\end{matrix}\right.\)

Nếu \(a+6=0\Leftrightarrow 2x^2-7x+11=0\)

\(\Leftrightarrow 2(x-\frac{7}{4})^2+\frac{39}{8}=0\Leftrightarrow 2(x-\frac{7}{4})^2=-\frac{39}{8}<0\) (vô lý)

Vậy...........

6.

PT\(\Leftrightarrow (x+1)(x+2)(x+3)(x+4)=24\)

\(\Leftrightarrow [(x+1)(x+4)][(x+2)(x+3)]=24\)

\(\Leftrightarrow (x^2+5x+4)(x^2+5x+6)=24\)

\(\Leftrightarrow (a+4)(a+6)=24\) (đặt \(x^2+5x=a\) )

\(\Leftrightarrow a^2+10a=0\Leftrightarrow a(a+10)=0\Rightarrow \left[\begin{matrix} a=0\\ a+10=0\end{matrix}\right.\)

Nếu \(a=0\Leftrightarrow x^2+5x=0\Leftrightarrow x(x+5)=0\Leftrightarrow \left[\begin{matrix} x=0\\ x=-5\end{matrix}\right.\)

Nếu \(a+10=0\Leftrightarrow x^2+5x+10=0\Leftrightarrow (x+\frac{5}{2})^2+\frac{15}{4}=0\)

\(\Leftrightarrow (x+\frac{5}{2})^2=-\frac{15}{4}\) (vô lý)

Vậy...........

a) \(\frac{4}{2x+3}+\frac{7}{3x-5}=0\)

ĐKXĐ: \(x\ne\frac{-3}{2};x\ne\frac{5}{3}\)

\(\Leftrightarrow\frac{4\left(3x-5\right)-7\left(2x+3\right)}{\left(3x-5\right)\left(2x+3\right)}=0\)

\(\Leftrightarrow12x-20-14x-21=0\)

\(\Leftrightarrow-2x-41=0\)

\(\Leftrightarrow x=\frac{-41}{2}\) thỏa mãn điều kiện

Vậy...

a/\(\left(4x-1\right)\left(x+5\right)=x^2-25\Leftrightarrow4x^2+20x-x-5=x^2-25\Leftrightarrow3x^2+19x+20\)\(\Leftrightarrow\left[{}\begin{matrix}\frac{-4}{3}\\-5\end{matrix}\right.\)

b/

\(2x^3-6x^2=x^2-3x\Leftrightarrow2x^3-6x^2-x^2+3x=0\Leftrightarrow2x^2\left(x-3\right)-x\left(x-3\right)=0\Leftrightarrow\left(2x^2-x\right)\left(x-3\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}2x^2-x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{1}{2}\\3\\0\end{matrix}\right.\)

c/\(x\left(x+3\right)^3-\frac{x}{4}\left(x+3\right)=0\Leftrightarrow\left(x+3\right)\left[\left(x+3\right)^2x-\frac{x}{4}\right]=0\Leftrightarrow\left(x+3\right)\left[\left(x^2+6x+9\right)x-\frac{x}{4}\right]=0\Leftrightarrow\left(x+3\right)\left(x^3+6x^2+9x-\frac{x}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^3+6x^2+\frac{35}{4}x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{5}{2}\\x=-\frac{7}{2}\end{matrix}\right.\)

d/\(\left(x-1\right)^2=\left(2x+5\right)^2\Leftrightarrow\left(x-1\right)^2-\left(2x+5\right)^2=0\Leftrightarrow\left(x-1+2x+5\right)\left(x-1-2x-5\right)=0\Leftrightarrow\left(3x+4\right)\left(-x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}3x+4=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{-4}{3}\\0\\-6\end{matrix}\right.\)

\(x^3+6x^2+3x-10=0\)

\(\Leftrightarrow x^3-x^2+7x^2-7x+10x-10=0\)

\(\Leftrightarrow x^2\left(x-1\right)+7x\left(x-1\right)+10\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+7x+10\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+2x+5x+10\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x\left(x+2\right)+5\left(x+2\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=-5\end{matrix}\right.\)

Vậy \(S=\left\{1;-2;-5\right\}\)

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

\(\Leftrightarrow x^3-x^2+5x^2-5x+6x-6=0\)

\(\Leftrightarrow x^2\left(x-1\right)+5x\left(x-1\right)+6\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+5x+6\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+2x+3x+6\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x+3\right)=0\)

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

Vậy \(S=\left\{1;-2;-3\right\}\)

\(\Leftrightarrow x^3+x^2-2x+5x^2+5x-10=0\)

\(\Leftrightarrow x\left(x^2+x-2\right)+5\left(x^2+x-2\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x^2+x-2\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x+2\right)\left(x-1\right)=0\)

b/ \(\Leftrightarrow x^3+5x^2+6x-x^2-5x-6=0\)

\(\Leftrightarrow x\left(x^2+5x+6\right)-\left(x^2+5x+6\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+5x+6\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x+3\right)=0\)