( x² - 1)² - x( x² - 1) - 2x² = 0 ( 1 )

Đặt : x² - 1 = a , ta có :

( 1) ⇔ a² - ax - 2x² = 0

⇔ a² + ax - 2ax - 2x² = 0

⇔ a( a + x) -2x( a + x) = 0

⇔ ( a + x)( a - 2x ) = 0

TH1 : Với : a + x = 0

⇔ x² + x - 1 = 0

⇔ x² +\(2.\dfrac{1}{2}x+\dfrac{1}{4}-1-\dfrac{1}{4}=0\)

⇔ \(\left(x+\dfrac{1}{2}\right)^2-\dfrac{5}{4}\) = 0

⇔ \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{5}{4}\)

* ) \(x+\dfrac{1}{2}=\dfrac{\sqrt{5}}{2}\)

⇔ \(x=\dfrac{\sqrt{5}-1}{2}\)

*) \(x+\dfrac{1}{2}=-\dfrac{\sqrt{5}}{2}\)

⇔\(x=-\dfrac{\sqrt{5}+1}{2}\)

TH2 . a - 2x = 0

⇔ x² - 2x - 1 = 0

⇔ x² - 2x + 1 - 2 = 0

⇔ ( x - 1)² = 2

*) x - 1 = \(\sqrt{2}\)

⇔ x = \(\sqrt{2}\) + 1

*) x - 1 = - \(\sqrt{2}\)

⇔ x = 1 - \(\sqrt{2}\)

KL.....

p/s : Mk nghĩ zậy

\(\left\{{}\begin{matrix}u=x^2\\dv=cos2xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=2xdx\\v=\dfrac{1}{2}sin2x\end{matrix}\right.\)

\(\Rightarrow I=\dfrac{1}{2}x^2sin2x|^{\pi}_0-\int\limits^{\pi}_0x.sin2xdx\)

Bài 1:

1.

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

\(\Leftrightarrow (x^2-6x)^2-2(x^2-6x+9)+2=0\)

\(\Leftrightarrow (x^2-6x)^2-2(x^2-6x)-16=0\)

Đặt $x^2-6x=a$ thì pt trở thành:

$a^2-2a-16=0$

$\Leftrightarrow a=1\pm \sqrt{17}$

Nếu $a=1+\sqrt{17}$

$\Leftrightarrow x^2-6x=1+\sqrt{17}$

$\Leftrightarrow (x-3)^2=10+\sqrt{17}$

$\Rightarrow x=3\pm \sqrt{10+\sqrt{17}}$

Nếu $a=1-\sqrt{17}$

$\Rightarrow x=3\pm \sqrt{10-\sqrt{17}}$

Vậy.........

2.

$x^4-2x^3+x=2$

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

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

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

Thấy rằng $x^2-x+1=(x-\frac{1}{2})^2+\frac{3}{4}>0$ nên $(x-2)(x+1)=0$

$\Rightarrow x=2$ hoặc $x=-1$

Vậy.......

Bài 2:

1.

ĐKXĐ: $x\neq 1$. Ta có:

\(x^2+(\frac{x}{x-1})^2=8\)

\(\Leftrightarrow x^2+(\frac{x}{x-1})^2+\frac{2x^2}{x-1}=8+\frac{2x^2}{x-1}\)

\(\Leftrightarrow (x+\frac{x}{x-1})^2=8+\frac{2x^2}{x-1}\)

\(\Leftrightarrow (\frac{x^2}{x-1})^2=8+\frac{2x^2}{x-1}\)

Đặt $\frac{x^2}{x-1}=a$ thì pt trở thành:

$a^2=8+2a$

$\Leftrightarrow (a-4)(a+2)=0$

Nếu $a=4\Leftrightarrow \frac{x^2}{x-1}=4$

$\Rightarrow x^2-4x+4=0\Leftrightarrow (x-2)^2=0\Rightarrow x=2$ (tm)

Nếu $a=-2\Leftrightarrow \frac{x^2}{x-1}=-2$

$x^2+2x-2=0\Rightarrow x=-1\pm \sqrt{3}$ (tm)

Vậy........

2. ĐKXĐ: $x\neq 0; 2$

$(\frac{x-1}{x})^2+(\frac{x-1}{x-2})^2=\frac{40}{49}$

$\Leftrightarrow (\frac{x-1}{x}+\frac{x-1}{x-2})^2-\frac{2(x-1)^2}{x(x-2)}=\frac{40}{49}$

$\Leftrightarrow 4\left[\frac{(x-1)^2}{x(x-2)}\right]^2-\frac{2(x-1)^2}{x(x-2)}=\frac{40}{49}$

Đặt $\frac{(x-1)^2}{x(x-2)}=a$ thì pt trở thành:

$4a^2-2a=\frac{40}{49}$

$\Rightarrow 2a^2-a-\frac{20}{49}=0$

$\Rightarrow a=\frac{7\pm \sqrt{209}}{28}$

$\Leftrightarrow 1+\frac{1}{x(x-2)}=\frac{7\pm \sqrt{209}}{28}$

$\Leftrightarrow \frac{1}{x(x-2)}=\frac{-21\pm \sqrt{209}}{28}$

$\Rightarrow x(x-2)=\frac{28}{-21\pm \sqrt{209}}$

$\Rightarrow (x-1)^2=\frac{7\pm \sqrt{209}}{-21\pm \sqrt{209}}$.

Dễ thấy $\frac{7+\sqrt{209}}{-21+\sqrt{209}}< 0$ nên vô lý

Do đó $(x-1)^2=\frac{7-\sqrt{209}}{-21-\sqrt{209}}$

$\Leftrightarrow x=1\pm \sqrt{\frac{7-\sqrt{209}}{-21-\sqrt{209}}}$

Vậy........

Bài 3:

ĐKXĐ: $x\neq 0$

PT $\Leftrightarrow (x-\frac{1}{x})^2+2=4-(x-\frac{1}{x})$

Đặt $x-\frac{1}{x}=a$ thì pt trở thành:

$a^2+2=4-a$

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

$\Leftrightarrow (a-1)(a+2)=0\Rightarrow a=1$ hoặc $a=-2$

Nếu $a=1\Leftrightarrow x^2-1=x$

$\Leftrightarrow x^2-x-1=0\Rightarrow x=\frac{1\pm \sqrt{5}}{2}$

Nếu $a=-2\Leftrightarrow x^2-1=-2x$

$\Leftrightarrow x^2+2x-1=0\Rightarrow x=-1\pm \sqrt{2}$

Vậy............

2. ĐKXĐ: $x\neq 0$

\(x^2+\frac{1}{4x^2}=2x-\frac{1}{x}+1\)

$\Rightarrow 4x^2+\frac{1}{x^2}=8x-\frac{4}{x}+4$

$\Rightarrow (2x-\frac{1}{x})^2+4=4(2x-\frac{1}{x})+4$

Đặt $2x-\frac{1}{x}=a$ thì pt trở thành:

$a^2+4=4a+4$

$\Leftrightarrow a(a-4)=0\Rightarrow a=0$ hoặc $a=4$

Nếu $a=0\Leftrightarrow 2x-\frac{1}{x}=0$

$\Rightarrow 2x=\frac{1}{x}\Rightarrow x=\pm \sqrt{\frac{1}{2}}$

Nếu $a=4\Rightarrow 2x^2-4x-1=0\Rightarrow x=\frac{2\pm \sqrt{6}}{2}$

Bài 1:

Để \(x^2-\left(m+1\right)x+4⋮x-1\Leftrightarrow4-m=0\Leftrightarrow m=4\)

Bài 2:

Sửa đề bài chút: g(x) = x^2 -3x +2

Để f(x) chia hết g(x) <=> a - 4 = 0 <=> a = 4

g: \(3\left(x-y\right)-5x\left(y-x\right)=\left(x-y\right)\left(5x+3\right)\)

f: \(4x^2\left(x+1\right)+2x^2\left(x+1\right)\)

\(=6x^2\left(x+1\right)\)

f: \(4x^2\left(x+1\right)+2x^2\left(x+1\right)=6x^2\left(x+1\right)\)

g: \(3\left(x-y\right)-5x\left(y-x\right)=\left(x-y\right)\left(5x+3\right)\)

câu f có ( x+1) là nhân tử chung

câu g đổi dấu - thành + thì (y-x) sẽ thành (x-y)

e: \(8\left(x+3y\right)-16x\left(x+3y\right)=8\left(x+3y\right)\left(1-2x\right)\)

f: \(4x^2\left(x+1\right)+2x^2\left(x+1\right)=6x^2\left(x+1\right)\)