Đặt \(\sqrt{x}=t\Rightarrow t^2=x\)

Mình nghĩ đề câu a là: \(x+\sqrt{5}+\sqrt{x}-1=-6\)

Đặt \(\sqrt{x}=t\Rightarrow t^2=x\)

\(Ta\)\(được\): \(t^2+\sqrt{5}+t-1=-6\)

\(\Leftrightarrow t^2-5+t+\sqrt{5}=0\)

\(\Leftrightarrow\left(t-\sqrt{5}\right).\left(t+\sqrt{5}\right)+\left(t+\sqrt{5}\right)=0\)

\(\Leftrightarrow\left(t+\sqrt{5}\right).\left(t-\sqrt{5}+1\right)=0\)

\(\Rightarrow\hept{\begin{cases}t=-\sqrt{5}\\t=\sqrt{5}-1\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=5\\x=6-2\sqrt{5}\end{cases}}\)

Gợi ý:

a) Đặt  \(x^2+3x+1=a\)

b)  \(\left(x^2+8x+7\right)\left(x+3\right)\left(x+5\right)+15\)

\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

Đặt     \(x^2+8x+11=a\)

c)  \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Đặt    \(x^2+7x+11=a\)

d) \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)

Đặt   \(12x^2+11x-1=a\)

Câu hỏi của Nguyễn Tấn Phát - Toán lớp 8 - Học toán với OnlineMath

Em tham khảo câu e nhé!

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........

a/ \(2\left(x^2-3x+2\right)=3\sqrt{x^3+8}\)

\(\Rightarrow2x^2-6x+4=3\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}\)

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

Chia 2 vế cho x² - 2x + 4 ta được:

\(\left(-2\right).\frac{x+2}{x^2-2x+4}+2=3\sqrt{\frac{x+2}{x^2-2x+4}}\)

Đặt \(a=\sqrt{\frac{x+2}{x^2-2x+4}}\left(a\ge0\right)\) ta được:

\(-2a^2-3a+2=0\Rightarrow\left(1-2a\right)\left(a+2\right)=0\Rightarrow\orbr{\begin{cases}a=\frac{1}{2}\left(n\right)\\a=-2\left(l\right)\end{cases}}\)

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

\(\Rightarrow x^2-6x-4=0\Rightarrow\orbr{\begin{cases}x=3+\sqrt{13}\\x=3-\sqrt{13}\end{cases}}\) (cái này tính denta là ra kết quả thôi)

                                                        Vậy có 2 nghiệm trên

câu b, c tương tự thôi

\(\Leftrightarrow4\left|x-2\right|=\left(x-2\right)^2+4\)

Đặt \(\left|x-2\right|=t\ge0\)

\(\Rightarrow4t=t^2+4\Rightarrow t^2-4t+4=0\)

\(\Rightarrow\left(t-2\right)^2=0\Rightarrow t=2\)

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

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)

a/ Đặt \(x^2+2x+1=\left(x+1\right)^2=t\ge0\)

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

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

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

Đặt \(x^2-x=t\Rightarrow t\left(t+1\right)-6=0\Rightarrow t^2+t-6=0\)

\(\Rightarrow\left[{}\begin{matrix}t=-3\\t=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2-x=-3\\x^2-x=2\end{matrix}\right.\)

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

\(\frac{x^2+5}{25-x^2}=\frac{3}{x+5}+\frac{x}{x-5}\)

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

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

\(\Rightarrow x^2+5=3\left(5-x\right)-x\left(5+x\right)\)

\(\Leftrightarrow x^2+5=15-3x-5x-x^2\)

\(\Leftrightarrow15-3x-5x-x^2-x^2-5=0\)

\(\Leftrightarrow10-8x-2x^2=0\)

\(\Leftrightarrow2x^2+8x-10=0\)

\(\Leftrightarrow2\left(x^2+4x-5\right)=0\)

\(\Leftrightarrow2\left(x^2+5x-x-5\right)=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}}\)