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

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

Đáp án C

Mấy cái này chỉ đơn giản là sử dụng các phép biến đổi đơn giản của biểu thức chứa căn bậc hai thôi nên bạn chú ý xem lại các bài trong SGK là làm được rồi! Chúc bạn học tốt nhé!

a) \(x^2.\sqrt{7}=\frac{\sqrt{6363}}{\sqrt{707}}=\sqrt{9}=3\)

\(\Rightarrow x^2=\frac{3}{\sqrt{7}}=\frac{3\sqrt{7}}{7}\Rightarrow x=\)\(\left[{}\begin{matrix}-\sqrt{\frac{3\sqrt{7}}{7}}\\\sqrt{\frac{3\sqrt{7}}{7}}\end{matrix}\right.\)

b) ĐKXĐ: \(x\ge\frac{1}{2}\)

PT \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2=2x-1\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\end{matrix}\right.\)(thỏa mãn)

\(\left\{{}\begin{matrix}x+y=3m\\x-2y=-3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=m+1\\x=2m-1\end{matrix}\right.\)

\(x^2+xy=30\)

\(\Leftrightarrow\left(2m-1\right)^2+\left(2m-1\right)\left(m+1\right)-30=0\)

\(\Leftrightarrow6m^2-3m-30=0\Rightarrow\left[{}\begin{matrix}m=-2\\m=\frac{5}{2}\end{matrix}\right.\)

Bài 2:

\(a+b+c=1-m+m-1=0\Rightarrow\left\{{}\begin{matrix}x_1=1\\x_2=m-1\end{matrix}\right.\)

a/ TH1: \(x_1=2x_2\Rightarrow1=2\left(m-1\right)\Rightarrow m=\frac{3}{2}\)

Th2: \(x_2=2x_1\Rightarrow m-1=2\Rightarrow m=3\)

b/ \(A=x_1^2+x_2^2-6x_1x_2\)

\(A=1+\left(m-1\right)^2-6\left(m-1\right)=8\)

\(\Rightarrow\left(m-1\right)^2-6\left(m-1\right)-7=0\)

\(\Rightarrow\left[{}\begin{matrix}m-1=-1\\m-1=7\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}m=0\\m=8\end{matrix}\right.\)

\(A=\left(m-1\right)^2-6\left(m-1\right)+1=\left(m-1\right)^2-6\left(m-1\right)+9-8\)

\(A=\left(m-1-3\right)^2-8=\left(m-4\right)^2-8\ge-8\)

\(\Rightarrow A_{min}=-8\) khi \(m=4\)

f) \(4x\left(x+1\right)=8\left(x+1\right)\)

\(\Leftrightarrow4x\left(x+1\right)-8\left(x+1\right)=0\)

\(\Leftrightarrow4\left(x-2\right)\left(x+1\right)=0\)

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

h) \(x^2-4x=0\)

\(\Leftrightarrow x\left(x-4\right)=0\)

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

i) \(2x\left(x-2\right)-\left(2-x\right)^2=0\)

\(\Leftrightarrow2x\left(x-2\right)-\left(x-2\right)^2=0\)

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

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

\(\Leftrightarrow x=\pm2\)

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

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

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

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

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

l) \(\left(x-3\right)^3+3-x=0\)

\(\Leftrightarrow\left(x-3\right)^3-\left(x-3\right)=0\)

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

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

m) \(x+6x^2=0\)

\(\Leftrightarrow x\left(1+6x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{6}\end{matrix}\right.\)

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

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

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

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

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

f ) \(4x\left(x+1\right)=8\left(x+1\right)\)

\(\Leftrightarrow4x\left(x+1\right)-8\left(x+1\right)=0\)

\(\Leftrightarrow4\left(x-2\right)\left(x+1\right)=0\)

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

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

Vậy ...

h ) \(x^2-4x=0\)

\(\Leftrightarrow x\left(x-4\right)=0\)

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

Vậy ...

I ) \(2x\left(x-2\right)-\left(2-x\right)^2=0\)

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

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

\(\Leftrightarrow\left(-2-x\right)\left(2-x\right)=0\)

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

Vậy ...

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

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

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

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

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

Vậy ...

i ) \(\left(x-3\right)^3+3-x=0\)

\(\Leftrightarrow\left(x-3\right)^3-\left(x-3\right)=0\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\\x=2\end{matrix}\right.\)

Vậy ...

m ) \(x+6x^2=0\)

\(\Leftrightarrow x\left(1+6x\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{6}\end{matrix}\right.\)

Vậy ...

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

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

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

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

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

Vậy ...

1.a)\(2a^2b\left(x+y\right)-4a^3b\left(-x-y\right)\)

\(=2a^2b\left(x+y\right)+4a^3b\left(x+y\right)\)

\(=2a^2b\left(1+2a\right)\left(x+y\right)\)

b)\(x^{m+1}+x^m\)

\(=x^m.x+x^m\)

\(=x^m\left(x+1\right)\)

c)\(x^{m+2}-x^{m+1}\)

\(=x^{m+1}.x-x^{m+1}\)

\(=x^{m+1}\left(x-1\right)\)

a, Đặt (x² +x ) = t ta có:

=> t² + 4t - 12 = 0

=> ( t + 2)² - 16 = 0

=> ( t + 2)² - 4² = 0

=> ( t -2)( t + 6) = 0

=>\(\left[{}\begin{matrix}t-2=0\\t+6=0\end{matrix}\right.\)

Thay t = x² + x

- x² + x -2 = 0 => (x+2)(x-1) = 0 => \(\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)

- x² + x + 6 = 0 => (x+3)(x-2) = 0 => \(\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)