a)  x 2  + 2x + 1.            b) x + 3.         c)  x 2  – x + 1.

\(a,=\left(6x^3+3x^2-10x^2-5x+4x+2\right):\left(2x+1\right)\\ =\left(2x+1\right)\left(3x^2-5x+2\right):\left(2x+1\right)=3x^2-5x+2\\ b,=\left(x^4-2x^3+3x^2+x^3-2x^2+3x\right):\left(x^2-2x+3\right)\\ =\left(x^2-2x+3\right)\left(x^2+x\right):\left(x^2-2x+3\right)=x^2+x\)

Thu gọn, sắp xếp đa thức theo lũy thừa giảm của biến:

* Ta có: f(x) = x5 – 3x2 + x3 – x2 – 2x + 5

= x5 – (3x2 + x2 ) + x3 - 2x + 5

= x5 – 4x2 + x3 – 2x + 5

= x5 + x3 – 4x2 – 2x + 5

Và g(x) = x2 – 3x + 1 + x2 – x4 + x5

= (x2 + x2 ) – 3x + 1 – x4 + x5

= 2x2 – 3x + 1 – x4 + x5

= x5 – x4 + 2x2 – 3x + 1

* f(x) + g(x):

\(f\left(x\right)-g\left(x\right)=\left(x^5-3x^2+x^3-x^2-2x+5\right)-\left(x^2-3x+1+x^2-x^4+x^5\right)\)

\(f\left(x\right)-g\left(x\right)=x^5-3x^2+x^3-x^2-2x+5-x^2+3x-1-x^2+x^4-x^5\)

\(f\left(x\right)-g\left(x\right)=\left(x^5-x^5\right)+\left(-3x^2-x^2-x^2-x^2\right)+x^3+\left(-2x+3x\right)+\left(5-1\right)+x^4\)

\(f\left(x\right)-g\left(x\right)=-6x^2+x^3+x+4+x^4\)

\(f\left(x\right)-g\left(x\right)=x^4+x^3-6x^2+x+4\)

`@` `\text {Ans}`

`\downarrow`

^{*Máy tớ cam hơi mờ, cậu thông cảm ._.*}

Cậu viết lại rõ đề câu c, nhé.

Sắp xếp hai đa thức theo lũy thừa giảm dần của x rồi đặt phép chia. Thương tìm được là: 3 x 2  – 2x + 2.

\(a.x^2-11x+15=-15.\Leftrightarrow x^2-11x+30=0.\)

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

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

\(c.x^3-4=4.\Leftrightarrow x^3=8.\Leftrightarrow x^3=2^3.\Rightarrow x=2.\)

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

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

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

1: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\)

hay \(x\in\left\{3;\dfrac{1}{4}\right\}\)

2: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2x+16\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x^2+2x-16\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x-15\right)=0\)

hay \(x\in\left\{1;5\right\}\)

3: \(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)

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

hay \(x\in\left\{1;\dfrac{1}{2};-\dfrac{1}{2}\right\}\)

4: \(\Leftrightarrow x^2\left(x+4\right)-9\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-3\right)\left(x+3\right)=0\)

hay \(x\in\left\{-4;3;-3\right\}\)

5: \(\Leftrightarrow\left[{}\begin{matrix}3x+5=x-1\\3x+5=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-6\\4x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)

6: \(\Leftrightarrow\left(6x+3\right)^2-\left(2x-10\right)^2=0\)

\(\Leftrightarrow\left(6x+3-2x+10\right)\left(6x+3+2x-10\right)=0\)

\(\Leftrightarrow\left(4x+13\right)\left(8x-7\right)=0\)

hay \(x\in\left\{-\dfrac{13}{4};\dfrac{7}{8}\right\}\)

1.

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

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

\(\Leftrightarrow4x=5\Leftrightarrow x=\dfrac{5}{4}\)

2.

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

\(\Leftrightarrow x^2+x+1=x^2-2x+16\)

\(\Leftrightarrow3x=15\Leftrightarrow x=5\)

3.

\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)

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

7.

\(\Leftrightarrow x^2+2x-15=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

8.\(\Leftrightarrow x^4+x^3+4x^3+4x^2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0;x=-4\end{matrix}\right.\)

9.\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=\left(x-2\right)\left(3-2x\right)\)

\(\Leftrightarrow x+2=3-2x\)

\(\Leftrightarrow3x=1\Leftrightarrow x=\dfrac{1}{3}\)