\(a,\\ \left(x+2\right)^2-x.\left(x-1\right)=10\\ \Leftrightarrow x^2+4x+4-x^2+x=10\\ \Leftrightarrow\left(x^2-x^2\right)+4x+x=10-4\\ \Leftrightarrow5x=6\\ \Leftrightarrow x=\dfrac{6}{5}\\ b,\\ x^3-6x^2+9x=0\\ \Leftrightarrow x.\left(x^2-6x+9\right)=0\\ \Leftrightarrow x.\left(x-3\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

\(5x^2-10x=5x.\left(x-2\right)\\ x^2-y^2-2x+2y=\left(x^2-y^2\right)-\left(2x-2y\right)\\ =\left(x-y\right).\left(x+y\right)-2.\left(x-y\right)\\ =\left(x+y-2\right).\left(x-y\right)\\ x^2+10x-y^2+25=\left(x^2+10x+25\right)-y^2\\ =\left(x+5\right)^2-y^2=\left(x+5-y\right).\left(x+5+y\right)\)

\(a,\\ \left(6x-7\right).\left(7x-1\right)=6x.7x-7x.7-6x.1-7.\left(-1\right)\\ =42x^2-49x-6x+7=42x^2-55x+7\\ b,\\ \left(4x-1\right)^2+\left(2x-5\right).\left(2x+5\right)=16x^2-8x+1+4x^2-25\\ =20x^2-8x-24\)

\(c,\\ \dfrac{x+5}{x}+\dfrac{x}{x-5}+\dfrac{25}{x^2-5x}\\ =\dfrac{\left(x-5\right).\left(x+5\right)}{x.\left(x-5\right)}+\dfrac{x.x}{x.\left(x-5\right)}+\dfrac{25}{x.\left(x-5\right)}\\ =\dfrac{x^2-25+x^2+25}{x.\left(x-5\right)}=\dfrac{2x^2}{x.\left(x-5\right)}=\dfrac{2x}{\left(x-5\right)}\left(ĐK:x\ne0;x\ne5\right)\)

Em muốn hỏi bài nào vậy?

dạ bài 3 ạ

Lời giải:
$A=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$

$A=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$

$=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)$

$=(2^8-1)(2^8+1)(2^{16}+1)=(2^{16}-1)(2^{16}+1)=2^{32}-1$

P/s: Lần sau bạn lưu ý ghi đầy đủ yêu cầu đề bài.

Lời giải:
a. ĐKXĐ: $x\neq \pm 1$

b.

\(A=\left[\frac{x+1}{(x-1)(x+1)}+\frac{x^2}{(x-1)(x+1)}-\frac{x(x-1)}{(x+1)(x-1)}\right].\frac{x+1}{2x+1}\)

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

c.

Khi $x=2$ thì $A=\frac{1}{2-1}=1$

x²+3x+3y+xy

=x(x+3)+y(3+x)

=(x+3)(x+y)

     x² + 3x + 3y + xy

=   ( x² + xy) + ( 3x + 3y)

= x( x + y) + 3 ( x + y)

= ( x + y) ( x + 3) 

Theo Bezout  đa thức  F(n) = 2n² + n - 18 chia hết cho đa thức n - 3 

⇔  F(3) ⋮ n- 3 ⇔ 2.3² + 3 - 18  ⋮ n - 3 ⇔ 3 ⋮ n - 3

n - 3 ⋮ Ư(3) = {  -3; -1; 1; 3} ⇔ n ϵ { 0; 2; 4; 6}

bạn có chép sai đề bài không ạ?

\(\dfrac{x^3\left(x-1\right)^3}{\left(x-1\right)^3}+\dfrac{x^3}{\left(x-1\right)^3}+\dfrac{3x^2\left(x-1\right)^2}{\left(x-1\right)^3}=28\)

ĐK: \(x\ne1\)

\(x^3+\dfrac{x^3}{\left(x-1\right)^3}+\dfrac{3x^2}{x-1}-28=0\)

\(x^3\left(x-1\right)^3+x^3+3x^2\left(x-1\right)^2-28\left(x-1\right)^3=0\)

\(\left(x^2-x\right)^3+3\left(x^2-x\right)^2+x^3-28\left(x^3-3x^2+3x-1\right)=0\)

\(\left(x^2-x\right)^3+3\left(x^2-x\right)^2+3\left(x^2-x\right)+1-\left(27x^3-81x^2+81x-27\right)=0\)

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

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

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

\(x=2\) (TMĐK)

Ta có:

+ ⇒ MTC = 2x( x + 3 )

Khi đó ta có: