5x(x-2011)-x+2011=0

5x(x-2011)-(x-2011)=0

(5x-1)(x-2011)=0

TH1 : 5x-1= 0

          5x=1

          x=1/5

TH2 : x-2011=0

          x=2011

\(\left(x-1\right)^2+x\left(5-x\right)=0\)

\(x^2-2x+1+5x-x^2=0\)

\(3x=-1\)

\(\Rightarrow x=-\frac{1}{3}\)

\(\left(x-1\right)^2+x\left(5-x\right)=0\)

\(\Rightarrow x^2-2x+1+5x-x^2=0\)

\(\Rightarrow-2x+1+5x=0\)

\(\Rightarrow3x=-1\)

\(\Rightarrow x=-\frac{1}{3}\)

\(\frac{1}{x-1}-\frac{1}{x+1}-\frac{2}{x^2+1}-\frac{4}{x^4+1}-\frac{8}{x^5+1}-\frac{16}{x^{16}+1}\)

\(=\frac{x+1-x+1}{\left(x+1\right)\left(x-1\right)}-\frac{2}{x^2+1}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)

\(=\frac{2}{x^2-1}-\frac{2}{x^2+1}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)

\(=\frac{2\left(x^2+1\right)-2.\left(x^2-1\right)}{x^2-1}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)

\(=\frac{2x^2+2-2x^2+2}{\left(x^2+1\right)\left(x^2-1\right)}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)

\(=\frac{4}{x^4-1}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)

\(=\frac{4\left(x^4+1\right)-4\left(x^4-1\right)}{\left(x^4-1\right)\left(x^4+1\right)}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)

\(=\frac{8}{x^8-1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)

\(=\frac{8.\left(x^8+1\right)-8\left(x^8-1\right)}{\left(x^8-1\right)\left(x^8+1\right)}-\frac{16}{x^{16}+1}\)

\(=\frac{16}{x^{16}-1}-\frac{16}{x^{16}+1}\)

\(=\frac{16.\left(x^{16}+1\right)-16.\left(x^{16}-1\right)}{\left(x^{16}-1\right)\left(x^{16}+1\right)}\)

\(=\frac{32}{x^{32}-1}\)

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

Dễ thấy đây là hằng đẳng thức thứ hai với 5x + 3 là A và x + 3 là B

Do đó : \(\left(5x+3\right)^2-2\left(5x+3\right)\left(x+3\right)+\left(x+3\right)^2\)

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

\(=\left(4x\right)^2\)

\(=16x^2\)

\(x^2-xy-2x+2y.\)

\(=\left(x^2-xy\right)-\left(2x-2y\right)\)

\(=x\left(x-y\right)-2\left(x-y\right)\)

\(=\left(x-y\right)\left(x-2\right)\)

Hok tốt ạ

x² - xy - 2x + 2y

=(x² - 2x) - (xy - 2y)

=x(x - 2) - y(x - 2)

=( x - y )(x - 2)

\(\frac{x^2+y^2}{xy}=\frac{25}{12}\)

\(\Rightarrow12x^2+12y^2=25xy\)

\(\Rightarrow12x^2+12y^2+24xy=49xy\)

\(\Rightarrow12\left(x^2+2xy+y^2\right)=49xy\)

\(\Rightarrow\left(x+y\right)^2=\frac{49xy}{12}\)

\(\Rightarrow x+y=\sqrt{\frac{49xy}{12}}\)

Lại có :\(12\left(x^2-2xy+y^2\right)=xy\)

\(\Rightarrow x-y=\sqrt{\frac{xy}{12}}\)

\(\Rightarrow A=\sqrt{\frac{\frac{xy}{12}}{\frac{49xy}{12}}}\)

\(\Rightarrow A=\sqrt{\frac{1}{49}}=\pm\frac{1}{7}\)

Phạm Tuấn Đạt Chỉ kiến thức lớp 7 là đủ rồi bạn ey!À mà \(\sqrt{\frac{1}{49}}=-\frac{1}{7}???\) không có căn bậc 2 của số âm nha bạn!

\(\frac{x^2+y^2}{xy}=\frac{25}{12}\Leftrightarrow\frac{x^2+y^2}{25}=\frac{xy}{12}\)

Đặt \(\frac{x^2+y^2}{25}=\frac{xy}{12}=k\Rightarrow x^2+y^2=25k;xy=12k\)

\(A^2=\frac{\left(x-y\right)^2}{\left(x+y\right)^2}=\frac{x^2-2xy+y^2}{x^2+2xy+y^2}=\frac{25k-2.12k}{25k+2.12k}=\frac{25k-24k}{25k+24k}=\frac{1k}{49k}=\frac{1}{49}\)

\(\Rightarrow A=\sqrt{\frac{1}{49}}=\frac{1}{7}\)

3(x+3) - x² + 9

= 3(x +3) - ( x² - 9 )

=3(x+3) - (x² - 3²)

= 3(x + 3) - (x - 3)(x+3)

= (x+3)(3 - x + 3)

= x(x+3)

Chúc học tốt^^