chị cả

ĐKXĐ : \(\left\{{}\begin{matrix}4x^2-1\ne0\\8x^3+1\ne0\end{matrix}\right.\Leftrightarrow x\ne\pm\dfrac{1}{2}\)

\(P=\dfrac{2x^5-x^4-2x+1}{4x^2-1}+\dfrac{8x^2-4x+2}{8x^3+1}\)

\(=\dfrac{\left(x^4-1\right)\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}+\dfrac{2\left(4x^2-2x+1\right)}{\left(2x+1\right)\left(4x^2-2x+1\right)}\)

\(=\dfrac{x^4-1}{2x+1}+\dfrac{2}{2x+1}=\dfrac{x^4+1}{2x+1}\)

https://sg.docworkspace.com/l/sIM-LioBEocfloAY

a) \(A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2x+10}{\left(x+5\right)\left(x-5\right)}\)

\(A=\dfrac{x-5+2x+10-2x-10}{\left(x+5\right)\left(x-5\right)}=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}=\dfrac{1}{x+5}\)

b) \(A=-3\Rightarrow\dfrac{1}{x+5}=-3\)

\(\Leftrightarrow x+5=-\dfrac{1}{3}\Leftrightarrow x=-\dfrac{1}{3}-5=\dfrac{-16}{3}\)

\(9x^2-42x+49=\left(3x-7\right)^2=\left(3.\dfrac{-16}{3}-7\right)^2=\left(-23\right)^2=529\) \(\left(x=\dfrac{-16}{3}\right)\)

a, Với \(5-x\ge0\Leftrightarrow x\le5\)

Biểu thức có dạng : \(P=5-x+3x-2=3+2x\)(*)

Với \(5-x< 0\Leftrightarrow x>5\)

Biểu thức có dạng : \(P=x-5+3x-2=4x-7\)(**)

b, Với biểu thức (*) \(P=3+2x=6\Leftrightarrow x=\frac{3}{2}\)

Với biểu thức (**) \(P=4x-7=6\Leftrightarrow x=\frac{13}{4}\)

a)B = ( 2 + 4 + 6 + 8 +........+ 2014 ) - ( 3 + 5 + 7 + 9 +.......+ 2011 )

= 1015056 - 1012035

= 3021

Mk nhanh nhất đó 

Cảm ơn nha nguyen thi lan huong

3021

ai tk mk

mk tk lại

mk hứa

yên tâm

thank nhiều

\(a,5\left(x-2\right)\left(x+2\right)-5.x^{28}\)

\(=\left(5x-10\right)\left(x+2\right)-5x^{28}\)

\(=5x^2+10x-10x-20-5x^{28}\)

\(=5x^2-2x-5x^{28}\)

\(b,\left(a-b\right)^2+4ab\)

\(=a^2-2ab+b^2+4ab\)

\(=a^2+2ab+b^2\)

a) 5(x - 2)(x + 2) - 5.x²⁸

= 5(x² - 4) - 5.x²⁸

= 5.x² - 5.4 - 5.x²

= 5x² - 20 - 5x²⁸ 

= -5x²⁸ + 5x² - 20

b) (a - b)² + 4ab

= a² - 2ab + b² + 4ab

= a² - 2ab + b²

\(a,5\left(x-2\right)\left(x+2\right)-5x^{28}\)

\(=\left(5x-10\right)\left(x+2\right)-5x^{28}\)

\(=5x^2+10x-10x-20-5x^{28}\)

\(=5x^2-2x-5x^{28}\)

\(b,\left(a-b\right)^2+4ab\)

\(=a^2-2ab+b^2+4ab\)

\(=a^2+2ab+b^2\)

câu 2

\(...=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2-\sqrt{5}\right|-\left|2+\sqrt{5}\right|=-4\)

câu 1

\(P=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{1}{\sqrt{x}}\right)\)

\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)

\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

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

\(P< -1\Leftrightarrow\frac{-3\sqrt{x}}{2\sqrt{x}+4}+1< 0\Leftrightarrow-\sqrt{x}+4< 0\Leftrightarrow\sqrt{x}>4\Leftrightarrow x>16\)

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

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

=4

a,M=2^0-2^1+2^2-2^3+2^4-2^5+.....+2^2012

2M=2^1-2^2+2^3-2^4+2^5-2^5+......-2^2012+2^2013

3M=2^0+2^2013

M=(2^0+2^2013)÷3

Vậy.......

b,N=3-3^2+3^3-3^4+3^5-3^6+.....+3^2011-3^2012

3N=3^2-3^3+3^4-3^5+3^6-3^7+......+3^2012-3^2013

4N=3-3^2013

N=(3-3^2013)÷4

Vậy........

K tao nhé ko lên lớp tao đánh m😈😈😈

Bt dễ thế mà ko làm dc😂😂😂😂😂

phần c đâu