a: \(A=\dfrac{x+2}{x}\cdot\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x-2}\)

b: \(B=\dfrac{2x-6+3x+9-5x+2}{\left(x-3\right)\left(x+3\right)}=\dfrac{5}{x^2-9}\)

e) Ta có: \(E=\left(3x+2\right)\left(3x-5\right)\left(x-1\right)\left(9x+10\right)+24x^2\)

\(=\left(9x^2-15x+6x-10\right)\left(9x^2+10x-9x-10\right)+24x^2\)

\(=\left(9x^2-10-9x\right)\left(9x^2-10+x\right)+24x^2\)

\(=\left(9x^2-10\right)^2-8x\left(9x^2-10\right)-9x^2+24x^2\)

\(=\left(9x^2-10\right)^2-8x\left(9x^2-10\right)+15x^2\)

\(=\left(9x^2-10\right)^2-3x\left(9x^2-10\right)-5x\left(9x^2-10\right)+15x^2\)

\(=\left(9x^2-10\right)\left(9x^2-3x-10\right)-5x\left(9x^2-10-3x\right)\)

\(=\left(9x^2-3x-10\right)\left(9x^2-5x-10\right)\)

\(a,\Leftrightarrow9x^2=-36\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)=0\\ \Leftrightarrow\left(3-x\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x^2-x-2x^2+3x+2=0\\ \Leftrightarrow2x=-2\Leftrightarrow x=-1\\ d,\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\\ \Leftrightarrow-3\left(4x-3\right)=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ e,\Leftrightarrow\dfrac{1}{3}x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ f,\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Đặt \(A=1\cdot2+2\cdot3+\cdots+99\cdot100\)

\(=1\left(1+1\right)+2\left(2+1\right)+\cdots+99\left(99+1\right)\)

\(=\left(1^2+2^2+\cdots+99^2\right)+\left(1+2+\cdots+99\right)\)

\(=\frac{99\left(99+1\right)\left(2\cdot99+1\right)}{6}+\frac{99\cdot\left(99+1\right)}{2}\)

\(=\frac{99\cdot100\cdot199}{6}+\frac{99\cdot100}{2}=\frac{99\cdot100\cdot199+3\cdot99\cdot100}{6}\)

\(=\frac{99\cdot100\cdot202}{6}=33\cdot100\cdot101=333300\)

Ta có: \(x^2+\left(x^2+1\right)+\left(x^2+2\right)+\cdots+\left(x^2+99\right)\)

\(=\left(x^2+0\right)+\left(x^2+1\right)+\cdots+\left(x^2+99\right)\)

\(=\left(x^2+x^2+\cdots+x^2\right)+\left(0+1+\cdots+99\right)=100x^2+\frac{99\cdot100}{2}=100x^2+99\cdot50=100x^2+4950\)

Ta có: \(\frac{1\cdot2+2\cdot3+\cdots+99\cdot100}{x^2+\left(x^2+1\right)+\left(x^2+2\right)+\cdots+\left(x^2+99\right)}=50\frac{116}{131}\)

=>\(\frac{333300}{100x^2+4950}=50+\frac{116}{131}\)

=>\(\frac{6666}{2x^2+99}=\frac{6666}{131}\)

=>\(2x^2+99=131\)

=>\(2x^2=32\)

=>\(x^2=16\)

=>\(\left[\begin{array}{l}x=4\\ x=-4\end{array}\right.\)

3^xx2^x=36

3^xx2^x=6²

6^x       =6²

x      =2

Vậy x = 2

a) (x-y)²-(x²-2xy)

=y²-2xy+x²-x²+2xy

=y²-(-2xy+2xy)+(x²-x²)

=y²

b)(x-y)²+x²+2xy-(x+y)²

=y²-2xy+x²+x²+2xy-y²-2xy-x²

=(y²-y²)-(2xy+2xy-2xy)+(x²+x²-x²)

=x²-2xy

xét delta phẩy có

1+1-m = 2-m vậy điều kiện để phương trình có 2 nghiệm x1;x2 là m ≤2 

theo Vi-ét ta có:

\(\left\{{}\begin{matrix}x1+x2=2\\x1x2=m-1\end{matrix}\right.\)

theo bài ra ta có: 

2x1 + x2 = 5 

x1 + 2 = 5 => x1 = 3 => x2 = -1 

ta có x1x2 = m - 1 => m - 1 = -3 

=> m = -2 vậy m = -2 để phương trình có 2 nghiệm x1;x2 thỏa mãn 2x1 + x2 = 5.

a: ĐKXĐ: \(x\notin\left\{-3;2\right\}\)

b: \(A=\dfrac{x^2-4-5+x+3}{\left(x-2\right)\left(x+3\right)}=\dfrac{x^2+x-6}{\left(x-2\right)\left(x+3\right)}=\dfrac{x+2}{x-2}\)

c: Để A=3/4 thì 4x-8=3x+6

=>x=14

d: Để A nguyên thì \(x-2\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(x\in\left\{3;1;4;0;6;-2\right\}\)