\(\Leftrightarrow3^x\cdot82=3^{50}+3^{54}=3^{50}\cdot82\)

hay x=50

\(2+\dfrac{1}{2}-\dfrac{4}{3}+\left(-\dfrac{1}{3}\right)^4\)

=\(2+\dfrac{1}{2}-\dfrac{4}{3}+\dfrac{1}{81}\)

=\(\dfrac{5}{2}-\dfrac{4}{3}+\dfrac{1}{81}\)

=\(\dfrac{7}{6}+\dfrac{1}{81}\)

=\(\dfrac{191}{162}\)

\(2+\dfrac{1}{2}-\dfrac{4}{3}+\left(-\dfrac{1}{3}\right)^4\)

\(=\dfrac{5}{2}-\dfrac{4}{3}+\dfrac{1}{81}\)

\(=\dfrac{5}{2}-\dfrac{108}{81}+\dfrac{1}{81}\)

\(=\dfrac{5}{2}-\dfrac{107}{81}\)

\(=\dfrac{405}{162}-\dfrac{214}{162}=\dfrac{191}{162}\)

B = 2-4-6+8 + 10-12-14+16 + ... + 90-92-94+96 + 98 - 100

B =  (2-4-6+8) + (10-12-14+16) + ... + (90-92-94+96) + (98 - 100)

B = 0 + 0 + ... + 0 + (-2)

B = -2

hello mọi người

\(\dfrac{3}{5}+\dfrac{1}{6}.\dfrac{3}{4}=\dfrac{3}{5}+\dfrac{1}{8}=\dfrac{29}{40}\)

\(A=2x^3+6x^2-3x+\dfrac{1}{2}=2\cdot\dfrac{1}{3}^3+6\cdot\dfrac{1}{3}^2-3\cdot\dfrac{1}{3}+\dfrac{1}{2}\)

=13/54

1: (3x+2)(x+2)(2x-1)

=(3x^2+6x+2x+4)(2x-1)

=(3x^2+8x+4)(2x-1)

=6x^3-3x^2+16x^2-8x+8x-4

=6x^3+13x^2-4

2: (5x+1)(x-1)+3x(2x+2)

=5x^2-5x+x-1+6x^2+6x

=11x^2+10x-1

3: 4x(2x+1)(x-1)+(x+5)(x-3)

=4x(2x^2-2x+x-1)+x^2+2x-15

=8x^3-4x^2-4x+x^2+2x-15

=8x^3-3x^2-2x-15

4: (2x-1)(x+2)(x-2)+(3x-1)(x-1)

=(2x-1)(x^2-4)+3x^2-4x+1

=2x^3-8x-x^2+4+3x^2-4x+1

=2x^3+2x^2-12x+5

a. \(\Leftrightarrow\dfrac{x+2}{98}+1+\dfrac{x+4}{96}+1=\dfrac{x+6}{94}+1+\dfrac{x+8}{92}+1\)

\(\Leftrightarrow\dfrac{x+100}{98}+\dfrac{x+100}{96}=\dfrac{x+100}{94}+\dfrac{x+100}{92}\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\right)=0\)

\(\Leftrightarrow x+100=0\Leftrightarrow x=-100\)

c. \(\Leftrightarrow3x^2+3x-x-1=0\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\Leftrightarrow\left[\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)