3x+1 + 3x+2 = (96 : 94) . 4
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\(\Leftrightarrow3^x\cdot82=3^{50}+3^{54}=3^{50}\cdot82\)
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
\(\dfrac{3}{5}+\dfrac{1}{6}.\dfrac{3}{4}=\dfrac{3}{5}+\dfrac{1}{8}=\dfrac{29}{40}\)
\(A=2x^3+6x^2-3x+\dfrac{1}{2}=2\cdot\dfrac{1}{3}^3+6\cdot\dfrac{1}{3}^2-3\cdot\dfrac{1}{3}+\dfrac{1}{2}\)
=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. \(\Leftrightarrow3x^2+3x-x-1=0\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\Leftrightarrow\left[\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)
=> 3x . 31 + 3x . 32 = 92 . 4
=> 3x . ( 31 + 32 ) = 81.4
=> 3x . 12 = 324
=> 3x = 324 : 12
=> 3x = 27
=> 3x = 33
=> x = 3
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