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\(2\cdot\left(2x-6\right)+\left(x-1\right)=2\)
\(\Leftrightarrow4x-12+x-1-2=0\)
\(\Leftrightarrow5x-15=0\)
\(\Leftrightarrow5x=15\)
\(\Leftrightarrow x=3\)
\(3\left(2x-6\right)-4\left(1+2x\right)-2\left(x-4\right)=4-3\left(1+2x\right)-5\left(1-2x\right).\)
\(\Leftrightarrow6x-18-4-8x-2x+8=4-3-6x-5+10x\)
\(\Leftrightarrow-4x-14=4x-4\)
\(\Leftrightarrow-4x-4x=-4+14\)
\(\Leftrightarrow-8x=10\)
\(\Leftrightarrow x=-\frac{5}{4}\)
Câu 1:
\(\Leftrightarrow6x-18-8x-4-2x+8=4-3\left(2x+1\right)+5\left(2x-1\right)\)
=>-4x-14=4-6x-3+10x-5
=>-4x-14=4x-4
=>-8x=10
hay x=-5/4
\(x-5=\left(-1\frac{11}{37}\right).4\frac{3}{7}-\left(1,5-6\frac{1}{3}.\frac{2}{19}\right)\)
\(x-5=-\frac{42}{31}.\frac{31}{7}-\left(\frac{3}{2}-\frac{19}{3}.\frac{2}{19}\right)\)
\(x-5=-6-\left(\frac{3}{2}-\frac{2}{3}\right)=-6-\left(\frac{9}{6}-\frac{4}{6}\right)=-\frac{36}{6}-\frac{5}{6}=-\frac{41}{6}\)
\(x=-\frac{41}{6}+5=-\frac{41}{6}+\frac{30}{6}=-\frac{11}{6}\)
(2x - 15)⁵ = (2x - 15)³
(2x - 15)⁵ - (2x - 15)³ = 0
(2x - 15)³.[(2x - 15)² - 1] = 0
(2x - 15)³.[(2x - 15)(2x - 15) - 1] = 0
(2x - 15)³.(4x² - 30x - 30x + 225 - 1) = 0
(2x - 15)³.(4x² - 60x + 225 - 1) = 0
(2x - 15)³.(4x² - 60x + 224) = 0
4.(2x - 15)³.(x² - 15x + 56) = 0
4.(2x - 15)³.(x² - 7x - 8x + 56) = 0
4.(2x - 15)³.[(x² - 7x) - (8x - 56)] = 0
4.(2x - 15)³.[x(x - 7) - 8(x - 7)] = 0
4.(2x - 15)³.(x - 7)(x - 8) = 0
(2x - 15)³ = 0 hoặc x - 7 = 0 hoặc x - 8 = 0
*) (2x - 15)³ = 0
2x - 15 = 0
2x = 15
x = 15/2
*) x - 7 = 0
x = 7
*) x - 8 = 0
x = 8
Vậy x = 7; x = 15/2; x = 8
tớ nè
x = 2
(2x+1)^3=125
=>(2x+1)=5 =>5^3=125
=>x=2