A, bạn ơi mình biết làm hết r

B,cài này dễ lằm bạn nghĩ đi okeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

a: \(\Leftrightarrow\left[{}\begin{matrix}2x-5=3-8x\\2x-5=8x-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}10x=8\\-6x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

b) ta có: x-1 = 0 => x= 0+1 = 1

            x-3 = 0 => x= 0+3 = 3

vậy x =1 và x = 3

b)<=>3x-x³=-x(x²-3)

=>-x(x²-3)=0

Th1:-x=0

Th2:x²-3=0

=>x²=3

=>x=\(\pm\sqrt{3}\)

c)(x-1).(x-3)=0

Th1:x-1=0

=>x=0

Th2:x-3=0

=>x=3

d)Ix+1I + Ix+2I+I2x+3I=2016x

<=>Ix+1I + Ix+2I+I2x+3I=|2x+3|+|x+2|+|x+1|

=>|2x+3|+|x+2|+|x+1|=2016x

=>x\(\approx\)0.00298210735586481

b) x=1

c) x=1;3

d) chịu

a ) 2|x - 3| - 5 = 3 <=> 2|x - 3| = 8 <=> |x - 3| = 4 => x - 3 = ± 4

TH1 : x - 3 = 4 => x = 7

TH2 : x - 3 = - 4 => x = - 1

Vậy x = { - 1; 7 }

b ) 2|2x + 3| + |2x + 3| = 6 <=> 3|2x + 3| = 6 => |2x + 3| = 2 => 2x + 3 = ± 2

=> x = { - 5/2 ; - 1/2 }

c ) 3|x + 1|² + |x + 1|² = 16

4|x + 1|² = 16

=> |x + 1|² = 4 = 2² ( ko xét TH |x + 1| = - 2 vì |x + 1| ≥ 0 )

=> |x + 1| = 2 => x + 1 = ± 2 => x = { - 3; 1 }