Объяснение:
{3*(x-1)-2*(1+x)<1 {3x-3-2-2x<1 {x-5<1 {x<6</p>
{3x-4≥0 {3x≥4 |÷3 {x≥4/3 {x≥4/3
Ответ: x∈[4/3;6).
2.
(√6+√3)*√12-2*√6*√3=√(6*12)+√(3*12)-√(4*6*3)=√72+√36-√72=√36=6.
3.
a) 3x²-11x+6=0 D=49 √D=7 x₁=3 x₂=2/3.
б) 4x²-1=0 (2x)²-1²=0 (2x+1)*(2x-1)=0 x₁=-1/2 x₂=1/2.
4.

5.
a) 4¹¹*4⁻⁹=4⁽¹¹⁻⁹⁾=4²=16.
б) (2⁻²)³:2⁻⁵=2⁽⁻²*³⁾:2⁻⁵=2⁻⁶:2⁻⁵=2⁽⁻⁶⁻⁽⁻⁵⁾⁾=2⁽⁻⁶⁺⁵⁾=2⁻¹=1/2=0,5.