![2x^2-7x-4 \leqslant 0 \\ D=49+32=81 \\ x_1=\frac{7-9}{4}=-0,5 \\ x_2=\frac{7+9}{4}=4 \\ 2(x+0,5)(x-4)\leqslant 0](https://tex.z-dn.net/?f=2x%5E2-7x-4+%5Cleqslant+0+%5C%5C+D%3D49%2B32%3D81+%5C%5C+x_1%3D%5Cfrac%7B7-9%7D%7B4%7D%3D-0%2C5+%5C%5C+x_2%3D%5Cfrac%7B7%2B9%7D%7B4%7D%3D4+%5C%5C+2%28x%2B0%2C5%29%28x-4%29%5Cleqslant+0)
--(+)--(●-0,5)---(-)-----(●4)---(+)--->
![x\in[-0,5 ; 4]](https://tex.z-dn.net/?f=x%5Cin%5B-0%2C5+%3B+4%5D+)
-(x+1)²+9=0
Смотр. рис.
Ответ: -4 и 2.
Y`=2x/π+2cos2x+1/2cosx-1/2sinx
y`(π/2)=1+2*(-1)+1/2*0-1/2*1=1-2+0-0,5=-1,5
Ответ:
Объяснение:
1) a(n) = n/(√n + 1)
a(1) = 1/(√1 + 1) = 1/2; a(2) = 2/(√2 + 1); a(3) = 3/(√3 + 1)
a(4) = 4/(√4 + 1) = 4/3; a(5) = 5/(√5 + 1)
2) a(n) = 2n/(√3n - 1)
a(1) = 2/(√3 - 1); a(2) = 4/(√6 - 1); a(3) = 6/(√9 - 1) = 6/(3 - 1) = 3
a(4) = 8/(√12 - 1); a(5) = 10/(√15 - 1)
3) a(n) = (2n - 1)/(√n + 2)
a(1) = 1/(√2 + 2); a(2) = 3/(√2 + 2); a(3) = 5/(√3 + 2)
a(4) = 7/(√4 + 2) = 7/4; a(5) = 9/(√5 + 2)
4) a(n) = 3n/(√(2n-1) + 1)
a(1) = 3/(√1 + 1) = 3/2; a(2) = 6/(√3 + 1); a(3) = 9/(√5 + 1)
a(4) = 12/(√7 + 1); a(5) = 15/(√9 - 1) = 15/2
9х^{2}=81
х^{2}=81:9
х^{2}=9
х=√9
х=-3 или х=3