Сворачиваем 2 скобку , получим
((x+9)^2/x-9)*1/x+9=(x+9)^2/(x+9)*(x-9)=x+9/x-9=-8/26=-4/13
ответ : -4/13
1)cosx<0⇒x∈(π/2+2πn;3π/2+2πn,n∈z)
-cosx+√3sinx=0
2(√3/2sinx-1/2cosx)=0
2sin(x-π/6)=0
x-π/6=πn
x=π/6+πn U x∈(π/2+2πn;3π/2+2πn,n∈z)⇒x=7π/6+2πn
2π≤7π/6+2πn≤7π/2
12≤7+12n≤21
5≤12n≤14
5/12≤n≤7/6
n=1⇒x=7π/6+2π=19π/6
2)cosx≥0⇒x∈[-π/2+2πk;π/2+2πk,k∈z]
cosx+√3sinx=0
2sin(x+π/6)=0
x+π/6=πk
x=-π/6+πk U x∈[-π/2+2πk;π/2+2πk,k∈z]⇒x=π/6+2πk
2π≤π/6+2πk≤7π/2
12≤1+12k≤21
11≤12k≤20
11/12≤k≤5/3
k=1⇒x=π/6+2π=13π/6
1,7*2,5+1,7*3,5-2=1,7*(2,5+3,5)-2=1,7*6-2=10,2-2=8,2
2,3*1,5-2,3*0,5-1=2,3*(1,5-0,5)-1=2,3*1-1=2,3-1=1,3
1. ∫(eˣ-1/x)dx=eˣ-ln|x|+c
2. ∫(sinx+4cosx-5√x)dx=-cosx+4sinx-5*3√x*x/2=-cosx+4sinx-7.5x√x+c
3. ∫(6-3x)⁷dx 6-3x=z 3x=6-z x=2-z/3 dx=-dz/3
∫(6-3x)⁷dx=-1/3∫z⁷dz=-1/3*8*z⁸=z⁸/24=(6-3x)⁸/24