Установить соответствие чисел а) 75% б) 100% в) 10% г) 50% д) 1 25%

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(-∞;-1) ∪ (;+∞)

Объяснение:

0\\D=49+4*12*19=961\\x1= \frac{7-31}{24} =-1\\x2=\frac{7+31}{24} =\frac{38}{24} =\frac{19}{12}" class="latex-formula" id="TexFormula2" src="https://tex.z-dn.net/?f=12x%5E2-7x-19%3E0%5C%5CD%3D49%2B4%2A12%2A19%3D961%5C%5Cx1%3D%20%5Cfrac%7B7-31%7D%7B24%7D%20%3D-1%5C%5Cx2%3D%5Cfrac%7B7%2B31%7D%7B24%7D%20%3D%5Cfrac%7B38%7D%7B24%7D%20%3D%5Cfrac%7B19%7D%7B12%7D" title="12x^2-7x-19>0\\D=49+4*12*19=961\\x1= \frac{7-31}{24} =-1\\x2=\frac{7+31}{24} =\frac{38}{24} =\frac{19}{12}">

(x+1)(x-)>0

(-∞;-1) ∪ (;+∞)

1tgx=2tgx/(1-tg²x) {tgx-tg³x-2tgx=0⇒tg³x+tgx=0⇒tgx(tg²x+1)=0⇒tgx=0⇒x=πn,n∈z (tg²x+1> 0) {1-tg²x≠0 ответ x=πn,n∈z 2 2sinxcosx-2√3cos²x=0 2cosx(sinx-√3cosx)=0 cosx=0⇒x=π/2+πn,n∈z sinx-√3cosx=0/cosx⇒tgx-√3=0⇒tgx=√3⇒x=π/3+πn,n∈z 3 sin²x-2cosx=0 1-cos²x-2cosx=0 cosx=a a²+2a-1=0 d=4+4=8 a1=(-2-2√2)/2=-1-√2⇒cosx=-1-⇒2< -1 нет решения a2=-1+√2⇒cosx=√2-1⇒x=+-arccos(√2-1)+2πn,n∈z 4 2cos(5x/2)*cos(3x/2)=0 cos(5x/2)=0⇒5x/2=π/2+πn⇒x=π/5+2πn/5,n∈z cos(3x/2)=0⇒3x/2=π/2+πn⇒x=π/3+2πn/3,n∈z 5 2*1-3cosx-2=0 -3cosx=0 cosx=0 x=π/2+πn,n∈z 6 2sin²x-sinx=0 sinx(2sinx-1)=0 sinx=0⇒x=πn,n∈z sinx=1/2⇒x=(-1)^n*π/6+πn,n∈z