a: \(cosα=∣3⋅12+(−4)⋅(−5)∣32+(−4)2⋅122+(−5)2=∣36+20∣5⋅13=5665cosα=32+(−4)2​⋅122+(−5)2​∣3⋅12+(−4)⋅(−5)∣​=5⋅13∣36+20∣​=6556​\)

b: (d)//Δ nên (d): 3x-4y+c=0 và c<>7

(C): \((x+3)2+(y−2)2=36(x+3)2+(y−2)2=36\)

=>I(-3;2); R=6

(d) tiếp xúc với (C)

=>\(d(I;(d))=6d(I;(d))=6\)

=>\(∣3⋅(−3)+(−4)⋅2+c∣32+(−4)2=632+(−4)2​∣3⋅(−3)+(−4)⋅2+c∣​=6\)

=>\(∣c−17∣=6⋅5=30∣c−17∣=6⋅5=30\)

=>\[[c−17=30c−17=−30⇔[c=47(nhận)c=−13(nhận)[c−17=30c−17=−30​⇔[c=47(nhận)c=−13(nhận)​\]

Vậy: (d): 3x-4y+47=0 hoặc (d): 3x-4y-13=0

a: \(−2x2+18x+20>=0−2x2+18x+20>=0\)

=>\(−x2+9x+10>=0−x2+9x+10>=0\)

=>\(x2−9x−10<=0x2−9x−10<=0\)

=>(x-10)(x+1)<=0

=>-1<=x<=10

b: 

ĐKXĐ: \(2x2−8x+4>=02x2−8x+4>=0\)

=>\(x2−4x+2>=0x2−4x+2>=0\)

=>\(x2−4x+4−2>=0x2−4x+4−2>=0\)

=>\((x−2)2>=2(x−2)2>=2\)

=>\([x−2>=2x−2<=−2⇔[x>=2+2x<=−2+2[x−2>=2​x−2<=−2​​⇔[x>=2+2​x<=−2​+2​\)

\(2x2−8x+4=x−22x2−8x+4​=x−2\)

=>\({2x2−8x+4=(x−2)2x−2>=0{2x2−8x+4=(x−2)2x−2>=0​\)

=>\({2x2−8x+4−x2+4x−4=0x>=2⇔{x2−4x=0x>=2{2x2−8x+4−x2+4x−4=0x>=2​⇔{x2−4x=0x>=2​\)

=>x=4