HOC24
Lớp học
Môn học
Chủ đề / Chương
Bài học
Tuyển Cộng tác viên Hoc24 nhiệm kì 28 tại đây: https://forms.gle/GrfwFgzveoKLVv3p6
21.a) `2sin(x-30^@)-1=0``<=>sin(x-30^@)=1/2``<=> sin(x-30^@)=sin30^@``<=>[(x-30^@=30^@+k360^@),(x-30^@=180^@-30^@+k360^@):}``<=> [(x=60^@+k360^@),(x=180^@+k360^@):}`b) `5sin^2x+3cosx+3=0``<=>5(1-cos^2x)+3cosx+3=0``<=>-5cos^2x+3cosx+8=0``<=>(cosx+1)(cosx=8/5)=0``<=>[(cosx=-1),(cosx=8/5\ (VN)):}``<=>x=180^@+k360^@`22.`-1<=sin2x<=1``<=>2<=3+sin2x<=4``=> y_(min)=2 ; y_(max)=4`
1. B2. A3. A4. C5. B6. A7. A
18. The cook told me to mix the eggs with the flour.
19. Luke told me to bring a bottle of wine to the party.
20. Mon said to me that I should not eat out too often.
`\sqrt((-1)/(2x)) .\sqrt(-8x^5)`
`=\sqrt( (-1)/(2x) . (-8x^5))`
`=\sqrt( 8x^5)/(2x))`
`=\sqrt(4x^2)`
`=|2x|=-2x`
`sin^2 α+cos^2 α =1`
`=> sinα =\sqrt(1-cos^2α)=\sqrt(1-(3/4)^2) = \sqrt7/4`
`=> tanα=(sinα)/(cosα)=(3\sqrt7)/7`
`=> cotα=1/(tanα)=\sqrt7/3`
`(3-\sqrt5)(\sqrt10+\sqrt2)(3-\sqrt5)`
`=(3-\sqrt5)^2 .(\sqrt10+\sqrt2)`
`=(3-6\sqrt5+5)(\sqrt10+\sqrt2)`
`=(8-6\sqrt5)(\sqrt10+\sqrt2)`
`=2\sqrt10 -22\sqrt2`
`=(3x-1)^2+2(2x-1)(2x+5)+(2x+5)^2-(2x-3)(4x^2+6x+9):(2x-3)`
`=(3x-1+2x+5)^2-(4x^2+6x+9)`
`=(5x+4)^2-(4x^2+6x+9)`
`=25x^2+40x+16-4x^2-6x-9`
`=21x^2+34x+7`
`2(x-4)-3(x+1)=4`
`2x-8-3x-3=4`
`2x-3x=4+8+3`
`-x=15`
`x=-15`
`3(x+5)-3=2(x+1)+7`
`3x+15-3=2x+2+7`
`x=-3`
.
`15-(x+2)^2=-1`
`(x+2)^2=16`
`(x+2)^2=4^2=(-4)^2`
`[(x+2=4),(x+2=-4):}`
`[(x=2),(x=-6):}`
`75-(2x+1)^3=11`
`(2x+1)^3=64`
`(2x+1)^2=4^3`
`2x+1=4`
`x=3/2`
Áp dụng hệ thức trong tam giác vuông:
`MH^2 =NH.PH`
`=>PH=MH^2 : NH = 2,4^2 : 1,8=3,2(cm)`
`=> NP=NH+PH=5(cm)`
`=> S= 1/2 . MH .NP =6(cm^2)`