mọi người ơi giúp em vớiiii
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b: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=8^x\)
=>\(8^x=\dfrac{4\cdot4^5}{3\cdot3^5}\cdot\dfrac{6\cdot6^5}{2\cdot2^5}\)
=>\(8^x=\dfrac{4^6}{2^6}\cdot\dfrac{6^6}{3^6}=2^6\cdot2^6=2^{12}=\left(2^3\right)^4=8^4\)
=>x=4
1: \(5^{x+4}-3\cdot5^{x+3}=2\cdot5^{11}\)
=>\(5^{x+3}\cdot5-3\cdot5^{x+3}=2\cdot5^{11}\)
=>\(2\cdot5^{x+3}=2\cdot5^{11}\)
=>x+3=11
=>x=8
2: \(\dfrac{1}{2}\cdot2^x+4\cdot2^x=9\cdot2^5\)
=>\(2^x\cdot\left(\dfrac{1}{2}+4\right)=9\cdot2^5\)
=>\(2^x\cdot\dfrac{9}{2}=9\cdot2^5\)
=>\(2^x=2^6\)
=>x=6
3: \(9^{2x+1}=27^3\)
=>\(3^{4x+2}=3^9\)
=>4x+2=9
=>4x=7
=>\(x=\dfrac{7}{4}\)
4: \(2^{-1}\cdot2^x+4\cdot2^x=9\cdot2^5\)
=>\(2^x\left(4+\dfrac{1}{2}\right)=9\cdot2^5\)
=>\(2^x\cdot\dfrac{9}{2}=9\cdot2^5\)
=>\(2^x=9\cdot2^5:\dfrac{9}{2}=2^6\)
=>x=6
5: \(\left(2x-1\right)^3=\dfrac{8}{27}\)
=>\(\left(2x-1\right)^3=\left(\dfrac{2}{3}\right)^3\)
=>\(2x-1=\dfrac{2}{3}\)
=>\(2x=\dfrac{2}{3}+1=\dfrac{5}{3}\)
=>\(x=\dfrac{5}{3}:2=\dfrac{5}{6}\)
Gọi số điểm cho trước là x(điểm)
(Điều kiện: \(x\in Z^+;x>3\))
Số điểm không thẳng hàng là x-3(điểm)
TH1: vẽ 1 đường thẳng đi qua 3 điểm thẳng hàng
=>Có 1 đường thẳng
TH2: Chọn 2 điểm bất kì trong x-3 điểm còn lại
Số đường thẳng là \(C^2_{x-3}=\dfrac{\left(x-3\right)!}{\left(x-3-2\right)!\cdot2!}=\dfrac{\left(x-4\right)\left(x-3\right)}{2}\)(đường)
TH3: Chọn 1 điểm trong 3 điểm thẳng hàng, 1 điểm trong x-3 điểm còn lại
=>Có 3(x-3) đường thẳng
Tổng số đường thẳng là 120 đường nên ta có:
\(1+\dfrac{\left(x-4\right)\left(x-3\right)}{2}+3\left(x-3\right)=120\)
=>\(\dfrac{2+\left(x-4\right)\left(x-3\right)+6\left(x-3\right)}{2}=120\)
=>2+(x-4)(x-3)+6(x-3)=240
=>\(2+x^2-7x+12+6x-18=240\)
=>\(x^2-x-244=0\)
=>\(\left[{}\begin{matrix}x=\dfrac{1+\sqrt{977}}{2}\left(loại\right)\\x=\dfrac{1-\sqrt{977}}{2}\left(loại\right)\end{matrix}\right.\)
Exercise 3 : Choose the best answer A, B, C or D to complete the sentences.
1. Oliver and Joe are from Toronto. They come from __________ city.
A. different B. the same C. same D. same as
2. Kevin likes painting, but hes not good at it. - Im__________ .
A. not too B. either C.also not D. not either
3. I dont agree with you . Your ideas are ___________ .
A. same as me B. the same as I C. different from mine D. similar to mine
4. The gallery is _______ it was 10 years ago. There is no change at all.
A. different from B. different as C. same as D. the same as
5. This years musical festival is not_______ it was last year.
. as exciting than B. same exciting as C. as exciting as D. as exciting more
6. I am waiting for the ticket office to open, and my cousin _______.
A. is too B. is same C. is either D. is so
7. It was noisy because he turned the volume up _______ it would go.
A. as loud as B. as loud than C. as loud like D. loud as
8. This film is _______the film I watched with my friend last week.
A. so interesting as B. not as interesting as C. as interesting so D. not as interesting to
9. My feelings about this song are _______ what I have imagined before.
A. different from quite B. quite different from C. quite from different D. different quite from
10. Listening to pop music is my favourite hobby. -___________. A. Im not either B. I do too C. Me too D. I dont either
1 is raining
2 warms - gives
3 is running
4 goes
5 work
6 rains - is raining
7 is cooking - cook s
8 is he working
9 are you going - am going
10 don't usually go
11 is swimming
12 does he take
13 is studying
14 is not
15 is not reading
16 gives
17 catches
Ta có số hạt ko mang điện = 1/2 số hạt mang điện và = 1/3 tất cả ( bao gồm cả 2 loại hạt)
Số hạt ko mang điện là 60 nhân 1/3=20
Số hạt mang điện là 60 - 20=40
Tick cho mình nha



Bài 6: Oz là phân giác của góc xOy
=>\(\widehat{xOz}=\dfrac{\widehat{xOy}}{2}=\dfrac{142^0}{2}=71^0\)
Ta có: \(\widehat{xOz}+\widehat{x'Oz}=180^0\)(hai góc kề bù)
=>\(\widehat{x'Oz}+71^0=180^0\)
=>\(\widehat{x'Oz}=109^0\)
Bài 7:
Ta có: Oz là phân giác của góc xOy
=>\(\widehat{xOz}=\widehat{yOz}=\dfrac{\widehat{xOy}}{2}=\dfrac{180^0}{2}=90^0\)
Ot là phân giác của góc xOz
=>\(\widehat{zOt}=\dfrac{\widehat{xOz}}{2}=\dfrac{90^0}{2}=45^0\)
Ov là phân giác của góc yOz
=>\(\widehat{vOz}=\dfrac{90^0}{2}=45^0\)
\(\widehat{vOt}=\widehat{zOv}+\widehat{zOt}=45^0+45^0=90^0\)