Sửa đề: \(\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}\)

\(=\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\dfrac{5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}\)

\(=\dfrac{3\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}+\dfrac{3\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{5\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}\)

\(=\dfrac{3}{-5}+\dfrac{3}{5}=0\)

lộn tìm x xin lỗi nha

780-(x-45):2=120

=>(x-45):2=780-120=660

=>\(x-45=660\times2=1320\)

=>x=1320+45=1365

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

Mấy chiếc thuyền nữa đang chạy

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

=>\(2+x^2-7x+12+6x-18=240\)

=>\(x^2-x-244=0\)

=>\(\left[{}\begin{matrix}x=\dfrac{1+\sqrt{977}}{2}\left(loại\right)\\x=\dfrac{1-\sqrt{977}}{2}\left(loại\right)\end{matrix}\right.\)

   27.75 + 25.27 - 150

= 27.(75 + 25) - 150

= 27.100 - 150

= 2700 - 150

= 2550

b;   142 - [50 - (23 x 10 - 23.5)]

   =  142 - [50 - 23x (10 - 5)]

   = 142 - [50 - 23 x 5]

   = 142 -  [50 - 115]

  = 142 - [ - 65]

  = 142 + 65

   = 207

\(F=\dfrac{\dfrac{2010}{1}+\dfrac{2009}{2}+...+\dfrac{1}{2010}+2010}{1+\dfrac{1}{2}+...+\dfrac{1}{2010}}\)

\(=\dfrac{\left(\dfrac{2010}{1}+1\right)+\left(\dfrac{2009}{2}+1\right)+...+\left(\dfrac{1}{2010}+1\right)}{1+\dfrac{1}{2}+...+\dfrac{1}{2010}}\)

\(=\dfrac{\dfrac{2011}{1}+\dfrac{2011}{2}+...+\dfrac{2011}{2010}}{1+\dfrac{1}{2}+...+\dfrac{1}{2010}}=2011\)

Help me có ai làm nhanh hum

a: 15:(x+2)=3

=>\(x+2=\dfrac{15}{3}=5\)

=>x=5-2=3

b: 20(1+x)=2

=>\(x+1=\dfrac{2}{20}=\dfrac{1}{10}\)

=>\(x=\dfrac{1}{10}-1=-\dfrac{9}{10}\)

f: \(12x-33=3^2\cdot3^3\)

=>\(12x-33=3^5=243\)

=>12x=243+33=276

=>\(x=\dfrac{276}{12}=23\)

g: 541+(218-x)=73

=>218-x=73-541=-468

=>x=218+468=686