\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}\)\(=\frac{24}{50}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x.1}\)=\(\frac{24}{50}\)

=\(\frac{1}{2}-\frac{1}{x.1}=\frac{24}{50}\)

=\(\frac{1}{x.1}=\frac{1}{2}-\frac{24}{50}\)

=\(\frac{1}{x.1}=\frac{1}{50}\)

\(\Rightarrow\)\(x.1=50\)

\(\Rightarrow x=50\)

Bài làm

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)

Bài làm

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)

Tìm x :

x - 0,27 = \(\frac{73}{100}\)

x           = \(\frac{73}{100}+0,27\)

x           = 1

Cậu P khó quá mik chưa nghĩ ra cách tính nhanh nhất !

Cậu tự giải nhé !

Hok tốt

\(x-0,27=\frac{\frac{73}{77}+\frac{73}{165}+\frac{73}{285}}{25\times\left(\frac{5}{84}+\frac{3}{180}+\frac{4}{285}\right)}.\)

\(x-0,27=\frac{\frac{146}{105}+\frac{73}{285}}{25\times\left(\frac{8}{105}+\frac{4}{285}\right)}\)

\(x-0,27=\frac{\frac{219}{133}}{25\times\frac{12}{133}}\)

\(x-0,27=\frac{\frac{219}{133}}{\frac{300}{133}}\)

\(x-0,27=0,73\)

\(x=0,73+0,27\)

\(x=1\)

\(\frac{1}{3}\) + \(\frac{5}{6}\): \(\left(x-2\frac{1}{5}\right)\)= \(\frac{3}{4}\)

<=> \(\frac{5}{6}\):\(\left(x-2\frac{1}{5}\right)\)= \(\frac{3}{4}\)- \(\frac{1}{3}\)

<=> \(\frac{5}{6}\) : \(\left(x-2\frac{1}{5}\right)\) = \(\frac{5}{12}\)

<=> \(\left(x-2\frac{1}{5}\right)\) =    \(\frac{5}{6}\) : \(\frac{5}{12}\)

,<=> \(\left(x-2\frac{1}{5}\right)\)=   2 

<=. x = 2 + \(\frac{11}{5}\)

<=> x = \(\frac{21}{5}\)

(1-1/3).(1-1/5).(1-1/7).(1-1/9).(1-1/11).(1-1/13).(1-1/2).(1-1/4).(1-1/6).(1-1/8).(1-1/10)

=2/3.4/5.6/7.8/9.10/11.12/13.1/2.3/4.5/6.7/8.9/10

=8/15.48/63.120/143.3/8.35/48.9/10

=384/945.360/1144.315/480

=138240/1081080.315/480

=43545600/518918400=84/1001

khó quá

\(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right)\)    \(\left(1-\frac{1}{11}\right)\)\(\left(1-\frac{1}{13}\right)\left(1-\frac{1}{2}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{6}\right)\)
 \(=1.\left(\frac{1}{3}-\frac{1}{2}\right).\left(\frac{1}{5}-\frac{1}{4}\right).\left(\frac{1}{7}-\frac{1}{6}\right)\)\(.\left(\frac{1}{11}-\frac{1}{10}\right).\frac{1}{13}\)
  \(=1.1.1.1.1.\frac{1}{13}\)
   \(=\frac{1}{13}\)
Cái này là chỉ do mình nghĩ thôi, ko biết đúng ko nữa
Có sai cko mình sorry nha!
Tk và kb hộ mình  với! thanks 

\(a_{n-1}=\frac{1}{1+2+3+...+n}=\frac{2}{n\left(n+1\right)}\)=>\(1-a_{n-1}=1-\frac{2}{n\left(n+1\right)}=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)

\(A=\left(1-\frac{2}{2.3}\right)\left(1-\frac{2}{3.4}\right)........\left(1-\frac{2}{2006.2007}\right)\)

\(=\left(\frac{1.4}{2.3}\right)\left(\frac{2.5}{3.4}\right)\left(\frac{3.6}{4.5}\right)........\left(\frac{2005.2008}{2006.2007}\right)\)\(=\frac{\left(1.2.3......2005\right)\left(4.5.6.....2008\right)}{\left(2.3.4.....2006\right)\left(3.4.5....2007\right)}=\frac{1.2008}{2006.3}=\frac{1004}{3009}\)

=\(\frac{9^{2016}}{16^{2016}}.\frac{16^{2015}}{9^{2015}}.\frac{4}{3}\)

=\(\frac{9}{16}.\frac{4}{3}\)

=\(\frac{3}{4}\)

k cho mk nhoa

\(\left(\frac{9}{16}\right)^{2016}.\left(\frac{16}{9}\right)^{2015}.\frac{4}{3}\)

\(=\left[\frac{9}{16}\left(\frac{9}{16}\right)^{2015}\right].\left(\frac{16}{9}\right)^{2015}.\frac{4}{3}\)

\(=\frac{9}{16}\left[\left(\frac{9}{16}\right)^{2015}.\left(\frac{16}{9}\right)^{2015}\right].\frac{4}{3}\)

\(=\frac{9}{16}\left[\left(\frac{9}{16}.\frac{16}{9}\right)^{2015}\right].\frac{4}{3}\)

\(=\frac{9}{16}.1^{2015}.\frac{4}{3}\)

\(=\frac{9}{16}.\frac{4}{3}\)

\(=\frac{3}{4}\)

\(\left(\frac{9}{16}\right)^{2016}.\left(\frac{16}{9}\right)^{2015}.\frac{4}{3}\)

\(=\frac{9^{2016}}{16^{2016}}.\frac{16^{2015}}{9^{2015}}.\frac{4}{3}\)

\(=\frac{9}{6}.\frac{4}{3}\)

\(=\frac{36}{48}\)

\(=\frac{3}{4}\)

OK chờ tí nhé

câu a tách ra nha

câu b thì tính trong ngoặc rồi tách

hok tốt

...

Câu a tham khảo link này nha bn

https://olm.vn/hoi-dap/detail/186073292336.html

\(\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}x=0\)

\(x.\left(\frac{1}{3}+\frac{2}{5}\right)-\frac{2}{5}=0\)

\(x.\frac{11}{15}=0+\frac{2}{5}=\frac{2}{5}\)

\(x=\frac{2}{5}:\frac{11}{15}=\frac{6}{11}\)

Mình không biết! Mình mới lớp 5 thôi!

=1/3*x+2/5*x-2/5.

=x*(1/3+2/5)-2/5.

=x*11/15-2/5=0.

=>x*11/15=2/5

=>x=2/5:11/15=6/11.

k nha có j kb vs mk