\(D=\frac{1}{6}+\frac{1}{66}+\frac{1}{176}+\frac{1}{336}+\frac{1}{546}\)
\(x+\frac{5}{3}=\frac{1}{81}\)
\(-52+\frac{2}{3}x=-46\)
\(\frac{x}{45}=\frac{5}{6}+\frac{-29}{30}\)
Câu 6. Giải các phương trình sau:
a, x+\(\frac{2x+\frac{x-1}{5}}{3}=1-\frac{3x-\frac{1-2x}{3}}{5}\); b, \(\frac{3x-1-\frac{x-1}{2}}{3}-\frac{2x+\frac{1-2x}{3}}{2}=\frac{\frac{3x-1}{2}}{5}-6\)
Câu 7. Giải các phương trình sau:
a, \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\); b, \(\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4+++==}{96}+1\right)+\left(\frac{x+5}{95}+1\right)\)
c, \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\); d, \(\frac{201-6}{99}+\frac{203-6}{97}=\frac{205-x}{95}+3=0\)
e, \(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\); f, \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
g, \(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\); h, \(\frac{2-x}{2002}-1=\frac{1-x}{2003}-\frac{x}{2004}\)
i, \(\frac{x^2-10x-29}{1971}+\frac{x^2-10x-27}{1973}=\frac{x^2-10x-1971}{29}+\frac{x^2-10x-1973}{27}\);
Câu 6 :
a, Ta có : \(x+\frac{2x+\frac{x-1}{5}}{3}=1-\frac{3x-\frac{1-2x}{3}}{5}\)
=> \(\frac{15x}{15}+\frac{5\left(2x+\frac{x-1}{5}\right)}{15}=\frac{15}{15}-\frac{3\left(3x-\frac{1-2x}{3}\right)}{15}\)
=> \(15x+5\left(2x+\frac{x-1}{5}\right)=15-3\left(3x-\frac{1-2x}{3}\right)\)
=> \(15x+10x+\frac{5\left(x-1\right)}{5}=15-9x+\frac{3\left(1-2x\right)}{3}\)
=> \(15x+10x+x-1=15-9x+1-2x\)
=> \(15x+10x+x-1-15+9x-1+2x=0\)
=> \(37x-17=0\)
=> \(x=\frac{17}{37}\)
Vậy phương trình trên có nghiệm là \(S=\left\{\frac{17}{37}\right\}\)
Bài 7 :
a, Ta có : \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
=> \(\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)
=> \(\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)
=> \(x-23=0\)
=> \(x=23\)
Vậy phương trình trên có nghiệm là \(S=\left\{23\right\}\)
c, Ta có : \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)
=> \(\frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\)
=> \(\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)
=> \(\frac{x+2005}{2004}+\frac{x+2005}{2003}-\frac{x+2005}{2002}-\frac{x+2005}{2001}=0\)
=> \(\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
=> \(x+2005=0\)
=> \(x=-2005\)
Vậy phương trình trên có nghiệm là \(S=\left\{-2005\right\}\)
e, Ta có : \(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
=> \(\frac{x-45}{55}-1+\frac{x-47}{53}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)
=> \(\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)
=> \(\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)
=> \(\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)
=> \(x-100=0\)
Vậy phương trình trên có nghiệm là \(S=\left\{100\right\}\)
\(\frac{1}{6}+\frac{1}{66}+\frac{1}{176}+\frac{1}{336}+...+\frac{1}{1886}\)
\(\frac{1}{6}+\frac{1}{66}+\frac{1}{176}+\frac{1}{336}+...+\frac{1}{1886}\)
\(=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+\frac{1}{16.21}+...+\frac{1}{41.46}\)
\(=\frac{5}{5}\left(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+\frac{1}{16.21}+...+\frac{1}{41.46}\right)\)
\(=\frac{1}{5}\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}+...+\frac{5}{41.46}\right)\)
\(=\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{41}-\frac{1}{46}\right)\)
\(=\frac{1}{5}\left(1-\frac{1}{46}\right)\)
\(=\frac{1}{5}.\frac{45}{46}=\frac{9}{46}\)
1/(1.6) + 1/(6.11) + 1/(11.16) + … + 1/[(5n-4)(5n+1)]
=(1/1 – 1/6)/5 + (1/6 – 1/11)/5 + (1/11 – 1/16)/5 +…+ [1/(5n-4) – 1/(5n+1)]/5
=[1/1 – 1/6 + 1/6 – 1/11 + 1/11 – 1/16 + … + 1/(5n-4) – 1/(5n+1)]/5
=[1 – 1/(5n+1)]/5
=[1 – 1/(5.100+1)]/5 = 100/501
\(B=\frac{1}{6}+\frac{1}{66}+\frac{1}{176}+\frac{1}{336}+...+\frac{1}{496.501}\)
Quy luật:
6 = 1.6
66 = 6.11
176 = 11.16
336 = 16.21
...
1/(1.6) + 1/(6.11) + 1/(11.16) + … + 1/[(5n-4)(5n+1)]
=(1/1 – 1/6)/5 + (1/6 – 1/11)/5 + (1/11 – 1/16)/5 +…+ [1/(5n-4) – 1/(5n+1)]/5
=[1/1 – 1/6 + 1/6 – 1/11 + 1/11 – 1/16 + … + 1/(5n-4) – 1/(5n+1)]/5
=[1 – 1/(5n+1)]/5
Tổng 100 số đầu =[1 – 1/(5.100+1)]/5 = 100/501
1/1.6 + 1/6.11+ 1/11.16+ ....
số thứ 100 có dạng 1/(496.501)
do đó tổng trên bằng 1/5( 1/1- 1/501) = 100/ 501
hc tốt
\(B=\frac{1}{6}+\frac{1}{66}+\frac{1}{176}+\frac{1}{336}+...+\frac{1}{496.501}\)
\(B=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+\frac{1}{16.21}+...+\frac{1}{496.501}\)
\(B=\frac{1}{5}.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{496}-\frac{1}{501}\right)\)
\(B=\frac{1}{5}.\left(1-\frac{1}{501}\right)\)
\(B=\frac{1}{5}.\frac{500}{501}=\frac{100}{501}\)
\(a,\frac{3}{4}-x=1\) \(b,x+4=\frac{1}{5}\) \(c,x-\frac{1}{5}=2\)
\(d,x+\frac{5}{3}=\frac{1}{81}\) \(e,x-\frac{10}{3}=\frac{7}{15}.\frac{3}{5}\) \(g,x+\frac{3}{22}=\frac{27}{121}.\frac{11}{9}\)
\(h,1-x=\frac{49}{65}.\frac{5}{7}\) \(i,\frac{8}{23}.\frac{46}{24}-x=\frac{1}{3}\) \(k,\frac{62}{7}.x=\frac{29}{9}:\frac{3}{56}\)
- Thiếu đề bài bạn ơi !!
- Nếu là giải phương trình thì nghĩ bạn nên tự làm bởi toàn là chuyển vế cơ bản ỷ lại quá sẽ khó để làm đến phần thông hiểu, nâng cao ,.... nha
- Nếu cần kết quả để xem đúng hay sai thì ấn hỏi mình hoặc các bạn khác cũng được nha !
- Vì bạn giải xong nên mình đưa kết quả bạn soát nha :
\(a,x=-\frac{1}{4}.b,x=-\frac{19}{5}.c,x=\frac{11}{5}.d,x=-\frac{134}{81}.e,x=\frac{271}{75}\)
\(g,x=\frac{3}{22}.h,x=\frac{6}{13}.i,x=\frac{1}{3}.k,\frac{5684}{837}\)
B= \(\frac{1}{6}+\frac{1}{66}+\frac{1}{176}+\frac{1}{336}+....+\frac{1}{496.501}\)
\(5B=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}+...+\frac{5}{496.501}\)
\(5B=\frac{6-1}{1.6}+\frac{11-6}{6.11}+\frac{16-11}{11.16}+\frac{21-16}{16.21}+...+\frac{501-496}{496.501}\)
\(5B=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{496}-\frac{1}{501}\)
\(5B=1-\frac{1}{501}=\frac{500}{501}\Rightarrow B=\frac{100}{501}\)
tìm x biết
a,\(\left|2x+\frac{3}{4}\right|=\frac{1}{2}\)
b,\(\frac{3x}{2,7}=\frac{\frac{1}{4}}{2\frac{1}{4}}\)
c,\(\left|x-1\right|+4=6\)
d, \(\frac{x}{3}=\frac{y}{5}\) và y-x=24
e, \(\left(x^2-3\right)^2=16\)
f, \(\frac{3}{4}+\frac{2}{5}x=\frac{29}{60}\),
g, \(\left(-\frac{1}{3}\right)^3.x=\frac{1}{81}\)
k,\(\frac{3}{4}-\frac{2}{5}x=\frac{29}{60}\)
l,\(\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
a) \(\left|2x+\frac{3}{4}\right|=\frac{1}{2}\)
\(\orbr{\begin{cases}2x+\frac{3}{4}=\frac{1}{2}\\2x+\frac{3}{4}=\frac{-1}{2}\end{cases}}\) => \(\orbr{\begin{cases}2x=\frac{1}{2}-\frac{3}{4}\\2x=\frac{-1}{2}-\frac{3}{4}\end{cases}}\) => \(\orbr{\begin{cases}2x=\frac{-1}{4}\\2x=\frac{-5}{4}\end{cases}}\) => \(\orbr{\begin{cases}x=\frac{-1}{8}\\x=\frac{-5}{8}\end{cases}}\)
Vậy \(x=\left\{\frac{-1}{8},\frac{-5}{8}\right\}\)
b) \(\frac{3x}{2,7}=\frac{\frac{1}{4}}{2\frac{1}{4}}\)= \(\frac{3x}{2,7}=\frac{\frac{1}{4}}{\frac{9}{4}}\)
=> \(3x.\frac{9}{4}=2,7.\frac{1}{4}\)=> \(\frac{27x}{4}=\frac{27}{40}\)
\(27x.40=27.4\)
\(1080.x=108\)
\(x=\frac{1}{10}\)
Vậy \(x=\frac{1}{10}\)
c) \(\left|x-1\right|+4=6\)
\(\left|x-1\right|=6-4\)
\(\left|x-1\right|=2\)
\(\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}}\)=> \(\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
Vậy \(x=\left[3,-1\right]\)
d) \(\frac{x}{3}=\frac{y}{5}=>\frac{y}{5}=\frac{x}{3}=>\frac{y-x}{5-3}=\frac{24}{2}=12\)
e) \(\left(x^2-3\right)^2=16\)
\(\left(x^2-3\right)^2=4^2\)\(=>x^2-3=4\)
\(x^2=7=>x=\sqrt{7}\)
Vậy \(x=\sqrt{7}\)
f) \(\frac{3}{4}+\frac{2}{5}x=\frac{29}{60}\)
\(\frac{2}{5}x=\frac{29}{60}-\frac{3}{4}\)
\(\frac{2}{5}x=-\frac{4}{15}\)
\(x=-\frac{4}{15}:\frac{2}{5}=-\frac{4}{15}.\frac{5}{2}=-\frac{2}{3}\)
Vậy \(x=-\frac{2}{3}\)
g) \(\left(-\frac{1}{3}\right)^3.x=\frac{1}{81}\)
\(\left(-\frac{1}{27}\right).x=\frac{1}{81}\)
\(x=\left(-\frac{1}{27}\right):\frac{1}{81}=\left(-\frac{1}{27}\right).81=-3\)
Vậy \(x=-3\)
k)\(\frac{3}{4}-\frac{2}{5}x=\frac{29}{60}\)
\(\frac{2}{5}x=\frac{3}{4}-\frac{29}{60}\)
\(\frac{2}{5}x=\frac{4}{15}\)
\(x=\frac{2}{5}-\frac{4}{15}=>x=\frac{2}{15}\)
Vậy \(x=\frac{2}{15}\)
I) \(\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
\(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}\)
\(\frac{3}{5}x=\frac{5}{14}\)
\(x=\frac{5}{14}:\frac{3}{5}=\frac{5}{14}.\frac{5}{3}=\frac{25}{42}\)
Vậy \(x=\frac{25}{42}\)
Bài 1: Tìm x biết:
a. \(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right)\left(5x+6\right)}=\frac{2010}{2011}\)
b. \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+..+\frac{4}{41.45}=\frac{29}{45}\)
c. \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+2\right)}:2\)
d. (x-20) . \(\frac{\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2000}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}=\frac{1}{2000}\)
Bài 2:
\(\frac{a}{n\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\left(n,a\right)\in Nsao\)
Bài 3:
a)\(\frac{3}{x}+\frac{y}{3}=\frac{5}{6}\)
b) \(\frac{x}{3}-\frac{4}{y}=\frac{1}{5}\)
c) \(\frac{x}{2}+\frac{y}{3}=\frac{x+y}{2+3}\)
d) \(\frac{x-1}{9}+\frac{1}{3}=\frac{1}{y+2}\)
tung từng vế một thôi
bạn nhác quá éo chịu suy nghĩ
bài này dễ vl
Bài 1:
a, \(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right)\left(5x+6\right)}=\frac{2010}{2011}\)
\(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2010}{2011}\)
\(1-\frac{1}{5x+6}=\frac{2010}{2011}\)
\(\frac{1}{5x+6}=1-\frac{2010}{2011}\)
\(\frac{1}{5x+6}=\frac{1}{2011}\)
=> 5x + 6 = 2011
5x = 2011 - 6
5x = 2005
x = 2005 : 5
x = 401
b, \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)
\(\frac{7}{x}+\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}\right)=\frac{29}{45}\)
\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\frac{7}{x}=\frac{29}{45}-\frac{8}{45}\)
\(\frac{7}{x}=\frac{7}{15}\)
=> x = 15
c, ghi lại đề
d, ghi lại đề
Bài 2:
\(\frac{1}{n}-\frac{1}{n+a}=\frac{n+a}{n\left(n+a\right)}-\frac{n}{n\left(n+a\right)}=\frac{a}{n\left(n+a\right)}\)
tim x
a)\(2\frac{3}{4}x-1\frac{5}{8}x=1\)
b)\(2\frac{3}{4}x-1\frac{5}{8}x=1\)
c)\(\frac{3}{2}x-\frac{2}{5}=\frac{1}{3}x-\frac{1}{4}\)
d)\(2x-\frac{1}{4}=\frac{5}{6}-\frac{1}{2}x\)
e)\(\frac{-5}{6}+3x=\frac{2}{3}-\frac{1}{2}x\)
f)\(\frac{5}{2}x-\frac{3}{2}=x+\frac{29}{10}\)
g)\(\frac{2}{3}+\frac{7}{3}x=\frac{5}{4}x+\frac{1}{6}\)
h)\(\frac{1}{3}.x+\frac{2}{5}\left(x-1\right)=0\)
giup mk nha moi nguoi
cách tim số hạng thứ n của dãy
VD: \(\frac{1}{6};\frac{1}{66};\frac{1}{176;}\frac{1}{336;....}\)
Tím số hạng thứ 46 của dãy