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Giải:
\(\dfrac{\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{2}{7}\right)+\left(1-\dfrac{3}{8}\right)+...+\left(1-\dfrac{88}{93}\right)}{\dfrac{-1}{12}-\dfrac{1}{14}-\dfrac{1}{16}-...-\dfrac{1}{186}}\)
Gọi dãy là A,phần tử là B. Ta có:
B=\(\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{2}{7}\right)+\left(1-\dfrac{3}{8}\right)+...+\left(1-\dfrac{88}{93}\right)\)
B=\(\dfrac{5}{6}+\dfrac{5}{7}+\dfrac{5}{8}+...+\dfrac{5}{93}\)
B=5.\(\left(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+...+\dfrac{1}{93}\right)\)
B=5.\(\left[\dfrac{2}{2}.\left(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+...+\dfrac{1}{93}\right)\right]\)
B=5.\(\left[2.\left(\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+...+\dfrac{1}{186}\right)\right]\)
B=10.\(\left(\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+...+\dfrac{1}{186}\right)\)
⇒A=\(\dfrac{10.\left(\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+...+\dfrac{1}{186}\right)}{\dfrac{-1}{12}+\dfrac{-1}{14}+\dfrac{-1}{16}+...+\dfrac{-1}{186}}\)
⇒A=-10
Chúc bạn học tốt!
Lời giải:
a) Hiển nhiên vế trái $\geq 0$ do tính chất của trị tuyệt đối.
$\Rightarrow 4x\geq 0\Rightarrow x\geq 0$. Đến đây ta có thể phá bỏ dấu trị tuyệt đối
$|x+\frac{11}{17}|+|x+\frac{2}{17}|+|x+\frac{4}{17}|=4x$
$x+\frac{11}{17}+x+\frac{2}{17}+x+\frac{4}{17}=4x$
$3x+1=4x$
$x=1$
b) Hiển nhiên vế trái $\geq 0$ nên $11x\geq 0\Rightarrow x\geq 0$
Khi đó:
$|x+\frac{1}{2}|+|x+\frac{1}{6}|+|x+\frac{1}{12}|+...+|x+\frac{1}{110}|=x+\frac{1}{2}+x+\frac{1}{6}+x+\frac{1}{12}+...+x+\frac{1}{110}$
$=10x+(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110})$
$=10x+(1-\frac{1}{11})=10x+\frac{10}{11}=11x$
$\Rightarrow x=\frac{10}{11}$
trời đất dung hoa vạn vật sinh sôi con mẹ mày lôi thôi đầu xanh mỏ đỏ gặp cỏ thay cơm đầu tóc bờm sờm khạc đờm tung tóe mà TAO ĐỊT CON MẸ MÀY NHƯ LỒN TRAU LỒN CHÓ LỒN BÓ XI MĂNG LỒN CHẰNG MẠNG NHỆN MÀ LỒN BẸN LÁ KHOÁI LỒN KHAI LÁ MIT LỒN ĐÍT LỒN TƠM LỒN TƠM LỒN ĐẬM LỒN GIA MAI LỒN ỈA CHẢY LỒN NHẨY HIPHOP LỒN LÔ XỐP LỒN HÀNG HIỆU LỒN HÀNG TRIỆU CON SÚC VẬT MÀ NÓ ĐÂM VÀO CÁI CON ĐĨ MẸ MÀY TỪ TRÊN CAO MÀ LAO ĐẦU XUỐNG ĐẤT ĐỊT LẤT PHẤT NHƯ MƯA RƠI
a)Ta thấy:
\(\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{x+a}{x\left(x+a\right)}-\dfrac{x}{x\left(x+a\right)}\)
\(=\dfrac{\left(x+a\right)-x}{x\left(x+a\right)}\)
\(=\dfrac{a}{x\left(x+a\right)}\)
\(\Rightarrowđpcm\)
b)Ta thấy:
\(\dfrac{1}{x\left(x+1\right)}-\dfrac{1}{\left(x+1\right)\left(x+2\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+2\right)}{x\left(x+1\right)^2\left(x+2\right)}-\dfrac{x\left(x+1\right)}{x\left(x+1\right)^2\left(x+2\right)}\)
\(=\dfrac{x+2}{x\left(x+1\right)\left(x+2\right)}-\dfrac{x}{x\left(x+1\right)\left(x+2\right)}\)
\(=\dfrac{\left(x+2\right)-x}{x\left(x+1\right)\left(x+2\right)}=\dfrac{2}{x\left(x+1\right)\left(x+2\right)}\Rightarrowđpcm\)
c)Ta thấy:
\(\dfrac{1}{x\left(x+1\right)\left(x+2\right)}-\dfrac{1}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+2\right)\left(x+3\right)}{x\left(x+1\right)^2\left(x+2\right)^2\left(x+3\right)}-\dfrac{x\left(x+1\right)\left(x+2\right)}{x\left(x+1\right)^2\left(x+2\right)^2\left(x+3\right)}=\dfrac{x+3}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}-\dfrac{x}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\dfrac{x+3-x}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\dfrac{3}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}\Rightarrowđpcm\)
a/ \(\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{a}{x\left(x+a\right)}\)
Ta có: \(\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{x+a}{x\left(x+a\right)}-\dfrac{x}{x\left(x+a\right)}\)
\(=\dfrac{\left(x-x\right)+a}{x\left(x+a\right)}\) hay \(\dfrac{a}{x\left(x+a\right)}\)
\(\Rightarrow\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{a}{x\left(x+a\right)}\left(đpcm\right)\)
a: =>1/3x+2/5x-2/5=0
=>11/15x-2/5=0
=>11/15x=2/5
=>x=2/5:11/15=2/5*15/11=30/55=6/11
b: =>-5x-1-1/2x+1/3=x
=>-11/2x-2/3-x=0
=>-13/2x=2/3
=>x=-2/3:13/2=-2/3*2/13=-4/39
c: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=1/3 hoặc x=-1/2
d: 9(3x+1)^2=16
=>(3x+1)^2=16/9
=>3x+1=4/3 hoặc 3x+1=-4/3
=>3x=1/3 hoặc 3x=-7/3
=>x=1/9 hoặc x=-7/9
\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right).......\left(1-\dfrac{1}{10}\right)\)
\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\left(\dfrac{3}{3}-\dfrac{1}{3}\right).........\left(\dfrac{10}{10}-\dfrac{1}{10}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}......\dfrac{9}{10}\)
\(=\dfrac{1}{10}\)
Bài 1:
a) \(\left(\dfrac{3}{8}+\dfrac{-3}{4}+\dfrac{7}{12}\right):\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\left(\dfrac{9}{24}+\dfrac{-18}{24}+\dfrac{14}{24}\right):\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\dfrac{5}{24}:\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\dfrac{5}{24}.\dfrac{6}{5}+\dfrac{1}{2}\)
\(=\dfrac{1}{4}+\dfrac{1}{2}\)
\(=\dfrac{1}{4}+\dfrac{2}{4}\)
\(=\dfrac{3}{4}\)
b) \(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}-\dfrac{4}{5}\right)\)
\(=\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}\)
\(=\left(\dfrac{1}{2}+\dfrac{4}{5}\right)+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)\)
\(=\dfrac{1}{2}+\dfrac{4}{5}\)
\(=\dfrac{5}{10}+\dfrac{8}{10}\)
\(=\dfrac{9}{5}\)
c) \(6\dfrac{5}{12}:2\dfrac{3}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{77}{12}:\dfrac{11}{4}+\dfrac{42}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{77}{12}.\dfrac{4}{11}+\dfrac{42}{4}.\left(\dfrac{5}{15}+\dfrac{3}{15}\right)\)
\(=\dfrac{7}{3}+\dfrac{42}{4}.\dfrac{8}{15}\)
\(=\dfrac{7}{3}+\dfrac{14.2}{1.3}\)
\(=\dfrac{7}{3}+\dfrac{28}{3}\)
\(=\dfrac{35}{3}\)
d) \(\left(\dfrac{7}{8}-\dfrac{3}{4}\right).1\dfrac{1}{3}-\dfrac{2}{7}.\left(3,5\right)^2\)
\(=\left(\dfrac{7}{8}-\dfrac{6}{8}\right).\dfrac{4}{3}-\dfrac{2}{7}.12\dfrac{1}{4}\)
\(=\dfrac{1}{8}.\dfrac{4}{3}-\dfrac{2}{7}.\dfrac{49}{4}\)
\(=\dfrac{1}{6}-\dfrac{7}{2}\)
\(=\dfrac{1}{6}-\dfrac{21}{6}\)
\(=\dfrac{-10}{3}\)
e) \(\left(\dfrac{3}{5}+0,415-\dfrac{3}{200}\right).2\dfrac{2}{3}.0,25\)
\(=\left(\dfrac{3}{5}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)
\(=\left(\dfrac{120}{200}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)
\(=1.\dfrac{8}{3}.\dfrac{1}{4}\)
\(=\dfrac{2}{3}\)
f) \(\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)
\(=\dfrac{5}{16}:\dfrac{1}{8}-\left(\dfrac{9}{4}-\dfrac{3}{5}\right).\dfrac{10}{11}\)
\(=\dfrac{5}{16}.\dfrac{8}{1}-\left(\dfrac{45}{20}-\dfrac{12}{20}\right).\dfrac{10}{11}\)
\(=\dfrac{5}{2}-\dfrac{33}{20}.\dfrac{10}{11}\)
\(=\dfrac{5}{2}-\dfrac{3}{2}\)
\(=\dfrac{2}{2}=1\)
g) \(0,25:\left(10,3-9,8\right)-\dfrac{3}{4}\)
\(=\dfrac{1}{4}:\dfrac{1}{2}-\dfrac{3}{4}\)
\(=\dfrac{1}{4}.\dfrac{2}{1}-\dfrac{3}{4}\)
\(=\dfrac{1}{2}-\dfrac{3}{4}\)
\(=\dfrac{2}{4}-\dfrac{3}{4}\)
\(=\dfrac{-1}{4}\)
h) \(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+20\%\right):\dfrac{7}{3}\)
\(=\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{11}{20}+\dfrac{1}{5}\right):\dfrac{7}{3}\)
\(=\dfrac{7}{5}-\left(\dfrac{11}{20}+\dfrac{4}{20}\right):\dfrac{7}{3}\)
\(=\dfrac{7}{5}-\dfrac{3}{4}:\dfrac{7}{3}\)
\(=\dfrac{7}{5}-\dfrac{9}{28}\)
\(=\dfrac{196}{140}-\dfrac{45}{140}\)
\(=\dfrac{151}{140}\)
i) \(\dfrac{\left(\dfrac{1}{2-0,75}\right).\left(0,2-\dfrac{2}{5}\right)}{\dfrac{5}{9}-1\dfrac{1}{12}}\)
\(=\dfrac{\left(\dfrac{1}{1,25}\right).\left(\dfrac{1}{5}-\dfrac{2}{5}\right)}{\dfrac{5}{9}-\dfrac{13}{12}}\)
\(=\dfrac{\dfrac{1}{1,25}.\dfrac{-1}{5}}{\dfrac{20}{36}-\dfrac{39}{36}}\)
\(=\dfrac{\dfrac{-1}{6,25}}{\dfrac{-19}{36}}\)
k) \(\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{1}{14}}{-1-\dfrac{3}{7}+\dfrac{3}{28}}\)
\(=\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{28}}{-\dfrac{3}{3}-\dfrac{3}{7}+\dfrac{3}{28}}\)
\(=\dfrac{2\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}{\left(-3\right)\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}\)
\(=-\dfrac{2}{3}\)
\(A=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(A=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)
\(A=\left(\dfrac{7}{10}.\dfrac{5}{28}\right).\left(\dfrac{8}{3}.\dfrac{3}{8}\right).20\)
\(A=\dfrac{1}{8}.1.20\)
\(A=\dfrac{20}{8}=\dfrac{5}{2}\)
\(B=\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)
\(B=\left(9\dfrac{3}{8}+7\dfrac{5}{8}\right)+4,03\)
\(B=\left[\left(9+7\right)+\left(\dfrac{3}{8}+\dfrac{5}{8}\right)\right]+4,03\)
\(B=\left(16+1\right)+4,03\)
\(B=17+4,03\)
\(B=21,03\)
\(C=\left(9,75.21\dfrac{3}{7}+\dfrac{39}{4}.18\dfrac{4}{7}\right).\dfrac{15}{78}\)
\(C=\left(\dfrac{39}{4}.\dfrac{150}{7}+\dfrac{39}{4}.\dfrac{130}{7}\right).\dfrac{15}{78}\)
\(C=\dfrac{39}{4}.\left(\dfrac{150}{7}+\dfrac{130}{7}\right).\dfrac{15}{78}\)
\(C=\dfrac{39}{4}.40.\dfrac{15}{78}\)
\(C=390.\dfrac{15}{78}\)
\(C=75\)
1: \(\left(3x-\dfrac{1}{5}\right)^2=\left(-\dfrac{3}{25}\right)^2\)
=>3x-1/5=3/25 hoặc 3x-1/5=-3/25
=>3x=8/25 hoặc 3x=2/25
=>x=8/75 hoặc x=2/75
2: \(\left(2x-\dfrac{1}{3}\right)^2=\left(-\dfrac{2}{9}\right)^2\)
=>2x-1/3=2/9 hoặc 2x-1/3=-2/9
=>2x=5/9 hoặc 2x=1/9
=>x=5/18 hoặc x=1/18
x/186 = (1 - 303030/313131) + (616161/626262 - 1) + (929292/939393 - 1)
x/186 = 1 - 30/31 + 61/62 - 1 + 92/93 - 1
x/186 = 61/62 - 30/31 + 92/93 - 1
x/186 = 1/62 - 1/93
x/186 = 1/186
x = 1