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\(A=\dfrac{14}{8.11}+\dfrac{14}{11.14}+\dfrac{14}{14.17}+.....+\dfrac{14}{197.200}\)
\(A=\dfrac{14}{3}\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)
\(A=\dfrac{14}{3}.\left(\dfrac{1}{8}-\dfrac{1}{200}\right)\)
\(A=\dfrac{14}{3}.\dfrac{24}{200}=\dfrac{28}{25}\)
\(B=\dfrac{7}{15}+\dfrac{7}{35}+\dfrac{7}{63}+...+\dfrac{7}{399}\)
\(B=\dfrac{7}{3.5}+\dfrac{7}{5.7}+\dfrac{7}{7.9}+.....\dfrac{7}{19.21}\)
\(B=\dfrac{7}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+....+\dfrac{1}{19}-\dfrac{1}{21}\right)\)
\(B=\dfrac{7}{2}.\left(\dfrac{1}{3}-\dfrac{1}{21}\right)\)
\(B=\dfrac{7}{2}.\dfrac{6}{21}=1\)
a, \(2-\dfrac{14}{x}=\dfrac{-22}{3}\)
\(\dfrac{14}{x}=2-\dfrac{-22}{3}=\dfrac{28}{3}\)
\(\dfrac{14}{x}=\dfrac{28}{3}\)
=> \(x.28=14.3\)
\(x.28=42\)
\(x=42:28\)
\(x=\dfrac{3}{2}=1,5\)
b, \(\left(\dfrac{2x}{5}+1\right):\left(-7\right)=\dfrac{1}{35}\)
\(\dfrac{2x}{5}+1=\dfrac{1}{35}.\left(-7\right)=-\dfrac{1}{5}\)
\(\dfrac{2x}{5}=-\dfrac{1}{5}-1=\dfrac{-6}{5}\)
\(\dfrac{2x}{5}=\dfrac{-6}{5}\)
=> \(2x=-6\)
\(x=-6:2=-3\)
a)
\(2-\dfrac{14}{x}=-\dfrac{22}{3}\)
\(\Rightarrow\dfrac{14}{x}=2-\dfrac{-22}{3}=\dfrac{28}{3}\)
\(\Rightarrow x=\dfrac{14.3}{28}=\dfrac{3}{2}=1,5\)
b)
\(\left(\dfrac{2x}{5}+1\right):\left(-7\right)=\dfrac{1}{35}\)
\(\Rightarrow\dfrac{2x}{5}+1=\dfrac{1}{35}.\left(-7\right)\)
\(\Rightarrow\dfrac{2x}{5}+1=-\dfrac{1}{5}\)
\(\Rightarrow\dfrac{2x}{5}=-\dfrac{1}{5}-1=-\dfrac{6}{5}\)
Hay \(\dfrac{2x}{5}=-\dfrac{6}{5}\)
\(\Rightarrow2x=-6\)
\(\Rightarrow x=-\dfrac{6}{2}=-3\)
Chúc bạn học tốt!
a) \(\dfrac{1}{2}x-\dfrac{3}{4}x-\dfrac{7}{3}=-\dfrac{5}{6}\\ < =>-\dfrac{1}{4}x-\dfrac{7}{3}=-\dfrac{5}{6}\\ < =>-\dfrac{1}{4}x=-\dfrac{5}{6}+\dfrac{7}{3}=\dfrac{3}{2}\\ =>x=\dfrac{3}{2}:\dfrac{-1}{4}=-6\)
b) \(\left|x-\dfrac{1}{6}\right|+-\dfrac{5}{12}=\dfrac{4}{7}.\dfrac{14}{48}\\ < =>\left|x-\dfrac{1}{6}\right|=\left(\dfrac{4}{7}.\dfrac{14}{48}\right)-\left(-\dfrac{5}{12}\right)=\dfrac{1}{6}+\dfrac{5}{12}=\dfrac{7}{12}\\ \)
Xảy ra 2 trường hợp:
+) TH1: \(x-\dfrac{1}{6}=\dfrac{7}{12}\\ =>x=\dfrac{7}{12}+\dfrac{1}{6}=\dfrac{3}{4}->\left(a\right)\)
+) TH2" \(-\left(x-\dfrac{1}{6}\right)=\dfrac{7}{12}\\ < =>-x+\dfrac{1}{6}=\dfrac{7}{12}\\ < =>-x=\dfrac{7}{12}-\dfrac{1}{6}=\dfrac{5}{12}\\ =>x=-\dfrac{5}{12}->\left(b\right)\)
Từ (a) và (b) => \(x\in\left\{-\dfrac{5}{12};\dfrac{3}{4}\right\}\)
a, \(\dfrac{1}{2}x-\dfrac{3}{4}x-\dfrac{7}{3}=\dfrac{-5}{6}\)
\(\Rightarrow\dfrac{-1}{4}x=\dfrac{-5}{6}+\dfrac{7}{3}\)
\(\Rightarrow\dfrac{-1}{4}x=\dfrac{3}{2}\Rightarrow x=-6\)
Vậy \(x=-6\)
b, \(\left|x-\dfrac{1}{6}\right|+\dfrac{-5}{12}=\dfrac{4}{7}.\dfrac{14}{48}\)
\(\Rightarrow\left|x-\dfrac{1}{6}\right|-\dfrac{5}{12}=\dfrac{1}{6}\)
\(\Rightarrow\left|x-\dfrac{1}{6}\right|=\dfrac{7}{12}\)
\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{6}=\dfrac{-7}{12}\\x-\dfrac{1}{6}=\dfrac{7}{12}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-5}{12}\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{-5}{12};\dfrac{3}{4}\right\}\)
Chúc bạn học tốt!!!
d, Vì B=10^1993+1/10^1992+1 > 1 =>10^1993+1/10^1992+1>10^1993+1+9/10^1992+1+9 = 10^1993+10/10^1992+10= 10. (10^1992+1)/10. (10^1991+1) = 10^1992+1/10^1991+1=A Vậy A=B
cau d B>1 ta co tinh chat (\(\dfrac{a}{b}>\dfrac{a+m}{b+m}\) ) B> \(\dfrac{10^{1993}+1+9}{10^{1992}+1+9}\)\(=\dfrac{10^{1993}+10}{10^{1992}+10}\)=\(\dfrac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}\)=\(\dfrac{10^{1992}+1}{10^{1991}+1}\)=A
Suy ra B>A(chuc ban hoc goi nhe)
a, \(\left(2,8x-32\right):\dfrac{2}{3}=-90\)
\(\Rightarrow2,8x-32=-60\)
\(\Rightarrow2,8x=-28\)
\(\Rightarrow x=-10\)
Vậy x = -10
b, \(\left(4,5-2x\right):1\dfrac{4}{7}=\dfrac{11}{14}\)
\(\Rightarrow\left(4,5-2x\right):\dfrac{11}{7}=\dfrac{11}{14}\)
\(\Rightarrow4,5-2x=\dfrac{121}{98}\)
\(\Rightarrow2x=\dfrac{160}{49}\)
\(\Rightarrow x=\dfrac{80}{49}\)
Vậy \(x=\dfrac{80}{49}\)
\(a,\left(2,8x-32\right):\dfrac{2}{3}=-90\)
\(2,8x-32=-90.\dfrac{2}{3}\)
\(2,8x-32=-60\)
\(2,8x=-60+32\)
\(2,8x=-28\)
\(x=-28:2,8\)
\(x=-10\)
Vậy \(x=-10\)
\(b,\left(4,5-2x\right):1\dfrac{4}{7}=\dfrac{11}{14}\)
\(\left(4,5-2x\right):\dfrac{11}{7}=\dfrac{11}{14}\)
\(4,5-2x=\dfrac{11}{14}.\dfrac{11}{7}\)
\(4,5-2x=\dfrac{121}{98}\)
\(2x=4,5-\dfrac{121}{98}\)
\(2x=\dfrac{160}{49}\)
\(x=\dfrac{160}{49}:2\)
\(x=\dfrac{80}{49}\)
Vậy \(x=\dfrac{80}{49}\)
Thay dấu ba chấm bởi x rồi tìm x.
Chẳng hạn:
\(a) \) \(\dfrac{7}{9}-\dfrac{x}{3}=\dfrac{1}{9}\)
\(\Rightarrow\dfrac{x}{3}=\dfrac{7}{9}-\dfrac{1}{9}\)
\(\Rightarrow\dfrac{x}{3}=\dfrac{6}{9}=\dfrac{2}{3}\)
Vậy x = 2
Đáp số:
a) x = 2
b) x = 3
c) x = 7
d) x =19.
a)x + 3/5 = 1/5 <=> x = 1/5 - 3/5 <=> x = -2/5
b)-1/2 - x = 1/3 - 1/-4 <=> -x = 1/3 + 1/4 + 1/2 <=> -x = 8/24 + 6/24 + 12/24 <=> -x = 26/24 <=> x = -26/24
c)x/14 = 1/7 + -3/14 <=> x/14 = 2/14 + -3/14<=> x/14 = -1/14 <=> x = -1
có j thiếu sót các bạn sửa và bình luận cho mình nha
a) \(x+\dfrac{3}{5}=\dfrac{1}{5}\Rightarrow x=\dfrac{1}{5}-\dfrac{3}{5}=\dfrac{-2}{5}\)
b) \(\dfrac{-1}{2}-x=\dfrac{1}{3}-\dfrac{1}{-4}\rightarrow\dfrac{-1}{2}-x=\dfrac{7}{12}\)\(\Rightarrow x=\dfrac{-1}{2}-\dfrac{7}{12}=\dfrac{-13}{12}\)
c) \(\dfrac{x}{14}=\dfrac{1}{7}+\dfrac{-3}{14}\rightarrow\dfrac{x}{14}=\dfrac{-1}{14}\Rightarrow x=-1\)
Câu a :
Chưa nghĩ ra! Sorry nhé!!
Câu b :
Câu hỏi của Trần Thùy Linh - Toán lớp 6 | Học trực tuyến
Câu c :
Câu hỏi của Trần Thùy Linh - Toán lớp 6 | Học trực tuyến
Vào link đó mà xem, t ngại chép lại
a) \(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)...\left(1-\dfrac{1}{780}\right)\)
\(=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}.....\dfrac{779}{780}\)\(=\)
\(\dfrac{a}{7}+\dfrac{1}{14}=\dfrac{-1}{b}\)
=>\(\dfrac{2a+1}{14}=\dfrac{-1}{b}\)
=>\(\left(2a+1\right)\cdot b=-14\)
mà 2a+1 lẻ
nên \(\left(2a+1\right)\cdot b=1\cdot\left(-14\right)=\left(-1\right)\cdot14=7\cdot\left(-2\right)=\left(-7\right)\cdot2\)
=>\(\left(2a+1;b\right)\in\left\{\left(1;-14\right);\left(-1;14\right);\left(7;-2\right);\left(-7;2\right)\right\}\)
=>\(\left(a,b\right)\in\left\{\left(0;-14\right);\left(-1;14\right);\left(3;-2\right);\left(-4;2\right)\right\}\)