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\(2\left(\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1}{9}\)
\(\Leftrightarrow2\left(\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1}{9}\)
\(\Leftrightarrow2\left(\frac{1}{9}-\frac{1}{x+1}\right)=\frac{1}{9}\)
\(\Leftrightarrow\frac{1}{9}-\frac{1}{x+1}=\frac{1}{9}\div2\)
\(\Leftrightarrow\frac{1}{9}-\frac{1}{x+1}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{9}-\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{18}\)
\(\Leftrightarrow x+1=18\)
\(\Leftrightarrow x=18-1\)
\(\Leftrightarrow x=17\)
\(\left|x\right|-\frac{3}{4}=\frac{5}{3}\)
\(\Leftrightarrow\left|x\right|=\frac{5}{3}+\frac{3}{4}\)
\(\Leftrightarrow\left|x\right|=\frac{20}{12}+\frac{9}{12}\)
\(\Leftrightarrow\left|x\right|=\frac{29}{12}\)
\(\Leftrightarrow x=\pm\frac{29}{12}\)
a) \(1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}..1\frac{1}{99}=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}....\frac{100}{99}=\frac{2.2.3.3.4.4...10.10}{1.3.2.4.3.5...9.11}=\frac{\left(2.3.4...10\right)\left(2.3.4...10\right)}{\left(1.2.3...9\right)\left(3.4.5...11\right)}\)
\(\frac{10.2}{1.11}=\frac{20}{11}\)
b) \(\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right).\left(1-\frac{1}{25}\right).\left(1-\frac{1}{36}\right)=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}.\frac{35}{36}\)
\(=\frac{1.3.2.4.3.5.4.6.5.7}{2.2.3.3.4.4.5.5.6.6}=\frac{\left(1.2.3.4.5\right).\left(3.4.5.6.7\right)}{\left(2.3.4.5.6\right).\left(2.3.4.5.6\right)}=\frac{1.7}{6.2}=\frac{7}{12}\)
c) \(\frac{99}{98}-\frac{98}{97}+\frac{1}{97.98}=\frac{99}{98}-\frac{98}{97}+\frac{1}{97}-\frac{1}{98}=\left(\frac{99}{98}-\frac{1}{98}\right)+\left(-\frac{98}{97}+\frac{1}{97}\right)=1-1=0\)
d) \(3\frac{1}{11}.\frac{27}{36}.1\frac{6}{7}.2\frac{4}{9}=\frac{34}{11}.\frac{3}{4}.\frac{13}{7}.\frac{22}{9}=\frac{34.3.13.22}{11.4.7.9}=\frac{34.13}{11.2.7.3}=\frac{442}{462}=\frac{221}{231}\)
\(\frac{1}{10\cdot11}+\frac{1}{11\cdot12}+.......+\frac{1}{19\cdot20}\)\(=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+.....+\frac{1}{18}-\frac{1}{19}\)\(+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{10}-\frac{1}{20}=\frac{2}{20}-\frac{1}{20}=\frac{1}{20}\)
K CHO MÌNH NHA CÁC BẠN
a. Ta có :
B = 308/1 + 307/2 +306/3+....+1/308
B = (1+1+....+1) + 307/2 + ....+ 1/308
B = (1 + 307/2) + (1+306/3) + ...+ (1+ 1/308) + 1
B = 309/2 + 309/3 + ....+ 309/308 + 309/309
B = 309.(1/2 + 1/3 + ....+1/309)
Vậy A/B: 1/2 + 1/3 + ... + 1/309 / 308/1 + 307/2 +....+ 2/307+1/308
A/B = 1/2 + 1/3 +... + 1/309 / 309.(1/2 + 1/3 + ....+1/309)
A/B = 1/309
b.7/10.11 + 7/11.12 + .... +7 /69.70
= 7. (1/10.11+1/11.12 + ...+ 1/69.70)
= 7.(1/10-1/11+1/11-1/12+....+1/69-1/70)
= 7.(1/10 - 1/70)
= 7. 3/35
= 3/5
ở câu 1 ở mỗi phẫn số chúng ta cộng thêm 1, tổng là ta cộng thêm 5. Lấy 5 + -5=0. Rồi ta được tất cả tử là x+200,đặt chung ra ngoài,từ đó tính x=-200
a) Đặt \(A=\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+.....+\frac{1}{\left(x-2\right)x}+\frac{1}{x\left(x+2\right)}\)
=> \(3A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+.....+\frac{3}{\left(x-2\right)x}+\frac{3}{x\left(x+2\right)}\)
=> \(3A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+....+\frac{1}{\left(x-2\right)}-\frac{1}{x}+\frac{1}{x}-\frac{1}{x+2}\)
=> 3A = \(\frac{1}{5}-\frac{1}{x+2}\)
=> A = \(\frac{1}{15}-\frac{1}{3x+6}\)
Mà : A = \(\frac{101}{1540}\)
=> \(\frac{1}{15}-\frac{1}{3x+6}=\frac{101}{1540}\)
=> \(\frac{1}{3x+6}=\frac{1}{15}-\frac{101}{1540}=\frac{1}{924}\)
=> 3x + 6 = 924
=> 3(x + 2) = 924
=> x + 2 = 308
=> x = 306
a) Ta có: \({{1} \over x(x+2)}= {{1} \over 3}({{1} \over x}-{{1} \over x+2})\) \(\Rightarrow\) \({{1} \over 3}({{1} \over 5}-{{1} \over 8}+{{1} \over 8}-...+{{1} \over x}-{{1} \over x+2})={{101} \over 1540} \)\(\Leftrightarrow\) \({{1} \over 3}({{1} \over 5}-{{1} \over x+2})={{101} \over 1540}\)\(\Leftrightarrow\)x+2 = 308 \(\Leftrightarrow\) x=306 Lúc sau lm hơi tắt mọi người thông cảm
\(a,\frac{1}{2}+\frac{2}{3}x=\frac{4}{5}\)
=> \(\frac{2}{3}x=\frac{4}{5}-\frac{1}{2}=\frac{3}{10}\)
=> \(x=\frac{3}{10}:\frac{2}{3}=\frac{9}{20}\)
Vậy \(x\in\left\{\frac{9}{20}\right\}\)
\(b,x+\frac{1}{4}=\frac{4}{3}\)
=> \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\)
Vậy \(x\in\left\{\frac{13}{12}\right\}\)
\(c,\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
=> \(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}=\frac{5}{14}\)
=> \(x=\frac{5}{14}:\frac{3}{5}=\frac{25}{42}\)
Vậy \(x\in\left\{\frac{25}{42}\right\}\)
\(d,\left|x+5\right|-6=9\)
=> \(\left|x+5\right|=9+6=15\)
=> \(\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=15-5=10\\x=-15-5=-20\end{matrix}\right.\)
Vậy \(x\in\left\{10;-20\right\}\)
\(e,\left|x-\frac{4}{5}\right|=\frac{3}{4}\)
=> \(\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{4}\\x-\frac{4}{5}=-\frac{3}{4}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{3}{4}+\frac{4}{5}=\frac{31}{20}\\x=-\frac{3}{4}+\frac{4}{5}=\frac{1}{20}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{31}{20};\frac{1}{20}\right\}\)
\(f,\frac{1}{2}-\left|x\right|=\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{2}-\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{6}\)
=> \(\left[{}\begin{matrix}x=\frac{1}{6}\\x=-\frac{1}{6}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{6};-\frac{1}{6}\right\}\)
\(g,x^2=16\)
=> \(\left|x\right|=\sqrt{16}=4\)
=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
vậy \(x\in\left\{4;-4\right\}\)
\(h,\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
=> \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\)
=> \(x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)
Vậy \(x\in\left\{\frac{5}{6}\right\}\)
\(i,3^3.x=3^6\)
\(x=3^6:3^3=3^3=27\)
Vậy \(x\in\left\{27\right\}\)
\(J,\frac{1,35}{0,2}=\frac{1,25}{x}\)
=> \(x=\frac{1,25.0,2}{1,35}=\frac{5}{27}\)
Vậy \(x\in\left\{\frac{5}{27}\right\}\)
\(k,1\frac{2}{3}:x=6:0,3\)
=> \(\frac{5}{3}:x=20\)
=> \(x=\frac{5}{3}:20=\frac{1}{12}\)
Vậy \(x\in\left\{\frac{1}{12}\right\}\)
Lưu ý : . là dấu nhân
ai làm đúng , nhanh mình k