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\(a,5,5-\left|x-0,4\right|=-1\frac{1}{5}\)
\(\Rightarrow5,5-\left|x-0,4\right|=-\frac{6}{5}\)
\(\Rightarrow-\left|x-0,4\right|=-\frac{6}{5}-5,5=-6,7\)
\(\Rightarrow\left|x-0,4\right|=6,7\)
\(\Rightarrow x-0,4=\pm6,7\)
\(\Rightarrow\orbr{\begin{cases}x-0,4=6,7\\x-0,4=-6,7\end{cases}\Rightarrow\orbr{\begin{cases}x=7,1\\x=-6,3\end{cases}}}\)
\(a,5,5-\left|x-0,4\right|=-1\frac{1}{5}\)
=> \(\left|x-0,4\right|=5,5-\left[-\frac{6}{5}\right]=5,5+1,2=6,7\)
=> \(\left|x-0,4\right|=\pm6,7\)
Xét hai trường hợp :
TH1 : x - 0,4 = 6,7
=> x = 6,7 + 0,4 = 7,1
TH2 : x - 0,4 = -6,7
=> x = -6,7 + 0,4 =-6,3
\(b,\left[1-\frac{3}{4}\left|x\right|\right]^2=\frac{16}{25}\)
=> \(\left[1-\frac{3}{4}\left|x\right|\right]=\pm\sqrt{\frac{16}{25}}\)
=> \(\left[1-\frac{3}{4}\left|x\right|\right]=\pm\frac{4}{5}\)
=> \(\orbr{\begin{cases}1-\frac{3}{4}\left|x\right|=\frac{4}{5}\\1-\frac{3}{4}\left|x\right|=-\frac{4}{5}\end{cases}}\)=> \(\orbr{\begin{cases}x=\pm\frac{4}{15}\\x=\pm\frac{12}{5}\end{cases}}\)
\(c,\left[0,1\left|x\right|-\frac{1}{2}\right]\left[0,5-\left|x\right|\right]=0\)
=> \(\orbr{\begin{cases}0,1\left|x\right|-\frac{1}{2}=0\\0,5-\left|x\right|=0\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{1}{10}\left|x\right|=\frac{1}{2}\\\left|x\right|=0,5\end{cases}}\)
=> \(\orbr{\begin{cases}\left|x\right|=5\\\left|x\right|=0,5\end{cases}}\)=> \(\orbr{\begin{cases}x\in\left\{5;-5\right\}\\x\in\left\{0,5;-0,5\right\}\end{cases}}\)
d, Xét hai trường hợp rồi ra kết quả thôi
a) \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
=> \(\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)
=> \(\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}+\frac{x+1}{12}=0\)
=> \(\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)=0\)
=> x + 1 = 0
=> x = -1
b) \(\frac{x-1}{2020}+\frac{x-2}{2019}-\frac{x-3}{2018}=\frac{x-4}{2017}\)
=> \(\left(\frac{x-1}{2020}-1\right)+\left(\frac{x-2}{2019}-1\right)-\left(\frac{x-3}{2018}-1\right)=\left(\frac{x-4}{2017}-1\right)\)
=> \(\frac{x-2021}{2020}+\frac{x-2021}{2019}-\frac{x-2021}{2018}=\frac{x-2021}{2017}\)
=> \(\left(x-2021\right)\left(\frac{1}{2020}+\frac{1}{2019}-\frac{1}{2018}-\frac{1}{2017}\right)=0\)
=> x - 2021 = 0
=> x = 2021
c) \(\left(\frac{3}{4}x+3\right)-\left(\frac{2}{3}x-4\right)-\left(\frac{1}{6}x+1\right)=\left(\frac{1}{3}x+4\right)-\left(\frac{1}{3}x-3\right)\)
=> \(\frac{3}{4}x+3-\frac{2}{3}x+4-\frac{1}{6}x-1=\frac{1}{3}x+4-\frac{1}{3}x+3\)
=> \(-\frac{1}{12}x+6=7\)
=> \(-\frac{1}{12}x=1\)
=> x = -12
làm lần lượt nhá,dài dòng quá khó coi.ahihihi!
\(\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{7\left(\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)
\(=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}=\frac{1}{4}\)
Bài 1 :
\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2018}}{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}{\left(\frac{2017}{1}+1\right)+\left(\frac{2016}{2}+1\right)+\left(\frac{2015}{3}+1\right)+...+\left(\frac{1}{2017}+1\right)+1}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}{\frac{2018}{1}+\frac{2018}{2}+\frac{2018}{3}+....+\frac{2018}{2017}+\frac{2018}{2018}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}{2018.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}+\frac{1}{2018}\right)}\)
\(=\frac{1}{2018}\)
B=\(\frac{\frac{1}{51}+\frac{1}{53}+...+\frac{1}{149}}{\frac{1}{101.99}+\frac{1}{103.97}+...+\frac{1}{149.51}}\)
\(\)TA CÓ E=\(\frac{1}{101.99}+\frac{1}{103.97}+...+\frac{1}{149.51}\)
\(200E=\frac{200}{101.99}+\frac{200}{103.97}+..+\frac{200}{149.51}\)
\(200E=\frac{101+99}{101.99}+\frac{103+97}{103.97}+...+\frac{149+51}{149.51}\)
\(200E=\frac{1}{99}+\frac{1}{101}+\frac{1}{97}+\frac{1}{103}+...+\frac{1}{51}+\frac{1}{149}\)
\(200E=\frac{1}{51}+\frac{1}{53}+...+\frac{1}{147}+\frac{1}{149}\)
\(E=\left(\frac{1}{51}+\frac{1}{53}+...+\frac{1}{147}+\frac{1}{149}\right):200\)\(=\left(\frac{1}{51}+\frac{1}{53}+...+\frac{1}{147}+\frac{1}{149}\right).\frac{1}{200}\)
\(\Rightarrow B=\frac{1}{51}+\frac{1}{53}+...+\frac{1}{149}\)/\(\left(\frac{1}{51}+\frac{1}{53}+..+\frac{1}{149}\right).\frac{1}{200}\)
\(\Rightarrow B=\frac{1}{\frac{1}{200}}=200\)
VẬY B=200
2.
a) Ta có:
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)=\left(x+1\right)\left(\frac{1}{13}+\frac{1}{14}\right)\)
Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\ne\frac{1}{13}+\frac{1}{14}\)nên \(x+1=0\Leftrightarrow x=-1\)
Vậy x = -1
b) Ta có:
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Rightarrow\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)
\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Rightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}\right)=\left(x+2004\right)\left(\frac{1}{2002}+\frac{1}{2003}\right)\)
Vì \(\frac{1}{2000}+\frac{1}{2001}\ne\frac{1}{2002}+\frac{1}{2003}\)nên \(x+2004=0\Leftrightarrow x=-2004\)
Vậy, x = -2004
a)\(\frac{x+3}{x+5}=7\Leftrightarrow x+3=7\left(x+5\right)\)
\(\Leftrightarrow x+3=7x+35\)
\(\Leftrightarrow-6x=32\)
\(\Leftrightarrow x=-\frac{16}{3}\)
b)\(\frac{2x-1}{3x+5}=-\frac{2}{3}\)
\(\Leftrightarrow3\left(2x-1\right)=-2\left(3x+5\right)\)
\(\Leftrightarrow6x-3=-6x-10\)
\(\Leftrightarrow12x=-7\)
\(\Leftrightarrow x=-\frac{7}{12}\)
c)\(\frac{x+1}{4}=\frac{9}{x+1}\Leftrightarrow\left(x+1\right)^2=36\)
\(\Leftrightarrow\left(x+1\right)^2=6^2\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=6\\x+1=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-7\end{cases}}}\)
d)\(\frac{6x-1}{2x+3}=\frac{3x}{x+2}\)
\(\Leftrightarrow\left(6x-1\right)\left(x+2\right)=3x\left(2x+3\right)\)
\(\Leftrightarrow6x^2+12x-x-2=6x^2+9x\)
\(\Leftrightarrow2x=2\Leftrightarrow x=1\)
a: =>x-8/5=1/20-1/10=-1/20
=>x=-0,05+1,6=1,55
b: =>x-3/2=4/3 hoặc x-3/2=-4/3
=>x=17/6 hoặc x=1/6
c: =>\(\left|x-\dfrac{1}{3}\right|=\dfrac{5}{2}-\dfrac{1}{4}+\dfrac{2}{3}=\dfrac{35}{12}\)
=>x-1/3=35/12 hoặc x-1/3=-35/12
=>x=39/12=13/4 hoặc x=-31/12
d: =>|x-5/8|=3/4
=>x-5/8=3/4 hoặc x-5/8=-3/4
=>x=11/8 hoặc x=-1/8