Bài 1 :
\(\frac{x+4}{20}+\frac{x+3}{21}+\frac{x+2}{22}\)\(=\frac{x+1}{23}=-4\)
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HD
1
30 tháng 6 2019
Bài làm
x = \(\frac{20}{21}+\frac{21}{22}+\frac{22}{23}+\frac{23}{20}\)
x = 1 + 1 + 1 + 1 + \((\)\(\frac{3}{20}-\frac{1}{21}-\frac{1}{22}-\frac{1}{23})\)
Ta thấy 0 < \(\frac{3}{20}-\frac{1}{21}-\frac{1}{22}-\frac{1}{23}\)
\(\Rightarrow\) 1 + 1 + 1 + 1 + \((\frac{3}{20}-\frac{1}{21}-\frac{1}{22}-\frac{1}{23})\)> 4
\(\Rightarrow\)x > 4
3 tháng 7 2017
a ) Ta có : \(\frac{x+11}{10}+\frac{x+21}{20}+\frac{x+31}{30}=\frac{x+41}{40}+\frac{x+101}{5}\)
\(\Leftrightarrow\left(\frac{x+11}{10}-1\right)+\left(\frac{x+21}{10}-1\right)+\left(\frac{x+31}{30}-1\right)=\left(\frac{x+41}{40}-1\right)+\left(\frac{x+101}{50}-2\right)\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}=\frac{x+1}{40}+\frac{x+1}{50}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}-\frac{x+1}{40}-\frac{x+1}{50}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)\ne0\)
Nên x + 1 = 0
=> x = -1
DP
24 tháng 6 2017
\(a,\)\(x+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=-\frac{37}{45}\)
\(x+\left(\frac{9-5}{5.9}+\frac{13-9}{9.13}+\frac{17-13}{13.17}+...+\frac{45-41}{41.45}\right)=-\frac{37}{45}\)
\(x+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+....+\frac{1}{41}-\frac{1}{45}\right)-\frac{37}{45}\)
\(x+\left(\frac{1}{5}-\frac{1}{45}\right)=-\frac{37}{45}\)
\(x+\frac{8}{45}=-\frac{37}{45}\)
\(x=-\frac{37}{45}-\frac{8}{45}\)
\(x=-1\)
HK
2
N
11 tháng 11 2018
\(\left(\frac{x}{20}+1\right)+\left(\frac{x-1}{21}+1\right)=\left(\frac{x-2}{22}+1\right)+\left(\frac{x-3}{23}+1\right)\)
\(\frac{x+20}{20}+\frac{x+20}{21}-\frac{x+20}{22}-\frac{x+20}{23}=0\)
\(\left(x+20\right).\left(\frac{1}{20}+\frac{1}{21}-\frac{1}{22}-\frac{1}{23}\right)=0\)
mà \(\left(\frac{1}{20}+\frac{1}{21}-\frac{1}{22}-\frac{1}{23}\right)\ne0\)
=> x+20=0 => x=-20
vậy x=-20
11 tháng 11 2018
\(\frac{x}{20}+\frac{x-1}{21}=\frac{x-2}{22}+\frac{x-3}{23}\)
\(1+\frac{x}{20}+1+\frac{x-1}{21}=1+\frac{x-2}{22}+1+\frac{x-3}{23}\)
\(\frac{x+20}{20}+\frac{21+x-1}{21}=\frac{22+x-2}{22}+\frac{23+x-3}{23}\)
\(\frac{x+20}{20}+\frac{x+20}{21}=\frac{x+20}{22}+\frac{x+20}{23}\)
\(\frac{x+20}{20}+\frac{x+20}{21}-\frac{x+20}{22}-\frac{x+20}{23}=0\)
\(\left(x+20\right)\left(\frac{1}{20}+\frac{1}{21}-\frac{1}{22}-\frac{1}{23}\right)=0\)
Mà \(\frac{1}{20}+\frac{1}{21}-\frac{1}{22}-\frac{1}{23}\ne0\)
\(\Rightarrow x+20=0\)
\(\Rightarrow x=-20\)
Vậy x = -20
14 tháng 9 2019
\(e,\frac{22}{15}-x=-\frac{8}{27}\)
=> \(x=\frac{22}{15}-\left[-\frac{8}{27}\right]\)
=> \(x=\frac{22}{15}+\frac{8}{27}\)
=> \(x=\frac{198}{135}+\frac{40}{135}=\frac{198+40}{135}=\frac{238}{135}\)
\(g,\left[\frac{2x}{5}-1\right]:\left[-5\right]=\frac{1}{4}\)
=> \(\left[\frac{2x}{5}-\frac{1}{1}\right]=\frac{1}{4}\cdot\left[-5\right]\)
=> \(\left[\frac{2x}{5}-\frac{5}{5}\right]=-\frac{5}{4}\)
=> \(\frac{2x-5}{5}=-\frac{5}{4}\)
=> \(2x-5=-\frac{5}{4}\cdot5=-\frac{25}{4}\)
=> \(2x=-\frac{5}{4}\)
=> \(x=-\frac{5}{8}\)
\(h,-2\frac{1}{4}x+9\frac{1}{4}=20\)
=> \(-\frac{9}{4}x+\frac{37}{4}=20\)
=> \(-\frac{9}{4}x=20-\frac{37}{4}=\frac{43}{4}\)
=> \(x=\frac{43}{4}:\left[-\frac{9}{4}\right]=\frac{43}{4}\cdot\left[-\frac{4}{9}\right]=\frac{43}{1}\cdot\left[-\frac{1}{9}\right]=-\frac{43}{9}\)
\(i,-4\frac{3}{5}\cdot2\frac{4}{23}\le x\le-2\frac{3}{5}:1\frac{6}{15}\)
=> \(-\frac{23}{5}\cdot\frac{50}{23}\le x\le-\frac{13}{5}:\frac{21}{15}\)
=> \(-\frac{1}{1}\cdot\frac{10}{1}\le x\le-\frac{13}{5}\cdot\frac{15}{21}\)
=> \(-10\le x\le-\frac{13}{1}\cdot\frac{3}{21}\)
=> \(-10\le x\le-\frac{13}{1}\cdot\frac{1}{7}\)
=> \(-10\le x\le-\frac{13}{7}\)
Đến đây tìm x
TN
27 tháng 2 2020
\(\frac{-5}{9}x+1=\frac{2}{3}x-10\)
\(\frac{-5}{9}x+\frac{9}{9}=\frac{6}{9}x-\frac{90}{9}\)
\(-5x+9=6x-90\)
\(-5x-6x=-90-9\)
\(-11x=-99\)
\(x=\frac{-99}{-11}=9\)
b. \(\frac{x-22}{8}+\frac{x-21}{9}+\frac{x-20}{10}+\frac{x-19}{11}=4\)
\(\frac{x-22}{8}-1+\frac{x-21}{9}-1+\frac{x-20}{10}-1+\frac{x-19}{11}-1=0\)
\(\frac{x-30}{8}+\frac{x-30}{9}+\frac{x-30}{10}+\frac{x-30}{11}=0\)
\(\left(x-30\right)\left(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}\right)=0\)
x=30
Chúc bạn học tốt!!
K
1
LC
6 tháng 11 2017
-Xét \(x\ge y\ge z\). Dễ cm bđt đúng
-Xét \(x\ge z\ge y\)
Đặt x=z+a, z=y+b với \(a,b\ge0\)
=>x=y+a+b
BĐT\(< =>\frac{x-y}{y\left(y+1\right)}\ge\frac{x-z}{x\left(x+1\right)}+\frac{z-x}{z\left(z+1\right)}\)
<=>\(\frac{a+b}{y\left(y+1\right)}\ge\frac{a}{x\left(x+1\right)}+\frac{b}{z\left(z+1\right)}\)
Vì \(x\ge z\ge y=>x\left(x+1\right)\ge z\left(z+1\right)\ge y\left(y+1\right)\)
\(=>\frac{a}{y\left(y+1\right)}\ge\frac{a}{x\left(x+1\right)},\frac{b}{y\left(y+1\right)}\ge\frac{b}{z\left(z+1\right)}\)
=>\(\frac{a+b}{y\left(y+1\right)}\ge\frac{a}{x\left(x+1\right)}+\frac{b}{z\left(z+1\right)}\)=>bđt cần cm đúng=>đpcm
Bác viết nhộn đề gồi :v
\(.\frac{x+4}{20}+\frac{x+3}{21}+\frac{x+2}{22}+\frac{x+1}{23}=-4\)
\(\Rightarrow\frac{x+4}{20}+1+\frac{x+3}{21}+1+\frac{x+2}{22}+1+\frac{x+1}{23}+1=0\)
\(\Rightarrow\frac{x+24}{20}+\frac{x+24}{21}+\frac{x+24}{22}+\frac{x+24}{23}=0\)
\(\Rightarrow\left(x+24\right)\left(\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+\frac{1}{23}\right)=0\)
=> x=-24
\(\frac{x+4}{20}+\frac{x+3}{21}\frac{x+2}{22}+\frac{x+1}{23}\)\(=-4\)
\(\Rightarrow\left(\frac{x+4}{20}+1\right)+\left(\frac{x+3}{21}+1\right)+\left(\frac{x+2}{22}+1\right)\)\(+\left(\frac{x+1}{23}+1\right)=0\)
\(\Rightarrow\left(\frac{x+4}{20}+\frac{20}{20}\right)+\left(\frac{x+3}{21}+\frac{21}{21}\right)\)\(+\left(\frac{x+2}{22}+\frac{22}{22}\right)+\left(\frac{x+1}{23}+\frac{23}{23}\right)=0\)
\(\frac{\Rightarrow x+24}{20}+\frac{x+24}{21}+\frac{x+24}{22}+\frac{x+24}{23}=0\)
\(\Rightarrow\left(x+24\right)+\left(\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+\frac{1}{23}\right)=0\)
Vì \(\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+\frac{1}{23}\ne0\)
\(\Rightarrow x+24=0\)
\(\Rightarrow x=24\)
Chúc bạn học tốt ( -_- )