✅ Kết quả tính được:

\(x \approx - 512,78\)

\(\left(\frac{x}{20}+1\right)+\left(\frac{x-1}{21}+1\right)=\left(\frac{x-2}{22}+1\right)+\left(\frac{x-3}{23}+1\right)\)

\(\frac{x+20}{20}+\frac{x+20}{21}-\frac{x+20}{22}-\frac{x+20}{23}=0\)

\(\left(x+20\right).\left(\frac{1}{20}+\frac{1}{21}-\frac{1}{22}-\frac{1}{23}\right)=0\)

mà \(\left(\frac{1}{20}+\frac{1}{21}-\frac{1}{22}-\frac{1}{23}\right)\ne0\)

=> x+20=0 => x=-20

vậy x=-20

\(\frac{x}{20}+\frac{x-1}{21}=\frac{x-2}{22}+\frac{x-3}{23}\)

\(1+\frac{x}{20}+1+\frac{x-1}{21}=1+\frac{x-2}{22}+1+\frac{x-3}{23}\)

\(\frac{x+20}{20}+\frac{21+x-1}{21}=\frac{22+x-2}{22}+\frac{23+x-3}{23}\)

\(\frac{x+20}{20}+\frac{x+20}{21}=\frac{x+20}{22}+\frac{x+20}{23}\)

\(\frac{x+20}{20}+\frac{x+20}{21}-\frac{x+20}{22}-\frac{x+20}{23}=0\)

\(\left(x+20\right)\left(\frac{1}{20}+\frac{1}{21}-\frac{1}{22}-\frac{1}{23}\right)=0\)

Mà \(\frac{1}{20}+\frac{1}{21}-\frac{1}{22}-\frac{1}{23}\ne0\)

\(\Rightarrow x+20=0\)

\(\Rightarrow x=-20\)

Vậy x = -20

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

\(\frac{3x+25}{144}=\frac{2y-169}{25}=\frac{z+144}{169}=\frac{3x+2y+z}{338}=\frac{169}{338}=\frac{1}{2}\)

\(\Rightarrow3x+25=\frac{1}{2}.144=72\)

\(\Leftrightarrow x=\frac{47}{3}\)

\(2y-169=\frac{1}{2}.25=\frac{25}{2}\)

\(\Leftrightarrow y=\frac{363}{4}\)

\(z+144=\frac{1}{2}.169=\frac{169}{2}\)

\(\Leftrightarrow z=\frac{-119}{2}\)

Áp dụng tính chất dãy tỉ số bằng nhau

\(\frac{3x+25}{144}=\frac{2y-169}{25}=\frac{z+144}{169}=\frac{\left(3x+2y+z\right)+\left(25-169+144\right)}{144+25+169}=\frac{169+25-169+144}{144+25+169}=\)

\(\frac{1}{2}\)

Ta có

\(\frac{3x+25}{144}=\frac{1}{2}\Rightarrow6x+50=144\Rightarrow6x=94\Rightarrow x=\frac{47}{3}\)

\(\frac{2y-169}{25}=\frac{1}{2}\Rightarrow4y-338=25\Rightarrow4y=363\Rightarrow y=\frac{363}{4}\)

\(\frac{z+144}{169}=\frac{1}{2}\Rightarrow2z+288=169\Rightarrow2z=-119\Rightarrow z=\frac{-119}{2}\)

\(\dfrac{148-x}{25}+\dfrac{169-x}{23}+\dfrac{186-x}{21}+\dfrac{199-x}{19}=10\)

\(\Leftrightarrow\left(\dfrac{148-x}{25}-1\right)+\left(\dfrac{169-x}{23}-2\right)+\left(\dfrac{186-x}{21}-3\right)+\left(\dfrac{199-x}{19}-4\right)=0\)

\(\Leftrightarrow\dfrac{123-x}{25}+\dfrac{123-x}{23}+\dfrac{123-x}{21}+\dfrac{123-x}{19}=0\)

\(\Leftrightarrow\left(123-x\right)\left(\dfrac{1}{25}+\dfrac{1}{23}+\dfrac{1}{21}+\dfrac{1}{19}\right)=0\)

\(\Leftrightarrow123-x=0\Leftrightarrow x=123\)

Vậy x = 123

thanks you

Tính chất về giá trị tuyệt đối ứng dụng để tìm giá trị lớn nhất, giá trị nhỏ nhất: 

\(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) dấu bằng xảy ra khi và chỉ khi  \(ab\ge0\)

+) A=|x-2012|+|2011-x|\(\ge\)|x-2012+2011-x|=1

min A=1 <=> (x-2012)(2011-x)\(\ge0\)<=> \(2011\le x\le2012\)(lập bảng xét dấu )

+) B=\(\frac{21.\left|4x+6\right|+23}{3\left|4x+6\right|+5}=\frac{7\left(3\left|4x+6\right|+5\right)-35+23}{3\left|4x+6\right|+5}=7-\frac{12}{3\left|4x+6\right|+5}\)

\(\left|4x+6\right|\ge0\Rightarrow3\left|4x+6\right|+5\ge5\Rightarrow\frac{12}{3\left|4x+6\right|+5}\le\frac{12}{5}\)

\(\Rightarrow-\frac{12}{3\left|4x+6\right|+5}\ge-\frac{12}{5}\Rightarrow7-\frac{12}{3\left|4x+6\right|+5}\ge7-\frac{12}{5}=\frac{23}{5}\)

Vậy \(A\ge\frac{23}{5}\)

min A=23/5 <=> 4x+6=0 <=> x=-3/2

ai bít

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

còn b vs c thì sao ạ

Hnay đi học, cô giáo có sửa cho bạn bài đó hong dọ, do cô mình giao cái bài về nhà  y sì dãy í, mà mai nộp ròi, nhưng mình k biết làm, nếu bạn biết , chỉ mình với :((