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ở hàng thứ 3 tính cả đề, ở phân số thứ 2 trên tử là số 3 ak bn???
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
\(\Leftrightarrow3\left(x^2-4x+4\right)-\dfrac{5}{4}\left(9x^2+6x+1\right)=\dfrac{4}{3}\left(-x^2+4x-3\right)-\dfrac{7}{6}x\left(x-3\right)\)
\(\Leftrightarrow3x^2-12x+12-\dfrac{45}{4}x^2-\dfrac{15}{2}x-\dfrac{5}{4}=-\dfrac{4}{3}x^2+\dfrac{16}{3}x-4-\dfrac{7}{6}x^2+\dfrac{7}{2}x\)
\(\Leftrightarrow x^2\cdot\dfrac{-33}{4}-\dfrac{39}{2}x+\dfrac{43}{4}+\dfrac{5}{2}x^2-\dfrac{53}{6}x+4=0\)
\(\Leftrightarrow x^2\cdot\dfrac{-23}{4}-\dfrac{85}{3}x+\dfrac{59}{4}=0\)
\(\Leftrightarrow12\left(\dfrac{-23}{4}x^2-\dfrac{85}{3}x+\dfrac{59}{4}\right)=0\)
\(\Leftrightarrow-69x^2-340x+177=0\)
\(\Leftrightarrow69x^2+340x-177=0\)
\(\text{Δ}=340^2-4\cdot69\cdot\left(-177\right)=164452\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-170-\sqrt{41113}}{69}\\x_2=\dfrac{-170+\sqrt{41113}}{69}\end{matrix}\right.\)
a)
\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}\)
Vậy \(x = \frac{{ - 2}}{3}\).
b)
\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)
Vậy\(x = \frac{1}{12}\).
c)
\(\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)
Vậy \(x = \frac{7}{3}\).
d)
\(\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}\)
Vậy \(x = \frac{{ - 9}}{{10}}\).
khỏi chép lại đề ha
- 2 - 4x - 5x + \(\frac{3}{2}\)= \(\frac{7}{4}\)
\(\frac{7}{2}\)- 9x = \(\frac{7}{4}\)
-9x = \(\frac{7}{2}-\frac{7}{4}\)
-9x = \(\frac{7}{4}\)
x = \(\frac{7}{4}:\left(-9\right)\)
x = \(\frac{-7}{36}\)
- 3 - 2x - \(\frac{1}{3}=7x-\frac{1}{4}\)
-2x - 7x = \(\frac{-1}{4}-3+\frac{1}{3}\)
-9x = \(\frac{-35}{12}\)
x = \(\frac{-35}{12}:\left(-9\right)\)
x = \(\frac{35}{108}\)
- \(\frac{-15}{2}\)+ \(\frac{1}{4}\)+ 4x -2 = 1
4x = 1 + \(\frac{15}{2}-\frac{1}{4}+2\)
4x = \(\frac{41}{4}\)
x = \(\frac{41}{4}:4\)
x = \(\frac{41}{16}\)