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Với \(\forall x\) ta có :
+) \(\left|x+\dfrac{1}{2}\right|\ge0\)
+) \(\left|x+\dfrac{1}{6}\right|\ge0\)
..........................
+) \(\left|x+\dfrac{1}{110}\right|\ge0\)
\(\Leftrightarrow\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{6}\right|+.........+\left|x+\dfrac{1}{110}\right|\ge0\)
Mà \(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{6}\right|+........+\left|x+\dfrac{1}{110}\right|=11x\)
\(\Leftrightarrow11x\ge0\)
\(\Leftrightarrow x\ge0\)
Với \(x\ge0\) thì :
+) \(\left|x+\dfrac{1}{2}\right|=x+\dfrac{1}{2}\)
+) \(\left|x+\dfrac{1}{6}\right|=x+\dfrac{1}{6}\)
.....................................
+) \(\left|x+\dfrac{1}{110}\right|=x+\dfrac{1}{110}\)
\(\Leftrightarrow x+\dfrac{1}{2}+x+\dfrac{1}{6}+......+x+\dfrac{1}{110}=11x\)
\(\Leftrightarrow11x+\left(\dfrac{1}{2}+\dfrac{1}{6}+........+\dfrac{1}{110}\right)=11x\)
\(\Leftrightarrow0x=\dfrac{1}{2}+\dfrac{1}{6}+....+\dfrac{1}{110}\) (vô lí)
\(\Leftrightarrow x\in\varnothing\)
\(\frac{2}{1^2}.\frac{6}{2^2}.\frac{10}{3^2}.\frac{20}{4^2}.......\frac{110}{10^2}\left(x-2\right)=-20\left(x+1\right)+60\)
\(\Rightarrow\frac{1.2}{1.1}.\frac{2.3}{2.2}.\frac{3.4}{3.3}.\frac{4.5}{4.4}......\frac{10.11}{10.10}\left(x-2\right)=-20x-20+60\)
\(\Rightarrow\frac{1.2.3.4.....10}{1.2.3.4.....10}.\frac{2.3.4.5.....11}{1.2.3.4.....10}\left(x-2\right)=-20x+40\)
\(\Rightarrow11\left(x-2\right)=-20x+40\)
\(\Rightarrow11x-22=-20x+40\)
\(\Rightarrow11x+20x=22+40\)
\(\Rightarrow31x=62\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
chắc h có mấy thành cay r nên ko làm bn lên mạng tải phẩn mêm có cánh iair đó :D
a, \(\dfrac{3}{5}-4.\left|\dfrac{1}{5}-\dfrac{3}{4}x\right|=\dfrac{1}{3}\)
\(\Rightarrow4\left|\dfrac{1}{5}-\dfrac{3}{4}x\right|=\dfrac{4}{15}\)
\(\Rightarrow\left|\dfrac{1}{5}-\dfrac{3}{4}x\right|=\dfrac{1}{15}\)
\(\Rightarrow\dfrac{1}{5}-\dfrac{3}{4}x\in\left\{-\dfrac{1}{15};\dfrac{1}{15}\right\}\)
\(\Rightarrow\dfrac{3}{4}x\in\left\{\dfrac{4}{15};\dfrac{2}{15}\right\}\Rightarrow x\in\left\{\dfrac{16}{45};\dfrac{8}{45}\right\}\)
b, \(\left|2\dfrac{2}{9}-x\right|=\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\)
\(\Rightarrow\left|2\dfrac{2}{9}-x\right|=\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}\)
\(\Rightarrow\left|2\dfrac{2}{9}-x\right|=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{8}-\dfrac{1}{9}\)
(do \(\dfrac{1}{a.\left(a+1\right)}=\dfrac{1}{a}-\dfrac{1}{a+1}\) với mọi \(a\in N\)*)
\(\Rightarrow\left|2\dfrac{2}{9}-x\right|=\dfrac{1}{3}-\dfrac{1}{9}\)
\(\Rightarrow\left|2\dfrac{2}{9}-x\right|=\dfrac{2}{9}\Rightarrow2\dfrac{2}{9}-x\in\left\{-\dfrac{2}{9};\dfrac{2}{9}\right\}\)
\(\Rightarrow x\in\left\{\dfrac{22}{9};2\right\}\)
c,\(\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)=0\)
\(\Rightarrow\dfrac{1}{3}x+\dfrac{2}{5}x-\dfrac{2}{5}=0\)
\(\Rightarrow\dfrac{11}{15}x=\dfrac{2}{5}\Rightarrow x=\dfrac{6}{11}\)
d, \(60\%x+\dfrac{2}{3}x=\dfrac{1}{3}.6\dfrac{1}{3}\)
\(\Rightarrow\dfrac{3}{5}x+\dfrac{2}{3}x=\dfrac{1}{3}.\dfrac{19}{3}\)
\(\Rightarrow\dfrac{19}{15}x=\dfrac{19}{9}\Rightarrow x=\dfrac{5}{3}\)
Chúc bạn học tốt!!!
1: \(\Leftrightarrow3x+4=2\)
=>3x=-2
=>x=-2/3
2: \(\Leftrightarrow7x-7=6x-30\)
=>x=-23
3: =>\(5x-5=3x+9\)
=>2x=14
=>x=7
4: =>9x+15=14x+7
=>-5x=-8
=>x=8/5
Bạn tham khảo nhé!
Câu hỏi của Hoàng Văn Cam - Toán lớp 6 - Học toán với OnlineMath