Nguyễn Phương Ly
Giới thiệu về bản thân
a) Các tia chung gốc \(A\) là:
\(A B\) (hay \(A y\)); \(A M\) (hay \(A C\), \(A z\)); \(A x\).
b) Các điểm thuộc tia \(A z\) mà không thuộc tia \(A y\) là:
\(M\) và \(C\).
c) Tia \(A M\) và tia \(M A\) không chung gốc nên không phải hai tia đối nhau.
Số tiền \(15\) quyển vở trước khi giảm giá là:
\(15.7\) \(000 = 105\) \(000\) (đồng)
Số tiền \(15\) quyển vở sau khi giàm giá \(10 \%\) là:
\(105\) \(000.90 \% = 94\) \(500\) (đồng)
Vậy bạn An đem theo \(100\) \(000\) đồng nên đủ tiền mua \(15\) quyển vở.
a) \(\frac{3}{8} - \frac{1}{6} x\) \(= \frac{1}{4}\)
\(\frac{3}{8} - \frac{1}{6} x\) | \(= \frac{1}{4}\) | ||||||||||
\(\frac{1}{6} x\) | \(= \frac{3}{8} - \frac{2}{8}\) | ||||||||||
\(\frac{1}{6} x\) | \(= \frac{1}{8}\) | ||||||||||
\(x\) | \(= \frac{1}{8} : \frac{1}{6}\) | ||||||||||
\(x\) | a) \(\frac{3}{8} - \frac{1}{6} x\) \(= \frac{1}{4}\)
Vậy \(x = \frac{3}{4}\). b) \(\left(\left(\right. x - 1 \left.\right)\right)^{2} = \frac{1}{4}\) Suy ra \(\left[\right. & \left(\left(\right. x - 1 \left.\right)\right)^{2} = \left(\left(\right. \frac{1}{2} \left.\right)\right)^{2} \\ & \left(\left(\right. x - 1 \left.\right)\right)^{2} = \left(\left(\right. \frac{- 1}{2} \left.\right)\right)^{2}\) hay \(\left[\right. & x - 1 = \frac{1}{2} \&\text{nbsp}; \\ & x - 1 = \frac{- 1}{2} \&\text{nbsp};\) \(\left[\right. & x = \frac{1}{2} + 1 \&\text{nbsp}; \\ & x = \frac{- 1}{2} + 1 \&\text{nbsp};\) suy ra \(\left[\right. & x = \frac{3}{2} \&\text{nbsp}; \\ & x = \frac{1}{2} \&\text{nbsp};\) Vậy \(x \in \left{\right. \frac{3}{2} ; \frac{1}{2} \left.\right}\). c) \(\left(\right. x - \frac{- 1}{2} \left.\right) . \left(\right. x + \frac{1}{3} \left.\right) = 0\). Suy ra \(\left[\right. & x - \frac{- 1}{2} = 0 \\ & x + \frac{1}{3} = 0\) hay \(\left[\right. & x = \frac{- 1}{2} \&\text{nbsp}; \\ & x = \frac{- 1}{3} \&\text{nbsp};\) Vậy \(x \in \left{\right. \frac{- 1}{2} ; \frac{- 1}{3} \left.\right}\). |
a) \(\frac{3}{8} - \frac{1}{6} x\) \(= \frac{1}{4}\)
\(\frac{3}{8} - \frac{1}{6} x\) | \(= \frac{1}{4}\) | ||||||||||
\(\frac{1}{6} x\) | \(= \frac{3}{8} - \frac{2}{8}\) | ||||||||||
\(\frac{1}{6} x\) | \(= \frac{1}{8}\) | ||||||||||
\(x\) | \(= \frac{1}{8} : \frac{1}{6}\) | ||||||||||
\(x\) | a) \(\frac{3}{8} - \frac{1}{6} x\) \(= \frac{1}{4}\)
Vậy \(x = \frac{3}{4}\). b) \(\left(\left(\right. x - 1 \left.\right)\right)^{2} = \frac{1}{4}\) Suy ra \(\left[\right. & \left(\left(\right. x - 1 \left.\right)\right)^{2} = \left(\left(\right. \frac{1}{2} \left.\right)\right)^{2} \\ & \left(\left(\right. x - 1 \left.\right)\right)^{2} = \left(\left(\right. \frac{- 1}{2} \left.\right)\right)^{2}\) hay \(\left[\right. & x - 1 = \frac{1}{2} \&\text{nbsp}; \\ & x - 1 = \frac{- 1}{2} \&\text{nbsp};\) \(\left[\right. & x = \frac{1}{2} + 1 \&\text{nbsp}; \\ & x = \frac{- 1}{2} + 1 \&\text{nbsp};\) suy ra \(\left[\right. & x = \frac{3}{2} \&\text{nbsp}; \\ & x = \frac{1}{2} \&\text{nbsp};\) Vậy \(x \in \left{\right. \frac{3}{2} ; \frac{1}{2} \left.\right}\). c) \(\left(\right. x - \frac{- 1}{2} \left.\right) . \left(\right. x + \frac{1}{3} \left.\right) = 0\). Suy ra \(\left[\right. & x - \frac{- 1}{2} = 0 \\ & x + \frac{1}{3} = 0\) hay \(\left[\right. & x = \frac{- 1}{2} \&\text{nbsp}; \\ & x = \frac{- 1}{3} \&\text{nbsp};\) Vậy \(x \in \left{\right. \frac{- 1}{2} ; \frac{- 1}{3} \left.\right}\). |
a) \(1 - \frac{1}{2} + \frac{1}{3} = \frac{6 - 3 + 2}{6} = \frac{5}{6}\).
b) \(\frac{2}{5} + \frac{3}{5} : \frac{9}{10} = \frac{2}{5} + \frac{3}{5} \cdot \frac{10}{9} = \frac{2}{5} + \frac{2}{3} = \frac{16}{15}\).
c) \(\frac{7}{11} \cdot \frac{3}{4} + \frac{7}{11} \cdot \frac{1}{4} + \frac{4}{11} = \frac{7}{11} \left(\right. \frac{3}{4} + \frac{1}{4} \left.\right) + \frac{4}{11} = \frac{7}{11} + \frac{4}{11} = 1\).
d) \(\left(\right. \frac{3}{4} + 0 , 5 + 25 \% \left.\right) \cdot 2 \frac{2}{3} = \left(\right. \frac{3}{4} + \frac{1}{2} + \frac{1}{4} \left.\right) \cdot \frac{8}{3} = \frac{3}{2} \cdot \frac{8}{3} = 4\).