Question

Đã trả lời bởi chuyên gia

Answer 1

$\frac{- 1}{9} + {(x - \frac{5}{6})}^{2} = \frac{1}{3} \Leftrightarrow {(x - \frac{5}{6})}^{2} = \frac{1}{3} + \frac{1}{9} \Leftrightarrow {(x - \frac{5}{6})}^{2} = \frac{4}{9} \Leftrightarrow {(x - \frac{5}{6})}^{2} = {(\frac{2}{3})}^{2} \Leftrightarrow x - \frac{5}{6} = \frac{2}{3} h o ặ c x - \frac{5}{6} = \frac{- 2}{3} \Leftrightarrow x = \frac{3}{2} h o ặ c x = \frac{1}{6}$

$x + \frac{5}{8} x - \frac{12}{16} x = 1 \Leftrightarrow x (1 + \frac{5}{8} - \frac{12}{16}) = 1 \Leftrightarrow x . \frac{7}{8} = 1 \Leftrightarrow x = \frac{8}{7}$

$\frac{1}{3} - {(x + \frac{1}{2})}^{2} = \frac{- 6}{9} \Leftrightarrow {(x + \frac{1}{2})}^{2} = 1 \Leftrightarrow x + \frac{1}{2} = 1 h o ặ c x + \frac{1}{2} = - 1 \Leftrightarrow x = \frac{1}{2} h o ặ c x = \frac{- 3}{2}$ ..

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Answer 2

$\frac{- 1}{9} + {(x - \frac{5}{6})}^{2} = \frac{1}{3} \Leftrightarrow {(x - \frac{5}{6})}^{2} = \frac{1}{3} + \frac{1}{9} \Leftrightarrow {(x - \frac{5}{6})}^{2} = \frac{4}{9} \Leftrightarrow {(x - \frac{5}{6})}^{2} = {(\frac{2}{3})}^{2} \Leftrightarrow x - \frac{5}{6} = \frac{2}{3} h o ặ c x - \frac{5}{6} = \frac{- 2}{3} \Leftrightarrow x = \frac{3}{2} h o ặ c x = \frac{1}{6}$

$x + \frac{5}{8} x - \frac{12}{16} x = 1 \Leftrightarrow x (1 + \frac{5}{8} - \frac{12}{16}) = 1 \Leftrightarrow x . \frac{7}{8} = 1 \Leftrightarrow x = \frac{8}{7}$

$\frac{1}{3} - {(x + \frac{1}{2})}^{2} = \frac{- 6}{9} \Leftrightarrow {(x + \frac{1}{2})}^{2} = 1 \Leftrightarrow x + \frac{1}{2} = 1 h o ặ c x + \frac{1}{2} = - 1 \Leftrightarrow x = \frac{1}{2} h o ặ c x = \frac{- 3}{2}$ ..

Answer 3

d,

$\frac{- 1}{9} + {(x - \frac{5}{6})}^{2} = \frac{1}{3} \Leftrightarrow {(x - \frac{5}{6})}^{2} = \frac{4}{9} \Rightarrow \begin{matrix} x - \frac{5}{6} = \frac{2}{3} \\ h o ặ c \\ x - \frac{5}{6} = \frac{- 2}{3} \end{matrix} \Leftrightarrow \begin{matrix} x = \frac{3}{2} \\ h o ặ c \\ x = \frac{1}{6} \end{matrix} e, x + \frac{5}{8} x - \frac{12}{16} x = 1 \Leftrightarrow x . (1 + \frac{5}{8} - \frac{12}{16}) = 1 \Leftrightarrow x . \frac{7}{8} = 1 \Rightarrow x = \frac{8}{7} f, \frac{1}{3} - {(x + \frac{1}{2})}^{2} = \frac{- 6}{9} \Leftrightarrow {(x + \frac{1}{2})}^{2} = 1 \Rightarrow \begin{matrix} x + \frac{1}{2} = 1 \\ h o ặ c \\ x + \frac{1}{2} = - 1 \end{matrix} \Leftrightarrow \begin{matrix} x = \frac{1}{2} \\ h o ặ c \\ x = \frac{- 3}{2} \end{matrix}$

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Answer 4

d,

$\frac{- 1}{9} + {(x - \frac{5}{6})}^{2} = \frac{1}{3} \Leftrightarrow {(x - \frac{5}{6})}^{2} = \frac{4}{9} \Rightarrow \begin{matrix} x - \frac{5}{6} = \frac{2}{3} \\ h o ặ c \\ x - \frac{5}{6} = \frac{- 2}{3} \end{matrix} \Leftrightarrow \begin{matrix} x = \frac{3}{2} \\ h o ặ c \\ x = \frac{1}{6} \end{matrix} e, x + \frac{5}{8} x - \frac{12}{16} x = 1 \Leftrightarrow x . (1 + \frac{5}{8} - \frac{12}{16}) = 1 \Leftrightarrow x . \frac{7}{8} = 1 \Rightarrow x = \frac{8}{7} f, \frac{1}{3} - {(x + \frac{1}{2})}^{2} = \frac{- 6}{9} \Leftrightarrow {(x + \frac{1}{2})}^{2} = 1 \Rightarrow \begin{matrix} x + \frac{1}{2} = 1 \\ h o ặ c \\ x + \frac{1}{2} = - 1 \end{matrix} \Leftrightarrow \begin{matrix} x = \frac{1}{2} \\ h o ặ c \\ x = \frac{- 3}{2} \end{matrix}$

Answer 5

ta có:

$d . \frac{- 1}{9} + {(x - \frac{5}{6})}^{2} = \frac{1}{3} = > {(x - \frac{5}{6})}^{2} = \frac{1}{3} + \frac{1}{9} = > {(x - \frac{5}{6})}^{2} = \frac{4}{9} = > x - \frac{5}{6} = \frac{2}{3} h o ặ c x - \frac{5}{6} = \frac{- 2}{3} = > x = \frac{3}{2} h o ặ c x = \frac{1}{6} e . x + \frac{5}{8} x - \frac{12}{16} x = 1 = > \frac{7}{8} x = 1 = > x = 1 : \frac{7}{8} = > x = \frac{8}{7} f . \frac{1}{3} - {(x + \frac{1}{2})}^{2} = \frac{- 6}{9} = > {(x + \frac{1}{2})}^{2} = \frac{1}{3} + \frac{6}{9} = > {(x + \frac{1}{2})}^{2} = 1 = > x + \frac{1}{2} = 1 h o ặ c x + \frac{1}{2} = - 1 = > x = \frac{1}{2} h o ặ c x = \frac{- 3}{2}$

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Answer 6

ta có:

$d . \frac{- 1}{9} + {(x - \frac{5}{6})}^{2} = \frac{1}{3} = > {(x - \frac{5}{6})}^{2} = \frac{1}{3} + \frac{1}{9} = > {(x - \frac{5}{6})}^{2} = \frac{4}{9} = > x - \frac{5}{6} = \frac{2}{3} h o ặ c x - \frac{5}{6} = \frac{- 2}{3} = > x = \frac{3}{2} h o ặ c x = \frac{1}{6} e . x + \frac{5}{8} x - \frac{12}{16} x = 1 = > \frac{7}{8} x = 1 = > x = 1 : \frac{7}{8} = > x = \frac{8}{7} f . \frac{1}{3} - {(x + \frac{1}{2})}^{2} = \frac{- 6}{9} = > {(x + \frac{1}{2})}^{2} = \frac{1}{3} + \frac{6}{9} = > {(x + \frac{1}{2})}^{2} = 1 = > x + \frac{1}{2} = 1 h o ặ c x + \frac{1}{2} = - 1 = > x = \frac{1}{2} h o ặ c x = \frac{- 3}{2}$

Số thứ tự	Loại hàng	Số lượng (quyển)	Giá đơn vị (đồng)	Tổng số tiền (đồng)
1	Vở loại 1	35	2000	...
2	Vở loại 2	42	1500	...
3	Vở loại 3	38	1200	...
Cộng:				...

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