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Задача Задача №44272
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13 ноября 2023 19:23 · публичный вопрос
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4 7/32x-(3 3/4x - 1 7/32x)

Нейросеть13 ноября 2023 19:23

To simplify the expression \( 4 \frac{7}{32}x - (3 \frac{3}{4}x - 1 \frac{7}{32}x) \), we first need to handle the operations inside the parenthesis, and then we can simplify the remaining part of the expression. Let's break it down step by step:

Firstly, we need to convert the mixed numbers to improper fractions to make the calculations easier:

\( 4 \frac{7}{32}x \) is \( 4 \times 32 + 7 = 128 + 7 = 135/32 \) when converted to an improper fraction, so: \( 4 \frac{7}{32}x = \frac{135}{32}x \)

\( 3 \frac{3}{4}x \) is \( 3 \times 4 + 3 = 12 + 3 = 15/4 \) when converted to an improper fraction, so: \( 3 \frac{3}{4}x = \frac{15}{4}x \)

Since \( 1 \frac{7}{32}x \) is already a part of the mixed number with the denominator 32, we'll keep it as is: \( 1 \frac{7}{32}x = \frac{39}{32}x \)

Now, let's substitute these values back into the equation:

\( \frac{135}{32}x - (\frac{15}{4}x - \frac{39}{32}x) \)

We need to distribute the negative sign inside the parentheses:

\( \frac{135}{32}x - \frac{15}{4}x + \frac{39}{32}x \)

To combine these terms, we'll need a common denominator. The common denominator between 4 and 32 is 32, so we will convert \( \frac{15}{4} \) to an equivalent fraction with the denominator of 32:

\( \frac{15}{4} \) is equivalent to \( \frac{15 \times 8}{4 \times 8} = \frac{120}{32} \)

Now we have:

\( \frac{135}{32}x - \frac{120}{32}x + \frac{39}{32}x \)

Combining the terms we get:

\( (\frac{135}{32} - \frac{120}{32} + \frac{39}{32})x = (\frac{135 - 120 + 39}{32})x = (\frac{54}{32})x \)

Now we can simplify \( \frac{54}{32} \) by dividing both numerator and denominator by 2 (their greatest common divisor):

\( \frac{54}{32} = \frac{27}{16} \)

So, the simplified expression is:

\( \frac{27}{16}x \)

We can leave the expression in this form, or you could convert it back to a mixed number if required:

\( \frac{27}{16}x = 1 \frac{11}{16}x \)

So the final simplified expression can be written as either:

\( 1 \frac{11}{16}x \) or \( \frac{27}{16}x \)

4 7/32x-(3 3/4x - 1 7/32x)
To simplify the expression \( 4 \frac{7}{32}x - (3 \frac{3}{4}x - 1 \frac{7}{32}x) \), we first need to handle the operations inside the parenthesis, and then we can simplify the remaining part of the expression. Let's break it down step by step: Firstly, we need to convert the mixed numbers to improper fractions to make the calculations easier: \( 4 \frac{7}{32}x \) is \( 4 \times 32 + 7 = 128 + 7 = 135/32 \) when converted to an improper fraction, so: \( 4 \frac{7}{32}x = \frac{135}{32}x \) \( 3 \frac{3}{4}x \) is \( 3 \times 4 + 3 = 12 + 3 = 15/4 \) when converted to an improper fraction, so: \( 3 \frac{3}{4}x = \frac{15}{4}x \) Since \( 1 \frac{7}{32}x \) is already a part of the mixed number with the denominator 32, we'll keep it as is: \( 1 \frac{7}{32}x = \frac{39}{32}x \) Now, let's substitute these values back into the equation: \( \frac{135}{32}x - (\frac{15}{4}x - \frac{39}{32}x) \) We need to distribute the negative sign inside the parentheses: \( \frac{135}{32}x - \frac{15}{4}x + \frac{39}{32}x \) To combine these terms, we'll need a common denominator. The common denominator between 4 and 32 is 32, so we will convert \( \frac{15}{4} \) to an equivalent fraction with the denominator of 32: \( \frac{15}{4} \) is equivalent to \( \frac{15 \times 8}{4 \times 8} = \frac{120}{32} \) Now we have: \( \frac{135}{32}x - \frac{120}{32}x + \frac{39}{32}x \) Combining the terms we get: \( (\frac{135}{32} - \frac{120}{32} + \frac{39}{32})x = (\frac{135 - 120 + 39}{32})x = (\frac{54}{32})x \) Now we can simplify \( \frac{54}{32} \) by dividing both numerator and denominator by 2 (their greatest common divisor): \( \frac{54}{32} = \frac{27}{16} \) So, the simplified expression is: \( \frac{27}{16}x \) We can leave the expression in this form, or you could convert it back to a mixed number if required: \( \frac{27}{16}x = 1 \frac{11}{16}x \) So the final simplified expression can be written as either: \( 1 \frac{11}{16}x \) or \( \frac{27}{16}x \)