Question

1) 3/5+4 1/5×(4/21-3/14)-3/5=2) |7x-5|+4×|5-7x|=​

Answer 1

Ответ:

$\displaystyle \bold 1) \; -\dfrac{1}{10}$

2) 5·|5-7·x|

Пошаговое объяснение:

$\displaystyle \bold 1) \; \dfrac{3}{5} +4\dfrac{1}{5} \cdot (\dfrac{4}{21}-\dfrac{3}{14})-\dfrac{3}{5}=-\dfrac{1}{10}.$

Изменим расположения дробей и получим:

$\displaystyle \bold \; \dfrac{3}{5}-\dfrac{3}{5} +4\dfrac{1}{5} \cdot (\dfrac{4}{21}-\dfrac{3}{14})=4\dfrac{1}{5} \cdot (\dfrac{4}{21}-\dfrac{3}{14})=-\dfrac{1}{10}.$

$\displaystyle \bold 1. \; \dfrac{4}{21}-\dfrac{3}{14}=\dfrac{4 \cdot 2}{21 \cdot 2}-\dfrac{3 \cdot 3}{14 \cdot 3}=\dfrac{8}{42}-\dfrac{9}{42}=\dfrac{8-9}{42}=-\dfrac{1}{42};$

$\displaystyle \bold 2. \; 4\dfrac{1}{5} \cdot (-\dfrac{1}{42})=-\dfrac{21}{5} \cdot \dfrac{1}{21 \cdot 2}=-\dfrac{21 \cdot 1}{5 \cdot 21 \cdot 2}=-\dfrac{1}{5 \cdot 2}=-\dfrac{1}{10}.$

2) |7·x-5|+4·|5-7·x| = |7·x-5|+4·|-(7·x-5)| = |7·x-5|+4·|7·x-5| = 5·|7·x-5|.

Answer 2

Let's solve these two equations step by step.

1. Equation: \frac{3}{5} + \frac{4 \frac{1}{5} \cdot \left(\frac{4}{21} - \frac{3}{14}\right) - \frac{3}{5}

First, simplify the expressions inside the parentheses:

$\frac{4}{21} - \frac{3}{14}$ can be simplified to a common denominator, which is 42:

$\frac{4}{21} - \frac{3}{14} = \frac{8}{42} - \frac{9}{42} = \frac{-1}{42}$

Now we can substitute this value back into the main equation:

\frac{3}{5} + \frac{4 \frac{1}{5} \cdot \frac{-1}{42} - \frac{3}{5}

Next, calculate $4 \frac{1}{5}$ as an improper fraction:

$4 \frac{1}{5} = \frac{21}{5}$

Now, substitute this value into the equation:

\frac{3}{5} + \frac{\frac{21}{5} \cdot \frac{-1}{42} - \frac{3}{5}

Now, perform the multiplications inside the parentheses:

\frac{3}{5} + \frac{\frac{21 \cdot -1}{5 \cdot 42} - \frac{3}{5}

Simplify further:

$\frac{3}{5} + \frac{-21}{210} - \frac{3}{5}$

Now, add the fractions:

$\frac{3}{5} - \frac{3}{5} - \frac{21}{210}$

Subtract the fractions:

$0 - \frac{21}{210}$

Finally, calculate $0 - \frac{21}{210}$:

$-\frac{21}{210} = -\frac{1}{10}$

So, the solution to equation 1 is $-\frac{1}{10}$.

2. Equation: $|7x - 5| + 4 \cdot |5 - 7x|$

This is an absolute value equation. You need to consider two cases, one for each absolute value expression:

Case 1: $7x - 5$ is non-negative ($7x - 5 \geq 0$)

$7x - 5 \geq 0$ when $7x \geq 5$, so $x \geq \frac{5}{7}$

In this case, the equation becomes:

$7x - 5 + 4(5 - 7x)$

Now, simplify this equation:

$7x - 5 + 20 - 28x$

Combine like terms:

$-21x + 15$

Case 2: $7x - 5$ is negative ($7x - 5 < 0$)

$7x - 5 < 0$ when $7x < 5$, so $x < \frac{5}{7}$

In this case, the equation becomes:

$-(7x - 5) + 4(7x - 5)$

Now, simplify this equation:

$-7x + 5 + 28x - 20$

Combine like terms:

$21x - 15$

So, we have two cases for the solutions:

Case 1 ($x \geq \frac{5}{7}$): $-21x + 15$ Case 2 ($x < \frac{5}{7}$): $21x - 15$

You would solve each case separately.