Question

(х-7)в квадрате -х(6+х) при х=- одна двадцатая

Answer 1

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Answer 2

To simplify the expression, let's substitute $x$ with $-\frac{1}{20}$ and calculate it step by step.

Given expression: $(x - 7)^2 - x(6 + x)$

Substituting $x = -\frac{1}{20}$: $(- \frac{1}{20} - 7)^2 - (- \frac{1}{20})(6 - \frac{1}{20})$

First, let's simplify the expression inside the parentheses: $(- \frac{141}{20})^2 - (- \frac{1}{20})(\frac{119}{20})$

Next, let's calculate the squared term: $\frac{141^2}{20^2} - (- \frac{1}{20})(\frac{119}{20})$

Simplifying further: $\frac{19881}{400} + \frac{119}{400}$

Combining the fractions: $\frac{19881 + 119}{400}$

Adding the numerator: $\frac{20000}{400}$

Finally, dividing the numerator by the denominator: $\frac{50}{1}$

Therefore, when $x = -\frac{1}{20}$, the expression $(x - 7)^2 - x(6 + x)$ evaluates to $50$.