Question

(2x-3)^2>25. (2x+7)^2≤169 ​

Answer 1

1)$(2x-3)^{2} >25\rightarrow \left[\begin{array}{cc}(2x-3)>5\\(2x-3)8\\2x4\\x$

Ответ: $x\in(-\infty;-1)\cup(4;+\infty)$

2)$(2x+7)^{2}\leq169\rightarrow \left[\begin{array}{cc}(2x+7)\leq13\\(2x+7)\geq-13\end] \rightarrow\left[\begin{array}{cc}2x\leq6\\2x\geq-20\end] \rightarrow\left[\begin{array}{cc}x\leq3\\x\geq-10\end]$

Ответ:$x\in[-10;3]$

Answer 2

To solve the inequalities, let's break them down step by step:

Inequality 1: (2x - 3)^2 > 25

Taking the square root of both sides of the inequality, we get:

2x - 3 > 5 or 2x - 3 < -5

Solving the first inequality: 2x - 3 > 5 2x > 5 + 3 2x > 8 x > 4

Solving the second inequality: 2x - 3 < -5 2x < -5 + 3 2x < -2 x < -1

Therefore, the solution to inequality 1 is x < -1 or x > 4.

Inequality 2: (2x + 7)^2 ≤ 169

Taking the square root of both sides of the inequality, we get:

2x + 7 ≤ 13 or 2x + 7 ≥ -13

Solving the first inequality: 2x + 7 ≤ 13 2x ≤ 13 - 7 2x ≤ 6 x ≤ 3

Solving the second inequality: 2x + 7 ≥ -13 2x ≥ -13 - 7 2x ≥ -20 x ≥ -10

Therefore, the solution to inequality 2 is -10 ≤ x ≤ 3.

To summarize:

Inequality 1: x < -1 or x > 4 Inequality 2: -10 ≤ x ≤ 3