# how to determine the LDU in examples of matsheets?

how to determine the LDU in examples of matsheets?

- These are the values in which the function exists. That is, if the function is a root of x, then x is positive, otherwise the function does not exist. The denominator of a fraction must not be zero. Descend from these conditions.
- The range of valid values (LDL). The range of allowable values (LDL) of an algebraic expression is the set of all admissible sets of values of letters included in this expression.
The equations can be solved according to the scheme:

Find the ODZ, that is, solve the corresponding inequalities and write out in an explicit form, on which numerical set the given equation makes sense.

Solve the equation with the help of various transformations.

Check if the roots of this equation belong to the LDU.Let us consider examples in which LDB is ambiguously represented

Example 1. Solve the equation $$ sqrt {- x ^ 5 + x ^ 2 + 1} = x + 1 $$

Solution: The DSA is determined from the condition: $$ - x ^ 5 + x ^ 2 + 1 ge 0 Leftrightarrow x ^ 5 - x ^ 2 - 1 le 0 $$. Such an inequality can only be solved approximately. Therefore, without determining the LDZ, we start solving differently: $$ sqrt {- x ^ 5 + x ^ 2 + 1} = x + 1 Leftrightarrow left {begin {array} {l} - x ^ 5 + x ^ 2 + 1 = left ( {x + 1} right) ^ 2, x + 1 ge 0 end {array} right.quad Leftrightarrow left {begin {array} {l} x ^ 5 + 2x = 0, x ge - 1 end {array} right. Leftrightarrow left {begin {array} {l} xleft ({x ^ 4 + 2} right) = 0,

x ge - 1 end {array} right. $$Answer: 0

Note: Sometimes it is more difficult to find the TLD, than to solve the equation, and maybe it is completely impossible. The solution of the equation was possible without finding the LDL and the root of the equation is verified by simple substitution.Example 2. Solve the equation $$ sqrt {2x - 1} = - x $$

The solution: ODZ is obtained from $$ 2x - 1 ge 0 Rightarrow x ge frac {1} {2} $$. By raising both sides of the equation to the square, we get the solution: $$ sqrt {2x - 1} = - x Leftrightarrow 2x - 1 = x ^ 2 Leftrightarrow x ^ 2 - 2x + 1 = 0 Rightarrow x = 1 $$. The obtained root enters the ODZ equation, but it is established by check that this number is not the root of the given equation.

The answer is: no horses

Note: The fact that the roots found are included in the DHS does not guarantee at all that they satisfy the original equation, even if all the transformations are correct. Finding the TLD in the decision was not useless. If we change the reasoning, we get: because $$ 2x - 1 ge 0 Leftrightarrow - x le - frac {1} {2} lt; 0 Rightarrow sqrt {2x - 1} lt; 0 $$, which contradicts the definition of the arithmetic square root.Example 3. Solve the equation $$ sqrt {5 - x} = sqrt4 {{x - 5}} + lg left ({x - 3} right) $$

Solution: The LDL of this equation is determined by the conditions $$ left {begin {array} {l} x ge 5 x le 5 x gt; 3 end {array} right.quad Leftrightarrow x = 5 $$. A check shows that x = 5 is the root of the equation.

Answer: 5

Note: Sometimes the use of a TLD is useful because it allows you to quickly solve an equation. In this example, the solution was obtained solely from the definition of a DHS. - ODZ - area of admissible values

It is necessary to look at the denominator, to determine at what values of the variable the denominator vanishes and write, for example, x unequally (sign) *, * - TLD is the value of the unknown in which the expression makes sense.

For example, if in the expression Vx, then ODZ will be the next x gt; = 0, since the square root can be taken from a positive number.

If the expression contains several functions of the restrictive DSA, then the DSA expression will be the set on which all functions are meaningful.

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How to determine the LDU in examples of matsheets

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How to determine the LDU in examples of matsheets

How to determine the LDU in examples of matsheets

How to determine the LDU in examples of matsheets