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Answer 1

$\large\underline{\bold{Solution-}}$

$\tt \: \sqrt[3]{2 - x} + \sqrt{x - 1} = 1 - - - (1)$

To solve this question, we use method of Substitution,

$\tt \: Put \: \sqrt{x - 1} = y$

On squaring both sides, we get

$\tt \: x - 1 = {y}^{2}$

$\therefore \: \tt \: x \: = \: {y}^{2} + 1$

Substituting these values in equation (1),

$\tt \: \sqrt[ 3]{2 - ( {y}^{2} - 1)} + y = 1$

$\tt \: \sqrt[3]{2 - {y}^{2} - 1} = 1 - y$

$\tt \: \sqrt[3 ]{1 - {y}^{2} } = 1 - y$

On Cubing both sides, we get

$\tt \: 1 - {y}^{2} = {(1 - y)}^{3}$

$\tt \: (1 - y)(1 + y) - {(1 - y)}^{3} = 0$

$\tt \: (1 - y)\bigg((1 + y) - {(1 - y)}^{2} \bigg) = 0$

$\tt \: (1 - y)(1 + y - 1 - {y}^{2} + 2y) = 0$

$\tt \: (1 - y)(3y - {y}^{2} ) = 0$

$\tt \: (1 - y)y(3 - y) = 0$

$\therefore \: \: \tt \: y = 0 \: \: or \: \: \: 1 \: \: \: or \: \: \: 3$

Hence,

Correspondence values of x are

$\begin{gathered}\boxed{\begin{array}{c|c} \bf y & \bf x = {y}^{2} + 1 \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 1 \\ \\ \sf 1 & \sf 2 \\ \\ \sf 3 & \sf 10 \end{array}} \\ \end{gathered}$

Answer 2

Answer:

hope it helps

Step-by-step explanation:

thanks you