Đáp án:
\(a,\ P=\sqrt{1-a}\\ b,\ P=\left |\dfrac{5}{7} \right |\\ c,\ a=(-3)\)
Giải thích các bước giải:
\(a,\ ĐK:\ a\neq ±1\\ P=\left (\dfrac{2}{\sqrt{1+a}}+\sqrt{1-a} \right ):\left (\dfrac{2}{\sqrt{1-a^{2}}}+1 \right )\\ =\left (\dfrac{2}{\sqrt{1+a}}+\dfrac{\sqrt{1-a}.\sqrt{1+a}}{\sqrt{1+a}} \right ):\left (\dfrac{2}{\sqrt{1-a^{2}}}+\dfrac{\sqrt{1-a^{2}}}{\sqrt{1-a^{2}}} \right )\\ =\dfrac{2+\sqrt{1-a^{2}}}{\sqrt{1+a}}:\dfrac{2+\sqrt{1-a^{2}}}{\sqrt{1-a^{2}}}\\ =\dfrac{2+\sqrt{1-a^{2}}}{\sqrt{1+a}}\times \dfrac{\sqrt{1+a}.\sqrt{1-a}}{2+\sqrt{1-a^{2}}}\\ =\dfrac{(2+\sqrt{1-a^{2}}).\sqrt{1+a}.\sqrt{1-a}}{\sqrt{1+a}.(2+\sqrt{1-a^{2}})}\\ =\sqrt{1-a}\\ b,\ \text{Thay a = $\dfrac{24}{49}$ vào P ta có:}\\ P=\sqrt{1-\dfrac{24}{49}}=\sqrt{\dfrac{25}{49}}=\sqrt{\left (\dfrac{5}{7} \right )^{2}}=\left |\dfrac{5}{7} \right |\\ c,\ \text{Để P = 2 thì:}\\ \sqrt{1-a}=2\\ ⇔(\sqrt{1-a})^{2}=2^{2}\\ ⇔1-a=4\\ ⇔a=1-4\\ ⇔a=(-3)\ (tmđk)\)
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