Toán P= 1+√a. 1-√a —-. -. —-. Với a≥ 0 ,a#1 1 -√a. 1+√a 24/07/2021 By Mary P= 1+√a. 1-√a —-. -. —-. Với a≥ 0 ,a#1 1 -√a. 1+√a
`P = (1+\sqrt{a})/(1-\sqrt{a}) – (1-\sqrt{a})/(1+\sqrt{a})` `(a\geq0;a\ne1)` `P=((1+\sqrt{a})^2 – (1-\sqrt{a})^2)/((-\sqrt{a}+1)(\sqrt{a}+1))` `=frac{[1+\sqrt{a}-(1-\sqrt{a})].[1+\sqrt{a}+(1+\sqrt{a})]}{(1-\sqrt{a})(1+\sqrt{a})}` `=frac{(1+\sqrt{a}-1+\sqrt{a}).(1+\sqrt{a}+1+\sqrt{a})}{(1-\sqrt{a})(1+\sqrt{a})}` `=frac{2\sqrt{a}.(2+2\sqrt{a})}{(1-\sqrt{a})(1+\sqrt{a})}` `=frac{4\sqrt{a}+4a}{(1-\sqrt{a})(1+\sqrt{a})}` `=frac{4\sqrt{a}.(\sqrt{a}+1)}{(1-\sqrt{a})(1+\sqrt{a})}` `=frac{4\sqrt{a}}{1-\sqrt{a}}` Trả lời
Đáp án: + Giải thích các bước giải: Với `a \ge 0 , a \ne 1` `P = (1+\sqrt{a})/(1-\sqrt{a}) – (1-\sqrt{a})/(1+\sqrt{a})` `= ((1+\sqrt{a})^2 – (1-\sqrt{a})^2)/((-\sqrt{a}+1)(\sqrt{a}+1))` `= (4\sqrt{a})/((-\sqrt{a}+1)(\sqrt{a}+1))` `= (4\sqrt{a})/(1-a)` Trả lời
`P = (1+\sqrt{a})/(1-\sqrt{a}) – (1-\sqrt{a})/(1+\sqrt{a})` `(a\geq0;a\ne1)`
`P=((1+\sqrt{a})^2 – (1-\sqrt{a})^2)/((-\sqrt{a}+1)(\sqrt{a}+1))`
`=frac{[1+\sqrt{a}-(1-\sqrt{a})].[1+\sqrt{a}+(1+\sqrt{a})]}{(1-\sqrt{a})(1+\sqrt{a})}` `=frac{(1+\sqrt{a}-1+\sqrt{a}).(1+\sqrt{a}+1+\sqrt{a})}{(1-\sqrt{a})(1+\sqrt{a})}`
`=frac{2\sqrt{a}.(2+2\sqrt{a})}{(1-\sqrt{a})(1+\sqrt{a})}`
`=frac{4\sqrt{a}+4a}{(1-\sqrt{a})(1+\sqrt{a})}`
`=frac{4\sqrt{a}.(\sqrt{a}+1)}{(1-\sqrt{a})(1+\sqrt{a})}`
`=frac{4\sqrt{a}}{1-\sqrt{a}}`
Đáp án: + Giải thích các bước giải:
Với `a \ge 0 , a \ne 1`
`P = (1+\sqrt{a})/(1-\sqrt{a}) – (1-\sqrt{a})/(1+\sqrt{a})`
`= ((1+\sqrt{a})^2 – (1-\sqrt{a})^2)/((-\sqrt{a}+1)(\sqrt{a}+1))`
`= (4\sqrt{a})/((-\sqrt{a}+1)(\sqrt{a}+1))`
`= (4\sqrt{a})/(1-a)`