cho P = ( x√x+1/x+√x – x√x-1/x-√x ) : 2(x+4√x+4)/x-4 a. Rút gọn P b. Tính P tại x = 9+4√5 23/09/2021 Bởi Alexandra cho P = ( x√x+1/x+√x – x√x-1/x-√x ) : 2(x+4√x+4)/x-4 a. Rút gọn P b. Tính P tại x = 9+4√5
Đáp án: `a)“ P=-(sqrtx-2)/((sqrtx+2))` `b)` `-(sqrt5)/((4+sqrt5)) ` Giải thích các bước giải: a) `P=( (xsqrtx+1)/(x+sqrtx) – (xsqrtx-1)/(x-sqrtx) ) : (2(x+4sqrtx+4))/(x-4)` `=( ((xsqrtx+1)(x-sqrtx))/((x+sqrtx)(x-sqrtx)) – ((xsqrtx-1)(x+sqrtx))/((x+sqrtx)(x-sqrtx)) ) : (2(sqrtx+2)^2)/((sqrtx-2)(sqrtx+2))` `=( (x^2sqrtx-x^2+x-sqrtx)/((x+sqrtx)(x-sqrtx)) – ((x^2sqrtx+x^2-x-sqrtx)/((x+sqrtx)(x-sqrtx)) ) : (2(sqrtx+2))/(sqrtx-2)` `= (-2x^2+2x)/((x+sqrtx)(x-sqrtx)) . (sqrtx-2)/(2(sqrtx+2))` `= (-2x^2+2x)/(x^2-x) . (sqrtx-2)/(2(sqrtx+2))` `= (-2x(x-1))/(x.(x-1)) . (sqrtx-2)/(2(sqrtx+2))= -2 . (sqrtx-2)/(2(sqrtx+2))` `= -(sqrtx-2)/((sqrtx+2))` . b) Ta có `x=9+4sqrt5` `<=> x = 4 + 4sqrt5 + 5 = 2^2 + 2.2.sqrt5 + sqrt(5)^2 = (2+sqrt5)^2` `=> sqrtx = sqrt( (2+sqrt5)^2) = 2+sqrt5 ` Thay `x = 9+4sqrt5` vào P ta có: `P =-(sqrtx-2)/((sqrtx+2))=-( 2+sqrt5-2)/(( 2+sqrt5+2))=-(sqrt5)/((4+sqrt5)) ` Bình luận
Đáp án: `a)“ P=-(sqrtx-2)/((sqrtx+2))`
`b)` `-(sqrt5)/((4+sqrt5)) `
Giải thích các bước giải:
a) `P=( (xsqrtx+1)/(x+sqrtx) – (xsqrtx-1)/(x-sqrtx) ) : (2(x+4sqrtx+4))/(x-4)`
`=( ((xsqrtx+1)(x-sqrtx))/((x+sqrtx)(x-sqrtx)) – ((xsqrtx-1)(x+sqrtx))/((x+sqrtx)(x-sqrtx)) ) : (2(sqrtx+2)^2)/((sqrtx-2)(sqrtx+2))`
`=( (x^2sqrtx-x^2+x-sqrtx)/((x+sqrtx)(x-sqrtx)) – ((x^2sqrtx+x^2-x-sqrtx)/((x+sqrtx)(x-sqrtx)) ) : (2(sqrtx+2))/(sqrtx-2)`
`= (-2x^2+2x)/((x+sqrtx)(x-sqrtx)) . (sqrtx-2)/(2(sqrtx+2))`
`= (-2x^2+2x)/(x^2-x) . (sqrtx-2)/(2(sqrtx+2))`
`= (-2x(x-1))/(x.(x-1)) . (sqrtx-2)/(2(sqrtx+2))= -2 . (sqrtx-2)/(2(sqrtx+2))`
`= -(sqrtx-2)/((sqrtx+2))`
.
b) Ta có `x=9+4sqrt5`
`<=> x = 4 + 4sqrt5 + 5 = 2^2 + 2.2.sqrt5 + sqrt(5)^2 = (2+sqrt5)^2`
`=> sqrtx = sqrt( (2+sqrt5)^2) = 2+sqrt5 `
Thay `x = 9+4sqrt5` vào P ta có:
`P =-(sqrtx-2)/((sqrtx+2))=-( 2+sqrt5-2)/(( 2+sqrt5+2))=-(sqrt5)/((4+sqrt5)) `
…….