Cho bt M = (1/ √x-1+ √x/x-1).x- √x/2 √x+1 vs x 0
≥
; x
≠
1
1. rút gọn M
2. tìm x để M =1/3
Cho bt M = (1/ √x-1+ √x/x-1).x- √x/2 √x+1 vs x 0 ≥ ; x ≠ 1 1. rút gọn M 2. tìm x để M =1/3
By Peyton
By Peyton
Cho bt M = (1/ √x-1+ √x/x-1).x- √x/2 √x+1 vs x 0
≥
; x
≠
1
1. rút gọn M
2. tìm x để M =1/3
C1 : M=(√x+1+√x/(√x-1)(√x+1).√x(√x-1)/(2√x+1)
=(2√x+1).√x(√x-1)/(√x-1)(√x+1)(2√x+1)
=√x/√x+1
C2 : M=1/3<=>√x/√x+1=3√x=√x+1=> 2√x=1=>√x=1/2
=>1/4(TMĐKXĐ)
Vậy M=1/3<=>x=1/4
Đáp án+Giải thích các bước giải:
`1)`
Với `x≥0;x\ne1`
Ta có:
`M=(1/(\sqrtx-1)+\sqrtx/(x-1)).(x-\sqrtx)/(2\sqrtx+1)`
`=(1/(\sqrtx-1)+\sqrtx/((\sqrtx-1)(\sqrtx+1))).(x-\sqrtx)/(2\sqrtx+1)`
`=(\sqrtx+1+\sqrtx)/((\sqrtx-1)(\sqrtx+1)).(\sqrtx(\sqrtx-1))/(2\sqrtx+1)`
`=(2\sqrtx+1)/(\sqrtx+1).(\sqrtx)/(2\sqrtx+1)`
`=\sqrtx/(\sqrtx+1)`
Vậy với `x≥0;x\ne1` thì `M=\sqrtx/(\sqrtx+1)`
`2)`
`M=\sqrtx/(\sqrtx+1)`
`\to M=1/3`
`⇔ \sqrtx/(\sqrtx+1)=1/3`
`⇔3\sqrtx=\sqrtx+1`
`⇔2\sqrtx=1`
`⇔\sqrtx=1/2`
`⇔x=1/4(tm)`
Vậy `x=1/4` để `M=1/3`