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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 9: tìm gtln 1) 2√x -x 2)6√x -x 3)1/√(x ²+1) 4)1/2x- √x -3

Toán Lớp 9: tìm gtln
1) 2√x -x
2)6√x -x
3)1/√(x ²+1)
4)1/2x- √x -3

Comments ( 1 )

  1. Giải đáp+Lời giải và giải thích chi tiết:
    1)2\sqrtx-x(x>=0)
    =-(x-2\sqrtx)
    =-(x-2\sqrtx+1-1)
    =-(\sqrtx-1)^2+1
    Vì -(\sqrtx-1)^2<=0
    =>-(\sqrtx-1)^2+1<=1
    Dấu “=” xảy ra khi \sqrtx=1<=>x=1.
    Vậy GTLN_1=1<=>x=1.
    2)6\sqrtx-x(x>=0)
    =-(x-6\sqrtx)
    =-(x-6\sqrtx+9-9)
    =-(\sqrtx-3)^2+9
    Vì -(\sqrtx-3)^2<=0
    =>-(\sqrtx-3)^2+9<=9
    Dấu “=” xảy ra khi \sqrtx=3<=>x=9.
    Vậy GTLN_2=9<=>x=9.
    3)1/\sqrt{x^2+1}
    Vì x^2+1>=1>0AAx
    =>\sqrt{x^2+1}>=1>0AAx
    =>1/\sqrt{x^2+1}<=1/1=1
    Dấu “=” xảy ra khi x^2=0<=>x=0.
    Vậy GTLN_3=1<=>x=0.
    4)1/(2x-\sqrtx+3)(x>=0)
    Ta có:
    2x-\sqrtx+3
    =2(x-\sqrtx/2)+3
    =2(x-2*\sqrtx*1/4+1/16-1/16)+3
    =2(\sqrtx-1/4)^2+3-1/8
    =2(\sqrtx-1/4)^2+23/8>=23/8
    =>1/(2x-\sqrtx+3)<=1:23/8=8/23
    Dấu “=” xảy ra khi \sqrtx=1/4<=>x=1/16
    Vậy GTLN_4=8/23<=>x=1/16.

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222-9+11+12:2*14+14 = ? ( )

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