Đáp án: $m^{2}$ + $M^{2}$ = 2
Giải thích các bước giải:
Áp dụng bất đăng thức Cô-si , ta có ∀x ∈ R :
$x^{2}$ +1 $\geq$ 2|x| ⇒ $\frac{2|x|}{x^2+1}$ $\leq$ 1 ⇒-1 $\leq$ $\frac{2x}{x^2+1}$ $\leq$ 1 ;
$\frac{2x}{x^2+1}$ =-1 ⇔ x=-1;$\frac{2x}{x^2+1}$=1⇔ x=1
Vậy m = min y = -1 ; M = max y =1 ⇒ $m^{2}$ + $M^{2}$ =2
Đáp án:
Xét `y-1`
`=(2x-x^2-1)/(x^2+1)`
`=(-(x-1)^2)/(x^2+1)`
Vì `x^2+1>=1>0`
Mà `-(x-1)^2<=0`
`=>y-1<=0`
`=>y<=1`
Dấu "=' xảy ra khi `x=1`.
Xét `y+1`
`=(2x+x^2+1)/(x^2+1)`
`=(x+1)^2/(x^2+1)`
Mà `(x+1)^2>=0`
`=>y+1>=0`
`=>y>=-1`
Vậy `min_y=-1<=>x=-1,max_y=1<=>x=1`.
`=>m^2+M^2=1+1=2`
Hãy tìm nội dung chính trong đoạn văn sau "Khi mặt trời ... trong đất liền" (Trong văn bản Cô Tô)
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vật dao động điều hòa trên quỹ đạo dài 20cm. lúc bật cashs gốc tọa độ 6cm thì vận tốc của vật là 16π cm/s. Chu kì dao động là A. 1s B. 2s C. 0,5s D. 0,25s
cho tam giác ABC vuông tại A, gọi AD là đường phân giác của góc BAC; AE là đường phân giác góc ngoài của tam giác ABC tại đỉnh A. C/m $\dfrac{1}{AB^2}$ + $\dfrac{1}{AC^2}$ + $\dfrac{1}{AD^2}$ + $\dfrac{1}{AE^2}$
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