chung minh rang
|x+y| <hoac = |x| + |y|
|x-y|>hoac bang |x|-|y|
cho x+y =2. chung minh rang x^2015+y^2015 ba hon hoac bang x^2016+y^2016
chung minh rang ( a^2+b^2)(x^2+y^2) lon hon hoac bang (ax+by)^2
giúp vớiiiiiiiiiiiiiiiiiiiiiiiii
Cái này là BĐT Bunhiacopxki đó bạn
\(\left(a^2+b^2\right)\left(x^2+y^2\right)\ge\left(ax+by\right)^2\)
\(\Leftrightarrow a^2x^2+b^2y^2+b^2x^2+a^2y^2\ge a^2x^2+b^2y^2+2axby\)
\(\Leftrightarrow b^2x^2+a^2y^2\ge2axby\)
\(\Leftrightarrow\left(bx-ay\right)^2\ge0\) ( luôn đúng )
\(\Rightarrowđpcm\)
\(\left(a^2+b^2\right)\left(x^2+y^2\right)\ge\left(ax+by\right)^2\)
\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2\ge a^2x^2+b^2y^2+2axby\)
\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2-a^2x^2-b^2y^2-2axby\ge0\)
\(\Leftrightarrow a^2y^2+b^2y^2-2axby\ge0\)
\(\Leftrightarrow\left(ay-bx\right)^2\ge0\) ( bất đẳng thức luôn đúng )
Vậy ................
Chung minh rang l x l lon hon hoac bang 0
nếu:\(|x|=0\Rightarrow x=0\)
\(|x|>0\Rightarrow x>0\)
vây \(|x|\ge0\)
Cho 3 so duong thoa man\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\) . Chung minh rang \(\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\)lon hon hoac bang\(\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}\)
Từ giả thiết : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\Rightarrow xy+yz+zx=xyz\)
Ta có : \(\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}\)
Vì hai vế luôn dương nên ta bình phương hai vế được :
\(\left(\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\right)^2\ge\left(\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2\)
Xét \(\left(\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\right)^2\)
\(=\left(x+y+z\right)+\left(xy+yz+zx\right)+2\left(\sqrt{x+yz}.\sqrt{y+zx}+\sqrt{y+zx}.\sqrt{z+xy}+\sqrt{z+xy}.\sqrt{x+yz}\right)\)
Xét \(\left(\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2\)
\(=xyz+\left(x+y+z\right)+2\left(x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)\)
Suy ra : \(\sqrt{x+yz}.\sqrt{y+zx}+\sqrt{y+zx}.\sqrt{z+xy}+\sqrt{z+xy}.\sqrt{x+yz}\ge\)
\(\ge x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\) (*)
Mà theo bất đẳng thức Bunhiacopxki , ta có :
\(\sqrt{\left(x+yz\right)}.\sqrt{y+zx}\ge\sqrt{xy}+\sqrt{yz.zx}=\sqrt{xy}+z\sqrt{xy}\) (1)
\(\sqrt{y+zx}.\sqrt{z+xy}\ge\sqrt{yz}+x\sqrt{yz}\)(2)
\(\sqrt{z+xy}.\sqrt{x+yz}\ge\sqrt{xz}+y\sqrt{xz}\)(3)
Cộng (1) , (2) và (3) theo vế ta được (*) đúng
Vậy bđt ban đầu được chứng minh.
chung minh rang : -3x^2+6x-y^2+6y-12 nho hon hoac bang 0
(cac ban giai nhanh nhe 5 like , minh dang can gap)
-3x2 + 6x -y2 +6y -12
<=> (-3x2 +6x -3) +( -y2 + 6y - 9)
<=> -3(x2 -2x +1) - (y2 -6y +9)
<=> -3(x-1)2 - (y-3)2
mà -3(x-1)2>=0
-(y-3)2>=0
=> dpcm
= -3(x^2-2x+1)-(y^2-6y+9)
=-3(x-2)^2-(y-3)^2 nhỏ hơn hoặc bằng 0
cho x, y,z >0 chung minh rang\(\frac{x}{2x+y+z}+\frac{y}{2y+x+z}+\frac{z}{2z+x+y}< hoac=\frac{3}{ }4\)3/4
cho ham so \(y=f\left(x\right)=ax^2+bx+c\)(a;b;c thuoc Q)
chung to rang f(-2)*f(3) nho hon hoac bang 0, biet rang 13a+b+2c=0
Ta có: \(f\left(x\right)=ax^2+bx+c\)
\(\Rightarrow f\left(-2\right)=4a-2b+c\)
\(f\left(3\right)=9a+3b+c\)
\(\Rightarrow f\left(-2\right)+f\left(3\right)=13a+b+2c=0\)(vì 13a+b+2c=0)
\(\Rightarrow f\left(-2\right)=-f\left(3\right)\)
\(\Rightarrow f\left(-2\right).f\left(3\right)=-\left[f\left(-2\right)\right]^2\le0\)( đpcm)
cho 70 so nguyen duong phan biet a1;a2;a3;.......;a70,moi so khong vuot qua 200. Chung minh rang ton tai2 trong 70 so da cho sao cho hieu chung bang 4 hoac bang 5 hoac bang 9
cho x,y thuoc Q . chung to rang:
a) lx+yl < hoac = lxl +lyl
b)lx-yl> hoac = lxl - lyl
a)\(\left|x+y\right|\le\left|x\right|+\left|y\right|\left(1\right)\)
Bình phương 2 vế của (1) ta được:
\(\left(\left|x+y\right|\right)^2\le\left(\left|x\right|+\left|y\right|\right)^2\)
\(\Leftrightarrow x^2+2xy+y^2\le x^2+2\left|xy\right|+y^2\)
\(\Leftrightarrow xy\le\left|xy\right|\) (Đpcm)
Dấu = khi \(xy\ge0\)
b)\(\left|x-y\right|\ge\left|x\right|-\left|y\right|\)
\(\Rightarrow\left|x-y\right|+\left|y\right|\ge\left|x\right|\)
Áp dụng câu a ta có:
\(\Rightarrow\left|x-y\right|+\left|y\right|\ge\left|x-y+y\right|=\left|x\right|\) (luôn đúng)
Suy ra đpcm