Cho x;y>0
CMR: \(\dfrac{\left(x^3+8\right)\left(y^2-y+1\right)}{\left(x^2+x\right)\left(xy^2+2\right)}\ge\dfrac{1}{2}\)
1, x,y,z∈N*. CMR x+3z-y là hợp số biết `x^2+y^2=z^2`
2,Tìm n∈N* để \(\left(4n^3+n+3\right)⋮\left(2n^2+n+1\right)\)
3, CMR:\(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{4}{xy}\forall x\ne y,xy\ne0\)
2.
\(4n^3+n+3=4n^3+2n^2+2n-2n^2-n-1+4=2n\left(2n^2+n+1\right)-\left(2n^2+n+1\right)+4\)-Để \(\left(4n^3+n+3\right)⋮\left(2n^2+n+1\right)\) thì \(4⋮\left(2n^2+n+1\right)\)
\(\Leftrightarrow2n^2+n+1\in\left\{1;-1;2;-2;4;-4\right\}\) (do n là số nguyên)
*\(2n^2+n+1=1\Leftrightarrow n\left(2n+1\right)=0\Leftrightarrow n=0\) (loại) hay \(n=\dfrac{-1}{2}\) (loại)
*\(2n^2+n+1=-1\Leftrightarrow2n^2+n+2=0\) (phương trình vô nghiệm)
\(2n^2+n+1=2\Leftrightarrow2n^2+n-1=0\Leftrightarrow n^2+n+n^2-1=0\Leftrightarrow n\left(n+1\right)+\left(n+1\right)\left(n-1\right)=0\Leftrightarrow\left(n+1\right)\left(2n-1\right)=0\)
\(\Leftrightarrow n=-1\) (loại) hay \(n=\dfrac{1}{2}\) (loại)
\(2n^2+n+1=-2\Leftrightarrow2n^2+n+3=0\) (phương trình vô nghiệm)
\(2n^2+n+1=4\Leftrightarrow2n^2+n-3=0\Leftrightarrow2n^2-2n+3n-3=0\Leftrightarrow2n\left(n-1\right)+3\left(n-1\right)=0\Leftrightarrow\left(n-1\right)\left(2n+3\right)=0\)\(\Leftrightarrow n=1\left(nhận\right)\) hay \(n=\dfrac{-3}{2}\left(loại\right)\)
-Vậy \(n=1\)
1. \(x^2+y^2=z^2\)
\(\Rightarrow x^2+y^2-z^2=0\)
\(\Rightarrow\left(x-z\right)\left(x+z\right)+y^2=0\)
-TH1: y lẻ \(\Rightarrow x-z;x+z\) đều lẻ.
\(x+3z-y=x+z-y+2x\) chia hết cho 2. \(\Rightarrow\)Hợp số.
-TH2: y chẵn \(\Rightarrow\)1 trong hai biểu thức \(x-z;x+z\) chia hết cho 2.
*Xét \(\left(x-z\right)⋮2\):
\(x+3z-y=x-z+4z-y\) chia hết cho 2. \(\Rightarrow\)Hợp số.
*Xét \(\left(x+z\right)⋮2\):
\(x+3z-y=x+z+2z-y\) chia hết cho 2 \(\Rightarrow\)Hợp số.
1. a,b,c>0 và a+b+c=2017
\(CM:\Sigma\dfrac{2017a-a^2}{bc}\ge\sqrt{2}\left(\Sigma\sqrt{\dfrac{2017-a}{a}}\right)\)
2. cho x,y,z tm: \(x^2+y^2+z^2=3\)
\(CM:8\left(2-x\right)\left(2-y\right)\left(2-z\right)\ge\left(x+yz\right)\left(y+zx\right)\left(z+xy\right)\)
3. a,b,c>0 và \(a^2+b^2+c^2\ge6\)
\(CM:\Sigma\dfrac{1}{1+ab}\ge\dfrac{3}{2}\)
Tương tự, ta được:
\(\left(2-y\right)\left(2-z\right)>=\dfrac{\left(x+1\right)^2}{4}\)
và \(\left(2-z\right)\left(2-x\right)>=\left(\dfrac{y+1}{2}\right)^2\)
=>8(2-x)(2-y)(2-z)>=(x+1)(y+1)(z+1)
(x+yz)(y+zx)<=(x+y+yz+xz)^2/4=(x+y)^2*(z+1)^2/4<=(x^2+y^2)(z+1)^2/4
Tương tự, ta cũng co:
\(\left(y+xz\right)\left(z+y\right)< =\dfrac{\left(y^2+z^2\right)\left(x+1\right)^2}{2}\)
và \(\left(z+xy\right)\left(x+yz\right)< =\dfrac{\left(z^2+x^2\right)\left(y+1\right)^2}{2}\)
Do đó, ta được:
\(\left(x+yz\right)\left(y+zx\right)\left(z+xy\right)< =\left(x+1\right)\left(y+1\right)\left(z+1\right)\)
=>ĐPCM
Cho x,y,z>0. CM: \(\dfrac{xy}{z^2\left(x+y\right)}+\dfrac{yz}{x^2\left(y+z\right)}+\dfrac{zx}{y^2\left(z+x\right)}\ge\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
\(VT=\dfrac{\left(\dfrac{1}{z}\right)^2}{\dfrac{1}{x}+\dfrac{1}{y}}+\dfrac{\left(\dfrac{1}{x}\right)^2}{\dfrac{1}{y}+\dfrac{1}{z}}+\dfrac{\left(\dfrac{1}{y}\right)^2}{\dfrac{1}{x}+\dfrac{1}{z}}\ge\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}{2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)}=\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Dâu "=" xảy ra khi \(x=y=z\)
cho x,y,z>0 thỏa mãn \(\left(x^2+y^2\right)\left(y^2+z^2\right)\left(z^2+x^2\right)=8\)
Tìm giá trị nhỏ nhất của S=\(xyz\left(x+y+z\right)^3\)
(có thể dùng BDT \(\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\))
tks mn<3
Cho các số thực dương x,y. CMR: \(\dfrac{1}{\left(1+x\right)^2}+\dfrac{1}{\left(1+y\right)^2}\ge\dfrac{1}{1+xy}\)
\(\left(1+x\right)^2=\left(1.1+\sqrt{xy}.\sqrt{\dfrac{x}{y}}\right)^2\le\left(1+xy\right)\left(1+\dfrac{x}{y}\right)=\dfrac{\left(1+xy\right)\left(x+y\right)}{y}\)
\(\Rightarrow\dfrac{1}{\left(1+x\right)^2}\ge\dfrac{y}{\left(1+xy\right)\left(x+y\right)}\)
Tương tự ta có: \(\dfrac{1}{\left(1+y\right)^2}\ge\dfrac{x}{\left(1+xy\right)\left(x+y\right)}\)
Cộng vế với vế:
\(\dfrac{1}{\left(1+x\right)^2}+\dfrac{1}{\left(1+y\right)^2}\ge\dfrac{x+y}{\left(1+xy\right)\left(x+y\right)}=\dfrac{1}{1+xy}\)
Dấu "=" xảy ra khi \(x=y=1\)
Cho x, y, z thỏa mãn : \(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\). Cmr :
\(\dfrac{x}{\sqrt{yz\left(1+x^2\right)}}+\dfrac{y}{\sqrt{zx\left(1+y^2\right)}}+\dfrac{z}{\sqrt{xy\left(1+z^2\right)}}\ge\dfrac{3}{2}\).
Sửa lại đề: cho x, y, z dương thỏa mãn \(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}=1\)
Chứng minh \(A=\dfrac{x}{\sqrt{yz\left(1+x^2\right)}}+\dfrac{y}{\sqrt{xz\left(1+y^2\right)}}+\dfrac{z}{\sqrt{xy\left(1+z^2\right)}}\le\dfrac{3}{2}\)
Giải:
Đặt \(a=\dfrac{1}{x};b=\dfrac{1}{y};c=\dfrac{1}{z}\Rightarrow ab+bc+ac=1\)
\(\Rightarrow A=\dfrac{\dfrac{1}{a}}{\sqrt{\dfrac{1}{bc}\left(1+\dfrac{1}{a^2}\right)}}+\dfrac{\dfrac{1}{b}}{\sqrt{\dfrac{1}{ac}\left(1+\dfrac{1}{b^2}\right)}}+\dfrac{\dfrac{1}{a}}{\sqrt{\dfrac{1}{ab}\left(1+\dfrac{1}{c^2}\right)}}\)
\(\Rightarrow A=\sqrt{\dfrac{bc}{a^2+1}}+\sqrt{\dfrac{ac}{b^2+1}}+\sqrt{\dfrac{ab}{c^2+1}}\)
\(\Rightarrow A=\sqrt{\dfrac{bc}{a^2+ab+bc+ac}}+\sqrt{\dfrac{ac}{b^2+ab+bc+ac}}+\sqrt{\dfrac{ab}{c^2+ab+bc+ac}}\)
\(\Rightarrow A=\sqrt{\dfrac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\dfrac{ac}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\dfrac{ab}{\left(a+c\right)\left(b+c\right)}}\)
\(\Rightarrow A\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{c}{a+c}+\dfrac{a}{a+b}+\dfrac{c}{b+c}+\dfrac{a}{a+c}+\dfrac{b}{b+c}\right)\)
\(\Rightarrow A\le\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{a+c}{a+c}\right)=\dfrac{3}{2}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{\sqrt{3}}{3}\) hay \(x=y=z=\sqrt{3}\)
Đề bài này có rất nhiều vấn đề, đầu tiên không có điều kiện x, y, z gì cả? Dương? Â? Bằng 0? Khác 0?
Sau nữa là chiều của BĐT cũng có vấn đề nốt, mình thử với \(x=y=2;z=\dfrac{4}{3}\) thì vế trái ra \(\dfrac{2+\sqrt{30}}{5}\) mà theo casio cho biết thì số này nhỏ hơn \(\dfrac{3}{2}\) , vậy BĐT cũng sai luôn
Cho x, y, z > 0 và \(x+y\le z\) . CMR :
\(\left(x^2+y^2+z^2\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)\ge\dfrac{27}{2}\)
\(VT=\left(x^2+y^2+z^2\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)=3+\dfrac{x^2+y^2}{z^2}+z^2\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\)
\(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}>=2\cdot\sqrt{\dfrac{y^2}{x^2}\cdot\dfrac{x^2}{y^2}}=2\)
=>\(VT>=5+\left(\dfrac{x^2}{z^2}+\dfrac{z^2}{16x^2}\right)+\left(\dfrac{y^2}{z^2}+\dfrac{z^2}{16y^2}\right)+\dfrac{15}{16}z^2\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\)
\(\dfrac{x^2}{z^2}+\dfrac{z^2}{16x^2}>=2\cdot\sqrt{\dfrac{x^2}{z^2}\cdot\dfrac{z^2}{16x^2}}=\dfrac{1}{2}\)
\(\dfrac{y^2}{z^2}+\dfrac{z^2}{16y^2}>=\dfrac{1}{2}\)
và \(\dfrac{1}{x^2}+\dfrac{1}{y^2}>=\dfrac{2}{xy}>=\dfrac{2}{\left(\dfrac{x+y}{2}\right)^2}=\dfrac{8}{\left(x+y\right)^2}\)
=>\(\dfrac{15}{16}z^2\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)>=\dfrac{15}{16}z^2\cdot\dfrac{8}{\left(x+y\right)^2}=\dfrac{15}{2}\left(\dfrac{z}{x+y}\right)^2=\dfrac{15}{2}\)
=>VT>=5+1/2+1/2+15/2=27/2
Cho các số x, y, z dương thỏa mãn: \(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=3\)
Cmr: \(\dfrac{1}{\left(2x+y+z\right)^2}+\dfrac{1}{\left(2y+z+x\right)^2}+\dfrac{1}{\left(2z+x+y\right)^2}\ge\dfrac{3}{16}\)
Thay $x=\sqrt{\frac{1}{2,5}}; y=z=\sqrt{\frac{1}{0,25}}$ ta thấy đề sai bạn nhé!
1Cho x,y,z >0 và xy+yz+zx=1. Chứng minh rằng \(3\left(\dfrac{1}{x^2+1}+\dfrac{1}{y^2+1}+\dfrac{1}{z^2+1}\right)+\left(1+x^2^x\right)\left(1+y^2\right)\left(1+z^2\right)\ge\dfrac{985}{108}\) 2 Cho p,q là hai số nguyên tố thoả mãn \(p-1⋮p\) và \(p^3-1p⋮\) Chứng minh rằng p+q là số chính phương
cái chỗ math processing error kia là \(3\left(\dfrac{1}{x^2+1}+\dfrac{1}{y^2+1}+\dfrac{1}{z^2+1}\right)+\left(1+x^2\right)\left(1+y^2\right)\left(1+z^2\right)\ge\dfrac{985}{108}\)