1) Cho \(2.\overline{xy}+1v\text{à}3\overline{xy}+1l\text{à}c\text{ác}s\text{ố}ch\text{ính}ph\text{ươ}ng.T\text{ìm}\overline{xy}\)
1) \(Cho2.\overline{xy}+1v\text{à}3.\overline{xy}+1l\text{à}S\text{ố}ch\text{ính}ph\text{ươ}ng.T\text{ìm}\overline{xy}\)
* 2xy + 1 =n2(1)
3xy+1=m2(2)
(1) => 2xy chia hết cho 8 => xy chia hết cho 4
(2)=>3xy chia hết cho 8 mà (3;8)=1 => xy chia hết cho 8
*(1)+(2)
=> 5xy +2=m2+n2
VP chia 5 dư 2 => m2+n2 chia 5 dư 2 => m2 và n2 chia 5 dư 1
=>xy chia hết cho 5
(8;5)=1
=>xy chia hết cho 40
\(T\text{ìm}\) \(s\text{ố}.nguy\text{ê}n.d\text{ư}\text{ơ}ng.nh\text{ỏ}.nh\text{ất}.th\text{ỏa}.m\text{ãn}:\frac{1}{2}s\text{ố}.\text{đ}\text{ó}.l\text{à}.s\text{ố}.ch\text{ính}.ph\text{ư}\text{ơ}ng\) \(\frac{1}{3}s\text{ố}.\text{đ}\text{ó}.l\text{à}.l\text{ập}.ph\text{ư}\text{ơ}ng.c\text{ủa}.1.s\text{ố}.nguy\text{ên}\) \(\)
\(\frac{1}{5}s\text{ố}.\text{đ}\text{ó}.l\text{à}.l\text{ũy}.th\text{ừa}.5.c\text{ủa}.1.s\text{ố.nguy\text{ê}n}\)
Cho x, y, z > 0 thỏa mãn 1/x + 1/y + 1/z = 1
C/m Vx+yz + V y+zx + Vz+xy ≥ Vxyz + Vx + Vy + Vz
Các bạn giúp mình đi
cái V x là căn đó nghen
Theo gt \(xyz=xy+yz+xz\) ta có:
\(\sqrt{x+yz}=\sqrt{\frac{x^2+xyz}{x}}=\sqrt{\frac{x^2+xy+yz+xz}{x}}=\sqrt{\frac{\left(x+y\right)\left(x+z\right)}{x}}\)
Theo BĐT Cauchy-Schwarz có: \(\sqrt{\left(x+y\right)\left(x+z\right)}\ge x+\sqrt{yz}\) do đó:
\(\sqrt{x+yz}=\sqrt{\frac{\left(x+y\right)\left(x+z\right)}{x}}\ge\frac{x+\sqrt{yz}}{x}=\sqrt{x}+\sqrt{\frac{yz}{x}}\)
Tương tự cho 2 BĐT còn lại ta có:
\(\sqrt{y+xz}\ge\sqrt{y}+\sqrt{\frac{xz}{y}};\sqrt{z+xy}\ge\sqrt{z}+\sqrt{\frac{xy}{z}}\)
Cộng 3 vế của BĐT lại ta có:
\(\sqrt{x+yz}+\sqrt{y+xz}+\sqrt{z+xy\ge}\sqrt{x}+\sqrt{\frac{yz}{x}}+\sqrt{y}+\sqrt{\frac{xz}{y}}+\sqrt{z}+\sqrt{\frac{xy}{z}}\)
\(\Leftrightarrow\sqrt{x+yz}+\sqrt{y+xz}+\sqrt{z+xy}\ge\sqrt{x}+\sqrt{y}+\sqrt{z}+\frac{xy+yz+xz}{\sqrt{xyz}}\)
\(\Leftrightarrow\sqrt{x+yz}+\sqrt{y+xz}+\sqrt{z+xy}\ge\sqrt{x}+\sqrt{y}+\sqrt{z}+\sqrt{xyz}\)
1) \(Ch\text{ứng}t\text{ỏ}:\overline{ababab}chiah\text{ết}cho13v\text{à}7\)
\(\overline{ababab}=\overline{ab}.10000+\overline{ab}.100+\overline{ab}\\ =\overline{ab}\left(10000+100+1\right)\\ =\overline{ab}.10101⋮13v\text{à}7\)
1)\(T\text{ìm}\overline{ab}bi\text{ết}:\overline{ab}^2=\left(a+b\right)^3\\ aigi\text{úp}v\text{ới}chu\text{ẩn}b\text{ị}\text{đ}ih\text{ọc}r\text{ồi}\)
\(Cho\)\(c\text{ác}\)\(s\text{ố}\)\(a,b,c,d,p\)\(l\text{à}\)\(c\text{ác}\)\(s\text{ố}\)\(nguy\text{ê}n\)\(t\text{ố}\)\(th\text{ỏa}\)\(m\text{ãn:}\)\(p=a+b=c-d.\)\(T\text{ìm}\)\(c\text{ác}\)\(s\text{ố}\)\(a,b,c,d\)\(v\text{à}\)\(p\)
1) Tìm \(\overline{xy}th\text{ỏa}:\overline{xxyy}=\overline{xx}^2+\overline{yy}^2\)
\(^{n^2+2019}l\text{à}s\text{ố}ch\text{ình}ph\text{ương}\)
1) \(T\text{ìm}:\overline{abc}bi\text{ết}:\frac{\overline{abc}}{1000}=\frac{1}{a+b+c}\)
Ai jup cái