*\(2\overline{xy}+1=n^2\left(1\right)\\ 3\overline{xy+1=m^2\left(2\right)\left(1\right)=>2\overline{xy}chia}h\text{ết}cho8=>\overline{xy}chiah\text{ết}cho4\\ \left(2\right)=>3\overline{xy}chiah\text{ết}cho8,\left(8;3\right)=1=>\overline{xy}chiah\text{ết}cho8\)

*\(\left(1\right)+\left(2\right)\\ =>5\overline{xy}+2=m^2+n^2\\ VPchia5d\text{ư}2=>m^2+n^2chia5d\text{ư}2=>m^2v\text{à}n^2chia5d\text{ư}1\\ =>\overline{xy}chiah\text{ết}cho5\\ \left(8;5\right)=1=>\overline{xy}\)

\(=>\overline{xy}chiah\text{ết}cho40\\ =>\overline{xy}\left(40;80\right)=>\overline{xy}=40\)

Các bạn giúp mình đi

cái V x là căn đó nghen

dùng bất đẳng thức Côsi nha bạn

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:

\(\sqrt{x+yz}+\sqrt{y+xz}+\sqrt{z+xy\ge}\sqrt{x}+\sqrt{\frac{yz}{x}}+\sqrt{y}+\sqrt{\frac{xz}{y}}+\sqrt{z}+\sqrt{\frac{xy}{z}}\)

\(\Leftrightarrow\sqrt{x+yz}+\sqrt{y+xz}+\sqrt{z+xy}\ge\sqrt{x}+\sqrt{y}+\sqrt{z}+\frac{xy+yz+xz}{\sqrt{xyz}}\)

\(\Leftrightarrow\sqrt{x+yz}+\sqrt{y+xz}+\sqrt{z+xy}\ge\sqrt{x}+\sqrt{y}+\sqrt{z}+\sqrt{xyz}\)

\(\overline{ababab}=\overline{ab}.10000+\overline{ab}.100+\overline{ab}\\ =\overline{ab}\left(10000+100+1\right)\\ =\overline{ab}.10101⋮13v\text{à}7\)

Giup cai j!!??

| 3x + 1/2 | - 2/3 = 1

| 3x + 1/2 | = 1 + 2 /3 

| 3x + 1/2 | = 5/3

| 3x | = 5/3 - 1/2

| 3x | = 7/6

| x | = 7/6 : 3 

| x | = 7/18

| x | = 0,3

x = 0,3

đề:...

x=-13/18