\(10\equiv1\left(mod9\right)\Rightarrow10^{12}\equiv1^{12}\equiv1\left(mod9\right)\Rightarrow10^{12}-1\equiv0\left(mod9\right)\Rightarrow10^{12}-1⋮9\)

\(10\equiv1\left(mod3\right)\Rightarrow10^{12}\equiv1^{12}\equiv1\left(mod3\right)\Rightarrow10^{12}-1\equiv0\left(mod3\right)\Rightarrow10^{12}-1⋮3\)

\(b,10\equiv1\left(mod3\right)\Rightarrow10^{10}\equiv1^{10}\equiv1\left(mod3\right)\Rightarrow10^{10}+2\equiv1+2\equiv3\equiv0\left(mod3\right)\Rightarrow10^{10}⋮3\) \(10^{10}+2=1000....02\left(\text{9 chu sô 0}\right).\text{ Tông các chu so là:}1+0.9+2=3\text{ không chia hết cho 9}\)

\(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=\sqrt{9^2-17}=\sqrt{64}=8\)

\(2\sqrt{2}\left(\sqrt{3}-2\right)+9+4\sqrt{2}-2\sqrt{6}=2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}-2\sqrt{6}=9\) \(\sqrt{7-2\sqrt{10}}+\sqrt{2}=\sqrt{2-2\sqrt{10}+5}+\sqrt{2}=\sqrt{\left(\sqrt{5}\right)^2-2.\sqrt{2}.\sqrt{5}+\left(\sqrt{5}\right)^2}+\sqrt{2}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{2}=\left|\sqrt{5}-\sqrt{2}\right|+\sqrt{2}=\sqrt{5}-\sqrt{2}+\sqrt{2}=\sqrt{5}\) \(\sqrt{\sqrt{3}+\sqrt{2}}.\sqrt{\sqrt{3}-\sqrt{2}}=\sqrt{\left(\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2}=\sqrt{3-2}=\sqrt{1}=1\) \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}=\sqrt{2}\left(\sqrt{4+\sqrt{15}}\right)\left(\sqrt{5}-\sqrt{3}\right)\left[\left(\sqrt{4+\sqrt{15}}\right)\left(\sqrt{4-\sqrt{15}}\right)\right]=\sqrt{2}\left(\sqrt{4+\sqrt{15}}\right)\left(\sqrt{5}-\sqrt{3}\right);\left[\sqrt{2}\left(\sqrt{4+\sqrt{15}}\right)\left(\sqrt{5}-\sqrt{3}\right)\right]^2=2.\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)=4\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=4\Rightarrow\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}=\sqrt{4}=2\left(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}>0\right)\)

abba : 176 = ba

=> abba : ba = 176

=> ab = 176

bằng 0

số lớn là : 216/ (5-3)* 5 = 540

số bé là 540-216=324

                 ĐS : SL:540

                         SB:324

Thank anh nhìu nhìu ạ

\(\text{Ta co:}\sqrt{3}+1=\sqrt{4+2\sqrt{3}}.Dêthay:4>\sqrt{3}\Rightarrow4+2\sqrt{3}>3\sqrt{3}>0\Rightarrow\sqrt{3\sqrt{3}}< \sqrt{3}+1\)

số bi đỏ là

56 : ( 3 +4 ) x 4=32 viên 

Đ/S ..........

TÍCH NHÁ !