Bài này là bài của lớp 9 nha!! có chỗ nào ko hiểu ib

\(a,A=\sqrt{18}+\sqrt{50}-\frac{1}{2}\sqrt{98}.\)

\(=3\sqrt{2}+5\sqrt{2}-\frac{7}{2}\sqrt{2}\)

\(=\sqrt{2}\left(3+5-\frac{7}{2}\right)\)

\(=\frac{9}{2}\sqrt{2}\)

\(b,B=\left(2\sqrt{3}+7\right)\left(2\sqrt{3}-7\right)\)

\(=2^2\sqrt{3^2}-7^2\)

\(=12-49=-37\)

a ) 

\(A=\sqrt{18}+\sqrt{50}-\frac{1}{2}\sqrt{98}\)

\(A=3\sqrt{2}+5\sqrt{2}-\frac{7}{2}\sqrt{2}\)

\(A=(3+5-\frac{7}{2})\sqrt{2}\)

\(A=\frac{9}{2}\sqrt{2}=\frac{9\sqrt{2}}{2}\)

b)

\(B=\left(2\sqrt{3}+7\right)\left(2\sqrt{3}-7\right)=\left(2\sqrt{3}\right)^2-7^2=12-49=-37\)

Xin mạo muội sửa đề nha! có j sai thì bỏ qua cho

\(c,C=\sqrt{7-2\sqrt{10}}\)

\(=\sqrt{5-2\sqrt{5.2}+2}\)

\(=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\)

\(=\sqrt{5}-\sqrt{2}\)

a) \(3\sqrt{3}-3\sqrt{4^2\cdot3}+2\sqrt{6^2\cdot3}-\left(2-\sqrt{3}\right)\)

\(3\sqrt{3}-3\cdot4\sqrt{3}+2\cdot6\sqrt{3}-2+\sqrt{3}\)

\(3\sqrt{3}-12\sqrt{3}+12\sqrt{3}-2+\sqrt{3}\)

\(4\sqrt{3}-2\)

b) \(3\sqrt{2}\left(\sqrt{5^2\cdot2}-2\sqrt{3^2\cdot2}+\sqrt{7^2\cdot2}\right)\)

\(3\sqrt{2}\left(5\sqrt{2}-6\sqrt{2}+7\sqrt{2}\right)\)

\(3\sqrt{2}\left(6\sqrt{2}\right)\) \(=36\)

\(a=\sqrt{27}-3\sqrt{48}+2\sqrt{108}-\sqrt{\left(2-\sqrt{3}\right)^2}\)

\(a=3\sqrt{3}-3\sqrt{48}+\sqrt{216}-2+\sqrt{3}\)

\(a=3\sqrt{3}-3\sqrt{48}+3\sqrt{24}-2+\sqrt{3}\)

\(a=3\left(\sqrt{3}-\sqrt{48}+\sqrt{24}+1\right)-2\)

Tính cái trong ngoặc là \(ok\).Em lười đi lấy máy tính lắm

\(b=3\sqrt{2}\left(\sqrt{50}-2\sqrt{18}+\sqrt{98}\right)\)

\(b=3\sqrt{100}-3\sqrt{72}+3\sqrt{196}\)

\(b=3\left(\sqrt{100}-\sqrt{72}+\sqrt{196}\right)\)(Tính trong ngoặc)

\(a.3\sqrt{2}\left(\sqrt{50}-2\sqrt{18}+\sqrt{98}\right)=30-36+42=36\)

\(b.B=\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}=\sqrt{13+30\sqrt{2}+30}=\sqrt{25+2.5\sqrt{18}+18}=\sqrt{\left(5+\sqrt{18}\right)^2}=5+3\sqrt{2}\)

a) \(2\sqrt{98}-3\sqrt{18}+\dfrac{1}{2}\sqrt{32}=14\sqrt{2}-9\sqrt{2}+2\sqrt{2}=7\sqrt{2}\)

b) \(\left(5\sqrt{2}+2\sqrt{5}\right).\sqrt{5}-\sqrt{250}=5\sqrt{10}+10-5\sqrt{10}=10\)

c) \(\left(2\sqrt{3}-5\sqrt{2}\right).\sqrt{3}-\sqrt{36}=6-5\sqrt{6}-6=5\sqrt{6}\)

d) \(3\sqrt{48}+2\sqrt{27}-\dfrac{1}{3}\sqrt{243}=12\sqrt{3}+6\sqrt{3}-3\sqrt{3}=15\sqrt{3}\)

e) \(6\sqrt{\dfrac{1}{3}}+\dfrac{9}{\sqrt{3}}-\dfrac{2}{\sqrt{3}-1}=2\sqrt{3}+3\sqrt{3}=\left(\sqrt{3}+1\right)=4\sqrt{3}-1\)

f) \(4\sqrt{\dfrac{1}{2}}-\dfrac{6}{\sqrt{2}}.\dfrac{2}{\sqrt{2}+1}=2\sqrt{2}-\left(12-6\sqrt{2}\right)=8\sqrt{2}-12\)

a) Ta có: \(D=\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)\cdot\left(-\sqrt{2}\right)\)

\(=-2+\sqrt{6-2\sqrt{5}}\)

\(=-2+\sqrt{5-2\cdot\sqrt{5}\cdot1+1}\)

\(=-2+\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(=-2+\left|\sqrt{5}-1\right|\)

\(=-2+\sqrt{5}-1\)(Vì \(\sqrt{5}>1\))

\(=-3+\sqrt{5}\)

b) Ta có: \(2\sqrt{3}\left(\sqrt{27}+2\sqrt{48}\right)-\sqrt{75}\)

\(=2\sqrt{81}+4\sqrt{144}-5\sqrt{3}\)

\(=18+48-5\sqrt{3}\)

\(=66-5\sqrt{3}\)

c) Ta có: \(E=\left(\sqrt{10}+\sqrt{6}\right)\sqrt{8-2\sqrt{15}}\)

\(=\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\)

\(=\sqrt{2}\cdot\left(\sqrt{5}+\sqrt{3}\right)\cdot\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

\(=\sqrt{2}\cdot\left(\sqrt{5}+\sqrt{3}\right)\cdot\left|\sqrt{5}-\sqrt{3}\right|\)

\(=\sqrt{2}\cdot\left(\sqrt{5}+\sqrt{3}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\)(Vì \(\sqrt{5}>\sqrt{3}\))

\(=\sqrt{2}\cdot\left(5-3\right)\)

\(=2\sqrt{2}\)

d) Ta có: \(P=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)

\(=\sqrt{\frac{3}{2}+2\cdot\sqrt{\frac{3}{2}}\cdot\sqrt{\frac{1}{2}}+\frac{1}{2}}+\sqrt{\frac{3}{2}-2\cdot\sqrt{\frac{3}{2}}\cdot\sqrt{\frac{1}{2}}+\frac{1}{2}}\)

\(=\sqrt{\left(\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}\right)^2}+\sqrt{\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}\right)^2}\)

\(=\left|\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}\right|+\left|\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}\right|\)

\(=\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}+\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}\)(Vì \(\sqrt{\frac{3}{2}}>\sqrt{\frac{1}{2}}>0\))

\(=2\sqrt{\frac{3}{2}}=\sqrt{4\cdot\frac{3}{2}}=\sqrt{6}\)

e) Ta có: \(M=-3\sqrt{50}+2\sqrt{98}-7\sqrt{72}\)

\(=\sqrt{2}\cdot\left(-3\cdot\sqrt{25}+2\cdot\sqrt{49}-7\cdot\sqrt{36}\right)\)

\(=\sqrt{2}\cdot\left(-15+14-42\right)\)

\(=-43\sqrt{2}\)

a) \(\text{2}\sqrt{\text{18}}-9\sqrt{50}+3\sqrt{8}\)

= \(\text{6}\sqrt{\text{2}}-45\sqrt{2}+6\sqrt{2}\)

= \(-33\sqrt{2}\)

b) = \(7-2.\sqrt{7}.\sqrt{3}+3+7.2\sqrt{21}\)

= \(10-2\sqrt{21}+14\sqrt{21}\)

= \(10+12\sqrt{21}\)

\(E=(\sqrt{18}-3\sqrt{6}+\sqrt{2}).\sqrt{2}+6\sqrt{3} \\ = (3\sqrt{2}-3\sqrt{6}+\sqrt{2}).\sqrt{2} + 6\sqrt{3} \\ = 6 - 6\sqrt{3}+2 + 6\sqrt{3} \\ = 8\)

\(G=(2\sqrt2-\sqrt5+\sqrt{18}).(\sqrt{50}+\sqrt5) \\ =(2\sqrt2-\sqrt5+3\sqrt2).\sqrt5(\sqrt{10}+1) \\ = (5\sqrt2-\sqrt5). \sqrt5 (\sqrt{10}+1) \\ = (5\sqrt{10}-5)(\sqrt{10}+1) \\ = 5(\sqrt{10}-1)(\sqrt{10}+1)=5.9=45\)