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a) \(\sqrt{3+2\sqrt{2}}+\sqrt{\left(\sqrt{2}-2\right)^2}\)
\(=\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{2}\cdot1+1^2}+\left|\sqrt{2}-2\right|\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}-\left(\sqrt{2}-2\right)\)
\(=\left|\sqrt{2}+1\right|-\sqrt{2}+2\)
\(=\sqrt{2}+1-\sqrt{2}+2\)
\(=3\)
b) \(\dfrac{1}{5}\sqrt{50}-2\sqrt{96}-\dfrac{\sqrt{30}}{\sqrt{15}}+12\sqrt{\dfrac{1}{6}}\)
\(=\dfrac{1}{5}\cdot5\sqrt{2}-2\cdot4\sqrt{6}-\sqrt{\dfrac{30}{15}}+\sqrt{\dfrac{144}{6}}\)
\(=\sqrt{2}-8\sqrt{6}-\sqrt{2}+2\sqrt{6}\)
\(=-8\sqrt{6}+2\sqrt{6}\)
\(=-6\sqrt{6}\)
c) \(\left(\dfrac{5-\sqrt{5}}{\sqrt{5}}-2\right)\left(\dfrac{4}{1+\sqrt{5}}+4\right)\)
\(=\left[\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}}-2\right]\left[\dfrac{4\left(1-\sqrt{5}\right)}{\left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right)}+4\right]\)
\(=\left(\sqrt{5}-1-2\right)\left(\dfrac{4\left(1-\sqrt{5}\right)}{1-5}+4\right)\)
\(=\left(\sqrt{5}-3\right)\left(\sqrt{5}-1+4\right)\)
\(=\left(\sqrt{5}-3\right)\left(\sqrt{5}+3\right)\)
\(=\left(\sqrt{5}\right)^2-3^2\)
\(=-4\)
a) \(\sqrt[]{3+2\sqrt[]{2}}+\sqrt[]{\left(\sqrt[]{2}-2\right)^2}\)
\(=\sqrt[]{2+2\sqrt[]{2}.1+1}+\left|\sqrt[]{2}-2\right|\)
\(=\sqrt[]{\left(\sqrt[]{2}+1\right)^2}+\left(2-\sqrt[]{2}\right)\) \(\left(\left(\sqrt[]{2}\right)^2=2< 2^2=4\right)\)
\(=\left|\sqrt[]{2}+1\right|+2-\sqrt[]{2}\)
\(=\sqrt[]{2}+1+2-\sqrt[]{2}\)
\(=3\)
a: \(=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=12\sqrt{2x}\)
b: \(=6-4\sqrt{3}+4\sqrt{3}-8=-2\)
c: \(=\sqrt{2}+1+2-\sqrt{2}=3\)
d: \(=\dfrac{1}{\sqrt{2}}\cdot\left(\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}+2\sqrt{3}\right)=0\)
f: \(=\sqrt{2}-8\sqrt{6}-\sqrt{2}+2\sqrt{6}=-6\sqrt{6}\)
Bài 1:
a: \(5\sqrt{8}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}\)
\(=5\cdot2\sqrt{2}-4\cdot3\sqrt{3}-2\cdot5\sqrt{3}+6\sqrt{3}\)
\(=10\sqrt{2}-12\sqrt{3}-10\sqrt{3}+6\sqrt{3}\)
\(=10\sqrt{2}-16\sqrt{3}\)
b: \(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(1-\sqrt{6}\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|1-\sqrt{6}\right|\)
\(=3-\sqrt{6}+\sqrt{6}-1\)
=3-1=2
c: \(\dfrac{5\sqrt{3}-3\sqrt{5}}{\sqrt{5}-\sqrt{3}}+\dfrac{1}{4+\sqrt{15}}\)
\(=\dfrac{\sqrt{15}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}+\dfrac{1\left(4-\sqrt{15}\right)}{16-15}\)
\(=\sqrt{15}+4-\sqrt{15}=4\)
d: \(\dfrac{2\sqrt{3-\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{10}-\sqrt{2}}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{12}+2}\)
\(=\dfrac{\sqrt{3-\sqrt{5}}\cdot\sqrt{2}\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}\left(\sqrt{3}+1\right)}{2\left(\sqrt{3}+1\right)}\)
\(=\dfrac{\sqrt{6-2\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}}{2}\)
\(=\sqrt{\left(\sqrt{5}-1\right)^2}\cdot\dfrac{\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}}{2}\)
\(=3+\sqrt{5}-\dfrac{\sqrt{5}}{2}=3+\dfrac{\sqrt{5}}{2}\)
Bài 2:
Vẽ đồ thị:
Phương trình hoành độ giao điểm là:
\(\dfrac{1}{2}x-4=-3x+3\)
=>\(\dfrac{1}{2}x+3x=3+4\)
=>\(\dfrac{7}{2}x=7\)
=>x=2
Thay x=2 vào y=-3x+3, ta được:
\(y=-3\cdot2+3=-3\)
Vậy: (d1) cắt (d2) tại A(2;-3)
a: \(=\sqrt{5}-3\sqrt{5}-4\sqrt{3}+15\sqrt{3}=-2\sqrt{5}+11\sqrt{3}\)
b: \(=3\sqrt{10}-\sqrt{5}+6-\sqrt{2}\)
c; \(=15\sqrt{2}-10\sqrt{3}-12\sqrt{2}-\sqrt{3}=-11\sqrt{3}+3\sqrt{2}\)
d: \(=3-\sqrt{3}+\sqrt{3}-1=2\)
f: \(=\sqrt{10}-\sqrt{10}-2-2\sqrt{10}=-2-2\sqrt{10}\)
a) Ta có: \(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right)\cdot\sqrt{3}\)
\(=\left(7\cdot4\sqrt{3}+3\cdot3\sqrt{3}-2\cdot2\sqrt{3}\right)\cdot\sqrt{3}\)
\(=33\sqrt{3}\cdot\sqrt{3}\)
=99
b) Ta có: \(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)
\(=\left(12\cdot5\sqrt{2}-8\cdot10\sqrt{2}+7\cdot15\sqrt{2}\right):\sqrt{10}\)
\(=\dfrac{85\sqrt{2}}{\sqrt{10}}=\dfrac{85}{\sqrt{5}}=17\sqrt{5}\)
c) Ta có: \(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\cdot2\sqrt{2}\right)\cdot3\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+3\sqrt{2}\right)\cdot3\sqrt{6}\)
\(=36-36\sqrt{2}+18\sqrt{3}\)
d) Ta có: \(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)
\(=3\cdot\sqrt{75\sqrt{2}}+5\cdot\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)
\(=3\cdot5\sqrt{2}\cdot\sqrt{\sqrt{2}}+4\sqrt{3}\sqrt{\sqrt{2}}\)
\(=15\sqrt{\sqrt{8}}+4\sqrt{\sqrt{18}}\)
a,=\(\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right).\sqrt{3}\)
\(=28.3+9.3-4.3=99\)
b,\(=\left(60\sqrt{2}-80\sqrt{2}+175\sqrt{2}\right):\sqrt{10}\)
\(=155\sqrt{2}:\sqrt{10}=\dfrac{155}{\sqrt{5}}\)
a: \(=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}=3\)
b: \(=\left(-\sqrt{5}-2+\sqrt{5}-\sqrt{3}\right)\cdot\left(2\sqrt{3}+3\right)\)
\(=-\sqrt{3}\left(2+\sqrt{3}\right)\cdot\left(2+\sqrt{3}\right)\)
\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)
c: \(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}=\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1\)
Đề không khó, mỗi tội dài
vậy thì bn làm hộ mik vs , mik cần gấp