TÍNH
a,\(\sqrt{9-4\sqrt{5}}\)
b,\(\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)
c,\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7-4\sqrt{3}}}}\)
d,\(\left(\sqrt{5}+\sqrt{2}\right).\left(3\sqrt{2}-1\right)\)
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\(A=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{3+1+2\sqrt{3.1}}-\sqrt{3+1-2\sqrt{3.1}}\)
\(=\sqrt{(\sqrt{3}+1)^2}-\sqrt{(\sqrt{3}-1)^2}=|\sqrt{3}+1|-|\sqrt{3}-1|=2\)
\(B=\sqrt{4+5-2\sqrt{4.5}}+\sqrt{4+5+2\sqrt{4.5}}=\sqrt{(\sqrt{4}-\sqrt{5})^2}+\sqrt{(\sqrt{4}+\sqrt{5})^2}\)
\(=|\sqrt{4}-\sqrt{5}|+|\sqrt{4}+\sqrt{5}|=2\sqrt{5}\)
\(C\sqrt{2}=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7+1-2\sqrt{7.1}}-\sqrt{7+1+2\sqrt{7.1}}\)
\(=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}\)
\(=|\sqrt{7}-1|-|\sqrt{7}+1|=-2\Rightarrow C=-\sqrt{2}\)
----------------------------
\(7+4\sqrt{3}=(2+\sqrt{3})^2\Rightarrow 10\sqrt{7+4\sqrt{3}}=10(2+\sqrt{3})\)
\(\Rightarrow \sqrt{48-10\sqrt{7+4\sqrt{3}}}=\sqrt{28-10\sqrt{3}}=\sqrt{(5-\sqrt{3})^2}=5-\sqrt{3}\)
\(\Rightarrow 3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}=3+5(5-\sqrt{3})=28-5\sqrt{3}\)
\(\Rightarrow D=\sqrt{5\sqrt{28-5\sqrt{3}}}\)
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) Ta có: \(9+4\sqrt{5}\)
\(=5+2\cdot\sqrt{5}\cdot2+4\)
\(=\left(\sqrt{5}+2\right)^2\)(đpcm)
b) Ta có: \(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}\)
\(=\sqrt{5}-2-\sqrt{5}\)
=-2(ddpcm)
c) Ta có: \(\left(4-\sqrt{7}\right)^2\)
\(=16-2\cdot4\cdot\sqrt{7}+7\)
\(=23-8\sqrt{7}\)(đpcm)
d) Ta có: \(\sqrt{17-12\sqrt{2}}+2\sqrt{2}\)
\(=\sqrt{9-2\cdot3\cdot2\sqrt{2}+8}+2\sqrt{2}\)
\(=\sqrt{\left(3-2\sqrt{2}\right)^2}+2\sqrt{2}\)
\(=3-2\sqrt{2}+2\sqrt{2}=3\)(đpcm)
\(a.VT=4+4\sqrt{5}+5=2^2+4\sqrt{5}+\sqrt{5}^2=\left(2+\sqrt{5}\right)^2=VP\)
\(b.\) Dựa vào câu a ta có: \(9-4\sqrt{5}=\left(\sqrt{5}-2\right)^2\)
\(VT=\left|\sqrt{5}-2\right|-\sqrt{5}=\sqrt{5}-2-\sqrt{5}=-2=VP\)
\(c.VT=16-8\sqrt{7}+7=4^2-8\sqrt{7}+\sqrt{7}^2=\left(4-\sqrt{7}\right)^2=VP\)
\(d.\)
Ta có: \(17-12\sqrt{2}=8-12\sqrt{2}+9=\left(2\sqrt{2}\right)^2-12\sqrt{2}+3^2=\left(2\sqrt{2}-3\right)^2\)
\(VT=\left|2\sqrt{2}-3\right|+2\sqrt{2}=3-2\sqrt{2}+2\sqrt{2}=3=VP\)
\(a,=\sqrt{17}-4-\sqrt{17}-2=-6\\ b,=7\left(\sqrt{3}+\sqrt{2}\right)-7\sqrt{3}-6\sqrt{2}\\ =7\sqrt{3}+7\sqrt{2}-7\sqrt{3}-6\sqrt{2}=\sqrt{2}\\ c,=\dfrac{6\sqrt{5}+12-6\sqrt{5}+12}{3}+2\sqrt{2}-\dfrac{4\sqrt{7}}{7}\\ =8+2\sqrt{2}-\dfrac{4\sqrt{7}}{7}=\dfrac{56+14\sqrt{2}-4\sqrt{7}}{7}\\ d,=\left(\dfrac{\sqrt{2}}{4}-\dfrac{6\sqrt{2}}{4}+8\sqrt{2}\right):\dfrac{1}{8}=\dfrac{-5\sqrt{2}+32\sqrt{2}}{4}\cdot8=54\sqrt{2}\)
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: \(2\sqrt{8\sqrt{3}}-\sqrt{2\sqrt{3}}-\sqrt{9\sqrt{12}}\)
\(=2\sqrt{4\cdot2\sqrt{3}}-\sqrt{2\sqrt{3}}-\sqrt{9\cdot2\sqrt{3}}\)
\(=4\sqrt{2\sqrt{3}}-\sqrt{2\sqrt{3}}-3\sqrt{2\sqrt{3}}\)
=0
b: \(\sqrt{3}+\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\sqrt{3}+\left|2-\sqrt{3}\right|\)
\(=\sqrt{3}+2-\sqrt{3}\)
=2
c: \(\sqrt{\left(\sqrt{7}-4\right)^2}-\sqrt{28}+\sqrt{63}\)
\(=\left|\sqrt{7}-4\right|-2\sqrt{7}+3\sqrt{7}\)
\(=4-\sqrt{7}+\sqrt{7}\)
=4
d: \(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)
\(=\dfrac{\sqrt{10}\left(15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\right)}{\sqrt{10}}\)
\(=15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\)
\(=15\sqrt{5}+5\cdot2\sqrt{5}-3\cdot3\sqrt{5}\)
\(=16\sqrt{5}\)
e: \(\sqrt{3}-2\sqrt{48}+3\sqrt{75}-4\sqrt{108}\)
\(=\sqrt{3}-2\cdot4\sqrt{3}+3\cdot5\sqrt{3}-4\cdot6\sqrt{3}\)
\(=\sqrt{3}-8\sqrt{3}+15\sqrt{3}-24\sqrt{3}\)
\(=-16\sqrt{3}\)
a: \(A=\left(1-\sqrt{7}\right)\cdot\left(1+\sqrt{7}\right)=1-7=-6\)
b: \(B=3\sqrt{3}+8\sqrt{3}-15\sqrt{3}=-4\sqrt{3}\)
c: \(C=4\sqrt{2}-5\sqrt{2}+3\sqrt{2}=2\sqrt{2}\)
a)
= căn 5 -2
b)
= \(\sqrt{4+\sqrt{8}}.\sqrt{4-2-\sqrt{2}}\\ \)
= \(\sqrt{4+\sqrt{8}}.\sqrt{2-\sqrt{2}}\)
= \(2.\sqrt{2+\sqrt{2}}.\sqrt{2-\sqrt{2}}\)
=2.(4-2)
=4.