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g, h. Câu hỏi của Nữ hoàng sến súa là ta - Toán lớp 9 - Học toán với OnlineMath
e) \(E=A-\sqrt{2}\)
\(A=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
\(A^2=8-2\sqrt{16-7}=8-6=2\)
\(A>0=>A=\sqrt{2}\)
\(E=A-\sqrt{2}=0\)
a)\(\left(\sqrt{10}+\sqrt{2}\right)\left(6-2\sqrt{5}\right)\sqrt{3+\sqrt{5}}\)
=\(\left(6\sqrt{10}+6\sqrt{2}-10\sqrt{2}-2\sqrt{10}\right)\sqrt{3+\sqrt{5}}\)
=\(\left(4\sqrt{10}-4\sqrt{2}\right)\sqrt{3+\sqrt{5}}=\left(4\sqrt{10}-4\sqrt{2}\right)\dfrac{\sqrt{5}+1}{2}\)
=\(\dfrac{20\sqrt{2}+4\sqrt{10}-4\sqrt{10}-4\sqrt{2}}{2}\)
=\(\dfrac{16\sqrt{2}}{2}=8\sqrt{2}\)
b)\(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}\)
=\(\dfrac{\sqrt{5}+1-\sqrt{5}+1-2}{\sqrt{2}}=0\)
c)\(\sqrt{3,5-\sqrt{6}}+\sqrt{3,5+\sqrt{6}}\)
=\(\dfrac{\sqrt{6}-1+\sqrt{6}+1}{\sqrt{2}}=2\sqrt{3}\)
d)\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{7}\)
=\(\dfrac{\sqrt{7}-1-\sqrt{7}-1+\sqrt{14}}{\sqrt{2}}=\sqrt{7}-1\)
e)\(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)
=\(\dfrac{\sqrt{7}+1-\sqrt{7}+1-2}{\sqrt{2}}=0\)
1) \(\sqrt{12}\)+\(5\sqrt{3}-\sqrt{48}\)
= \(2\sqrt{3}+5\sqrt{3}-4\sqrt{3}\)
= (2+5-4).\(\sqrt{3}\)
= \(3\sqrt{3}\)
2)\(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)
= \(5\sqrt{5}+2\sqrt{5}-3.3\sqrt{5}\)
= \(5\sqrt{5}+2\sqrt{5}-9\sqrt{5}\)
= \(\left(5+2-9\right).\sqrt{5}\)
= -2\(\sqrt{2}\)
3)\(3\sqrt{32}+4\sqrt{8}-5\sqrt{18}\)
= \(3.4\sqrt{2}+4.2\sqrt{2}-5.3\sqrt{2}
\)
= 12\(\sqrt{2}\) \(+8\sqrt{2}\) \(-15\sqrt{2}\)
= \(\left(12+8-15\right).\sqrt{2}\)
= \(5\sqrt{2}\)
4)\(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)
= \(3.2\sqrt{3}-4.3\sqrt{3}+5.4\sqrt{3}\)
= \(6\sqrt{3}-12\sqrt{3}+20\sqrt{3}\)
= \(\left(6-12+20\right).\sqrt{3}\)
= \(14\sqrt{3}\)
5)\(\sqrt{12}+\sqrt{75}-\sqrt{27}\)
= \(2\sqrt{3}+5\sqrt{3}-3\sqrt{3}\)
= \(\left(2+5-3\right).\sqrt{3}\)
= \(4\sqrt{3}\)
6) \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\)
= \(2.3\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= 6\(\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= \(\left(6-7+9\right).\sqrt{2}\)
= 8\(\sqrt{2}\)
7)\(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)
= \(3.2\sqrt{5}-2.3\sqrt{5}+4\sqrt{5}\)
= \(6\sqrt{5}-6\sqrt{5}+4\sqrt{5}\)
= \(4\sqrt{5}\)
8)\(\left(\sqrt{2}+2\right).\sqrt{2}-2\sqrt{2}\)
= \(\left(\sqrt{2}\right)^2+2\sqrt{2}-2\sqrt{2}\)
= 2
\(A=\sqrt{8}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =2\sqrt{2}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =4\sqrt{2}+4\sqrt{7}\)
\(B=\left(3+2\sqrt{6}+2\right)\left(25-20\sqrt{6}+24\right)\sqrt{3-2\sqrt{6}+2}\\ =\left(\sqrt{3}+\sqrt{2}\right)^2\left(5-2\sqrt{6}\right)^2\left(\sqrt{3}-\sqrt{2}\right)\\ =\left(\sqrt{3}+\sqrt{2}\right)\left(3-2\sqrt{6}+2\right)^2\\ =\left(\sqrt{3}-\sqrt{2}\right)^3\\ =9\sqrt{3}-11\sqrt{2}\)
\(a,\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)
\(=-13\sqrt{3}\)
\(b,2\sqrt{3}.\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)=2\sqrt{3}\left(3\sqrt{3}+8\sqrt{3}-5\sqrt{3}\right)\)
\(=2\sqrt{3}.6\sqrt{3}=36\)
\(c,\left(2\sqrt{2}-\sqrt{3}\right)^2=8-4\sqrt{6}+3\)
\(=11-4\sqrt{6}\)
\(d,\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)=1+2\sqrt{3}+3-2\)
\(=2+2\sqrt{3}\)
\(a.\sqrt{\left(1-\sqrt{5}\right)^2}+1=\left|1-\sqrt{5}\right|+1=\sqrt{5}-1+1=\sqrt{5}\)
\(b.\sqrt{3+2\sqrt{2}}-2=\sqrt{\left(\sqrt{2}+1\right)^2}-2=\sqrt{2}+1-2=\sqrt{2}-1\)
\(c.\sqrt{b^2-b+\dfrac{1}{4}}-\left(2b-\dfrac{1}{2}\right)=\sqrt{\left(b-\dfrac{1}{2}\right)^2}-2b+\dfrac{1}{2}=b-\dfrac{1}{2}-2b+\dfrac{1}{2}=-2b\)
\(d.\sqrt{7+2\sqrt{10}}=\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}=\sqrt{5}+\sqrt{2}\)
\(e.\sqrt{11-4\sqrt{7}}=\sqrt{\left(\sqrt{7}-2\right)^2}=\sqrt{7}-2\)
\(g.3x+\sqrt{x^2-2x+1}=3x+\sqrt{\left(x-1\right)^2}\)
* \(x\ge1\Rightarrow3x+\left|x-1\right|=3x+x-1=4x-1\)
* \(x< 1\Rightarrow3x+\left|x-1\right|=3x+1-x=2x+1\)
\(h.\sqrt{y+2\sqrt{y^2-2y+1}}=\sqrt{y+2\sqrt{\left(y-1\right)^2}}=\sqrt{y+2y-2}=\sqrt{3y-2}\left(y\ge1\right)\) hoặc: \(\sqrt{y+2-2y}=\sqrt{-y+2}\left(y< 1\right)\)
\(H=\sqrt{17-2\sqrt{32}}+\sqrt{17+2\sqrt{32}}\)
\(H^2=17-2\sqrt{32}+17+2\sqrt{32}+2\sqrt{\left(17-2\sqrt{32}\right)\left(17+2\sqrt{32}\right)}=34+2\sqrt{161}\)
\(H=\sqrt{34+2\sqrt{161}}\)
\(k.\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)
\(\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(8+\sqrt{5}\right)\\ =\sqrt{3-\sqrt{5}}.\sqrt{2}\left(\sqrt{5}-1\right)\left(8+\sqrt{5}\right)\\ =\sqrt{6-2\sqrt{5}}\left(\sqrt{5}-1\right)\left(8+\sqrt{5}\right)\\ =\sqrt{\left(\sqrt{5}-1\right)^2}\left(\sqrt{5}-1\right)\left(8+\sqrt{5}\right)\\ =\left(\sqrt{5}-1\right)^2\left(8+\sqrt{5}\right)\\ =\left(6-2\sqrt{5}\right)\left(8+\sqrt{5}\right)\)
Thấy lạ ._. Bạn xem lại thử đề có đúng k giúp mình nhé!
a)\(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\sqrt{10\left(4-\sqrt{15}\right)}+\sqrt{6\left(4-\sqrt{15}\right)}\)
\(=\sqrt{40-10\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)
\(=\sqrt{\left(5-\sqrt{15}\right)^2}+\sqrt{\left(3-\sqrt{15}\right)^2}\)
\(=5-\sqrt{15}+\sqrt{15}-3\)
\(=2\)
b) \(2\left(\sqrt{10}-\sqrt{2}\right)\left(4+\sqrt{6-2\sqrt{5}}\right)\)
\(=\left(2\sqrt{10}-2\sqrt{2}\right)\left(4+\sqrt{\left(1-\sqrt{5}\right)^2}\right)\)
\(=\left(2\sqrt{10}-2\sqrt{2}\right)\left(4+\sqrt{5}-1\right)\)
\(=\left(2\sqrt{10}-2\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
\(=6\sqrt{10}+2\sqrt{50}-6\sqrt{2}-2\sqrt{10}\)
\(=6\sqrt{10}+10\sqrt{2}-6\sqrt{2}-2\sqrt{10}\)
\(=4\sqrt{10}+4\sqrt{2}\)
c) \(\left(\sqrt{7}+\sqrt{14}\right)\sqrt{9-2\sqrt{14}}\)
\(=\left(\sqrt{7}+\sqrt{14}\right)\sqrt{\left(\sqrt{2}-\sqrt{7}\right)^2}\)
\(=\left(\sqrt{7}+\sqrt{14}\right)\left(\sqrt{7}-\sqrt{2}\right)\)
\(=7\sqrt{7}-7\sqrt{2}+\sqrt{98}-\sqrt{28}\)
\(=7\sqrt{7}-7\sqrt{2}+7\sqrt{2}-2\sqrt{7}\)
\(=5\sqrt{7}\)
d) \(\sqrt{\dfrac{289+4\sqrt{72}}{16}}\)
\(=\sqrt{\dfrac{289+42\sqrt{2}}{16}}\)
\(=\dfrac{\sqrt{289+42\sqrt{2}}}{\sqrt{4^2}}\)
\(=\dfrac{\sqrt{\left(1+12\sqrt{2}\right)^2}}{4}\)
\(=\dfrac{1+12\sqrt{2}}{4}\)
e) \(\left(\sqrt{21}+7\right)\sqrt{10-2\sqrt{21}}\)
\(=\left(\sqrt{21}+\sqrt{7}\right)\sqrt{\left(\sqrt{3}-\sqrt{7}\right)^2}\)
\(=\left(\sqrt{21}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
\(=\sqrt{147}-\sqrt{63}+7-\sqrt{21}\)
\(=7\sqrt{3}-\sqrt{63}+7-\sqrt{21}\)
f) bạn xem đề lại nhé
đó là số 2 ko phải chữ s mik xin lỗi