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mjk cần gấp lắm !!
b và c.... ok!
b) \(\sqrt{\left(\sqrt{3}-2\right)^2}-\sqrt{\left(\sqrt{3}+2\right)^2}=\left(\sqrt{3}-2\right)-\left(\sqrt{3}+2\right)=-4\)
nãy nhìn không kĩ nên mới nói là bình phương lên,sorry nhak
c) Đặt \(C=\sqrt{3-2\sqrt{2}}-\sqrt{3+2\sqrt{2}}\)
ta có: \(C^2=3-2\sqrt{2}+3+2\sqrt{2}-2=4\)
=> \(C=-\sqrt{2}\) (vì \(\sqrt{3-2\sqrt{2}}< \sqrt{3+2\sqrt{2}}\))
a) hằng đẳng thức số 3 (hiệu 2 bình phương)
b) bình phương cả cái biểu thức đó lên, tính bình thường
c) bình phương cả lên như câu b
d) giống câu a
e) hẳng đẳng thức số 1
f) phá căn ra (biến đổi biểu thức trong căn thành hằng đẳng thức số 1 hoặc 2)
h) nghi là hằng đẳng thức số 1 hoặc số 2, từ từ lát nữa tớ xem
khó hiểu chỗ nào thì hỏi nhé
b,\(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\) \(=\sqrt{8\sqrt{3}}-2\sqrt{50\sqrt{3}}+4\sqrt{8\sqrt{3}}\)
\(=2\sqrt{2\sqrt{3}}-10\sqrt{2\sqrt{3}}+8\sqrt{2\sqrt{3}}\)
\(=0\)
d,\(A=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
\(\sqrt{2}A=\sqrt{2}(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}})\)
\(\sqrt2A=\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}\)
\(\sqrt2A=\sqrt{(\sqrt5-1)^2}\) \(+\sqrt{(\sqrt5+1)^2}\) \(=\sqrt5-1 +\sqrt5+1=2\sqrt5\)
\(\Rightarrow A=\dfrac{2\sqrt5}{\sqrt2}\) \(=\sqrt{10}\)
a. \(\frac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}=\frac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}\)
\(=\frac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)
\(=\frac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\frac{3\sqrt{5}-3+5-\sqrt{5}}{2\left(\sqrt{5}+1\right)}\)
\(=\frac{2\sqrt{5}+2}{2\left(\sqrt{5}+1\right)}=\frac{2\left(\sqrt{5}+1\right)}{2\left(\sqrt{5}+1\right)}=1\)
\(\sqrt{10-4\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{2^2-2.2.\sqrt{6}+\left(\sqrt{6}\right)^2}+\sqrt{3^2-2.3.2\sqrt{6}+\left(2\sqrt{6}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{6}\right)^2}+\sqrt{\left(3-2\sqrt{6}\right)^2}\)
\(=-\left(2-\sqrt{6}\right)-\left(3-2\sqrt{6}\right)\)
\(=-2+\sqrt{6}-3+2\sqrt{6}\)
\(=-5+3\sqrt{6}\)
\(\sqrt{16-6\sqrt{7}}+\sqrt{32-8\sqrt{7}}\)
\(=\sqrt{3^2-2.3.\sqrt{7}+\left(\sqrt{7}\right)^2}+\sqrt{2^2-2.2.2\sqrt{7}+\left(2\sqrt{7}\right)^2}\)
\(=\sqrt{\left(3-\sqrt{7}\right)^2}+\sqrt{\left(2-2\sqrt{7}\right)^2}\)
\(=3-\sqrt{7}-\left(2-2\sqrt{7}\right)\)
\(=3-\sqrt{7}-2+2\sqrt{7}\)
\(=1+\sqrt{7}\)
e) Ta có: \(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{2}+1-\sqrt{2}+1\)
=2
a: \(=\dfrac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\dfrac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)
\(=\dfrac{3\sqrt{5}+5-3-\sqrt{5}}{2\left(\sqrt{5}+1\right)}=\dfrac{2\sqrt{5}+2}{2\left(\sqrt{5}+1\right)}=1\)
b: \(=\sqrt{\sqrt{3}}\left(2\sqrt{2}-2\cdot5\sqrt{2}+4\cdot8\sqrt{2}\right)\)
\(=\sqrt{\sqrt{3}}\cdot24\sqrt{2}\)
d: \(=\dfrac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\dfrac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)
a: \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{9-2\cdot3\cdot\sqrt{6}+6}+\sqrt{24-2\cdot2\sqrt{6}\cdot3+9}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
b: \(\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\)
\(=\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(3-\sqrt{5}\right)^2}\)
\(=\left|3+\sqrt{5}\right|+\left|3-\sqrt{5}\right|\)
\(=3+\sqrt{5}+3-\sqrt{5}=6\)
c: \(\dfrac{3}{2\sqrt{3}+3}+\dfrac{3}{2\sqrt{3}-3}\)
\(=\dfrac{3\left(2\sqrt{3}-3\right)+3\left(2\sqrt{3}+3\right)}{12-9}\)
\(=2\sqrt{3}-3+2\sqrt{3}+3=4\sqrt{3}\)
d: \(\sqrt{\left(\sqrt{3}+4\right)\cdot\sqrt{19-8\sqrt{3}}+3}\)
\(=\sqrt{\left(4+\sqrt{3}\right)\cdot\sqrt{\left(4-\sqrt{3}\right)^2}+3}\)
\(=\sqrt{\left(4+\sqrt{3}\right)\cdot\left(4-\sqrt{3}\right)+3}\)
\(=\sqrt{16-3+3}=\sqrt{16}=4\)
e: \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}\)
\(=\dfrac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}+\dfrac{3\left(3-\sqrt{6}\right)}{3}\)
\(=\dfrac{\sqrt{6}}{2}+3-\sqrt{6}=3-\dfrac{\sqrt{6}}{2}\)
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. \(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\) \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
= \(3-\sqrt{6} +2\sqrt{6}-3\) = \(\sqrt{6}\)
b. \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)
= \(\sqrt{8\sqrt{3}}-2.5\sqrt{12}+4\sqrt{8\sqrt{3}}\)
= \(5\sqrt{8\sqrt{3}}-5\sqrt{4.\sqrt{12}}=5\sqrt{8\sqrt{3}}-5\sqrt{4.2\sqrt{3}}\)
= \(5\sqrt{8\sqrt{3}}-5\sqrt{8\sqrt{3}}=0\)
c. \(\sqrt{2-\sqrt{3}}.\left(\sqrt{6}+\sqrt{2}\right)\) = \(\sqrt{2}.\sqrt{2-\sqrt{3}}\left(\sqrt{3}+1\right)=\sqrt{4-2\sqrt{3}}.\left(\sqrt{3}+1\right)\)
=\(\sqrt{\left(\sqrt{3}-1\right)^2}\left(\sqrt{3}+1\right)\)
= 3 - 1 = 2
d. \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
= \(\dfrac{\sqrt{2}\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)}{\sqrt{2}}=\dfrac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)
= \(\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{2}}\) = \(\dfrac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}\)
= \(\dfrac{2\sqrt{5}}{\sqrt{2}}\)= \(\sqrt{10}\)
e. \(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3=\left(\sqrt{2}+1-\sqrt{2}+1\right)\left(\left(\sqrt{2}+1\right)^2+\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)+\left(\sqrt{2}-1\right)^2\right)\)\(2.\left(3+2\sqrt{2}+2-1+3-2\sqrt{2}\right)=2.7=14\)