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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
a)\(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\)
= \(2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\)
= \(4\sqrt{5}\)
b) \(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\)
= \(\sqrt{3\left(5-2\sqrt{6}\right)}-\sqrt{33-12\sqrt{6}}\)
= \(\sqrt{3\left(5-2\sqrt{6}\right)}-\sqrt{3\left(11-4\sqrt{6}\right)}\)
\(a,2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\)
\(=2\sqrt{5}-\sqrt{5^2.5}-\sqrt{4^2.5}+\sqrt{11^2.5}\)
\(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\)
\(=4\sqrt{5}\)
\(b,\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3-2\sqrt{6}\right)^2}\)
\(=|3-\sqrt{6}|+|3-2\sqrt{6}|\)
\(=3-\sqrt{6}+2\sqrt{6}-3\)
\(=\sqrt{6}\)
mjk cần gấp lắm !!
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. \(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3=\left(\sqrt{2}+1-\sqrt{2}+1\right)\left(\left(\sqrt{2}+1\right)^2+\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)+\left(\sqrt{2}-1\right)^2\right)\)\(2.\left(3+2\sqrt{2}+2-1+3-2\sqrt{2}\right)=2.7=14\)
\( 2)2\sqrt {\dfrac{{16}}{3}} - 3\sqrt {\dfrac{1}{{27}}} - 6\sqrt {\dfrac{4}{{75}}} \\ = 2.\dfrac{4}{{\sqrt 3 }} - 3.\dfrac{1}{{\sqrt {27} }} - 6\dfrac{2}{{\sqrt {75} }}\\ = \dfrac{8}{{\sqrt 3 }} - \dfrac{3}{{3\sqrt 3 }} - \dfrac{{12}}{{5\sqrt 2 }}\\ = \dfrac{{8\sqrt 3 }}{3} - \dfrac{{\sqrt 3 }}{3} - \dfrac{{4\sqrt 3 }}{5}\\ = \dfrac{{23\sqrt 3 }}{{15}}\\ 3)2\sqrt {27} - 6\sqrt {\dfrac{4}{3}} + \dfrac{3}{5}\sqrt {75} \\ = 6\sqrt 3 - \dfrac{{12}}{{\sqrt 3 }} + 3\sqrt 3 \\ = 9\sqrt 3 - 4\sqrt 3 \\ = 5\sqrt 3 \)
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: \(A=\sqrt{\left(1-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=\sqrt{3}-1-2-\sqrt{3}\)
=-3
b: Ta có: \(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)
\(=2-\sqrt{3}+\sqrt{3}-1\)
=1
c: Ta có: \(C=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=3-\sqrt{6}+2\sqrt{6}-3\)
\(=\sqrt{6}\)
a) \(\sqrt{3-2\sqrt{2}}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=\sqrt{\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot1+1^2}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{2}-1\right|+\left|2-\sqrt{2}\right|\)
\(=\sqrt{2}-1+2-\sqrt{2}\)
\(=1\)
b) \(\sqrt{33-12\sqrt{6}}-\sqrt{\left(5-2\sqrt{6}\right)^2}\)
\(=\sqrt{\left(2\sqrt{6}\right)^2-2\cdot2\sqrt{6}\cdot3+3^2}-\sqrt{\left(5-2\sqrt{6}\right)^2}\)
\(=\sqrt{\left(2\sqrt{6}-3\right)^2}-\sqrt{\left(5-2\sqrt{6}\right)^2}\)
\(=\left|2\sqrt{6}-3\right|-\left|5-2\sqrt{6}\right|\)
\(=2\sqrt{6}-3-5+2\sqrt{6}\)
\(=4\sqrt{6}-8\)
c) \(\sqrt{7-2\sqrt{6}}+\sqrt{15-6\sqrt{6}}\)
\(=\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot1+1^2}+\sqrt{3^2-2\cdot3\cdot\sqrt{6}+\left(\sqrt{6}\right)^2}\)
\(=\sqrt{\left(\sqrt{6}-1\right)^2}+\sqrt{\left(3-\sqrt{6}\right)^2}\)
\(=\left|\sqrt{6}-1\right|+\left|3-\sqrt{6}\right|\)
\(=\sqrt{6}-1+3-\sqrt{6}\)
\(=2\)
\(a,\sqrt{3-2\sqrt{2}}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=\sqrt{\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot1+1}+\left|2-\sqrt{2}\right|\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}+2-\sqrt{2}\)
\(=\left|\sqrt{2}-1\right|+2-\sqrt{2}\)
\(=\sqrt{2}-1+2-\sqrt{2}\)
\(=1\)
\(---\)
\(b,\sqrt{33-12\sqrt{6}}-\sqrt{\left(5-2\sqrt{6}\right)^2}\)
\(=\sqrt{\left(2\sqrt{6}\right)^2-2\cdot2\sqrt{6}\cdot3+3^2}-\left|5-2\sqrt{6}\right|\)
\(=\sqrt{\left(2\sqrt{6}-3\right)^2}-5+2\sqrt{6}\)
\(=\left|2\sqrt{6}-3\right|-5+2\sqrt{6}\)
\(=2\sqrt{6}-3-5+2\sqrt{6}\)
\(=4\sqrt{6}-8\)
\(---\)
\(c,\sqrt{7-2\sqrt{6}}+\sqrt{15-6\sqrt{6}}\)
\(=\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot1+1^2}+\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot3+3^2}\)
\(=\sqrt{\left(\sqrt{6}-1\right)^2}+\sqrt{\left(\sqrt{6}-3\right)^2}\)
\(=\left|\sqrt{6}-1\right|+\left|\sqrt{6}-3\right|\)
\(=\sqrt{6}-1+3-\sqrt{6}\)
\(=2\)
#\(Toru\)
\(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}=\sqrt{6-6\sqrt{6}+9}+\sqrt{24-12\sqrt{6}+9}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(\sqrt{24}-3\right)^2}=\left|3-\sqrt{6}\right|+\left|\sqrt{24}-3\right|=3-\sqrt{6}+\sqrt{24}-3=2\sqrt{6}-\sqrt{6}=\sqrt{6}\)
\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}=-\dfrac{\sqrt{2}\left(\sqrt{6}-4\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}=\dfrac{-\sqrt{2}}{\sqrt{3}}-\dfrac{1}{\sqrt{6}}=\dfrac{-\sqrt{6}}{3}-\dfrac{\sqrt{6}}{6}=-\dfrac{\sqrt{6}}{2}\).