\(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\)
= \(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3-2\sqrt{6}\right)^2}\)
= \(3-\sqrt{6}+3-2\sqrt{6}\)
K
Khách
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Những câu hỏi liên quan
19 tháng 8 2021
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: \(C=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=3-\sqrt{6}+2\sqrt{6}-3\)
\(=\sqrt{6}\)
30 tháng 9 2023
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) \(\sqrt{7-2\sqrt{6}}+\sqrt{15-6\sqrt{6}}\)
\(=\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot1+1^2}+\sqrt{3^2-2\cdot3\cdot\sqrt{6}+\left(\sqrt{6}\right)^2}\)
\(=\sqrt{\left(\sqrt{6}-1\right)^2}+\sqrt{\left(3-\sqrt{6}\right)^2}\)
\(=\left|\sqrt{6}-1\right|+\left|3-\sqrt{6}\right|\)
\(=\sqrt{6}-1+3-\sqrt{6}\)
\(=2\)
T
30 tháng 9 2023
\(a,\sqrt{3-2\sqrt{2}}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=\sqrt{\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot1+1}+\left|2-\sqrt{2}\right|\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}+2-\sqrt{2}\)
\(=\left|\sqrt{2}-1\right|+2-\sqrt{2}\)
\(=\sqrt{2}-1+2-\sqrt{2}\)
\(=1\)
\(---\)
\(b,\sqrt{33-12\sqrt{6}}-\sqrt{\left(5-2\sqrt{6}\right)^2}\)
\(=\sqrt{\left(2\sqrt{6}\right)^2-2\cdot2\sqrt{6}\cdot3+3^2}-\left|5-2\sqrt{6}\right|\)
\(=\sqrt{\left(2\sqrt{6}-3\right)^2}-5+2\sqrt{6}\)
\(=\left|2\sqrt{6}-3\right|-5+2\sqrt{6}\)
\(=2\sqrt{6}-3-5+2\sqrt{6}\)
\(=4\sqrt{6}-8\)
\(---\)
\(c,\sqrt{7-2\sqrt{6}}+\sqrt{15-6\sqrt{6}}\)
\(=\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot1+1^2}+\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot3+3^2}\)
\(=\sqrt{\left(\sqrt{6}-1\right)^2}+\sqrt{\left(\sqrt{6}-3\right)^2}\)
\(=\left|\sqrt{6}-1\right|+\left|\sqrt{6}-3\right|\)
\(=\sqrt{6}-1+3-\sqrt{6}\)
\(=2\)
#\(Toru\)
26 tháng 10 2023
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: \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}\)
\(=\dfrac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}+\dfrac{3\left(3-\sqrt{6}\right)}{3}\)
\(=\dfrac{\sqrt{6}}{2}+3-\sqrt{6}=3-\dfrac{\sqrt{6}}{2}\)
CW
26 tháng 6 2017
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 \(\sqrt{3-2\sqrt{2}}< \sqrt{3+2\sqrt{2}}\))
CW
26 tháng 6 2017
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
1 tháng 10 2023
\(A=\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right)\left(2+\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}\right)\)
\(A=\left[2-\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}\right]\left[2+\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}+3}\right]\)
\(A=\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)\)
\(A=2^2-\left(\sqrt{5}\right)^2\)
\(A=4-5\)
\(A=-1\)
____
\(B=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3+\sqrt{6}\right)\left(3-\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)\)
\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right]\left(\sqrt{6}+11\right)\)
\(B=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(B=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
\(B=6-121\)
\(B=-115\)
27 tháng 7 2023
b: =căn 10-3+4-căn 10=1
a: \(=\sqrt{11-4\sqrt{6}+\sqrt{15}}\)
26 tháng 6 2021
`1)A=sqrt{4+sqrt{10+2sqrt5}}+sqrt{4-sqrt{10+2sqrt5}}`
`<=>A^2=4+sqrt{10+2sqrt5}+4-sqrt{10+2sqrt5}+2sqrt{16-10-2sqrt5}`
`<=>A^2=8+2sqrt{6-2sqrt5}`
`<=>A^2=8+2sqrt{(sqrt5-1)^2}`
`<=>A^2=8+2(sqrt5-1)`
`<=>A^2=6+2sqrt5=(sqrt5+1)^2`
`<=>A=sqrt5+1(do \ A>0)`
`b)B=sqrt{35+12sqrt6}-sqrt{35-12sqrt6}`
Vì `35+12sqrt6>35-12sqrt6`
`=>B>0`
`B^2=35+12sqrt6+35-12sqrt6-2sqrt{35^2-(12sqrt6)^2}`
`<=>B^2=70-2sqrt{361}`
`<=>B^2=70-2sqrt{19^2}=70-38=32`
`<=>B=sqrt{32}=4sqrt2(do \ B>0)`
`3)(4+sqrt{15})(sqrt{10}-sqrt6)sqrt{4-sqrt{15}}`
`=sqrt{4+sqrt{15}}.sqrt{4-sqrt{15}}.sqrt{4+sqrt{15}}(sqrt{10}-sqrt6)`
`=sqrt{16-15}.sqrt2(sqrt5-sqrt3).sqrt{4+sqrt{15}}`
`=sqrt{8+2sqrt{15}}(sqrt5-sqrt3)`
`=sqrt{5+2sqrt{5.3}+3}(sqrt5-sqrt3)`
`=sqrt{(sqrt5+sqrt3)^2}(sqrt5-sqrt3)`
`=(sqrt5+sqrt3)(sqrt5-sqrt3)`
`=5-3=2`
Ơ lám sao để ra bc 2 nhỉ . T quên rồi.
j đây ?