√3 - √53+4√90
1.(√2 + 1)^3 − (√2 − 1)^3
2.√13 − √160 − √53 + 4√90
3.(√3 + 1)^3 − (√3 − 1)^3
1)( \(\sqrt{2}\) +1)^3-( \(\sqrt{2}\) -1)^3=
( \(\sqrt{2}\) +1- \(\sqrt{2}\) +1)[( \(\sqrt{2}\) +1)^2+( \(\sqrt{2}\) -1)( \(\sqrt{2}\) +1)+(\(\sqrt{2}\) -1)^2]
=2( 2+\(2\sqrt{2}\)+1+2-1+2-\(2\sqrt{2}\)+1)=2.7=14
2) \(\sqrt{13}\)-\(\sqrt{160}\)-\(\sqrt{53}\)+\(4\sqrt{90}\)
=\(\sqrt{13}\)-\(4\sqrt{10}\)-\(\sqrt{53}\)+\(12\sqrt{10}\)=\(\sqrt{13}\)-\(\sqrt{53}\)+\(16\sqrt{10}\)=
hình như sai đề rồi
3) tương tự bài 1)
( \(\sqrt{3}\) +1)^3-( \(\sqrt{3}\) -1)^3=
( \(\sqrt{3}\) +1- \(\sqrt{3}\) +1)[( \(\sqrt{3}\) +1)^2+( √\(\sqrt{3}\) -1)( \(\sqrt{3}\) +1)+(\(\sqrt{3}\) -1)^2]
=2( 3+\(2\sqrt{3}\)+1+3-1+3-\(2\sqrt{3}\)+1)=2.10=20
Kí hiệu [x] là số nguyên lớn nhất ko vượt quá x, gọi là phân nguyên của x, chẳng hạn [2,6] = 2; [7] = 7; [ -3,5] = -4. Hãy tính [ 2/3 ], [ 53/4 ], [ -2021,01], [ -39/5 ]
Trả lời :
\(\left[\frac{2}{3}\right]=0\)
\(\left[\frac{53}{4}\right]=\left[13\frac{1}{4}\right]=13\)
\(\left[-2021,01\right]=-2022\)
\(\left[\frac{-39}{5}\right]=\left[-7\frac{4}{5}\right]=-8\)
\(\sqrt{13-\sqrt{160}}\) + \(\sqrt{53+4\sqrt{90}}\)
\(\sqrt{13-\sqrt{160}}+\sqrt{53+4\sqrt{90}}\)
\(=\sqrt{13-4\sqrt{10}}+\sqrt{53+12\sqrt{10}}\)
\(=2\sqrt{2}-\sqrt{5}+3\sqrt{5}-2\sqrt{2}\)
\(=2\sqrt{5}\)
`\sqrt{13-\sqrt{160}}+\sqrt{53+4\sqrt{90}}`
`=\sqrt{8-2.2\sqrt{2}.\sqrt{5}+5}+\sqrt{45+2.3\sqrt{5}.2\sqrt{2}+8}`
`=\sqrt{(2\sqrt{2}-\sqrt{5})^2}+\sqrt{(3\sqrt{5}+2\sqrt{2})^2}`
`=|2\sqrt{2}-\sqrt{5}|+3\sqrt{5}+2\sqrt{2}`
`=2\sqrt{2}-\sqrt{5}+3\sqrt{5}+2\sqrt{2}`
`=4\sqrt{2}+2\sqrt{5}`
\(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
\(=\sqrt{13-4\sqrt{10}}-\sqrt{53-12\sqrt{10}}\)
\(=\sqrt{13-2\cdot2\sqrt{2}\cdot\sqrt{5}}-\sqrt{53-2\cdot3\sqrt{5}\cdot2\sqrt{2}}\)
\(=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{5}-2\sqrt{2}\right)^2}\)
\(=2\sqrt{2}-\sqrt{5}-3\sqrt{5}+2\sqrt{2}\)
\(=4\sqrt{2}-4\sqrt{5}\)
\(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
\(=\sqrt{13-\sqrt{4^2\cdot10}}-\sqrt{53+4\sqrt{3^2\cdot10}}\)
\(=\sqrt{13-4\sqrt{10}}-\sqrt{53+12\sqrt{10}}\)
\(=\sqrt{\left(2\sqrt{2}\right)^2-2\cdot2\sqrt{2}\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{5}\right)^2+2\cdot2\sqrt{2}\cdot3\sqrt{5}+\left(2\sqrt{2}\right)^2}\)
\(=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{5}+2\sqrt{2}\right)^2}\)
\(=\left|2\sqrt{2}-\sqrt{5}\right|-\left|3\sqrt{5}+2\sqrt{2}\right|\)
\(=2\sqrt{2}-\sqrt{5}-\left(3\sqrt{5}+2\sqrt{2}\right)\)
\(=2\sqrt{2}-\sqrt{5}-3\sqrt{5}-2\sqrt{2}\)
\(=-4\sqrt{5}\)
\(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
\(=\sqrt{13-2\sqrt{40}}-\sqrt{53+2\sqrt{360}}\)
\(=\sqrt{\left(\sqrt{5}\right)^2-2.\sqrt{5}.\sqrt{8}+\left(\sqrt{8}\right)^2}-\sqrt{\left(\sqrt{45}\right)^2+2\sqrt{45}.\sqrt{8}+\left(\sqrt{8}\right)^2}\)
\(=\sqrt{\left(\sqrt{8}-\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{45}+\sqrt{8}\right)^2}\)
\(=\sqrt{8}-\sqrt{5}-\left(\sqrt{45}+\sqrt{8}\right)\)
\(=\sqrt{8}-\sqrt{5}-3\sqrt{5}-\sqrt{8}\)
\(=-4\sqrt{5}\)
√13-√160 -√53+4√90
\(=2\sqrt{2}-\sqrt{5}-3\sqrt{5}-2\sqrt{2}=-4\sqrt{5}\)
A= 13/30 + 28/45∙ 2 1/2 -[(1/2+1/3): 53/90]: 50/53
Tính P = √(13-√(160)) - √(53+4√(90))
\(P=\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
.....\(=\sqrt{13-4\sqrt{10}}-\sqrt{53+12\sqrt{10}}\)
.....\(=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}-\sqrt{\left(2\sqrt{2}+3\sqrt{5}\right)^2}\)
.....\(=\left|2\sqrt{2}-\sqrt{5}\right|-\left|2\sqrt{2}+3\sqrt{5}\right|\)
.....\(=2\sqrt{2}-\sqrt{5}-2\sqrt{2}-3\sqrt{5}\)
.....\(=-4\sqrt{5}\)
\(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
Bài này không sai đề , tớ làm lại cho :
\(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}=\sqrt{13-4\sqrt{10}}-\sqrt{53+12\sqrt{10}}=\sqrt{8-2.2\sqrt{2}.\sqrt{5}+5}-\sqrt{45+2.3\sqrt{5}.2\sqrt{2}+8}=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{5}+2\sqrt{2}\right)^2}=\text{ |}2\sqrt{2}-\sqrt{5}\text{ |}-\text{ |}3\sqrt{5}+2\sqrt{2}\text{ |}=4\sqrt{2}-4\sqrt{5}\)
Đề này mình làm không ra nên mình sẽ sửa đề.
Giải:
\(\sqrt{14-\sqrt{160}}-\sqrt{49+4\sqrt{90}}\)
\(=\sqrt{14-4\sqrt{10}}-\sqrt{49+12\sqrt{10}}\)
\(=\sqrt{10-4\sqrt{10}+4}-\sqrt{40+12\sqrt{10}+9}\)
\(=\sqrt{\left(\sqrt{10}\right)^2-2.\sqrt{10}.2+2^2}-\sqrt{\left(2\sqrt{10}\right)^2+2.2\sqrt{10}.3+3^2}\)
\(=\sqrt{\left(\sqrt{10}-2\right)^2}-\sqrt{\left(2\sqrt{10}+3\right)^2}\)
\(=\sqrt{10}-2-\left(2\sqrt{10}+3\right)\)
\(=\sqrt{10}-2-2\sqrt{10}-3\)
\(=-\sqrt{10}-5\)
Vậy ...
Nếu sai mong bạn thông cảm
\(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
\(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
\(=\sqrt{8-2.2\sqrt{2}.\sqrt{5}+5}-\sqrt{45+2.3\sqrt{5}.2\sqrt{2}+8}\)
\(=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{5}+2\sqrt{2}\right)^2}\)
\(=|2\sqrt{2}-\sqrt{5}|-|3\sqrt{5}+2\sqrt{2}|\)
\(=2\sqrt{2}-\sqrt{5}-3\sqrt{5}-2\sqrt{2}\)
\(=-4\sqrt{5}\)