✓17-3✓32x✓17+3✓32
Tính nhanh
a)47/53x(17/3-53/47)+17/3x(6/17-47/53)
b)15/37x(38/41-75/45)-38/41x(15/37-82/76)
c)-41/32x(15/8-16/41)+15/8x(41/32-8/3)
Tính hợp lí (nếu có thể)
a, 21 x 23 - 3 x 7 x (-17)
b, 42x3-7x[(-34)+18]
c, 71x64+32x(-7)-13x32
d, 13x(23-17)-13x(23+17)
giúp mình vớiiiii
Lời giải:
a.
$21\times 23-3\times 7\times (-17)$
$=21\times 23-21\times (-17)$
$=21\times 23+21\times 17=21\times (23+17)=21\times 40$
$=840$
b.
$42\times 3-7\times [(-34)+18]$
$=7\times 18-7\times (-16)$
$=7\times 18+7\times 16=7\times (18+16)=7\times 34=238$
c.
$=142\times 32+32\times (-7)-13\times 32$
$=32\times (142-7-32)=32\times 103=3296$
d.
$=13\times [(23-17)-(23+17)]$
$=13\times (23-17-23-17)=13\times (-34)=-442$
√17-3√32 -√17+3√32
\(\sqrt{17}-3\sqrt{32}-\sqrt{17}+3\sqrt{32}\\ =\left(\sqrt{17}-\sqrt{17}\right)+\left(3\sqrt{32}-3\sqrt{32}\right)\\ =0+0=0\)
\(=\sqrt{17-12\sqrt{2}}-\sqrt{17+12\sqrt{2}}\)
\(=\sqrt{17-2\cdot3\cdot2\sqrt{2}}-\sqrt{17+2\cdot3\cdot2\sqrt{2}}\)
\(=\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(3+2\sqrt{2}\right)^2}\)
=3-2căn 2-3-2căn 2
=-4căn 2
\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}=3-2\sqrt{2}+3+2\sqrt{2}=6\)
\(\sqrt{17-3\sqrt{32}}+\sqrt{17-3\sqrt{32}}\)
`\sqrt(17-3\sqrt32)+\sqrt(17+3\sqrt32)`
`=\sqrt(17-12\sqrt2)+\sqrt(17+12\sqrt2)`
`=\sqrt(9-12\sqrt2+8)+\sqrt(9+12\sqrt2+8)`
`=\sqrt(3^2-2.3.2\sqrt2 +(2\sqrt2)^2)+\sqrt(3^2+2.3.2\sqrt2+(2\sqrt2)^2)`
`=\sqrt((3-2\sqrt2)^2)+\sqrt((3+2\sqrt2)^2)`
`=3-2\sqrt2+3+2\sqrt2`
`=6`
\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
\(=3-2\sqrt{2}+3+2\sqrt{2}\)
=6
Tính
\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
\(=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(=\sqrt{9+2.3.\sqrt{8}+8}+\sqrt{9-2.3.\sqrt{8}+8}\)
\(=\sqrt{\left(3+\sqrt{8}\right)^2}+\sqrt{\left(3-\sqrt{8}\right)^2}=\left|3+\sqrt{8}\right|+\left|3-\sqrt{8}\right|\)
\(=3+\sqrt{8}+3-\sqrt{8}\) (do \(3>\sqrt{8}\))
\(=6\)
\(\sqrt{17-3\sqrt{32}}-\sqrt{17+3\sqrt{32}}\)
\(=\sqrt{17-12\sqrt{2}}-\sqrt{17+12\sqrt{2}}\)
\(=\sqrt{\left(2\sqrt{2}-3\right)^2}-\sqrt{\left(2\sqrt{2}+3\right)^2}\)
\(=-\left(2\sqrt{2}-3\right)-\left(2\sqrt{2}+3\right)=-2\sqrt{2}+3-2\sqrt{2}-3\)
\(=-4\sqrt{2}\)
Tính: \(\sqrt{17-3\sqrt{32}}+\sqrt{17-3\sqrt{32}}\)
\(\sqrt{17-3\sqrt{32}}+\sqrt{17-3\sqrt{32}}=2\sqrt{17-3\sqrt{32}}\)
\(=\sqrt{4\left(17-3\sqrt{32}\right)}=\sqrt{68-12\sqrt{32}}=\sqrt{36-12\sqrt{32}+32}\)
\(=\sqrt{6^2-2.6.\sqrt{32}+\left(\sqrt{32}\right)^2}=\sqrt{\left(6-\sqrt{32}\right)^2}=\left|6-\sqrt{32}\right|\)
\(=6-\sqrt{32}=6-4\sqrt{2}\)
\(\sqrt{17-12\sqrt{2}}-\sqrt{24-8\sqrt{8}}\)
\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
cần gấp
Cần gấp thì bạn cũng nên viết đầy đủ đề bài nhé.
** Bài toán rút gọn**
Lời giải:
\(\sqrt{17-12\sqrt{2}}=\sqrt{17-2\sqrt{72}}=\sqrt{9-2\sqrt{8.9}+8}=\sqrt{(\sqrt{9}-\sqrt{8})^2}\)
\(=\sqrt{9}-\sqrt{8}=3-2\sqrt{2}\)
\(\sqrt{24-8\sqrt{8}}=\sqrt{24-2\sqrt{128}}=\sqrt{16-2\sqrt{16.8}+8}=\sqrt{(\sqrt{16}-\sqrt{8})^2}\)
\(=\sqrt{16}-\sqrt{8}=4-2\sqrt{2}\)
\(\Rightarrow \sqrt{17-12\sqrt{2}}-\sqrt{24-8\sqrt{8}}=(3-2\sqrt{2})-(4-2\sqrt{2})=-1\)
--------------------
\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{17-12\sqrt{2}}+\sqrt{17+12\sqrt{2}}\)
\(=\sqrt{8-2\sqrt{8.9}+9}+\sqrt{8+2\sqrt{8.9}+9}\)
\(=\sqrt{(\sqrt{8}-\sqrt{9})^2}+\sqrt{(\sqrt{8}+\sqrt{9})^2}\)
\(=|\sqrt{8}-\sqrt{9}|+|\sqrt{8}+\sqrt{9}|=3-2\sqrt{2}+3+2\sqrt{2}=6\)
----------------------
\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{9+2\sqrt{9.2}+2}-\sqrt{9-2\sqrt{9.2}+2}\)
\(=\sqrt{(\sqrt{9}+\sqrt{2})^2}-\sqrt{(\sqrt{9}-\sqrt{2})^2}\)
\(=|\sqrt{9}+\sqrt{2}|-|\sqrt{9}-\sqrt{2}|=3+\sqrt{2}-(3-\sqrt{2})=2\sqrt{2}\)
\(\sqrt{17-12\sqrt{2}}-\sqrt{24-8\sqrt{8}}=\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(4-2\sqrt{2}\right)^2}\)
\(=\left|3-2\sqrt{2}\right|-\left|4-2\sqrt{2}\right|=3-2\sqrt{2}-4+2\sqrt{2}\)
\(=-1\)
\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}\)
\(=\left|3-2\sqrt{2}\right|+\left|3+2\sqrt{2}\right|=3-2\sqrt{2}+3+2\sqrt{2}\)
\(=6\)
\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=\left|3+\sqrt{2}\right|-\left|3-\sqrt{2}\right|=3+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}\)
c) Ta có: \(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(=3+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}\)