Bài 11: Số vô tỉ. Khái niệm về căn bậc hai
Các câu hỏi tương tự
a,sqrt{1}+sqrt{9}+sqrt{25}+sqrt{49}+sqrt{81} csqrt{0,04}+sqrt{0,09}+sqrt{0,16}
b,sqrt{dfrac{1}{4}}+sqrt{dfrac{1}{9}}+sqrt{dfrac{1}{36}}+sqrt{dfrac{1}{16}} esqrt{2^2}+sqrt{4^2}+sqrt{left(-6^2right)}+sqrt{left(-8^2right)}
j,sqrt{1,44}-sqrt{1,69}+sqrt{1,96}
g, sqrt{dfrac{4}{25}}+sqrt{dfrac{25}{4}}+sqrt{dfrac{81}{100}}+sqrt{dfrac{9}{16}}
dsqrt{81}-sqrt{64}+sq...
Đọc tiếp
a,\(\sqrt{1}+\sqrt{9}+\sqrt{25}+\sqrt{49}+\sqrt{81}\) c\(\sqrt{0,04}+\sqrt{0,09}+\sqrt{0,16}\)
b,\(\sqrt{\dfrac{1}{4}}+\sqrt{\dfrac{1}{9}}+\sqrt{\dfrac{1}{36}}+\sqrt{\dfrac{1}{16}}\) e\(\sqrt{2^2}+\sqrt{4^2}+\sqrt{\left(-6^2\right)}+\sqrt{\left(-8^2\right)}\)
j,\(\sqrt{1,44}-\sqrt{1,69}+\sqrt{1,96}\)
g, \(\sqrt{\dfrac{4}{25}}+\sqrt{\dfrac{25}{4}}+\sqrt{\dfrac{81}{100}}+\sqrt{\dfrac{9}{16}}\)
d\(\sqrt{81}-\sqrt{64}+\sqrt{49}\)
bài 1: tính
a) 3/4+(-5/2)+(-3/5)
b) \(\sqrt{\left(7\right)^2}+\sqrt{\dfrac{25}{16}-\dfrac{3}{2}}\)
c)\(\dfrac{1}{2}.\sqrt{100}-\sqrt{\dfrac{1}{16}+\left(\dfrac{1}{3}\right)^0}\)
\(\sqrt{4}+\sqrt{9}+\sqrt{16}+\sqrt{25}\)
Bài 1: Tính
a) \(\sqrt{49}+\sqrt{4}\)
b) \(\sqrt{0,25}-\sqrt{0,01}\)
c) \(\sqrt{\dfrac{16}{25}}-\sqrt{\dfrac{1}{81}}\)
d) \(\sqrt{64}-\sqrt{16}+\sqrt{\left(-3\right)^2}\)
e) \(2-\sqrt{0,36}\)
1)so sánh các số sau:
a)0,5\(\sqrt{100}\)-\(\sqrt{\dfrac{4}{25}}\) và (\(\sqrt{1\dfrac{1}{9}}\)-\(\sqrt{\dfrac{9}{16}}\)):5
b)\(\sqrt{25+9}\) và \(\sqrt{25}+\sqrt{9}\)
2) CMR: Với a,b dương thì \(\sqrt{a+b}< \sqrt{a}+\sqrt{b}\)
giúp mình với
1, tính
a, 7timessqrt{dfrac{6^2}{7^2}}-sqrt{25}+sqrt{dfrac{left(-3right)^2}{2}}
b, -sqrt{dfrac{64}{49}}-dfrac{3}{5}timessqrt{dfrac{25}{64}}+sqrt{0,25}
c, sqrt{dfrac{10000}{5}}-dfrac{1}{4}.sqrt{dfrac{16}{9}}+sqrt{dfrac{left(-3right)^2}{left(4right)}}
d, left|dfrac{1}{4}-sqrt{0,0144}right|-dfrac{3}{2}+sqrt{dfrac{81}{169}}
Đọc tiếp
giúp mình với
1, tính
a, \(7\times\sqrt{\dfrac{6^2}{7^2}}-\sqrt{25}+\sqrt{\dfrac{\left(-3\right)^2}{2}}\)
b, \(-\sqrt{\dfrac{64}{49}}-\dfrac{3}{5}\times\sqrt{\dfrac{25}{64}}+\sqrt{0,25}\)
c, \(\sqrt{\dfrac{10000}{5}}-\dfrac{1}{4}.\sqrt{\dfrac{16}{9}}+\sqrt{\dfrac{\left(-3\right)^2}{\left(4\right)}}\)
d, \(\left|\dfrac{1}{4}-\sqrt{0,0144}\right|-\dfrac{3}{2}+\sqrt{\dfrac{81}{169}}\)
\(\sqrt{\left(-7^2\right)}+\sqrt{\frac{25}{16}}-\frac{3}{2}\)
Ta có : \(\sqrt{25}=5;-\sqrt{25}=-5;\sqrt{\left(-5\right)^2}=\sqrt{25}=5\)
Theo mẫu trên, hãy tính :
a) \(\sqrt{36}\)
b) \(-\sqrt{16}\)
c) \(\sqrt{\dfrac{9}{25}}\)
d) \(\sqrt{3^2}\)
e) \(\sqrt{\left(-3\right)^2}\)
Tìm x: \(\sqrt{x^2\sqrt{x^4\sqrt{x^8\sqrt{x^{16}}}}}=\text{ }\sqrt{3^{14}}\)
Tìm x biết :a) \(\sqrt{x}\) = \(x\)
b) \(x-2\sqrt{x}=0\)
c) \(\sqrt{x+1}=1-x\)
d) \(\sqrt{3x^2+4}+\sqrt{2004x^2+1}=3-4x^2\)
e) \(\sqrt{3x^2+4}+\sqrt{2007x^2+25}=7-69x^2\)