Tính :
\(\sqrt{1}+\sqrt{49}\)
\(\sqrt{25}+\sqrt{100}\)
Những câu hỏi liên quan
\(\sqrt{49}+\sqrt{36}-\sqrt{25}+\sqrt{100}=?\)
\(\sqrt{49}+\sqrt{36}-\sqrt{25}+\sqrt{100}\)
\(=7+6-5+10\)
\(=18\)
có phải như vậy ko?
Đúng 0
Bình luận (0)
\(\sqrt{49}+\sqrt{36}-\sqrt{25}+\sqrt{100}\)
\(=7+6-5+10\)
\(=13-5+10\)
\(=8+10\)
\(=18\)
Đúng 0
Bình luận (0)
13 tháng 11 2021 lúc 10:32
Bài 2: Tính:
a.\(\sqrt{81}\)
b. \(\sqrt{8100}\)
c. \(\sqrt{64}\)
d. \(\sqrt{\dfrac{49}{100}}\)
e . \(\sqrt{\dfrac{4}{25}}\)
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\(\sqrt{81}-\sqrt{64}+\sqrt{49}\)
a)\(\sqrt{1}\)+\(\sqrt{9}\)+\(\sqrt{25}\)+\(\sqrt{49}\)+\(\sqrt{81}\)
=1+3+5+7+9
=25
b)=\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{4}\)
=\(\dfrac{6}{12}\)+\(\dfrac{4}{12}\)+\(\dfrac{2}{12}\)+\(\dfrac{3}{12}\)
=\(\dfrac{15}{12}\)
c) =0,2+0.3+0,4
= 0.9
d) =9-8+7
=8
j) =1,2-1,3+1.4
= (-0,1)+1,4
=1,4
g) \(\dfrac{2}{5}\)+\(\dfrac{5}{2}\)+\(\dfrac{9}{10}\)+\(\dfrac{3}{4}\)
= (\(\dfrac{4}{10}\)+\(\dfrac{15}{10}\)+\(\dfrac{9}{10}\))+\(\dfrac{3}{4}\)
= \(\dfrac{14}{5}\)+\(\dfrac{3}{4}\)
=\(\dfrac{56}{20}\)+\(\dfrac{15}{20}\)
= \(\dfrac{71}{20}\)
Nhớ tick cho mk nha~
Đúng 3
Bình luận (0)
a,\(\sqrt{49-\sqrt{4+\sqrt{25}}}\)
b,\(\left(\sqrt{100-\sqrt{1}}\right):\sqrt{\left(-3\right)^2}\)
c,\(\sqrt{16+9-\sqrt{25-9}}\)
d,\(\sqrt{\left(-7\right)^2-\sqrt{1^{40}}.\sqrt{16}}\)
tờ phắc??? toán lớp 7???
a, \(\sqrt{1\frac{24}{25}.5\frac{1}{16}.0,01}=\sqrt{\frac{49}{25}.\frac{81}{16}.\frac{1}{100}}\)
b, \(\sqrt{2,25.1,46-2,25.0,02}\)
B2 : Tính : a, left(sqrt{x}-3right).left(sqrt{x}+2right)b, left(sqrt{x}-sqrt{y}right).left(sqrt{x}+sqrt{y}right)c, left(sqrt{dfrac{25}{3}}-sqrt{dfrac{49}{3}}+sqrt{3}right).sqrt{3}d,left(1+sqrt{3}-sqrt{5}right).left(1+sqrt{3}+sqrt{5}right)
Đọc tiếp
B2 : Tính :
a, \(\left(\sqrt{x}-3\right)\)\(.\left(\sqrt{x}+2\right)\)
b, \(\left(\sqrt{x}-\sqrt{y}\right).\)\(\left(\sqrt{x}+\sqrt{y}\right)\)
c, \(\left(\sqrt{\dfrac{25}{3}}-\sqrt{\dfrac{49}{3}}+\sqrt{3}\right)\)\(.\sqrt{3}\)
d,\(\left(1+\sqrt{3}-\sqrt{5}\right)\)\(.\left(1+\sqrt{3}+\sqrt{5}\right)\)
a. \(\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)=x-3\sqrt{x} +2\sqrt{x}-6=x-\sqrt{x}-6\)
b. \(\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)=x-y\)
c. \(\left(\sqrt{\dfrac{25}{3}}-\sqrt{\dfrac{49}{3}}+\sqrt{3}\right).\sqrt{3}\)
\(=\left(\dfrac{5}{\sqrt{3}}-\dfrac{7}{\sqrt{3}}+\sqrt{3}\right).\sqrt{3}=\dfrac{5}{3}-\dfrac{7}{3}+9=\dfrac{25}{3}\)
d. \(\left(1+\sqrt{3}-\sqrt{5}\right)\left(1+\sqrt{3}+\sqrt{5}\right)\)
\(=\left(1+\sqrt{3}\right)^2-5=1+2\sqrt{3}+3-5=2\sqrt{3}-1\)
Đúng 2
Bình luận (1)
Tính
A=\(\sqrt{0,36}-\left(\sqrt{25}-\sqrt{49}:\sqrt{4}\right).\sqrt{0,4}-\sqrt{10}^2\)
Thực hiện phép tính:
a. \(\sqrt{25}+2\sqrt{49}\)
b. \(\sqrt{16}.\sqrt{25}+\sqrt{169}:\sqrt{49}\)
c. \(\sqrt{\left(3-\sqrt{7}\right)^2}+\sqrt{7}\)
d. \(2\sqrt{3}-\sqrt{75}+2\sqrt{12}\)
Tính :
\(a)\sqrt{16}.\sqrt{25}+\sqrt{196}:\sqrt{49}\)
\(b)\sqrt{\sqrt{81}}\)
a) \(\sqrt{16}.\sqrt{25}+\sqrt{196}:\sqrt{49}\)
=4.5+14:7
=20+2
=22
b) chưa học nhó:))
a) \(\sqrt{16}.\sqrt{25}+\sqrt{196}\div\sqrt{49}\)
\(=4.5+14\div7=20+2=22\)
b) \(\sqrt{\sqrt{81}}=\sqrt{9}=3\)
Xem thêm câu trả lời