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Bài1:
Ta có:
a)\(\sqrt{\dfrac{3^2}{5^2}}=\sqrt{\dfrac{9}{25}}=\dfrac{3}{5}\)
b)\(\dfrac{\sqrt{3^2}+\sqrt{42^2}}{\sqrt{5^2}+\sqrt{70^2}}=\dfrac{\sqrt{9}+\sqrt{1764}}{\sqrt{25}+\sqrt{4900}}=\dfrac{3+42}{5+70}=\dfrac{45}{75}=\dfrac{3}{5}\)
c)\(\dfrac{\sqrt{3^2}-\sqrt{8^2}}{\sqrt{5^2}-\sqrt{8^2}}=\dfrac{\sqrt{9}-\sqrt{64}}{\sqrt{25}-\sqrt{64}}=\dfrac{3-8}{5-8}=\dfrac{-5}{-3}=\dfrac{5}{3}\)
Từ đó, suy ra: \(\dfrac{3}{5}=\sqrt{\dfrac{3^2}{5^2}}=\dfrac{\sqrt{3^2}+\sqrt{42^2}}{\sqrt{5^2}+\sqrt{70^2}}\)
Bài 2:
Không có đề bài à bạn?
Bài 3:
a)\(\sqrt{x}-1=4\)
\(\Rightarrow\sqrt{x}=5\)
\(\Rightarrow x=\sqrt{25}\)
\(\Rightarrow x=5\)
b)Vd:\(\sqrt{x^4}=\sqrt{x.x.x.x}=x^2\Rightarrow\sqrt{x^4}=x^2\)
Từ Vd suy ra:\(\sqrt{\left(x-1\right)^4}=16\)
\(\Rightarrow\left(x-1\right)^2=16\)
\(\Rightarrow\left(x-1\right)^2=4^2\)
\(\Rightarrow x-1=4\)
\(\Rightarrow x=5\)
a: \(=2\cdot\dfrac{5}{4}-3\cdot\dfrac{7}{6}+4\cdot\dfrac{9}{8}=\dfrac{5}{2}-\dfrac{7}{2}+\dfrac{9}{2}=\dfrac{7}{2}\)
b: \(=18-16\cdot\dfrac{1}{2}+\dfrac{1}{16}\cdot\dfrac{3}{4}\)
=10+3/64
=643/64
c: \(=\dfrac{2}{3}\cdot\dfrac{9}{4}-\dfrac{3}{4}\cdot\dfrac{8}{3}+\dfrac{7}{5}\cdot\dfrac{5}{14}=\dfrac{3}{2}-2+\dfrac{1}{2}=2-2=0\)
a: \(=7\cdot\dfrac{6}{7}-5+\dfrac{3\sqrt{2}}{2}=1+\dfrac{3}{2}\sqrt{2}\)
b: \(=-\dfrac{8}{7}-\dfrac{3}{5}\cdot\dfrac{5}{8}+\dfrac{1}{2}=\dfrac{-16+7}{14}-\dfrac{3}{8}=\dfrac{-9}{14}-\dfrac{3}{8}\)
\(=\dfrac{-72-42}{112}=\dfrac{-114}{112}=-\dfrac{57}{56}\)
c: \(=20\sqrt{5}-\dfrac{1}{4}\cdot\dfrac{4}{3}+\dfrac{3}{2}=20\sqrt{5}+\dfrac{3}{2}-\dfrac{1}{3}=20\sqrt{5}+\dfrac{7}{6}\)
a)\(\sqrt{\dfrac{3^2}{7^2}}=\sqrt{\dfrac{9}{49}}=\sqrt{\dfrac{3}{7}}\)
b)\(\dfrac{\sqrt{3^2}+\sqrt{39^2}}{\sqrt{7^2}+\sqrt{91^2}}=\dfrac{\sqrt{9}+\sqrt{1521}}{\sqrt{49}+\sqrt{8281}}=\dfrac{3+39}{7+91}=\dfrac{42}{98}\)
c)Tương tự câu b, ta đc:
\(\dfrac{\sqrt{3^2}-\sqrt{39^2}}{\sqrt{7^2}-\sqrt{91^2}}=\dfrac{3-39}{7-91}=\dfrac{-36}{86}=\dfrac{3}{7}\)
d)Tương tự câu a, ta đc:
\(\dfrac{\sqrt{39^2}}{\sqrt{91^2}}=\dfrac{39}{91}\)
Chúc Bạn Học Tốt!!!
a) \(\sqrt{\dfrac{3^2}{7^2}}=\sqrt{\left(\dfrac{3}{7}\right)^2}=\left|\dfrac{3}{7}\right|=\dfrac{3}{7}\)
b) \(\dfrac{\sqrt{3}^2+\sqrt{39}^2}{\sqrt{7}^2+\sqrt{91}^2}=\dfrac{\left|3\right|+\left|39\right|}{\left|7\right|+\left|91\right|}=\dfrac{3+39}{7+91}=\dfrac{42}{98}=\dfrac{3}{7}\)
c) \(\dfrac{\sqrt{3}^2-\sqrt{39}^2}{\sqrt{7}^2-\sqrt{91}^2}=\dfrac{\left|3\right|- \left|39\right|}{\left|7\right|-\left|91\right|}=\dfrac{3-39}{7-91}=\dfrac{-36}{-84}=\dfrac{3}{7}\)
d) \(\sqrt{\dfrac{39^2}{91^2}}=\sqrt{\left(\dfrac{39}{91}\right)^2}=\left|\dfrac{39}{91}\right|=\dfrac{39}{91}=\dfrac{3}{7}\)
B1
a. = 7/3. ( 37/5 - 32/5)
= 7/3 . 1
= 7/3
Phần b có gì đó sai sao lại có 3:+
c. = 4 + 6 - 3 + 5
= 12
d. = -5/21 : -19/21 : 4/5
= 25/76
B2
a. 1/4 : x =1/2 - 3/4
x = -1/4
x = 1/4 : -1/4
x = -1
b. 2 . | 2x - 3 | = 4 - (-8)
2 . | 2x - 3| = 12
| 2x - 3 | = 12:2
| 2x - 3 | = 6
| x - 3 | = 6:2
| x - 3 | = 3
=> x - 3 = +- 3
* x - 3 = 3
x = 6
* x - 3 = -3
x = 0
Chúc bạn vui vẻ
1.
0,2 . \(\sqrt{100}\) - \(\sqrt{\dfrac{16}{25}}\)
= 0,2 . 10 - \(\dfrac{4}{5}\)
= 2 - \(\dfrac{4}{5}\)
= \(\dfrac{6}{5}\)
1/ \(0,2.\sqrt{100}-\sqrt{\dfrac{16}{25}}\)
\(=0,2.10-0,8\)
\(=2-0,8=1,2\)
2/ \(\dfrac{2^7.9^3}{6^5.8^2}\)
\(=\dfrac{93312}{497664}=\dfrac{3}{16}=0,1875\)
3/ \(\sqrt{0,01}-\sqrt{0,25}\)
\(=0,1-0,5\)
\(=-0,4\)
4/ \(0,5.\sqrt{100}-\sqrt{\dfrac{1}{4}}\)
\(=0,5.10-0,5\)
\(=5-0,5=4,5\)
5/ \(7.\sqrt{0,01}+2.\sqrt{0,25}\)
\(=7.0,1+2.0,5\)
\(=0,7+1=1,7\)
6/ \(0,5.\sqrt{100}-\sqrt{\dfrac{1}{25}}\)
\(=0,5.10-0,2\)
\(=5-0,2=4,8\)
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~
Tính:
C = \(3\dfrac{1}{2}.\dfrac{4}{49}-\left[2,\left(4\right).2\dfrac{5}{11}\right]:\left(\dfrac{-42}{5}\right)\)
= \(\dfrac{7}{2}.\dfrac{4}{49}-\left[\left(2+0,\left(1\right).4\right).\dfrac{27}{5}\right].\dfrac{-5}{42}\)
= \(\dfrac{1.2}{1.7}-\left[\left(2+\dfrac{1.4}{9}\right).\dfrac{27}{5}\right].\dfrac{-5}{42}\)
= \(\dfrac{2}{7}-\left[\dfrac{18+4}{9}.\dfrac{27}{5}\right].\dfrac{-5}{42}\)
= \(\dfrac{2}{7}-\left[\dfrac{22.9}{3.5}\right].\dfrac{-5}{42}\)
= \(\dfrac{2}{7}-\dfrac{198}{15}.\dfrac{-5}{42}=\dfrac{2}{7}-\dfrac{11}{3}.\dfrac{-1}{7}\)
= \(\dfrac{2}{7}+\dfrac{11}{21}\) = \(\dfrac{6+11}{21}\) = \(\dfrac{17}{21}\)
#Giải:
a)\(\sqrt{27}\)+\(\sqrt{75}\)-\(\sqrt{\dfrac{1}{3}}\)=8\(\sqrt{3}\)-\(\sqrt{\dfrac{1}{3}}\)=\(\dfrac{23\sqrt{3}}{3}\).
b)\(\sqrt{4+2\sqrt{3}}\)-\(\sqrt{4-2\sqrt{3}}\)=2.
c)\(\dfrac{3}{\sqrt{7}+\sqrt{2}}\)+\(\dfrac{2}{3+\sqrt{7}}\)+\(\dfrac{2-\sqrt{2}}{\sqrt{2}-1}\)=1,093+\(\dfrac{2-\sqrt{2}}{\sqrt{2}-1}\)=2,507.
a) = \(3\sqrt{3}+5\sqrt{3}-\dfrac{1}{\sqrt{3}}\)
= \(3\sqrt{3}+5\sqrt{3}-\dfrac{3}{\sqrt{3}}\)
= \(\dfrac{23\sqrt{3}}{3}\)
b) = \(\sqrt{\left(1+\sqrt{3}\right)^2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)
= \(1+\sqrt{3}-\left(\sqrt{3}-1\right)\)
= \(1+\sqrt{3}-\sqrt{3}+1\)
= 2
c) = \(\dfrac{3\left(\sqrt{7}-\sqrt{2}\right)}{5}+\dfrac{2\left(3-\sqrt{7}\right)}{2}+\left(2-\sqrt{2}\right)\left(\sqrt{2}+1\right)\)
= \(3\sqrt{7}-3\sqrt{2}+3-\sqrt{7}+2\sqrt{2}+2-2-\sqrt{2}\)
= \(\dfrac{3\sqrt{7}-3\sqrt{2}}{5}+3-\sqrt{7}+\sqrt{2}\)
= \(\dfrac{3\sqrt{7}-3\sqrt{2}-5\sqrt{7}+5\sqrt{2}}{5}+3\)
= \(\dfrac{-2\sqrt{7}+2\sqrt{2}}{5}+3\)
\(\approx2,5\)