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a)
Vì 5^2 =25
=> \(\sqrt{25}=5\)
Vì 7^2=49
=>\(\sqrt{49}=7\)
Vì 1^2=1
=>\(\sqrt{1}=1\)
Vì \(\left(\frac{2}{3}\right)^2=\frac{4}{9}\)
=> \(\sqrt{\frac{4}{9}}=\frac{2}{3}\)
a) Vì \(5^2=25\Rightarrow\sqrt{25}=5\)
b) Vì \(7^2=49\Rightarrow\sqrt{49}=7\)
c) Vì \(1^2=1\Rightarrow\sqrt{1}=1\)
d) Vì \(\left(\frac{2}{3}\right)^2=\frac{4}{9}\Rightarrow\sqrt{\frac{4}{9}}=\frac{2}{3}\)
a) Vì \(5^2\)= 25 nên \(\sqrt{25}\)= 5 ;
b) Vì \(7^2\)= 49 nên \(\sqrt{49}\)= 7 ;
c) Vì \(1^2\)= 1 nên \(\sqrt{1}\)= 1
d: \(D=-8\cdot\left(\dfrac{3}{4}-\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=-8\cdot\dfrac{1}{2}:\dfrac{27-14}{12}\)
\(=-4:\dfrac{13}{12}\)
\(=-4\cdot\dfrac{12}{13}=-\dfrac{48}{13}\)
e: \(E=5\cdot4-4\cdot3+5-0.3\cdot20\)
=20-12+5-6
=8+5-6
=13-6=7
f: \(F=\dfrac{9}{4}+\dfrac{5}{6}-\dfrac{3}{2}:6\)
\(=\dfrac{9}{4}+\dfrac{5}{6}-\dfrac{3}{12}\)
\(=\dfrac{27}{12}+\dfrac{10}{12}-\dfrac{3}{12}=\dfrac{34}{12}=\dfrac{17}{6}\)
\(a,4\frac{5}{9}:\frac{\left(-5\right)}{7}+\frac{4}{9}:\frac{-5}{7}\)
\(=\frac{41}{9}.\frac{-7}{5}+\frac{4}{9}.\frac{-7}{5}\)
\(=\frac{-7}{5}.\left(\frac{41}{9}+\frac{4}{9}\right)\)
\(=-\frac{7}{9}.5\)
\(=-7\)
a)Bn Kaito Kid làm rùi!
B)Không viết lại đề
\(=\frac{11}{7}\cdot\left(-\frac{3}{5}+\frac{4}{9}-\frac{2}{5}+\frac{5}{9}\right)=\frac{11}{7}\cdot0=0\)
c)Không viết lại đề
\(A=\left(2+4+...+100\right)\left(\frac{3}{5}\cdot\frac{10}{7}-\frac{6}{7}\right):\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(2+4+6+...+100\right)\cdot0\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)=0\)
\(=\frac{7}{6}\cdot\left(\frac{3}{26}-\frac{3}{13}+\frac{1}{10}-\frac{8}{5}\right)=\frac{7}{6}\left(\frac{-3}{26}+\frac{-17}{10}\right)=\frac{7}{6}\cdot\frac{236}{130}=\frac{413}{195}\)
D)
\(\frac{3}{4}+\frac{1}{4}:\left(-\frac{2}{3}\right)-\left(-5\right)\)
\(=\frac{3}{4}+\frac{1}{4}.\left(-\frac{3}{2}\right)+5\)
\(=\frac{3}{4}-\frac{3}{8}+5\)
\(=\frac{3}{8}+5=\frac{43}{8}\)
\(12.\left(\frac{2}{5}-\frac{5}{6}\right)^2=12.\left(-\frac{13}{30}\right)^2=12.\frac{169}{900}=\frac{169}{75}\)
\(\left(-2\right)^2+\sqrt{36}-\sqrt{9}+\sqrt{25}=4+6-3+5=12\)
\(\left(9\frac{3}{4}:3.4.2\frac{7}{34}\right):\left(-1\frac{9}{16}\right)=\left(\frac{39}{4}:3.4.\frac{75}{34}\right):\left(-\frac{25}{16}\right)=\frac{975}{34}.\left(-\frac{16}{25}\right)=-\frac{312}{17}\)
\(\frac{\sqrt{3^2}+\sqrt{39^2}}{\sqrt{91^2}-\sqrt{\left(-7\right)^2}}=\frac{3+39}{91-7}=\frac{42}{84}=\frac{1}{2}\)
a. \(25^3:5^2\)
\(=\left(5^2\right)^3:5^2\)
\(=5^6:5^2=5^4\)
b. \(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6\)
\(=\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6\)
\(=\left(\frac{3}{7}\right)^{21-\left(2+6\right)}=\left(\frac{3}{7}\right)^{21-12}=\left(\frac{3}{7}\right)^9\)
\(a,25^3:5^2\)
=\(\left(5^2\right)^3:5^2\)
=\(5^6:5^2\)
=\(5^4\)
\(b,\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6\)
=\(\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6\)
\(=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}\)
\(=\left(\frac{3}{7}\right)^9\)
\(c,3-\left(\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2\)
=\(3-1+\frac{1}{4}:2\)
\(=2+\frac{1}{4}\cdot\frac{1}{2}\)
\(=2+\frac{1}{8}\)
\(=\frac{17}{8}\)
\(d,\left(-\frac{7}{4}:\frac{5}{8}\right)\cdot\frac{11}{16}\)
\(=\left(-\frac{7}{4}\cdot\frac{8}{5}\right)\cdot\frac{11}{16}\)
\(=-\frac{14}{5}\cdot\frac{11}{16}\)
\(=-\frac{77}{40}\)
\(e,\frac{2}{3}+\frac{1}{3}\cdot\frac{-6}{10}\)
\(=\frac{2}{3}-\frac{1}{5}\)
\(=\frac{7}{15}\)
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
a) Vì 52 = 25 nên √25 = 5
b) Vì 72= 49 nên √49 = 7
c) Vì 12 = 1 nên √1 = 1
d) Vì (23)2=49(23)2=49 = nên √49=23
a) Vì 52=25 nên \(\sqrt{25}=5\).
b) Vì 72=49 nên \(\sqrt{49}=7\).
c) Vì 1n=1 nên \(\sqrt{1}=1\). (\(\forall n\in N\))
d) Vì \(\left(\dfrac{2}{3}\right)^2=\dfrac{4}{9}\) nên \(\sqrt{\dfrac{4}{9}}=\dfrac{\sqrt{4}}{\sqrt{9}}=\dfrac{2}{3}\).