Rút gọn các biểu thức sau :
a) \(A=\)\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
b) \(B=\)\(\frac{\left(\frac{2}{3}\right)^3.\left(\frac{-3}{4}\right)^2.\left(-1\right)^5}{\left(\frac{2}{5}\right)^2.\left(\frac{-5}{12}\right)^3}\)
Rút gọn các biểu thức sau:
1) A=\(\frac{4^59^4-2.6^9}{2^{10}3^8+6^8.20}\)
2) B=\((\frac{3}{5})^2.5^2-(2\frac{1}{4})^3:\left(\frac{3}{4}\right)^3+\frac{1}{2}\)
1) A = -1 / 3
2) B = -35 / 2
Chúc em làm bài tốt !
\(\frac{11^4.6-11^5}{11^4-11^5}:\frac{9^8.3-9^9}{9^8.5+9^8.7}\)
\(\frac{3}{5}:\left(\frac{-1}{5}-\frac{1}{6}\right)+\frac{3}{5}:\left(\frac{-1}{3}-1\frac{1}{15}\right)\)
\(\left(\frac{1}{2}-\frac{13}{14}\right):\frac{5}{7}-\left(-\frac{2}{21}+\frac{1}{7}\right):\frac{5}{7}\)
\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
a)\(\frac{11^4.6-11^5}{11^4-11^5}:\frac{9^8.3-9^9}{9^8.5+9^8.7}\)
\(=1.6:\frac{9^8.3-9^8.9}{9^8.\left(5+7\right)}\)
\(=6:\frac{9^8.\left(3-9\right)}{9^8.12}\)
\(=6:\frac{9^8.\left(-6\right)}{9^8.12}\)
\(=6:\left(-\frac{6}{12}\right)\)
\(=6:\left(-\frac{1}{2}\right)\)
\(=-12\)
b) 3/5 : ( -1/5-1/6)+3/5:(-1/3-16/15) ( mình chuyển về ps luôn )
=3/5: (-11/30) + 3/5 : (-7/5)
=3/5:[-11/30+(-7/5)]
=3/5:53/30
=18/53
c) (1/2-13/14):5/7-(-2/21+1/7):5/7
= -3/7:5/7-1/21:5/7
=(-3/7-1/21):5/7
=-10/21:5/7
=-2/3
câu b vá c mình làm tắt nha. chúc bạn học tốt
Bài 2: Thực hiện phép tính (tính hợp lí nếu có)
a) \(\left(-2\right)^3+\frac{1}{2}\div\frac{1}{8}-\sqrt{25}+\left|-64\right|\)
b) \(\left(\frac{-3}{4}+\frac{2}{7}\right):\frac{2}{3}+\left(\frac{-1}{4}+\frac{5}{7}\right):\frac{2}{3}\)
c)\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
Giúp mk với!!! Mk cần gấp lắm!!!
a) Ta có: \(\left(-2\right)^3+\frac{1}{2}:\frac{1}{8}-\sqrt{25}+\left|-64\right|\)
\(=-8+\frac{1}{2}\cdot8-5+64\)
\(=-8+4-5+64=55\)
b) Ta có: \(\left(\frac{-3}{4}+\frac{2}{7}\right):\frac{2}{3}+\left(\frac{-1}{4}+\frac{5}{7}\right):\frac{2}{3}\)
\(=\left(\frac{-3}{4}+\frac{2}{7}\right)\cdot\frac{3}{2}+\left(\frac{-1}{4}+\frac{5}{7}\right)\cdot\frac{3}{2}\)
\(=\left(\frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{5}{7}\right)\cdot\frac{3}{2}\)
\(=0\cdot\frac{3}{2}=0\)
c) Ta có: \(\frac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\frac{2^{10}\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot20}=\frac{2\left(2^9\cdot9^4-6^9\right)}{6^8\left(2^2+20\right)}=\frac{-1}{3}\)
a) ( -2 )3 + \(\frac{1}{2}:\frac{1}{8}\) - √25 + \(|-64|\)
= \(\frac{-8}{1}\) + \(\frac{1}{2}.\frac{8}{1}\) - \(\frac{5}{1}\) + \(\frac{64}{1}\)
= \(\frac{-16}{2}+\frac{1}{2}.\frac{8}{1}-\frac{10}{2}+\frac{128}{2}\)
= \(\frac{-16}{2}+\frac{8}{2}-\frac{10}{2}+\frac{128}{2}\)
= \(\frac{-16+8-10+128}{2}\) = \(\frac{110}{2}\) = 55
a/ = (-8) + \(\frac{1}{2}\) * 8 - 5 + 64
=(-8) + 4 - 5 + 64 = 55
b/ = ( \(\frac{-3}{4}\frac{ }{ }\) + \(\frac{2}{7}\) ) * 3 + ( \(\frac{-1}{4}+\frac{5}{7}\)) * 3
= 3*( \(\frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{5}{7}\) )
= 3*[\(\frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}\)]
= 3* (-1) + 1 = -3 +1 = -2
c/ = \(\frac{\text{(2^2 )^5 (3^2)^4 - 2 * (2*3)^9}}{2^{10}\cdot3^8+\left(2.3\right)^8\cdot2^2\cdot5}\) =\(\frac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}=\frac{2\cdot2\cdot3}{1\cdot1\cdot2^2\cdot5}=\frac{12}{20}=\frac{3}{5}\)
Rút gọn các biểu thức sau :
A=\(\frac{2.8^4.27^2+4.6^9}{2^7.6^7+2^7.40.9^4}\)
B=\(\frac{\left(\frac{2}{3}\right)^3.\left(-\frac{3}{4}\right)^2.\left(-1\right)^5}{\left(\frac{2}{5}\right)^2.\left(-\frac{5}{12}\right)^3}\)
a,\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
b,\(\frac{27^2.8^5}{6^6.32^3}+\frac{3^4.4^4}{2^2.6^2}\)
c,\(\left(1+\frac{1}{1.3}\right)\)\(\left(1+\frac{1}{2.4}\right)\)\(\left(1+\frac{1}{3.5}\right)\)........\(\left(1+\frac{1}{20.22}\right)\)
d,\(\frac{3}{2}\)+\(\frac{7}{6}\)+\(\frac{13}{12}+\frac{21}{20}+.....+\frac{91}{90}\)
e,\(\left(-2^2\right)+\sqrt{36}-\sqrt{9}+\sqrt{25}\) 20).20)
Bài 1:
a) \(A=\frac{4^5.9-2.6^9}{2^{10}.3^8+6^8.20}\)
b) \(B=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
Bài 2: Tìm x biết:
\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\left(-3,2\right)+\frac{2}{5}\right|\)
Bài 3: Tìm các số: a1, a2, a3,..., a9 biết:
\(\frac{a_1-1}{9}=\frac{a_2-2}{8}=...=\frac{a_9-9}{1}\)và \(a_1+a_2+a_3+...+a_9=90\)
Cứu với!!! ╥_╥
Bài 2
| x - \(\frac{1}{3}\)| + \(\frac{4}{5}\)= | ( -3,2) + \(\frac{2}{5}\)|
=> | x - \(\frac{1}{3}\)| + \(\frac{4}{5}\)= | -2,8|
=> | x - \(\frac{1}{3}\)| + \(\frac{4}{5}\)= -2,8
=> | x - \(\frac{1}{3}\)| = -2,8 - \(\frac{4}{5}\)
=> | x - \(\frac{1}{3}\)| = - 3,6
=> x - \(\frac{1}{3}\)= -3,6
=> x = -3,6 + \(\frac{1}{3}\)
=> x = \(\frac{-49}{15}\)
Bài 3 :
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{a_1-1}{9}=\frac{a_2-2}{8}=...=\frac{a_9-9}{1}=\frac{a_1-1+a_2-2+...+a_9-9}{9+8+...+1}\)
\(=\frac{\left[a_1+a_2+...+a_9\right]-\left[1+2+...+9\right]}{9+8+...+1}=\frac{90-45}{45}=1\)
Ta có : \(\frac{a_1-1}{9}=1\Rightarrow a_1=10\)
Tương tự : \(a_1=a_2=....=a_9=10\)
Tính giá trị của các biểu thức sau:
\(\begin{array}{l}a)\left( {\frac{2}{3} + \frac{1}{6}} \right):\frac{5}{4} + \left( {\frac{1}{4} + \frac{3}{8}} \right):\frac{5}{2}\\b)\frac{5}{9}:\left( {\frac{1}{{11}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{2}{7}} \right)\end{array}\)
\(\begin{array}{l}a)\left( {\frac{2}{3} + \frac{1}{6}} \right):\frac{5}{4} + \left( {\frac{1}{4} + \frac{3}{8}} \right):\frac{5}{2}\\ = \left( {\frac{4}{6} + \frac{1}{6}} \right).\frac{4}{5} + \left( {\frac{2}{8} + \frac{3}{8}} \right).\frac{2}{5}\\ = \frac{5}{6}.\frac{4}{5} + \frac{5}{8}.\frac{2}{5}\\ = \frac{2}{3} + \frac{1}{4}\\ = \frac{8}{{12}} + \frac{3}{{12}}\\ = \frac{{11}}{{12}}\\b)\frac{5}{9}:\left( {\frac{1}{{11}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{2}{7}} \right)\\ = \frac{5}{9}:\left( {\frac{2}{{22}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{4}{{14}}} \right)\\ = \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{7}{4}.\frac{{ - 3}}{{14}}\\ = \frac{5}{9}.\frac{{ - 22}}{3} + \frac{{ - 3}}{8}\\ = \frac{{ - 110}}{{27}} + \frac{{ - 3}}{8}\\ = \frac{{ - 880}}{{216}} + \frac{{ - 81}}{{216}}\\ = \frac{{ - 961}}{{216}}\end{array}\)
Rút gọn biểu thức:
a) \(\frac{2.8^4.27^2+4.6^9}{2^7.6^7+2^7.40.9^4}\)
b) \(\frac{\left(\frac{2}{3}\right)^3.\left(\frac{-3}{4}\right)^2.\left(-1\right)^5}{\left(\frac{2}{5}\right)^2.\left(\frac{-5}{12}\right)^3}\)
a, \(\dfrac{2\cdot8^4\cdot27^2+4\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)
=\(\dfrac{2\cdot\left(2^3\right)^4\cdot\left(3^3\right)^2+2^2\cdot2^9\cdot3^9}{2^7\cdot2^7\cdot3^7+2^7\cdot2^3\cdot5\cdot\left(3^2\right)^4}\)
=\(\dfrac{2\cdot2^{12}\cdot3^6+2^{11}\cdot3^9}{2^{14}\cdot3^7+2^{10}\cdot5\cdot3^8}\)
=\(\dfrac{2^{11}\cdot3^6\cdot\left(2^2+3^3\right)}{2^{10}\cdot3^7\cdot\left(2^4+5\cdot3\right)}\)
=\(\dfrac{2^{11}\cdot3^6\cdot31}{2^{10}\cdot3^7\cdot31}\)
=\(\dfrac{2}{3}\)
b, \(\dfrac{\dfrac{8}{27}\cdot\dfrac{9}{16}\cdot\left(-1\right)}{\dfrac{4}{25}\cdot\dfrac{-125}{1728}}\)
=\(\dfrac{\dfrac{8\cdot9\cdot\left(-1\right)}{27\cdot16}}{\dfrac{4\cdot\left(-125\right)}{25\cdot1728}}\)
=\(\dfrac{\dfrac{-1}{6}}{\dfrac{-5}{432}}\)
=\(\dfrac{-1}{6}\cdot\dfrac{-432}{5}\)
=\(\dfrac{72}{5}\)
Rút gọn biểu thức
a)\(\frac{\left(\frac{2}{3}\right)^3.\left(-\frac{3}{4}\right)^2.\left(-1\right)^5}{\left(\frac{2}{5}\right)^2.\left(-\frac{5}{12}\right)^2}\)
b)\(6^6+6^3.3^3+3^6\)/-73