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Bài 1:
a) Ta có: \(\frac{-5}{7}+\frac{2}{7}+\frac{4}{-9}+\frac{4}{9}\)
\(=-\frac{3}{7}+\frac{-4}{9}+\frac{4}{9}\)
\(=-\frac{3}{7}\)
b) Ta có: \(\left(\frac{1}{2}:\frac{3}{4}\right)^2\)
\(=\left(\frac{1}{2}\cdot\frac{4}{3}\right)^2\)
\(=\left(\frac{2}{3}\right)^2=\frac{4}{9}\)
c) Ta có: \(\frac{1}{2}+\frac{3}{4}-\left(\frac{4}{5}+\frac{3}{4}\right)\)
\(=\frac{1}{2}+\frac{3}{4}-\frac{4}{5}-\frac{3}{4}\)
\(=\frac{1}{2}-\frac{4}{5}\)
\(=\frac{5}{10}-\frac{8}{10}=\frac{-3}{10}\)
d) Ta có: \(5^6:5^4+2^3\cdot2^2-225:15^2\)
\(=5^2+2^5-\frac{15^2}{15^2}\)
\(=25+32-1\)
\(=56\)
e) Ta có: \(\frac{7}{23}+\frac{4}{17}-\frac{7}{23}+\frac{13}{17}\)
\(=\frac{4}{17}+\frac{13}{17}\)
\(=\frac{17}{17}=1\)
g) Ta có: \(19\frac{1}{4}\cdot\frac{7}{12}-15\frac{1}{4}\cdot\frac{7}{12}\)
\(=\frac{7}{12}\left(19+\frac{1}{4}-15-\frac{1}{4}\right)\)
\(=\frac{7}{12}\cdot4=\frac{7}{3}\)
a) \(19-\left(x-1\right)^3=18\)
\(\Leftrightarrow-\left(x-1\right)^3=-1\)
\(\Leftrightarrow x-1=1\Leftrightarrow x=2\)
b) \(\left(x-5\right)^2+17=17\)
\(\Leftrightarrow\left(x-5\right)^2=0\)
\(\Leftrightarrow x-5=0\Leftrightarrow x=5\)
c) \(3^{x+2}-8\cdot3^x+5=86\)
\(\Leftrightarrow3^x\cdot3^2-8\cdot3^x=81\)
\(\Leftrightarrow3^x\left(3^2-8\right)=81\)
\(\Leftrightarrow3^x=81\)
\(\Leftrightarrow x=4\)
Nguyễn Huy Thắng
Trần Việt Linh
Trần Quỳnh Mai
Phạm Nguyễn Tất Đạt
Nguyễn Huy Tú
Nguyễn Thị Thu An
Trần Hương Thoan
Trương Hồng Hạnh
Nguyễn Quốc Việt
soyeon_Tiểubàng giải
Hoàng Lê Bảo NgọcPhương An
Silver bullet
Võ Đông Anh Tuấn
Lê Nguyên Hạo
ai lm đc thì giúp nha
\(a,4^{21}:16^5.\)
\(=4^{21}:\left(4^2\right)^5.\)
\(=4^{21}:4^{10}.\)
\(=4^{21-10}=4^{11}.\)
Vậy.....
\(b,32^8:4^{19}.\)
\(=\left(2^5\right)^8:\left(2^2\right)^{19}.\)
\(=2^{40}:2^{38}.\)
\(=2^{40-38}.\)
\(=2^2=4.\)
Vậy.....
\(c,27^{15}:9^{22}.\)
\(=\left(3^3\right)^{15}:\left(3^2\right)^{22}.\)
\(=3^{45}:3^{44}.\)
\(=3^{45-44}.\)
\(=3^1=3.\)
Vậy.....
\(d,25^{10}:125^6.\)
\(=\left(5^2\right)^{10}:\left(5^3\right)^6.\)
\(=5^{20}:5^{18}.\)
\(=5^{20-18}.\)
\(=5^2=25.\)
Vậy.....
~ Hok tốt!!! ~ :))
a, 421 : 165
= 421 : (42 )5
= 421 : 410
= 411
b, 328 : 419
= (25)8 : (22 )19
= 240 : 238
= 22
c, 2715 : 922
= (33 ) 15 : (32 )22
= 345 : 344
= 3
d, 2510 : 1256
= (52)10 : (53)6
= 520 : 518
= 52
Bài 1 :
a) A = \(8^2\) . \(32^4\) = \(\)(2\(^3\))\(^2\) . ( \(2^5\))\(^4\) = 2\(^6\) . 2\(^{20}\) = 2\(^{26}\)
b) B = 27\(^3\) . 9\(^4\) . 243 = ( \(3^3\))\(^3\) . ( \(3^2\) )\(^4\) . 3\(^5\) = 3\(^9\) . \(3^8\) . 3\(^5\) = 3\(^{22}\)
Bài 2 : So sánh
a) A = 27\(^5\) và B =2433
Ta có : 27\(^5\) =(3\(^3\))\(^5\) = 3\(^8\) = 6561
Vì 6561 > 2433 nên A > B .
b) A = 2300 và B = 3\(^{200}\)
Ta có : B = \(3^{200}\) = 3\(^8\) . 3\(^{192}\) = 6561 . 3\(^{192}\)
Vậy chắc chắn rằng B > A .
\(.2.\)
\(a.\)
\(2x+\dfrac{1}{2}=-\dfrac{5}{3}\)
\(\Rightarrow2x=-\dfrac{5}{3}-\dfrac{1}{2}=-\dfrac{13}{6}\)
\(\Rightarrow x=-\dfrac{13}{6}:2=-\dfrac{13}{12}\)
Vậy : \(x=-\dfrac{13}{12}\)
\(b.\)
\(\dfrac{1}{7}-\dfrac{3}{5}x=\dfrac{3}{5}\)
\(\Rightarrow\dfrac{3}{5}x=\dfrac{1}{7}-\dfrac{3}{5}=-\dfrac{16}{35}\)
\(\Rightarrow x=-\dfrac{16}{35}:\dfrac{3}{5}=-\dfrac{16}{21}\)
Vậy : \(x=-\dfrac{16}{21}\)
\(c.\)
\(\dfrac{3}{4}x+\dfrac{1}{2}=-\dfrac{3}{5}\)
\(\Rightarrow\dfrac{3}{4}x=-\dfrac{3}{5}-\dfrac{1}{2}=-\dfrac{11}{10}\)
\(\Rightarrow x=-\dfrac{11}{10}:\dfrac{3}{4}=-\dfrac{22}{15}\)
Vậy : \(x=-\dfrac{22}{15}\)
\(d.\)
\(-\dfrac{2}{15}-x=-\dfrac{3}{10}\)
\(\Rightarrow x=-\dfrac{2}{15}-\left(-\dfrac{3}{10}\right)=\dfrac{1}{6}\)
Vậy : \(x=\dfrac{1}{6}\)
a) Ta có: 218 = 215.23 = 215.8
Vì 8 > 7 nên 7.2157.215 < 8.215
Vậy 7.1215 < 218
b) Ta có: 3245 - 3244 = 3244(32 - 1) = 3244.31
3244 - 3243 = 3243(32 - 1) = 3243.31
Vì 3244 > 3243 nên 3244.31 > 3244.31
Vậy: 3245−32443245−3244 và 3244−3243
Chúc bạn học tốt@@
Vậy 3245 - 3244 > 3244 - 3243 nhé! Mình cop đề xuống nên mới vậy.
a) Ta có: \(71+\frac{26-3x}{5}=75\)
\(\Leftrightarrow\frac{26-3x}{5}=4\)
\(\Leftrightarrow26-3x=20\)
\(\Leftrightarrow3x=6\)
hay x=2
Vậy: x=2
b) Ta có: \(\left|x-12\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}x-12=5\\x-12=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=17\\x=-5+12=7\end{matrix}\right.\)
Vậy: \(x\in\left\{17;7\right\}\)
c) Ta có: \(\frac{7}{8}+x=\frac{3}{5}\)
\(\Leftrightarrow x=\frac{3}{5}-\frac{7}{8}\)
\(\Leftrightarrow x=\frac{24}{40}-\frac{35}{40}\)
hay \(x=-\frac{11}{40}\)
Vậy: \(x=-\frac{11}{40}\)
d) Ta có: \(\frac{1}{2}x-\frac{2}{5}=\frac{1}{5}\)
\(\Leftrightarrow\frac{1}{2}\cdot x=\frac{1}{5}+\frac{2}{5}=\frac{3}{5}\)
\(\Leftrightarrow x=\frac{3}{5}:\frac{1}{2}=\frac{3}{5}\cdot2\)
hay \(x=\frac{6}{5}\)
Vậy: \(x=\frac{6}{5}\)
b) Đặt B = A : C ta có:
\(A=\frac{5^3}{6}+\frac{5^3}{12}+\frac{5^3}{20}+\frac{5^3}{42}+\frac{5^3}{56}+\frac{5^3}{72}+\frac{5^3}{90}\)
\(A=5^3.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(A=5^3.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=5^3.\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(A=\frac{5^3.2}{5}\)
\(A=5^2.2\)
\(\Rightarrow A=50\)
\(C=\frac{1124.2247-1123}{1124+1123.2247}\)
\(C=\frac{\left(1123+1\right).2274-1123}{1123.2247+1124}\)
\(C=\frac{1123.2247-2247-1123}{1123.2247+1124}\)
\(C=\frac{1123.2247+1124}{1123.2247+1124}=1\)
\(\Rightarrow B=50:1=50\)
Vậy B = 50
Tìm x :
a) \(x-5=49:7\)
\(\Leftrightarrow x-5=7\)
\(\Leftrightarrow x=7+5\)
\(\Leftrightarrow x=12\)
Vậy : \(x=12\)
b) \(2x+6=24\)
\(\Leftrightarrow2x=24-6=18\)
\(\Leftrightarrow x=18:2\)
\(\Leftrightarrow x=9\)
Vậy : \(x=9\)
c) \(\frac{1}{3}:x+\frac{1}{2}=5\)
\(\Leftrightarrow\frac{1}{3}:x=5-\frac{1}{2}=\frac{9}{5}\)
\(\Leftrightarrow x=\frac{1}{3}:\frac{9}{5}\)
\(\Leftrightarrow x=\frac{5}{27}\)
Vậy : \(x=\frac{5}{27}\)
d) \(\frac{1}{6}.x-\frac{1}{3}=2\)
\(\Leftrightarrow\frac{1}{6}.x=2-\frac{1}{3}=\frac{5}{3}\)
\(\Leftrightarrow x=\frac{5}{3}:\frac{1}{6}\)
\(\Leftrightarrow x=10\)
Vậy : \(x=10\)
e) \(\frac{x}{27}=\frac{3}{x}\)
\(\Leftrightarrow x.x=27.3\)
\(\Leftrightarrow x^2=81\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=9\end{matrix}\right.\) mà \(x\in N\)
\(\Rightarrow x=9\)
Vậy : \(x=9\)
g) \(1200:24-\left(17-x\right)=36\)
\(\Leftrightarrow50-17+x=36\)
\(\Leftrightarrow33+x=36\)
\(\Leftrightarrow x=36-33\)
\(\Leftrightarrow x=3\)
Vậy : \(x=3\)
h) \(674-\left(12+x\right)=427\)
\(\Leftrightarrow12+x=674-427=247\)
\(\Leftrightarrow x=247-12\)
\(\Leftrightarrow x=230\)
Vậy : \(x=230\)
k) \(36.\left(x-9\right)=900\)
\(\Leftrightarrow x-9=900:36\)
\(\Leftrightarrow x-9=25\)
\(\Leftrightarrow x=25+9\)
\(\Leftrightarrow x=34\)
m) \(1,2:x+3,8:x=2,5\)
\(\Leftrightarrow\left(1,2-3,8\right):x=2,5\)
\(\Leftrightarrow-2,6:x=2,5\)
\(\Leftrightarrow x=\frac{-2,6}{2,5}=-\frac{26}{25}\)
Vậy : \(x=-\frac{26}{25}\)
n) \(\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right).x=\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right).x=\frac{1}{3}\)
\(\Leftrightarrow\left[\frac{1}{2}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\right].x=\frac{1}{3}\)
\(\Leftrightarrow\left[\frac{1}{2}.\left(1-\frac{1}{10}\right)\right].x=\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{2}.\frac{9}{10}.x=\frac{1}{3}\)
\(\Leftrightarrow\frac{9}{20}.x=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{3}:\frac{9}{20}\)
\(\Leftrightarrow x=\frac{20}{27}\)
Vậy : \(x=\frac{20}{27}\)
a)5.22-18:32
=(18.5)-(32-22)
=90-10
=80
b)17.85+15.17-120
=(85+15).17-120
=100.17-120
=1700-120
=1580
f)2.52 + 3: 710 – 54: 33
=2.52 + 3 : 710 – 54 : 33
= 104 + 3/710 - 18/11
= 102,3678617
d) 20 – [ 30 – (5 – 1)2 ]
=20-[30-4.2]
=20[30-8]
=20.28
=560
h)5.32-32:42
=(42.5):32
=210:32
=6,5625
d)