K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
S
subjects
CTVHS
9 tháng 9 2025
\(1)2^8:2^4+3^2\cdot3=2^4+3^3=16+27=43\)
\(2)3^{24}:3^{21}+2^2\cdot2^3=3^3+2^5=27+32=59\)
\(3)5^9:5^7+12\cdot3+7^0=5^2+36+1=25+37=62\)
\(4)5^6:5^4+3^2-2021^0=5^2+3^2-1=25+9-1=33\)
\(5)3^{19}:3^{16}+5^2\cdot2^3-1^{2021}=3^3+25\cdot8-1=27+200-1=226\)
\(6)3^6:3^5+2\cdot2^3+2021^0=3^1+2^4+1=3+16+1=20\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
9 tháng 9 2025
1: \(3^2\cdot5^3+9^2\)
\(=9\cdot125+81\)
=1125+81
=1206
2: \(55+45:3^2\)
\(=55+45:9\)
=55+5
=60
3: \(8^3:4^2-5^2=64:16-25=4-25=-21\)
4: \(5\cdot3^2-32:2^2=5\cdot9-32:4=45-8=37\)
5: \(16:2^3+5^2\cdot4=16:8+25\cdot4\)
=2+100
=102
6: \(5\cdot2^2-18:3^2\)
\(=5\cdot4-18:9\)
=20-2
=18
7: \(3\cdot5^2-15\cdot2^2=3\cdot25-15\cdot4=75-60=15\)
8: \(2^3\cdot6-72:3^2=8\cdot6-72:9=48-8=40\)
9: \(5\cdot2^2-27:3^2\)
\(=5\cdot4-27:9\)
=20-3
=17
10: \(3\cdot2^4+81:3^2=3\cdot16+81:9=48+9=57\)
11: \(4\cdot5^3-32:2^5=4\cdot125-32:32=500-1=499\)
12: \(6\cdot5^2-32:2^4=6\cdot25-32:16=150-2=148\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
16 tháng 8 2025
Bài 5:
a: \(37\cdot146+46\cdot2-46\cdot37\)
\(=37\left(146-46\right)+46\cdot2\)
\(=37\cdot100+92=3700+92=3792\)
b: \(2\cdot5\cdot71+5\cdot18\cdot2+10\cdot11\)
\(=10\cdot71+10\cdot18+10\cdot11\)
\(=10\left(71+18+11\right)=10\cdot100=1000\)
c: \(135+360+65+40\)
=135+65+360+40
=200+400
=600
d: \(27\cdot75+25\cdot27-450\)
\(=27\left(75+25\right)-450\)
=2700-450
=2250
Bài 4:
a: \(32\cdot163+32\cdot837\)
\(=32\cdot\left(163+837\right)\)
\(=32\cdot1000=32000\)
b: \(2\cdot3\cdot4\cdot5\cdot25=2\cdot5\cdot4\cdot25\cdot3=3\cdot10\cdot100=3000\)
c: \(25\cdot27\cdot4=27\cdot100=2700\)
Bài 3:
a: \(128\cdot19+128\cdot41+128\cdot40\)
\(=128\cdot\left(19+41+40\right)=128\cdot100=12800\)
b: \(375+693+625+307\)
=375+625+693+307
=1000+1000
=2000
c: \(37+42-37+22\)
=37-37+42+22
=0+64
=64
d: \(21\cdot32+21\cdot68\)
\(=21\cdot\left(32+68\right)=21\cdot100=2100\)
Bài 2:
a: \(17\cdot85+15\cdot17-120\)
\(=17\left(85+15\right)-120\)
=1700-120
=1580
b: \(189+73+211+127\)
=189+211+73+127
=400+200
=600
c: \(38\cdot73+27\cdot38\)
\(=38\left(73+27\right)=38\cdot100=3800\)
Bài 1:
a: \(28\cdot76+23\cdot28-28\cdot13\)
\(=28\left(76+23-13\right)=28\cdot86=2408\)
b: \(39\cdot50+25\cdot39+75\cdot61\)
\(=39\left(50+25\right)+75\cdot61\)
\(=39\cdot75+75\cdot61=75\left(39+61\right)=75\cdot100=7500\)
c: \(32\cdot163+837\cdot32\)
\(=32\left(163+837\right)=32\cdot1000=32000\)
d: \(63+118+37+82\)
=63+37+118+82
=100+200
=300
Ta có: \(\left(-\frac57-\frac79+\frac{9}{11}-\frac{1}{13}\right)\cdot\left(3-\frac34\right)\)
\(=\left(-\frac57-\frac79+\frac{9}{11}-\frac{11}{13}\right)\cdot\left(\frac{12}{4}-\frac34\right)\)
\(=\left(-\frac57-\frac79+\frac{9}{11}-\frac{11}{13}\right)\cdot\frac94\)
TA có: \(\left(\frac{10}{21}+\frac{14}{27}-\frac{6}{11}+\frac{22}{39}\right):\left(2-\frac23\right)\)
\(=-\frac23\left(\frac{-5}{7}-\frac79+\frac{9}{11}-\frac{11}{13}\right):\frac43\)
\(=-\frac23\left(\frac{-5}{7}-\frac79+\frac{9}{11}-\frac{11}{13}\right)\cdot\frac34\)
\(=\left(\frac{-5}{7}-\frac79+\frac{9}{11}-\frac{11}{13}\right)\cdot\frac{-2}{4}=\left(-\frac57-\frac79+\frac{9}{11}-\frac{11}{13}\right)\cdot\frac{-1}{2}\)
Ta có: \(\frac{\left(-\frac57-\frac79+\frac{9}{11}-\frac{1}{13}\right)\cdot\left(3-\frac34\right)}{\left(\frac{10}{21}+\frac{14}{27}-\frac{6}{11}+\frac{22}{39}\right):\left(2-\frac23\right)}\)
\(=\frac{\left(-\frac57-\frac79+\frac{9}{11}-\frac{11}{13}\right)\cdot\frac94}{\left(-\frac57-\frac79+\frac{9}{11}-\frac{11}{13}\right)\cdot\frac{-1}{2}}=\frac94:\frac{-1}{2}=\frac94\cdot\left(-2\right)=-\frac92\)
Ta có: $A = \dfrac{\left(\dfrac{-5}{7} - \dfrac{7}{9} + \dfrac{9}{11} - \dfrac{11}{13}\right) \cdot \left(3 - \dfrac{3}{4}\right)}{\left(\dfrac{10}{21} + \dfrac{14}{27} - \dfrac{6}{11} + \dfrac{22}{39}\right) : \left(2 - \dfrac{2}{3}\right)}$
$A = \dfrac{\left(\dfrac{-5}{7} - \dfrac{7}{9} + \dfrac{9}{11} - \dfrac{11}{13}\right) \cdot \dfrac{9}{4}}{\left(\dfrac{10}{21} + \dfrac{14}{27} - \dfrac{6}{11} + \dfrac{22}{39}\right) : \dfrac{4}{3}}$
$A = \dfrac{\left(\dfrac{-5}{7} - \dfrac{7}{9} + \dfrac{9}{11} - \dfrac{11}{13}\right) \cdot \dfrac{9}{4}}{\left(\dfrac{10}{21} + \dfrac{14}{27} - \dfrac{6}{11} + \dfrac{22}{39}\right) \cdot \dfrac{3}{4}}$
$A = \dfrac{\left(\dfrac{-5}{7} - \dfrac{7}{9} + \dfrac{9}{11} - \dfrac{11}{13}\right) \cdot \dfrac{9}{4}}{-\dfrac{2}{3} \cdot \left(\dfrac{-5}{7} - \dfrac{7}{9} + \dfrac{9}{11} - \dfrac{11}{13}\right) \cdot \dfrac{3}{4}}$
$A = \dfrac{\dfrac{9}{4}}{-\dfrac{2}{3} \cdot \dfrac{3}{4}}$ $A = \dfrac{\dfrac{9}{4}}{-\dfrac{1}{2}}$
$A = \dfrac{9}{4} \cdot (-2)$
$A = \dfrac{-18}{4}$
$A = -4,5$
\(A=\frac{\left(\frac{-5}{7}-\frac{7}{9}+\frac{9}{11}-\frac{11}{13}\right)\left(3-\frac{3}{4}\right)}{\left(\frac{10}{21}+\frac{14}{27}-\frac{6}{11}+\frac{22}{39}\right):\left(2-\frac{2}{3}\right)}\)
- Nhóm ở tử số:
- Nhóm ở mẫu số:
Đặt ngoặc lớn ở tử số là \(N\):\(3-\frac{3}{4}=\frac{12}{4}-\frac{3}{4}=\frac{9}{4}\)
\(2-\frac{2}{3}=\frac{6}{3}-\frac{2}{3}=\frac{4}{3}\)
\(N=\frac{-5}{7}-\frac{7}{9}+\frac{9}{11}-\frac{11}{13}\)Đặt ngoặc lớn ở mẫu số là \(D\) và biến đổi bằng cách đưa nhân tử chung \(-\frac{2}{3}\) ra ngoài:
\(D=\frac{10}{21}+\frac{14}{27}-\frac{6}{11}+\frac{22}{39}\)
\(D=-\frac{2}{3}\cdot \left(\frac{-5}{7}\right)-\frac{2}{3}\cdot \left(\frac{-7}{9}\right)-\frac{2}{3}\cdot \left(\frac{9}{11}\right)-\frac{2}{3}\cdot \left(\frac{-11}{13}\right)\)
\(D=-\frac{2}{3}\cdot \left(\frac{-5}{7}-\frac{7}{9}+\frac{9}{11}-\frac{11}{13}\right)=-\frac{2}{3}\cdot N\)Thay \(N\), \(D\) và các giá trị đã tính ở Bước 1 vào biểu thức \(A\):
\(A=\frac{N\cdot \frac{9}{4}}{\left(-\frac{2}{3}\cdot N\right):\frac{4}{3}}\)Rút gọn biểu thức ở mẫu số trước:
\(\left(-\frac{2}{3}\cdot N\right):\frac{4}{3}=-\frac{2}{3}\cdot N\cdot \frac{3}{4}=-\frac{1}{2}\cdot N\)Khi đó biểu thức \(A\) trở thành:
\(A=\frac{N\cdot \frac{9}{4}}{-\frac{1}{2}\cdot N}\)Triệt tiêu \(N\) ở cả tử và mẫu (vì \(N \neq 0\)):
\(A=\frac{\frac{9}{4}}{-\frac{1}{2}}=\frac{9}{4}\cdot (-2)=-\frac{9}{2}\)Kết luận\(\text{Kt\ qu\ ca\ phép\ tính\ là:\ }-\frac{9}{2}\text{\ (hoc\ }-4.5\text{)}\)