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\(a.9\cdot3^2\cdot\frac{1}{81}=\frac{3^2.3^2.1}{3^4}=\frac{3^4}{3^4}=1\)
\(b.2\frac{1}{2}+\frac{4}{7}:\left(\frac{-8}{9}\right)\)
\(=\frac{5}{2}+\frac{4}{7}.\left(\frac{-9}{8}\right)\)
\(=\frac{5}{2}+\frac{-9}{14}=\frac{13}{7}\)
\(c.3,75.\left(7,2\right)+2,8.\left(3,75\right)\)
\(=3,75.\left(7,2+2,8\right)\)
\(=3,75.10=37,5\)
\(d.\left(\frac{-5}{13}\right).\frac{3}{7}+\left(\frac{-8}{13}\right).\frac{3}{7}+\left(\frac{-4}{7}\right)\)
\(=\frac{3}{7}.\left[\left(\frac{-5}{13}\right)+\left(\frac{-8}{13}\right)\right]+\left(\frac{-4}{7}\right)\)
\(=\frac{3}{7}.\left(-1\right)+\frac{-4}{7}\)
\(=\frac{-3}{7}+-\frac{4}{7}=-1\)
\(e.\sqrt{81}-\frac{1}{8}.\sqrt{64}+\sqrt{0,04}\)
\(=9-\frac{1}{8}.8+0,2\)
\(=9-1+0,2=8+0,2=8,2\)
Bài 1
\(a,\left(\frac{3}{5}\right)^2-\left[\frac{1}{3}:3-\sqrt{16}.\left(\frac{1}{2}\right)^2\right]-\left(10.12-2014\right)^0\)
\(=\frac{9}{25}-\left[\frac{1}{9}-4.\frac{1}{4}\right]-1\)
\(=\frac{9}{25}-\left(-\frac{8}{9}\right)-1\)
\(=\frac{9}{25}+\frac{8}{9}-1\)
\(=\frac{56}{225}\)
\(b,|-\frac{100}{123}|:\left(\frac{3}{4}+\frac{7}{12}\right)+\frac{23}{123}:\left(\frac{9}{5}-\frac{7}{15}\right)\)
\(=\frac{100}{123}:\left(\frac{4}{3}\right)+\frac{23}{123}:\frac{4}{3}\)
\(=\left(\frac{100}{123}+\frac{23}{123}\right):\frac{4}{3}\)
\(=1:\frac{4}{3}=\frac{3}{4}\)
Phần c đăng riêng vì mk chưa tìm đc cách giải bt mỗi đáp án :v
\(c,\frac{\left(-5\right)^{32}.20^{43}}{\left(-8\right)^{29}.125^{25}}\)
\(=\frac{\left(-5\right)^{32}.\left(4.5\right)^{43}}{\left[4.\left(-2\right)\right]^{29}.\left(-5^3\right)^{25}}\)
\(=\frac{-5^{32}.4^{43}.5^{43}}{4^{29}.\left(-2\right)^{29}.\left(5\right)^{75}}\)
\(=\frac{\left(-5^4\right)^8.4^{43}.5^{43}}{4^{29}.\left(-2\right)^{29}.\left(5^3\right)^{25}}\)
\(=-\frac{1}{2}\)
a, \(\frac{-5}{9}.\left(\frac{3}{10}-\frac{2}{5}\right)\)
\(=\frac{-5}{9}.\left(\frac{3}{10}-\frac{4}{10}\right)\)
\(=\frac{-5}{9}.\frac{-1}{10}\)
\(=\frac{5}{90}\)
\(=\frac{1}{18}\)
b,\(\frac{2}{3}+\frac{-1}{3}+\frac{7}{15}\)
\(=\frac{10}{15}-\frac{5}{15}+\frac{7}{15}\)
\(=\frac{12}{15}\)
\(=\frac{4}{5}\)
c, \(\frac{3}{8}.3\frac{1}{3}\)
\(=\frac{3}{8}.\frac{10}{3}\)
\(=\frac{10}{8}\)
\(=\frac{5}{4}\)
d, \(\frac{-3}{5}+0,8.\left(-7\frac{1}{2}\right)\)
\(=\frac{-3}{5}+\frac{4}{5}.\frac{-15}{2}\)
\(=\frac{-3}{5}+\frac{-60}{10}\)
\(=\frac{-3}{5}+\frac{-30}{5}\)
\(=\frac{-33}{5}\)
e, \(\frac{2}{5}.8\frac{1}{3}+1\frac{2}{3}.\frac{2}{5}\)
\(=\frac{2}{5}.\left(8\frac{1}{3}+1\frac{2}{3}\right)\)
\(=\frac{2}{5}.10\)
\(=4\)
f, \(\frac{3}{7}.19\frac{1}{3}-\frac{3}{7}.33\frac{1}{3}\)
\(=\frac{3}{7}.\left(19\frac{1}{3}-33\frac{1}{3}\right)\)
\(=\frac{3}{7}.-14\)
\(=-6\)
~Study well~
#KSJ
\(-\frac{5}{9}\left(\frac{3}{10}-\frac{2}{5}\right)=-\frac{5}{9}\left(\frac{3}{10}-\frac{4}{10}\right)=-\frac{5}{9}.\frac{-1}{10}=\frac{1}{18}\)
\(\frac{1}{2}\sqrt{64}-\sqrt{\frac{9}{25}}+1^{2016}=\frac{1}{2}.8-\frac{3}{5}+1=4+\frac{2}{5}=\frac{22}{5}\)
\(2^8:2^5+3^2.2-12=2^3+9.2-12=8+18-12=8+6=14\)
\(3^x+\sqrt{\frac{16}{81}}-\sqrt{9}+\frac{\sqrt{81}}{3}=9\frac{4}{9}\)
\(3^x+\frac{4}{9}-3+\frac{9}{3}=9\frac{4}{9}\)
\(3^x+\frac{4}{9}-3+3=9\frac{4}{9}\)
\(3^x+\frac{4}{9}=9+\frac{4}{9}\)
\(\Rightarrow3^x=9+\frac{4}{9}-\frac{4}{9}\)
\(3^x=9\)
\(3^x=3^2\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
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 : \(a_1=a_2=....=a_9=10\)
a) \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
=> \(\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)
=> \(\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}+\frac{x+1}{12}=0\)
=> \(\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)=0\)
=> x + 1 = 0
=> x = -1
b) \(\frac{x-1}{2020}+\frac{x-2}{2019}-\frac{x-3}{2018}=\frac{x-4}{2017}\)
=> \(\left(\frac{x-1}{2020}-1\right)+\left(\frac{x-2}{2019}-1\right)-\left(\frac{x-3}{2018}-1\right)=\left(\frac{x-4}{2017}-1\right)\)
=> \(\frac{x-2021}{2020}+\frac{x-2021}{2019}-\frac{x-2021}{2018}=\frac{x-2021}{2017}\)
=> \(\left(x-2021\right)\left(\frac{1}{2020}+\frac{1}{2019}-\frac{1}{2018}-\frac{1}{2017}\right)=0\)
=> x - 2021 = 0
=> x = 2021
c) \(\left(\frac{3}{4}x+3\right)-\left(\frac{2}{3}x-4\right)-\left(\frac{1}{6}x+1\right)=\left(\frac{1}{3}x+4\right)-\left(\frac{1}{3}x-3\right)\)
=> \(\frac{3}{4}x+3-\frac{2}{3}x+4-\frac{1}{6}x-1=\frac{1}{3}x+4-\frac{1}{3}x+3\)
=> \(-\frac{1}{12}x+6=7\)
=> \(-\frac{1}{12}x=1\)
=> x = -12
A=\(\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right).\left(\frac{1}{16}-1\right).............\left(\frac{1}{9801}-1\right).\left(\frac{1}{10000}-1\right)\)
A=\(\left(\frac{1-4}{4}\right).\left(\frac{1-9}{9}\right).\left(\frac{1-16}{16}\right).............\left(\frac{1-9801}{9801}\right).\left(\frac{1-10000}{10000}\right)\)
A=\(\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}.....................\frac{-9800}{9801}.\frac{-9999}{10000}\)
A=\(\frac{-1.3}{2^2}.\frac{-2.4}{3^2}.\frac{-3.5}{4^2}.....................\frac{-98.100}{99^2}.\frac{-99.101}{100^2}\)
A=\(\frac{\left[\left(-1\right).\left(-2\right).\left(-3\right)....................\left(-98\right).\left(-99\right)\right].\left(3.4.5............100.101\right)}{\left(2.3.4.........99.100\right).\left(2.3.4...............99.100\right)}\)
A=\(\frac{1.101}{100.2}\)=\(\frac{101}{200}\)
2
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.................+\frac{2}{x.\left(x+1\right)}=\frac{2015}{2017}\)
\(\frac{1}{3.2}+\frac{1}{6.2}+\frac{1}{10.2}+.................+\frac{2}{2.x.\left(x+1\right)}=\frac{1}{2}.\frac{2015}{2017}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.................+\frac{1}{x.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+..................+\frac{1}{x.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+..............+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{x+1}{2.\left(x+1\right)}-\frac{2}{2.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{\left(x+1\right)-2}{2.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{x-1}{2.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
=>\(\frac{x-1}{x+1}=\frac{2015}{2017}.\frac{1}{2}:\frac{1}{2}\)
\(\frac{x-1}{x+1}=\frac{2015}{2017}\)
=>x+1=2017
=>x=2018-1
=>x=2016
Vậy x=2016
Còn bài 3 em ko biết làm em ms lớp 6
Chúc anh học tốt
a/ \(\left(\frac{-2}{3}\right)^4:24=\frac{16}{81}:24=\frac{2}{243}\)
b/ \(\left(\frac{3}{4}\right)^3.4^4=\frac{27}{64}.256=108\)
c/ \(\frac{3.0,8^5}{2,4^4}=\frac{3.0,32768}{33,1776}=\frac{0,98304}{33,1776}=\frac{4}{135}\)
d/ \(\frac{3^3-0,9^5}{2,7^4}=\frac{27-0,59049}{53,1441}=\frac{26,40951}{53,1441}=0,4969415231\)
e/\(\left(\frac{-7}{2}\right)^2+\left(\frac{-3}{4}\right)^3.64-\left(\frac{-61}{5}\right)\)
\(=\frac{49}{4}+\frac{-27}{64}.64+\frac{61}{5}\)
\(=12,25-27+12,2\)
\(=-2,55\)
f/ \(\frac{2^4.2^6}{\left(2^5\right)^2}-\frac{2^5.15^3}{6^3.10^2}=\frac{2^{10}}{2^{10}}-\frac{2^5.5^3.3^3}{2^3.3^3.5^2.2^2}\)
\(=1-\frac{2^5.5^3.3^3}{2^5.3^3.5^2}=1-\frac{5}{1}=-4\)
\(\)
chúc bạn học tốt
a)\(\left(\frac{-1}{3}\right)^3\cdot x=\frac{1}{81}\) \(< =>\frac{-1}{27}x=\frac{1}{81}\)\(< =>x=\frac{-1}{3}\)
Dài quá nói chung là đăng kí trước
Họ và tên : Công
Lớp 7A ; Trường THCS Nam Cao
`
Giai cau 5 thôi làm biếng quá
.a1/a2=a2/a3=a3/a4=a4/a5
(=) a1/a2.(a2/a3).( a3/a4).( a4/a5)=(a1+…+a5)/(a2+…+a6)
Rut gon: a1/16= ((a1+…+a5)/(a2+…+a6))^5
uhm bạn ơi bài 4 í phần b là câu hỏi là gì nhỉ
Tên: Nguyễn Đặng Phương Thảo
Lớp: 7
Trường: THCS Dịch Vọng Hậu
Bài làm của mik nè:
https://lazi.vn/uploads/edu/exercise/1551881808_toan.PNG
Câu 5( trước do muộn òi)
Theo bài ra ta có:
\(\left(+\right)a_2^2=a_1.a_3\Rightarrow\frac{a_1}{a_2}=\frac{a_2}{a_3}\left(1\right)\)
\(\left(+\right)a_3^2=a_2.a_4\Rightarrow\frac{a_2}{a_3}=\frac{a_3}{a_4}\left(2\right)\)
\(\left(+\right)a_4^2=a_3.a_5\Rightarrow\frac{a_3}{a_4}=\frac{a_4}{a_5}\left(3\right)\)
\(\left(+\right)a_5^2=a_4.a_6\Rightarrow\frac{a_4}{a_5}=\frac{a_5}{a_6}\left(4\right)\)
Từ (1), (2),(3),(4) => \(\frac{a_1}{a_2}=\frac{a_2}{a_3}=\frac{a_3}{a_4}=\frac{a_4}{a_5}=\frac{a_5}{a_6}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=\frac{a_3}{a_4}=\frac{a_4}{a_5}=\frac{a_5}{a_6}=\frac{a_1+a_2+a_3+a_4+a_5}{a_2+a_3+a_4+a_5+a_6}\)
\(\Rightarrow\frac{a_1}{a_2}.\frac{a_2}{a_3}.\frac{a_3}{a_4}.\frac{a_4}{a_5}.\frac{a_5}{a_6}=\left(\frac{a_1+a_2+a_3+a_4+a_5}{a_2+a_3+a_4+a_5+a_6}\right)^5\)
\(\Rightarrow\frac{a_1}{a_6}=\left(\frac{a_1+a_2+a_3+a_4+a_5}{a_2+a_3+a_4+a_5+a_6}\right)^5\left(đpcm\right)\)
họ và tên : hiếu
trường thcs xuân thành
lớp 7A
đề dễ vkl ib với tui , tui chỉ từng câu cho
Câu 2
a)\(5x-7=\frac{1}{2}\) b)\(3|x-2|=1\) c)\(3^{3x+2}=9^{2x+1}\)
\(\Rightarrow5x=\frac{1}{2}+7\) \(\Rightarrow|x-2|=\frac{1}{3}\) \(\Rightarrow3^{3x+2}=3^{4x+2}\)
\(\Rightarrow5x=\frac{15}{2}\) \(\Rightarrow x\in\left\{\frac{7}{3};\frac{5}{3}\right\}\) \(\Rightarrow3x+2=4x+2\)
\(\Rightarrow x=\frac{3}{2}\) \(\Rightarrow3x=4x\)
\(\Rightarrow x=0\)
Họ và tên: Lê Tài Bảo Châu
Lớp:7
Trường: THCS Nguyễn Đăng Đạo
Tôi sẽ gửi bài làm của tôi bằng chat riêng nhé.
bạn kb với tôi để tôi gửi bài nhé.
Câu 1
a) \(\frac{4}{3}+\frac{1}{2}:0,5=\frac{4}{3}+\frac{1}{2}:\frac{1}{2}=\frac{4}{3}+1=\frac{7}{3}\)
b) \(\sqrt{16}+\left(\frac{1}{2}\right)^2-4=4+\frac{1}{4}-4=\frac{1}{4}\)
c) \(27\frac{51}{59}-\left(26\frac{51}{59}-\frac{1}{3}\right)=27\frac{51}{59}-26\frac{51}{59}+\frac{1}{3}=1+\frac{1}{3}=\frac{4}{3}\)
Họ và tên : Nguyễn Lâm Việt Hoàng
Lớp : 7a
Trường THCS nhân đạo
Bài 1 :
a) \(\frac{4}{3}+\frac{1}{2}:0,5\)
\(=\frac{4}{3}+\frac{1}{2}:\frac{1}{2}=\frac{4}{3}+1\)
\(=\frac{7}{3}\)
b) \(\sqrt{16}+\left(\frac{1}{2}\right)^2-4\)
\(=4+\frac{1}{4}-4\)
\(=\frac{1}{4}\)
c) \(27\frac{51}{59}-\left(26\frac{51}{59}-\frac{1}{3}\right)\)
\(=27\frac{51}{59}-26\frac{51}{59}+\frac{1}{3}\)
\(=1+\frac{1}{3}=\frac{4}{3}\)
Bài 2 :
a) \(5x-7=\frac{1}{2}\)
\(5x=\frac{1}{2}+7\)
\(5x=\frac{15}{2}\)
\(x=\frac{15}{2}:5\)
\(x=\frac{3}{2}\)
b) \(3|x-2|=1\)
\(|x-2|=\frac{1}{3}\)
\(\Rightarrow\hept{\begin{cases}x-2=\frac{1}{3}\\x-2=\frac{-1}{3}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{7}{3}\\x=\frac{5}{3}\end{cases}}\)
c) \(3^{3x+2}=9^{2x+1}\)
\(\Leftrightarrow3^{3x+2}=3^{4x+2}\)
\(\Leftrightarrow3x+2=4x+2\)
\(\Rightarrow3x=4x\)
\(\Rightarrow x=0\)
Bài 3 :
a) \(f\left(-3\right)=\left(-3\right)^2-2=9-2=7\)
\(f\left(\frac{4}{3}\right)=\left(\frac{4}{3}\right)^2-2=\frac{16}{9}-2=\frac{-2}{9}\)
b) Gọi số công nhân lúc dầu và lúc giảm đi 10 người lần lượt là x và y.
Vì năng suất mỗi công nhân như nhau và công việc không thay đổi nên số công nhân và số ngày hoàn thành công việc tỉ lệ nghịch.
Theo bài ra ta có:
\(\frac{x}{\frac{1}{30}}=\frac{y}{\frac{1}{40}}\)và \(x-y=10\)
Áp dụng tính chất của dãy tỉ số bằng nhau , ta có :
\(\frac{x}{\frac{1}{30}}=\frac{y}{\frac{1}{40}}=\frac{x-y}{\frac{1}{30}-\frac{1}{40}}=\frac{10}{\frac{1}{12}}=120\)
\(\Rightarrow x=120.\frac{1}{30}=40\)
Vậy số công nhân lúc đầu là 40 người.