Mg giải giúp mình ạ
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Mg giải giúp mình ạ
T
7 tháng 11 2023
a) \(14\dfrac{2}{5}\cdot\dfrac{7}{8}-6\dfrac{2}{5}\cdot\dfrac{7}{8}\)
\(=\dfrac{7}{8}\cdot\left(14\dfrac{2}{5}-6\dfrac{2}{5}\right)\)
\(=\dfrac{7}{8}\cdot\left(\dfrac{72}{5}-\dfrac{32}{5}\right)\)
\(=\dfrac{7}{8}\cdot\dfrac{40}{5}\)
\(=\dfrac{7}{8}\cdot8=7\)
b) \(\dfrac{1}{4}+\dfrac{3}{4}\cdot\dfrac{-5}{9}-\left(-\dfrac{2021}{2022}\right)^0\)
\(=\dfrac{3}{12}-\dfrac{5}{12}-1\)
\(=-\dfrac{2}{12}-\dfrac{12}{12}\)
\(=-\dfrac{14}{12}=-\dfrac{7}{6}\)
c) \(\dfrac{2}{3}\cdot\left(-6\right)+0,25:1\dfrac{1}{4}\)
\(=-4+\dfrac{1}{4}:\dfrac{5}{4}\)
\(=-4+\dfrac{1}{4}\cdot\dfrac{4}{5}\)
\(=\dfrac{-20}{5}+\dfrac{1}{5}=-\dfrac{19}{5}\)
d) \(\left(\left|-0,6\right|+\dfrac{4}{5}\right)\sqrt{\dfrac{9}{49}}+\left(\dfrac{-2}{5}\right)^3\)
\(=\left(0,6+\dfrac{4}{5}\right)\cdot\sqrt{\left(\dfrac{3}{7}\right)^2}+\dfrac{\left(-2\right)^3}{5^3}\)
\(=\dfrac{7}{5}\cdot\dfrac{3}{7}+\dfrac{-8}{125}\)
\(=\dfrac{3}{5}-\dfrac{8}{125}\)
\(=\dfrac{75}{125}-\dfrac{8}{125}=\dfrac{67}{125}\)
\(\text{#}Toru\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
7 tháng 11 2023
\(a,14\dfrac{2}{5}.\dfrac{7}{8}-6\dfrac{2}{5}.\dfrac{7}{8}=\left(14\dfrac{2}{5}-6\dfrac{2}{5}\right).\dfrac{7}{8}=8.\dfrac{7}{8}=7\\ b,\dfrac{1}{4}+\dfrac{3}{4}.\dfrac{-5}{9}-\left(-\dfrac{2021}{2022}\right)^0\\ =\dfrac{1}{4}+\dfrac{-15}{36}-1\\ =\dfrac{1}{4}-\dfrac{5}{12}-1=\dfrac{3-5-12}{12}=-\dfrac{14}{12}=-\dfrac{7}{6}\)
Mg giải giúp mình ạ
Mg giải giúp mình ạ
6 tháng 11 2023
Bài 2:
a: \(\dfrac{3}{5}x+\dfrac{2}{3}=\dfrac{4}{5}\)
=>\(\dfrac{3}{5}x=\dfrac{4}{5}-\dfrac{2}{3}=\dfrac{12-10}{15}=\dfrac{2}{15}\)
=>\(x=\dfrac{2}{15}:\dfrac{3}{5}=\dfrac{2}{15}\cdot\dfrac{5}{3}=\dfrac{10}{45}=\dfrac{2}{9}\)
b: \(\dfrac{1}{3}+\dfrac{2}{3}:x=-2\)
=>\(\dfrac{2}{3}:x=-2-\dfrac{1}{3}=-\dfrac{7}{3}\)
=>\(x=-\dfrac{2}{3}:\dfrac{7}{3}=\dfrac{-2}{3}\cdot\dfrac{3}{7}=-\dfrac{2}{7}\)
c: \(\left|x+\dfrac{1}{2}\right|=1,5\)
=>|x+0,5|=1,5
=>\(\left[{}\begin{matrix}x+0,5=1,5\\x+0,5=-1,5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
d: \(\left(2x-1\right)^2=-\dfrac{16}{25}\)
mà \(\left(2x-1\right)^2>=0\forall x\)
nên \(x\in\varnothing\)
e: \(\left(2x-\dfrac{4}{3}\right)\left(x+\dfrac{1}{2}\right)=0\)
=>\(\left[{}\begin{matrix}2x-\dfrac{4}{3}=0\\x+\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{3}=0\\x+\dfrac{1}{2}=0\end{matrix}\right.\)
=>x=2/3 hoặc x=-1/2
6 tháng 11 2023
a: \(\dfrac{13}{25}-\dfrac{31}{41}+\dfrac{12}{25}-\dfrac{10}{41}-0,5\)
\(=\left(\dfrac{13}{25}+\dfrac{12}{25}\right)-\left(\dfrac{31}{41}+\dfrac{10}{41}\right)-\dfrac{1}{2}\)
\(=1-1-\dfrac{1}{2}=-\dfrac{1}{2}\)
b: \(\left(-2\right)^3-\left(-\dfrac{1}{2}\right)^2:\dfrac{-1}{16}-2023^0\)
\(=-8-\dfrac{1}{4}\cdot\dfrac{-16}{1}-1\)
=-8+4-1
=-4-1
=-5
c: \(\left(-\dfrac{1}{3}\right)^2+\dfrac{4}{3}:2-0,6\)
\(=\dfrac{1}{9}+\dfrac{4}{3}\cdot\dfrac{1}{2}-\dfrac{3}{5}\)
\(=\dfrac{1}{9}+\dfrac{2}{3}-\dfrac{3}{5}\)
\(=\dfrac{10}{90}+\dfrac{60}{90}-\dfrac{54}{90}=\dfrac{16}{90}=\dfrac{8}{45}\)
d: \(\dfrac{2}{9}\cdot\dfrac{7}{5}+\dfrac{2}{9}\cdot\dfrac{-11}{5}+\dfrac{4}{5}\cdot\dfrac{2}{9}\)
\(=\dfrac{2}{9}\left(\dfrac{7}{5}-\dfrac{11}{5}+\dfrac{4}{5}\right)\)
\(=\dfrac{2}{9}\cdot0=0\)
TP
8 tháng 5 2022
\(5.a.V_{rượu}=\dfrac{46.25}{100}=11,5\left(l\right)\\ m_{rượu}=11,5.0,8=9,2\left(g\right)\\ b.C_2H_5OH+CH_3COOH⇌CH_3COOC_2H_5+H_2O\\n_{C_2H_5OH}=\dfrac{9,2}{46}=0,2\left(mol\right)\\ n_{CH_3COOC_2H_5}=n_{C_2H_5OH}=0,2\left(mol\right)\\ \Rightarrow m_{CH_3COOC_2H_5}=0,2.88=17,6\left(g\right)\\ VìH=30\%\Rightarrow m_{CH_3COOC_2H_5}=17,6.30\%=5,28\left(g\right)\)
\(6.a.C_2H_5OH+CH_3COOH⇌CH_3COOC_2H_5+H_2O\\ n_{C_2H_5OH}=\dfrac{23}{46}=0,5\left(mol\right)\\ n_{CH_3COOH}=0,5.60=30\left(g\right)\\ b.n_{CH_3COOC_2H_5}=n_{C_2H_5OH}=0,5\left(mol\right)\\ \Rightarrow m_{CH_3COOC_2H_5}=0,5.88=44\left(g\right)\\ VìH=70\%\Rightarrow m_{CH_3COOC_2H_5}=44.70\%=30,8\left(g\right)\)
Mg giải giúp mình ạ mình đang rất gấp
14 tháng 11 2023
1,2=6/5
Hiệu số phần bằng nhau là:
6-5=1(phần)
Điểm số của Leicester City là:
11:1*6=66(điểm)
Điểm số của Aston Villa là:
66-11=55(điểm)
23 tháng 10 2023
2:
a: \(A=2^4\cdot5-\left[131-\left(13-4\right)^2\right]\)
\(=16\cdot5-\left[131-81\right]\)
=80-50
=30
b: \(B=2^3+3\cdot\left(\dfrac{1}{2}\right)^0-1+\left[\left(-2\right)^2:\dfrac{1}{2}\right]-8\)
\(=8+3-1+\left[4\cdot2\right]-8\)
=8+2
=10
Bài 1:
a: \(A=32,125-\left(6,325+12,125\right)-\left(37+13,675\right)\)
\(=32,125-12,125-6,325-13,675-37\)
=20-20-37
=-37
b: \(B=4,75+\left(-\dfrac{1}{2}\right)^3+0,5^2-3\cdot\dfrac{-3}{8}\)
\(=4,75-0,125+0,25+1,125\)
=5+1
=6
H9
HT.Phong (9A5)
CTVHS
23 tháng 10 2023
Bài 4:
a) \(2^x=2^5\)
\(\Rightarrow x=5\)
b) \(\left(-7\right)^x=\left(-7\right)^9\)
\(\Rightarrow x=9\)
c) \(4^x=64\)
\(\Rightarrow4^x=4^3\)
\(\Rightarrow x=3\)
d) \(5^x=625\)
\(\Rightarrow5^x=5^4\)
\(\Rightarrow x=4\)
23 tháng 10 2023
3:
a: \(-\dfrac{3}{5}-x=-0,75\)
=>\(x=-\dfrac{3}{5}+\dfrac{3}{4}=\dfrac{3}{20}\)
b: \(x+\dfrac{1}{3}=\dfrac{2}{5}-\left(-\dfrac{1}{3}\right)\)
=>\(x+\dfrac{1}{3}=\dfrac{2}{5}+\dfrac{1}{3}\)
=>x=2/5
c: \(-0,15-x=1\dfrac{4}{5}\)
=>\(-0,15-x=1,8\)
=>x=-0,15-1,8=-1,95
d: \(-\dfrac{4}{7}-x=\dfrac{3}{5}\)
=>\(x=-\dfrac{4}{7}-\dfrac{3}{5}\)
=>\(x=\dfrac{-20-21}{35}=-\dfrac{41}{35}\)
30 tháng 10 2023
Diện tích đáy là
\(\dfrac{1}{2}\cdot8\cdot6=4\cdot6=24\left(cm^2\right)\)
Thể tích của lăng trụ là:
\(V=24\cdot10=240\left(cm^3\right)\)
a) Diện tích cần quét sơn:
(9 + 6).2.4 + 9.6 - 11,25 = 172,75 (m²)
b) Số tiền phải trả:
172,75 . 100000 = 17250000 (đồng)