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

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)

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: \(\dfrac{2}{9}\cdot\dfrac{7}{5}+\dfrac{2}{9}\cdot\dfrac{-11}{5}+\dfrac{4}{5}\cdot\dfrac{2}{9}\)

\(=\dfrac{2}{9}\left(\dfrac{7}{5}-\dfrac{11}{5}+\dfrac{4}{5}\right)\)

\(=\dfrac{2}{9}\cdot0=0\)

\(5.a.V_{rượu}=\dfrac{46.25}{100}=11,5\left(l\right)\\ m_{rượu}=11,5.0,8=9,2\left(g\right)\\ b.C_2H_5OH+CH_3COOH⇌CH_3COOC_2H_5+H_2O\\n_{C_2H_5OH}=\dfrac{9,2}{46}=0,2\left(mol\right)\\ n_{CH_3COOC_2H_5}=n_{C_2H_5OH}=0,2\left(mol\right)\\ \Rightarrow m_{CH_3COOC_2H_5}=0,2.88=17,6\left(g\right)\\ VìH=30\%\Rightarrow m_{CH_3COOC_2H_5}=17,6.30\%=5,28\left(g\right)\)

\(6.a.C_2H_5OH+CH_3COOH⇌CH_3COOC_2H_5+H_2O\\ n_{C_2H_5OH}=\dfrac{23}{46}=0,5\left(mol\right)\\ n_{CH_3COOH}=0,5.60=30\left(g\right)\\ b.n_{CH_3COOC_2H_5}=n_{C_2H_5OH}=0,5\left(mol\right)\\ \Rightarrow m_{CH_3COOC_2H_5}=0,5.88=44\left(g\right)\\ VìH=70\%\Rightarrow m_{CH_3COOC_2H_5}=44.70\%=30,8\left(g\right)\)

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)

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

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) \(5^x=625\)

\(\Rightarrow5^x=5^4\)

\(\Rightarrow x=4\)

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: \(-\dfrac{4}{7}-x=\dfrac{3}{5}\)

=>\(x=-\dfrac{4}{7}-\dfrac{3}{5}\)

=>\(x=\dfrac{-20-21}{35}=-\dfrac{41}{35}\)

Diện tích đáy là

\(\dfrac{1}{2}\cdot8\cdot6=4\cdot6=24\left(cm^2\right)\)

Thể tích của lăng trụ là:

\(V=24\cdot10=240\left(cm^3\right)\)