\(1,0,75-\dfrac{2}{3}-0,5=\dfrac{3}{4}-\dfrac{2}{3}-\dfrac{1}{2}=\dfrac{9}{12}-\dfrac{8}{12}-\dfrac{1}{2}=\dfrac{1}{12}-\dfrac{1}{2}\)

\(=\dfrac{2}{24}-\dfrac{12}{24}=\dfrac{-10}{24}=\dfrac{-5}{12}\)

\(2,\dfrac{1}{5}-0,125-\dfrac{5}{4}=\dfrac{1}{5}-\dfrac{1}{8}-\dfrac{5}{4}=\dfrac{8}{40}-\dfrac{5}{40}-\dfrac{5}{4}=\dfrac{3}{40}-\dfrac{5}{4}\)

\(=\dfrac{3}{40}-\dfrac{50}{40}=\dfrac{-47}{40}\)

\(3,1,25-\dfrac{3}{4}+\dfrac{4}{3}=\dfrac{5}{4}-\dfrac{3}{4}+\dfrac{4}{3}=\dfrac{2}{4}+\dfrac{4}{3}=\dfrac{6}{12}+\dfrac{16}{12}=\dfrac{22}{12}=\dfrac{11}{6}\)

\(4,0,15-\dfrac{1}{4}+\dfrac{2}{5}=\dfrac{3}{20}-\dfrac{1}{4}+\dfrac{2}{5}=\dfrac{3}{20}-\dfrac{5}{20}+\dfrac{2}{5}=\dfrac{-2}{20}+\dfrac{2}{5}\)

\(=\dfrac{-2}{20}+\dfrac{8}{20}=\dfrac{6}{20}=\dfrac{3}{10}\)

\(5,5-3,4+\dfrac{1}{5}=\dfrac{5}{1}-\dfrac{17}{5}+\dfrac{1}{5}=\dfrac{25}{5}-\dfrac{17}{5}+\dfrac{1}{5}=\dfrac{25-17+1}{5}=\dfrac{9}{5}\)

\(6,\dfrac{1}{4}-0,3+\dfrac{4}{3}=\dfrac{1}{4}-\dfrac{3}{10}+\dfrac{4}{3}=\dfrac{10}{40}-\dfrac{12}{40}+\dfrac{4}{3}=\dfrac{-2}{40}+\dfrac{4}{3}\)

\(=\dfrac{-1}{20}+\dfrac{4}{3}=\dfrac{-3}{60}+\dfrac{80}{60}=\dfrac{77}{60}\)

\(7,0,2-3,25+4,7=\dfrac{1}{5}-\dfrac{13}{4}+\dfrac{47}{10}=\dfrac{4}{20}-\dfrac{65}{20}+\dfrac{47}{10}=\dfrac{-61}{20}+\dfrac{47}{10}\)

\(=\dfrac{-61}{20}+\dfrac{94}{20}=\dfrac{33}{20}=1,65\)

\(8,5,4+\dfrac{-7}{3}-\dfrac{-5}{7}=\dfrac{27}{5}+\dfrac{-7}{3}-\dfrac{-5}{7}=\dfrac{81}{15}+\dfrac{-35}{15}-\dfrac{-5}{7}\)

\(=\dfrac{46}{15}-\dfrac{-5}{7}=\dfrac{322}{105}-\dfrac{-75}{105}=\dfrac{397}{105}\)

\(9,\dfrac{-4}{2}+\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{-12}{6}+\dfrac{2}{6}-\dfrac{1}{4}=\dfrac{-10}{6}-\dfrac{1}{4}=\dfrac{-5}{3}-\dfrac{1}{4}\)

\(=\dfrac{-20}{12}-\dfrac{3}{12}=\text{ }\dfrac{-23}{12}\)

\(10,5,4-1,5-\left(7,2-1\right)=3,9-6,2=-2,3\)

\(11,4,9-\left(1,5-7,7+3\right)=4,9-\left(-3,2\right)=8,1\)

\(12,7,8-4,7+\left(5,3-1,4\right)=3,1+3,9=7\)

\(14,\dfrac{1}{2}-0,4+\dfrac{1}{5}\text{=}0,5-0,4+0,2=0,3\)

\(15,4,2-\dfrac{4}{5}+\dfrac{1}{2}=4,2-0,8+0,5=3,9\)

a. \(3,5x+\left(-1,5\right)x+3,2=-5,4\)

\(\Rightarrow2x=-5,4-3,2\)

\(\Rightarrow2x=-8,6\)

\(\Rightarrow x=-8,6:2=-4,3\)

Vậy...................

b. \(\left(-7,2\right)x+3,7x+2,7=-7,8\)

\(\Rightarrow-3,5x=-7,8-2,7\)

\(\Rightarrow-3,5x=-10,5\)

\(\Rightarrow x=-10,5:\left(-3,5\right)=3\)

Vậy...........

câu 1 sai đề

a,Câu này bạn xem lại đề nhé!!!

b,\(\left(-7,2\right)x+3,7x+2,7=-7,8\)

\(-3,5x+2,7=-7,8\)

\(-3,5x=-10,5\)

\(x=3\)

a) Ta có: \(\frac{1}{2}+\frac{2}{3}:\left(x-1\right)=\frac{2}{3}\)

⇒\(\frac{2}{3}:\left(x-1\right)=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}\)

⇒\(x-1=\frac{2}{3}:\frac{1}{6}=\frac{2}{3}\cdot6=4\)

hay x=5

Vậy: x=5

b) \(5,4-3\left[x-120\%\right]=\frac{3}{10}\)

⇔\(\frac{27}{5}-3\cdot\left(x-\frac{6}{5}\right)=\frac{3}{10}\)

⇔\(3\left(x-\frac{6}{5}\right)=\frac{27}{5}-\frac{3}{10}=\frac{51}{10}\)

hay \(x-\frac{6}{5}=\frac{51}{10}\cdot\frac{1}{3}=\frac{17}{10}\)

⇔\(x=\frac{17}{10}+\frac{6}{5}=\frac{29}{10}\)

Vậy: \(x=\frac{29}{10}\)

c) \(10\cdot3^{x+2}-3^x=89\)

\(\Leftrightarrow10\cdot3^2\cdot3^x-3^x=89\)

\(\Leftrightarrow3^x\left(90-1\right)=89\)

\(\Leftrightarrow3^x=1\)

hay x=0

Vậy: x=0

d) \(5\cdot\left(x-0,2\right)=3x+\left(\frac{-2}{3}\right)^3\)

⇒\(5\cdot\left(x-\frac{1}{5}\right)=3x+\frac{-8}{27}\)

\(\Leftrightarrow5x-1-3x-\frac{-8}{27}=0\)

\(\Leftrightarrow2x-\frac{19}{27}=0\)

\(\Leftrightarrow2x=\frac{19}{27}\)

hay \(x=\frac{\frac{19}{27}}{2}=\frac{19}{27}\cdot\frac{1}{2}=\frac{19}{54}\)

Vậy: \(x=\frac{19}{54}\)

e) \(\left(2x+\frac{3}{4}\right)^2-1,5=2\frac{1}{2}\)

\(\Leftrightarrow\left(2x+\frac{3}{4}\right)^2=\frac{5}{2}+\frac{3}{2}=\frac{8}{2}=4\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{3}{2}=-2\\2x+\frac{3}{2}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-2-\frac{3}{2}\\2x=2-\frac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\frac{7}{2}\\2x=\frac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-7}{2}\cdot\frac{1}{2}\\x=\frac{1}{2}\cdot\frac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-7}{4}\\x=\frac{1}{4}\end{matrix}\right.\)

Vậy: \(x\in\left\{-\frac{7}{4};\frac{1}{4}\right\}\)

a) = ( 5,4 - 4,4 ) + ( 6,5 - 5,5 ) + ( 7,6 - 6,6 ) + ( 8,7 - 7,7 )

    =  1 + 1 + 1 + 1

    = 4

b) = 9/10

a,\(\dfrac{2}{3}x-\dfrac{4}{9}=-\dfrac{5}{27}\)

\(\dfrac{2}{3}x=\dfrac{17}{27}\)

\(x=\dfrac{17}{18}\)

b,\(\dfrac{2}{3}x+\dfrac{4}{9}=\dfrac{5}{27}\)

\(\dfrac{2}{3}x=-\dfrac{7}{27}\)

\(x=-\dfrac{7}{18}\)

c,\(x:1,2=5,4:6\)

\(x:1,2=0,9\)

\(x=1,08\)

d,\(1,68:1,2=5,4:x\)

\(1,4=5,4:x\)

\(x=\dfrac{27}{7}\)

e,\(\left(1-2x\right)^2+1=10\)

\(\left(1-2x\right)^2=9\)

\(1-2x=3\)

\(2x=-2\)

\(x=-1\)

f,\(\left(1-2x\right)^2-6=10\)

\(\left(1-2x\right)^2=16\)

\(1-2x=4\)

\(2x=-3\)

\(x=-\dfrac{3}{2}\)

Đáp án B

Khối lượng của niken được giải phóng ra ở điện cực của bình điện phân tuân theo định luật I Fa-ra-đây :

m = kq = kIt

trong đó k là đương lượng điện hoá của niken, q = It là điện lượng chuyển qua dung dịch điện phân.

Thay số, ta tìm được : m = 0,3. 10 - 3 .5,0.3600 = 5,4g.

a) 7.3x8.3+3.8x7.3-5x7.3x0.4

=7.3x(8.3+3.8-5x0.4)

=7.3x10.1

=73.73

b)\(\frac{1}{10}:0.1-\frac{1}{8}:0.125+\frac{1}{2}:0.5-\frac{1}{4}:0.25\)

=1-1+1-1=0

c)\(\left(\frac{2727}{3636}+\frac{1212}{5454}\right)\times\left(\frac{9}{70}\right)=\left(\frac{3}{4}+\frac{2}{9}\right)\times\left(\frac{9}{70}\right)\)

=\(\frac{35}{36}\times\frac{9}{70}\)=\(\frac{1}{8}\)

d) 120x999+\(\frac{120}{5\times2.5\times125\times4\times0.8}\)

=120x999+\(\frac{120}{5000}\)

=120x999+\(\frac{3}{125}\)

=119880.024

a 73,73

b1/8

c 119880,024

đúng nha 

k mình nha mn

sai câu d thì phải