toán 5 mà khó thế bn

\(A=\dfrac{1}{2}\times\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2011}-\dfrac{1}{2013}\right)=\dfrac{1}{2}\times\left(\dfrac{1}{5}-\dfrac{1}{2013}\right)=\dfrac{1004}{10065}\)

\(\dfrac{-7}{x+1}=\dfrac{6}{x+27}\)

\(\Rightarrow-7x-189=6x+6\)

\(\Rightarrow-7x-6x=6+189\)

\(\Rightarrow-13x=195\)

\(\Rightarrow x=-15\)

Vay x = -15

\(\dfrac{-7}{x+1}=\dfrac{6}{x+27}\)

\(\Rightarrow-7\left(x+27\right)=6\left(x+1\right)\)

\(\Rightarrow-7x-189=6x+6\)

\(\Rightarrow-7x-6x=6+189\)

\(\Rightarrow-13x=195\)

\(\Rightarrow x=-15\)

Vậy x=-15

15 - 3/7 rồi bạn tự tính nha

\(15-\dfrac{x}{2}=\dfrac{3}{7}\)

\(\dfrac{x}{2}=15-\dfrac{3}{7}=\dfrac{15\cdot7-3}{7}=\dfrac{102}{7}\)

\(x=\dfrac{102}{7}\cdot2=\dfrac{204}{7}\)

\(15-\dfrac{x}{2}=\dfrac{3}{7}\)

\(\dfrac{x}{2}=15-\dfrac{3}{7}\)

\(\dfrac{x}{2}=\dfrac{102}{7}\)

\(x\times7=102\times2\)

\(x\times7=204\)

\(x=204\div7\)

\(x=\dfrac{204}{7}\)

\(\left(3+5+7+...+2021\right)\times\left(6,2:0,25-2,48\times10\right)\)
\(=\left(3+5+7+...+2021\right)\times\left(24,8-24,8\right)\)
\(=\left(3+5+7+...+2021\right)\times0\)
\(=0\)

\(\left(3+5+7+...+2021\right)\times\left(6,2:0,25-2,48\times10\right)\)

\(=\left(3+5+7+...+2021\right)\times\left(24,8-2,48\times10\right)\)

\(=\left(3+5+7+...+2021\right)\times\left(24,8-24,8\right)\)

\(=\left(3+5+7+...+2021\right)\times0\)

\(=0\)

130

=130

200 - 120 + 50 = 80 + 50 = 130

1300

1300

1300

( 7 - \(x\))³ + (11 - 7)² = 141

(7 - \(x\))³ + 4² = 141

( 7 - \(x\))³ + 16  = 141

(7 - \(x\))³          = 141 - 16

( 7 - \(x\))³        = 125

 (7 - \(x\))³       = 5³

 7 - \(x\)          = 5

       \(x\)         = 7 - 5

       \(x\)          = 2

\(\left(7-x\right)^3+\left(11-7\right)^2=141\)

\(\left(7-x\right)^3+4^2=141\)

\(\left(7-x\right)^3+16=141\)

\(\left(7-x\right)^3=141-16\)

\(\left(7-x\right)^3=125\)

\(\left(7-x\right)=5^3\)

\(\Rightarrow7-x=5\)

\(x=7-5\)

\(x=2\)

\(\text{Vậy x=2}\) 

( 7 - $x$)3 + (11 - 7)2 = 141

(7 - $x$)3 + 42 = 141

( 7 - $x$)3 + 16  = 141

(7 - $x$)3          = 141 - 16

( 7 - $x$)3        = 125

 (7 - $x$)3       = 53

 7 - $x$          = 5

       $x$         = 7 - 5

       $x$          = 2

Câu 2 :

a. \(n_C=\dfrac{3.6}{12}=0,3\left(mol\right)\)

\(n_{O_2}=\dfrac{4.48}{22,4}=0,2\left(mol\right)\)

Ta thấy : 0,3 > 0,2 => C dư , O₂ đủ

PTHH : C + O₂ -> CO₂

            0,2   0,2     0,2

b. \(m_{CO_2}=0,2.44=8,8\left(g\right)\)

c.\(m_{O_2\left(dư\right)}\left(0,3-0,2\right).32=3,2\left(g\right)\)

=>3x+1-x+5=6

=>2x+6=6

=>x=0