Áp dụng tính chất dãy tỉ số bằng nhau:\(\dfrac{\left(5z-3y\right)+\left(3x-2z\right)+\left(2y-5x\right)}{2+5+3}\)

=\(\dfrac{\left(3x-5x\right)+\left(-3y+2y\right)+\left(5z-2z\right)}{2+5+3}\)

=\(\dfrac{-2x-y+3z}{2+5+3}\)(???!!!!)

=\(\dfrac{-2x}{2}=\dfrac{-y}{5}=\dfrac{3z}{3}\)

=\(\dfrac{2}{-2x}=\dfrac{5}{-y}=\dfrac{3}{3z}\)

tớ xin chịu trận vì ko chứng minh được :(((

nó lại ra như thế này

15 - 3/7 rồi bạn tự tính nha

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

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

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

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

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

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

\(x\times7=102\times2\)

\(x\times7=204\)

\(x=204\div7\)

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

130

=130

200 - 120 + 50 = 80 + 50 = 130

1300

1300

1300

\(\frac{\frac{2}{3}+\frac{2}{5}-\frac{2}{9}}{\frac{4}{3}+\frac{4}{5}-\frac{4}{9}}\)

\(=\frac{2\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{9}\right)}{4\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{9}\right)}\)

\(=\frac{2}{4}=\frac{1}{2}\)

= (2/3 + 2/5 -2/9) : 2 x ( 2/3 + 2/5 - 2/9) = 1/2

1/2 nha bạn

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

\(\dfrac{6}{5}\)

5/6

8/15 + 4/15 :2/5 = 8/15 + 2/5 = 14/15

`(2x^2 +5)(3x-2)=0`

`<=> 2x^2 +5=0` hoặc `3x-2=0`

`<=> 2x^2 =-5` (vô lí vì `2x^2 ≥0 ∀x`) hoặc `x=2/3`

vậy pt có tập nghiệm `S={2/3}`

\(\left(2x^2+5\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x^2=-5\\3x=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=-10\left(l\right)\\x=\dfrac{2}{3}\left(n\right)\end{matrix}\right.\)

Vậy \(S=\left\{\dfrac{2}{3}\right\}\)