Bài 5: 

CTPT: C_xH_yO

\(n_{CaCO_3}=\dfrac{40}{100}=0,4\left(mol\right)\)

PTHH: 2C_xH_yO + \(\dfrac{4x+y-2}{2}\)O₂ --t^o--> 2xCO₂ + yH₂O

              \(\dfrac{0,4}{x}\)<--\(\dfrac{0,4\left(4x+y-2\right)}{4x}\)<------0,4

             Ca(OH)₂ + CO₂ --> CaCO₃ + H₂O

                                0,4<-----0,4

=> \(M_{C_xH_yO}=\dfrac{7,4}{\dfrac{0,4}{x}}=18,5x\left(g/mol\right)\)

=> y + 16 = 6,5x (1)

Có \(n_{O_2}=\dfrac{19,2}{32}=0,6\left(mol\right)\)

=> \(\dfrac{0,4\left(4x+y-2\right)}{4x}=0,6\)

=> 0,8x = 0,4y - 0,8 (2)

(1)(2) => x = 4; y = 10

CTPT: C₄H₁₀O

                      MÌNH CHỌN ĐÁP ÁN:   C

Em tách nhỏ ra rồi hỏi nhe!! VD như 1 bài hỏi 1 lần á

3) Ta có: \(\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{8}+\sqrt{12}}-\dfrac{3\sqrt{5}}{4}\)

\(=\dfrac{\sqrt{5}\left(\sqrt{2}+\sqrt{3}\right)}{2\left(\sqrt{2}+\sqrt{3}\right)}-\dfrac{3\sqrt{5}}{4}\)

\(=\dfrac{2\sqrt{5}-3\sqrt{5}}{4}\)

\(=\dfrac{-\sqrt{5}}{4}\) 

1) Ta có: \(\dfrac{1}{\sqrt{3}-1}+\dfrac{1}{4+2\sqrt{3}}-\dfrac{2}{\sqrt{3}}-\dfrac{3}{2}\)

\(=\dfrac{\sqrt{3}+1}{2}+\dfrac{2-\sqrt{3}}{2}-\dfrac{2\sqrt{3}}{3}-\dfrac{3}{2}\)

\(=\dfrac{\sqrt{3}+1+2-\sqrt{3}-3}{2}-\dfrac{2\sqrt{3}}{3}\)

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

3) Ta có: \(\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{8}+\sqrt{12}}-\dfrac{3\sqrt{5}}{4}\)

\(=\dfrac{\sqrt{5}\left(\sqrt{2}+\sqrt{3}\right)}{2\left(\sqrt{2}+\sqrt{3}\right)}-\dfrac{3\sqrt{5}}{4}\)

\(=\dfrac{2\sqrt{5}-3\sqrt{5}}{4}\)

\(=\dfrac{-\sqrt{5}}{4}\) 

a: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{15}{3x+y}+\dfrac{12}{2x-3y}=-6\\\dfrac{15}{3x+y}-\dfrac{25}{2x-3y}=105\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{37}{2x-3y}=-111\\\dfrac{4}{2x-3y}+\dfrac{5}{3x+y}=-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=-\dfrac{1}{3}\\3x+y=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3y=-\dfrac{1}{3}\\9x+3y=\dfrac{3}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}11x=\dfrac{7}{6}\\3x+y=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{66}\\y=\dfrac{2}{11}\end{matrix}\right.\)

\(ĐK:x\ge0;x\ne1\\ 1,P=\dfrac{x-2\sqrt{x}+1-x-\sqrt{x}+5\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2}{\sqrt{x}-1}\\ 2,P< 0\Leftrightarrow\sqrt{x}-1< 0\left(2>0\right)\\ \Leftrightarrow\sqrt{x}< 1\Leftrightarrow0\le x< 1\)

Bạn cứ nhân trong ngoặc trc xong ngoài ngoặc sau

đây dạng toán tính nhanh mà :v

(56 x 182) : (13 x 8)

= ( 8 x 7 x 13 x 14) : (13 x 8)

Ta thấy có 2 lần 8 và 13 nên ta gạch đi 2 số thì ta còn:

(7 x 14) : (1 x 1)

= 98 : 1

= 98

Đúng thì like cho mik

cần gấp lắm ạ xin các bạn giúp đỡ . ___ .

Câu 1: 

0,9 x 218 x 2 + 0,18 x 4290 + 0,6 x 353 x 3

= 9/10 x 436 + 9/50 x 4290 + 6/10 x 1059

= 9 x 43,6 + 9 x 85,8 + 6 x 105,9

= 3 x 130,8 + 3 x 257,4 + 3 x 211,8

= 3 x ( 130,8 + 257,4 + 211,8 )

= 3 x 600 

= 1800

Câu 2: 

3/4 x X + 1/2 x X - 15 = 35

X x ( 3/4 + 1/2 ) - 15 = 35

X x ( 3/4 + 1/2 ) = 50

X x 5/4 = 50

X = 40

VẬy X = 40

1.  0.9x218x2+0.18x4290+0.6x353x3  

= 1.8x218+0.18x10x429+1.8x353

=1.8x218+1.8x429+1.8x353

= 1.8x( 218+429+353 )

= 1.8x1000

=1800

2. 

3/4 x X + 1/2 x X - 15 = 35

3/4 x X + 1/2 x X = 35+15

3/4 x X +1/2 x X = 50

X x ( 3/4 + 1/2 ) = 50 

X x 5/4 = 50 

X = 50 : 5/4

X =40