ptich da thuc thanh nhan tu
x2-y2 - 2x + 2y
x2 + 4y2 - 25 + 4xy
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\(n_S=\frac{m}{M}=\frac{16}{32}=0,5\left(mol\right)\)
a,PTHH: \(S+O_2\rightarrow SO_2\)(có nhiệt độ nữa nhé)
(mol) 1 1 1
(mol) 0,5 0,5 0,5
b) Theo pt, ta có: \(n_S=n_{SO_2}=0,5\left(mol\right)\)
\(\Rightarrow m_{SO_2}=n.M=0,5.64=32\left(gam\right)\)
c)Theo pt, ta có: \(n_S=n_{O_2}=n_{SO_2}=0,5\left(mol\right)\)
\(\Rightarrow V_{O_2}\left(đktc\right)=n.22,4=0,5.22,4=11,2\left(lít\right)\)
\(\Rightarrow V_{SO_2}\left(đktc \right)=n.22,4=0,5.22,4=11,2\left(lit\right)\)
TL
a)\(S+O_2\underrightarrow{t^0}SO_2\)
b)\(n_S=\frac{16}{32}=0,5\left(mol\right)\)
\(S+O_2\underrightarrow{t^0}SO_2\)\(\)
0,5->0,5 0,5 (mol)
\(m_{SO_2}=0,5.64=32\left(g\right)\)
c)
\(V_{SO_2}=0,5.22,4=11,2\left(l\right)\)
Có:\(A=\frac{1}{x^2+4x+5}\)
\(=\frac{1}{\left(x^2+4x+4\right)+1}\)
\(=\frac{1}{\left(x+2\right)^2+1}\)
Vì\(\left(x+2\right)^2\ge0\forall x\)
\(\Leftrightarrow\left(x+2\right)^2+1\ge1\forall x\)
\(\Leftrightarrow\frac{1}{\left(x+2\right)^2+1}\le1\forall x\)
\(\Leftrightarrow A\le1\forall x\)
Dấu "=" xảy ra <=> x + 2 = 0
<=> x = -2
Vậy A đạt GTLN bằng 1 tại x = -2.
Dumflinz
\(\frac{7}{4}-y.\frac{5}{6}=\frac{1}{2}+\frac{1}{3}\)
\(\Leftrightarrow\frac{5}{6}.y=\frac{7}{4}-\frac{1}{2}-\frac{1}{3}\)
\(\Leftrightarrow\frac{5}{6}.y=\frac{11}{12}\)
\(\Leftrightarrow y=\frac{11}{12}:\frac{5}{6}\)
\(\Leftrightarrow y=\frac{11}{12}.\frac{6}{5}\)
\(\Leftrightarrow y=\frac{11}{10}\)
Vậy\(y=\frac{11}{10}\)
Dumflinz
Có:\(x^4+64y^4\)
\(=\left(x^4+16x^2y^2+64y^4\right)-16x^2y^2\)
\(=\left(x^2+8y^2\right)^2-\left(4xy\right)^2\)
\(=\left(x^2+4xy+8y^2\right)\left(x^2-4xy+8y^2\right)\)
Linz
= 64y4 + 32xy3 + 8y2x2 - 32xy3 -16x2y2 - 4x3y + 8x2y2 +4x3y +x4
= 8y2 ( 8y2 + 4xy + x2 ) - 4xy ( 8y2 + 4xy + x2 ) + x2 ( 8y2 + 4xy + x2 )
= ( 8y2 - 4xy + x2 ) ( 8y2 + 4xy + x2 )
\(a)\)\(x^2-y^2-2x+2y\)
\(=\left(x^2-2x+1\right)-\left(y^2-2y+1\right)\)
\(=\left(x-1\right)^2-\left(y-1\right)^2\)
\(=\left(x-y\right)\left(x+y-2\right)\)
\(b)\)\(x^2+4y^2-25+4xy\)
\(=\left(x^2+4xy+4y^2\right)-25\)
\(=\left(x+2y\right)^2-25\)
\(=\left(x+2y-5\right)\left(x+2y+5\right)\)
Dumflinz