B

\(a,\dfrac{x+1}{3}+\dfrac{1}{2}=\dfrac{x-1}{6}-1\\ \Leftrightarrow\dfrac{2\left(x+1\right)}{6}+\dfrac{3}{6}-\dfrac{x-1}{6}+\dfrac{6}{6}=0\\ \Leftrightarrow2x+2+3-x+1+6=0\\ \Leftrightarrow x+12=0\\ \Leftrightarrow x=-12\\ b,4x^2-1-x\left(2x-1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1\right)-x\left(2x-1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-x\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\)

c, ĐKXĐ:\(x\ne\pm3\)

\(\dfrac{x+3}{x-3}+\dfrac{x-3}{x+3}=\dfrac{x^2+2x}{x^2-9}+1\\ \Leftrightarrow\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}+\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}-\dfrac{x^2+2x}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=0\\ \Leftrightarrow\dfrac{x^2+6x+9+x^2-6x+9-x^2-2x-x^2+9}{\left(x-3\right)\left(x+3\right)}=0\\ \Rightarrow-2x+27=0\\ \Leftrightarrow x=\dfrac{27}{2}\left(tm\right)\)

\(d,\left(x^2+x-1\right)\left(x^2+x+3\right)=5\\ \Leftrightarrow\left(x^2+x-1\right)\left[\left(x^2+x-1\right)+4\right]=5\left(x^2+x-1\right)^2+4\left(x^2+x-1\right)-5=0\\ \Leftrightarrow\left[\left(x^2+x-1\right)^2+5\left(x^2+x-1\right)\right]-\left[\left(x^2+x-1\right)+5\right]=0\Leftrightarrow\left(x^2+x-1\right)\left(x^2+x-1+5\right)-\left(x^2+x-1+5\right)=0\\ \Leftrightarrow\left(x^2+x-1+5\right)\left(x^2+x-1-1\right)=0\\ \Leftrightarrow\left(x^2+x+4\right)\left(x^2+x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{15}{4}=0\\\left(x^2+2x\right)-\left(x+2\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\left(vô.lí\right)\\\left(x-1\right)\left(x+2\right)=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

\(5x^2y-4xy^2+5x-3-xyz+4x^2y-xy^2-5x+\dfrac{1}{2}\\ =\left(5x^2y+4x^2y\right)-\left(4xy^2+xy^2\right)+\left(5x-5x\right)-xyz-\left(3-\dfrac{1}{2}\right)\\ =9x^2y-5xy^2-xyz+\dfrac{5}{2}\)

1, \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)

\(n_{O_2}=\dfrac{6,4}{32}=0,2\left(mol\right)\)

PTHH: 4P+5O₂ \(\underrightarrow{t^o}\) 2P₂O₅

Vì \(\dfrac{0,2}{4}>\dfrac{0,2}{5}\) nên P dư, Oxi hết

2,\(n_{O_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PTHH : 2KMnO₄ \(\underrightarrow{t^o}\) K₂MnO₄ + MnO₂ + O₂

                0,2                                            0,1       (mol)

\(m_{KMnO_4}=0,2.158=31,6\left(g\right)\)

KMnO4 -t-> K2MnO4 + MnO2 + O2 chx cân bằng bạn ơi

D

\(B=\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{x+2}{x-\sqrt{x}-2}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{x+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{x+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{x+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{x+\sqrt{x}-x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{1}{\sqrt{x}+1}\)

Vận tốc khi xe đi từ B về A là: 40+10=50(km/h)

Đổi 30'=`1/2`h

Gọi x (km) là quãng đường AB (x>0)
Thời gian khi đi từ A đến B là: `x/40`(h)

Thời gian khi đi từ B về A là: `x/50`(h)

Vì thời gian về hết ít hơn thời gian đi là 30' nên ta có pt:
\(\dfrac{x}{40}-\dfrac{x}{50}=\dfrac{1}{2}\\ \Leftrightarrow...\\ \Leftrightarrow x=100\left(tm\right)\)

Vậy quãng đường AB là 100km

`x xx 2+ x xx 3,5 +x xx 4,5 =100`

⇒`x xx (2+3,5+4,5)=100`

⇒`x xx 10 =100`

⇒ `x =100:10`

⇒ `x =10`

Chiều rộng thửa ruộng là:
`15904:142=112(m)`

Chu vi thửa ruộng là:

`(142+112)xx2=508(m)`

Đáp số: 508m