Câu 3:

a: A=1/2

=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}}=\dfrac{1}{2}\)

=>\(2\sqrt{x}-4=\sqrt{x}\)

=>\(2\sqrt{x}-\sqrt{x}=4\)

=>\(\sqrt{x}=4\)

=>x=16(nhận)

b: \(B=\dfrac{\sqrt{x}-3}{\sqrt{x}-1}+\dfrac{\sqrt{x}+3}{x-1}\)

\(=\dfrac{\sqrt{x}-3}{\sqrt{x}-1}+\dfrac{\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)+\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

c: \(P=A\cdot B=\dfrac{\sqrt{x}-2}{\sqrt{x}}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)

\(\sqrt{P}< \dfrac{1}{2}\)

=>\(0< =P< \dfrac{1}{4}\)

=>P>=0 và P<1/4

=>P>=0 và P-1/4<0

=>\(\left\{{}\begin{matrix}\dfrac{\sqrt{x}-2}{\sqrt{x}+1}>=0\\\dfrac{\sqrt{x}-2}{\sqrt{x}+1}-\dfrac{1}{4}< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\sqrt{x}-2>=0\\\dfrac{4\sqrt{x}-8-\sqrt{x}-1}{4\left(\sqrt{x}+1\right)}< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\sqrt{x}>=2\\3\sqrt{x}-9< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}>=2\\\sqrt{x}< 3\end{matrix}\right.\)

=>\(2< =\sqrt{x}< 3\)

=>4<=x<9

\(D=\left(\dfrac{3}{4}xy+3y\right)\left(x-12y\right)\\ =x\left(\dfrac{3}{4}xy+3y\right)-12y\left(\dfrac{3}{4}xy+3y\right)\\ =\dfrac{3}{4}x^2y+3xy-9xy^2-36y^2\)

\(CuO+CO\underrightarrow{t^o}Cu+CO_2\)

\(MgO+CO->\left(CO.ko.khử,đc\right)\)

\(Fe_2O_3+3CO\underrightarrow{t^o}2Fe+3CO_2\)

\(FeO+CO\underrightarrow{t^o}Fe+CO_2\)

\(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Từ các pthh trên thấy: \(n_{CO_2.sinh.ra}=n_{CO.pứ}=0,2\left(mol\right)\left(theo.tỉ.lệ.pthh\right)\)

Áp dụng ĐLBTKL có: \(m_{hh}+m_{CO}=m_{rắn}+mCO_2\)

=> \(m_{rắn}=m_{hh}+m_{CO}-m_{CO_2}=12,5+0,2.28-0,2.44=9,3\left(g\right)\)

\(a,VP=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\\ =\left(x+2y\right)\left[x^2-x.2y+\left(2y\right)^2\right]\\ =x^3+\left(2y\right)^3=x^3+8y^3=VT\left(đpcm\right)\\ b,VT=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\left(x-y\right)\\ =x^3-y^3-3xy\left(x-y\right)\\ =x^3-3x^2y+3xy^2-y^3\\ =\left(x-y\right)^3=VP\left(đpcm\right)\)

\(c,VT=\left(x-3y\right)\left(x^2+3xy+9y^2\right)-\left(3y+x\right)\left(9y^2-3xy+x^2\right)\\ =\left(x-3y\right)\left[x^2+x.3y+\left(3y\right)^2\right]-\left(x+3y\right).\left[x^2-x.3y+\left(3y\right)^2\right]\\ =x^3-27y^3-\left(x^3+27y^3\right)\\ =-54y^3=VP\left(đpcm\right)\)

Trà xanh đã tanh thì đừng có lên mặt tỏ vẻ mik là thanh cao

đúng rồi

Bài 14:
a) xx' // yy' vì :
Góc x'An = góc ABy = 60 độ ( 2 góc so le trong)
=> xx' // yy'
b) Ax // By vì:
Góc zBy = góc zAx = 50 độ ( 2 góc đồng vị )
=> Ax // By

kb với mk nha

k tớ...Nha!

okie hồ hương quế

$M_{hợp\ chất} = 1X + 1O = X + 16 = \dfrac{1}{4}M_{CuSO_4} = \dfrac{1}{4}.160 = 40(đvC)$

$\Rightarrow X = 24(đvC)$

Vậy X là nguyên tố Magie, số hiệu nguyên tử : 12. kí hiệu : Mg

a) \(\left(0,5\right)^{12}:\left(0,5\right)^{10}=\left(0,5\right)^{12-10}=\left(0,5\right)^2\)

b) \(\sqrt{36}=\pm6\)

c)\(\left(0,75\right)^{22}:\left(0,75\right)^{12}=\left(0,75\right)^{22-12}=\left(0,75\right)^{10}\)

d) \(\sqrt{49}=\pm7\)