bài 11:

a: \(\frac{15x}{7y^3}\cdot\frac{2y^2}{x^2}=\frac{15x\cdot2y^2}{7x^2y^3}=\frac{30xy^2}{7x^2y^3}=\frac{30}{7xy^{}}\)

b: \(\frac{5x+10}{4x-8}\cdot\frac{4-2x}{x+2}\)

\(=\frac{5\left(x+2\right)}{4\left(x-2\right)}\cdot\frac{-2\left(x-2\right)}{x+2}\)

\(=\frac54\cdot\left(-2\right)=-\frac52\)

c: \(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}\)

\(=\frac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}\cdot\frac{3x}{2\left(1-2x\right)}\)

\(=\frac{3\left(2x+1\right)}{2\left(x+4\right)}=\frac{6x+3}{2x+8}\)

d: \(\left(\frac{1}{x^2+x}-\frac{2-x}{x+1}\right):\left(\frac{1}{x}+x-2\right)\)

\(=\left(\frac{1}{x\left(x+1\right)}+\frac{x-2}{x+1}\right):\frac{1+x^2-2x}{x}\)

\(=\frac{1+x\left(x-2\right)}{x\left(x+1\right)}\cdot\frac{x}{x^2-2x+1}\)

\(=\frac{x^2-2x+1}{x+1}\cdot\frac{1}{x^2-2x+1}=\frac{1}{x+1}\)

ko spam nha bạn

đè môn sinh, sử, anh, đia là zì zạy chụp giúp mik đc khum

$M_O = 16(đvC)$

$PTK = Y + 16.2 = M_{Cu} = 64(đvC) \Rightarrow Y = 32$
Vậy Y là lưu huỳnh, KHHH : S

More progress in gender equality will be made by the Vietnamese government

Old men should be treated with respect by children

2 câu dưới mình làm rồi nha bạn =))

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne1\end{matrix}\right.\)

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

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

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

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

\(=\dfrac{2}{x+\sqrt{x}+1}\)

b: \(x=7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2\)

Khi \(x=\left(2-\sqrt{3}\right)^2\) thì 

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

\(P=\dfrac{2}{7-4\sqrt{3}+2-\sqrt{3}+1}\)

\(=\dfrac{2}{10-5\sqrt{3}}=\dfrac{4+2\sqrt{3}}{5}\)

c: P>=2/3

=>P-2/3>=0

=>\(\dfrac{2}{x+\sqrt{x}+1}-\dfrac{2}{3}>=0\)

=>\(\dfrac{1}{x+\sqrt{x}+1}-\dfrac{1}{3}>=0\)

=>\(\dfrac{3-x-\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}>=0\)

=>\(-x-\sqrt{x}+2>=0\)

=>\(x+\sqrt{x}-2< =0\)

=>\(\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)< =0\)

=>\(\sqrt{x}-1< =0\)

=>0<=x<=1

Kết hợp ĐKXĐ, ta được: 0<=x<1

Bài 8:

a) PTK(hc)= 2.NTK(X) + 3.NTK(Y)

<=> 4,25. NTK(Mg)= 2.NTK(X) + 3.NTK(Y)

<=> 2.NTK(X) + 3.NTK(Y)= 4,25. 24=102(đ.v.C)

=> PTK(hc)=102(đ.v.C)

b) Ta có:

\(\dfrac{2.NTK_X}{102}.100\%=52,94\%\\ \Leftrightarrow NTK_X=27\left(\dfrac{g}{mol}\right)\)

=> X là nhôm (Al=27)

2.27+3.NTK(Y)=102

<=>NTK(Y)=16(đ.v.C)

=>Y là Oxi (O=16)

ko 

vì chưa chắc người ta đã lấy mình