C1: 

Bảo toàn C: n_C = 0,4 (mol)

Bảo toàn H: n_H = 1,2 (mol)

=> m_hh = 12.0,4 + 1.1,2 = 6(g)

C2: 

Bảo toàn O: n_O2 = \(\dfrac{0,4.2+0,6}{2}=0,7\left(mol\right)\)

Theo ĐLBTKL: m_hh + m_O2 = m_CO2 + m_H2O

=> m_hh = 0,4.44 + 0,6.18 - 0,7.32 = 6(g)

C3:

Gọi công thức chung của hh là C_xH₄

PTHH: C_xH₄ + (x+1)O₂ --t^o--> xCO₂ + 2H₂O

               a------------------------>ax----->2a

=> \(\left\{{}\begin{matrix}ax=0,4\\2a=0,6\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}a=0,3\\x=\dfrac{4}{3}=>CTHH:C_{\dfrac{4}{3}}H_4\end{matrix}\right.\)

=> \(m_{hh}=0,3.20=6\left(g\right)\)

Bài 1 :

Mình nghĩ phải sửa đề ntn :

\(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

\(\Leftrightarrow\left[2\left(2x+7\right)\right]^2-\left[3\left(x+3\right)\right]^2=0\)

\(\Leftrightarrow\left[2\left(2x+7\right)-3\left(x+3\right)\right]\left[2\left(2x+7\right)+3\left(x+3\right)\right]=0\)

\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\7x+23=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=\frac{-23}{7}\end{cases}}}\)

Vậy....

b) \(A=\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

Đặt \(q=x^2+x+1\)ta có :

\(A=q\left(q+1\right)-12\)

\(A=q^2+q-12\)

\(A=q^2+4q-3q-12\)

\(A=q\left(q+4\right)-3\left(q+4\right)\)

\(A=\left(q+4\right)\left(q-3\right)\)

Thay \(q=x^2+x+1\)ta có :

\(A=\left(x^2+x+1+4\right)\left(x^2+x+1-3\right)\)

\(A=\left(x^2+x+5\right)\left(x^2+x-2\right)\)

\(A=\left(x^2+x+5\right)\left(x^2+2x-x-2\right)\)

\(A=\left(x^2+x+5\right)\left[x\left(x+2\right)-\left(x+2\right)\right]\)

\(A=\left(x^2+x+5\right)\left(x+2\right)\left(x-1\right)\)

Cảm ơn ạ><

Cách 2 của câu 2:

Đặt \(x^2+x+2=t\)

Ta có: \(A=\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

\(=t\left(t-1\right)-12=t^2-t-12\)

\(=\left(t-4\right)\left(t+3\right)=\left(x^2+x-2\right)\left(x^2+x+5\right)\)

\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+5\right)\)

đkxđ: \(\dfrac{x+3}{x-1}\ge0\)

Ptr ⇔\(\left(x-1\right)\left(x+3\right)+\dfrac{2\left(x-1\right)\sqrt{\left(x+3\right)\left(x-1\right)}}{x-1}=8\\ \Leftrightarrow\left(x-1\right)\left(x+3\right)+2\sqrt{\left(x-1\right)\left(x+3\right)}-8=0\) 

Đặt \(\sqrt{\left(x-1\right)\left(x+3\right)}=a\)      (a≥0)

Ptr ⇔ \(a^2+2a-8=0\) 

⇔a=2 (tm) hoặc a=-4 (loại)

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

⇔\(x^2+2x-3=4\)

\(\Leftrightarrow x^2+2x-7=0\)

⇔ \(x=-1+2\sqrt{2}\)        (tm)

hoặc \(x=-1-2\sqrt{2}\) (tm)

Vậy...

ĐK: \(x\ge-2\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x+2}=a\ge0\\\sqrt{x^2-2x+4}=b\ge0\end{matrix}\right.\Leftrightarrow a^2+b^2=x^2-x+6\), PTTT:

\(5ab=2\left(a^2+b^2\right)\\ \Leftrightarrow2a^2-5ab+2b^2=0\\ \Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=2b\\2a=b\end{matrix}\right.\)

Với \(a=2b\Leftrightarrow x+2=4\left(x^2-2x+4\right)\)

\(\Leftrightarrow4x^2-8x+16=x+2\\ \Leftrightarrow4x^2-9x+14=0\\ \Delta=81-224< 0\\ \Leftrightarrow x\in\varnothing\)

Với \(2a=b\Leftrightarrow4\left(x+2\right)=x^2-2x+4\)

\(\Leftrightarrow4x+8=x^2-2x+4\\ \Leftrightarrow x^2-6x-4=0\\ \Delta=36+16=52\\ \Leftrightarrow\left[{}\begin{matrix}x=3+\sqrt{13}\left(tm\right)\\x=3-\sqrt{13}\left(tm\right)\end{matrix}\right.\)

Vậy PT có nghiệm \(x=3\pm\sqrt{13}\)

Lời giải:

ĐKXĐ: $x\geq -2$

PT $\Leftrightarrow 5\sqrt{(x+2)(x^2-2x+4)}=2(x^2-x+6)$

Đặt $\sqrt{x+2}=a; \sqrt{x^2-2x+4}=b(a,b\geq 0)$ thì pt trở thành:

$5ab=2(a^2+b^2)$

$\Leftrightarrow 2a^2-5ab+2b^2=0$

$\Leftrightarrow (2a-b)(a-2b)=0$

$\Rightarrow 2a=b$ hoặc $a=2b$

Nếu $2a=b\Leftrightarrow 4a^2=b^2$

$\Leftrightarrow 4(x+2)=x^2-2x+4$

$\Leftrightarrow x=3\pm \sqrt{13}$ (tm)

Nếu $a=2b\Leftrightarrow a^2=4b^2$

$\Leftrightarrow x+2=4(x^2-x+6)$

$\Leftrightarrow 4x^2-5x+22=0$ (dễ thấy pt này vô nghiệm)
 

1/ Đặt \(\hept{\begin{cases}\sqrt{x-2013}=a\\\sqrt{x-2014}=b\end{cases}}\)

Thì ta có:

\(\frac{\sqrt{x-2013}}{x+2}+\frac{\sqrt{x-2014}}{x}=\frac{a}{a^2+2015}+\frac{b}{b^2+2014}\)

\(\le\frac{a}{2a\sqrt{2015}}+\frac{b}{2b\sqrt{2014}}=\frac{1}{2\sqrt{2015}}+\frac{1}{2\sqrt{2014}}\)

2/ \(\frac{x}{2x+y+z}+\frac{y}{x+2y+z}+\frac{z}{x+y+2z}\)

\(\le\frac{1}{4}\left(\frac{x}{x+y}+\frac{x}{x+z}+\frac{y}{y+x}+\frac{y}{y+z}+\frac{z}{z+x}+\frac{z}{z+y}\right)\)

\(=\frac{3}{4}\)

=>x^2-4x+2y^2-4y+6=0

=>x^2-4x+4+2y^2-4y+2=0

=>(x-2)^2+2(y-1)^2=0

=>x=2 và y=1