2CO+O₂ \(\rightarrow\)2CO₂

\(\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{8x}{3-6x}\)

\(=\frac{8}{4x^2-1}.\frac{8x}{3-6x}\)

\(=\frac{8.8}{\left(4x^2-1\right).\left(3-6x\right)}\)

\(=\frac{64}{\left(4x^2-1\right)\left(3-6x\right)}\)

CO + O = CO₂

#Học tốt!!!

~NTTH~

Áp dụng 

\(\left(x+y+z\right)^3=x^3+y^3+z^3+\left(x+y+z\right)\left(xy+yz+zx\right)-3xyz\)

Ta có: 

\(\left(a+b+c\right)^2=a^2+b^2+c^2\)

=> \(2ab+2ac+2bc=0\)

=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

KHi đó:

 \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^3=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)-\frac{3}{abc}\)

=> \(0=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+0-\frac{3}{abc}\)

=> \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)

a) \(\left(x-5\right)^2+\left(x^2-25\right)=0\)

\(\Leftrightarrow\left(x-5\right)^2+\left(x+5\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-5+x+5\right)=0\)

\(\Leftrightarrow2x\left(x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)

b) \(\frac{x-2}{4}+\frac{2x-3}{3}=\frac{x-18}{6}\)

\(\Rightarrow\frac{3x-6}{12}+\frac{8x-12}{12}=\frac{2x-36}{12}\)

\(\Rightarrow\frac{11x-18}{12}=\frac{2x-36}{12}\)

\(\Rightarrow11x-18=2x-36\)

\(\Rightarrow11x-2x=18-36\)

\(\Rightarrow9x=-18\Rightarrow x=-2\)

c) \(\frac{1}{x-3}+\frac{x-3}{x+3}=\frac{5x-6}{x^2-9}\)

\(\Rightarrow\frac{x+3}{\left(x+3\right)\left(x-3\right)}+\frac{\left(x-3\right)^2}{\left(x+3\right)\left(x-3\right)}=\frac{5x-6}{x^2-9}\)

\(\Rightarrow\frac{x+3}{\left(x+3\right)\left(x-3\right)}+\frac{x^2-6x+9}{\left(x+3\right)\left(x-3\right)}=\frac{5x-6}{x^2-9}\)

\(\Rightarrow\frac{x^2-5x+12}{x^2-9}=\frac{5x-6}{x^2-9}\)

\(\Rightarrow x^2-5x+12=5x-6\)

\(\Rightarrow x^2-10x+18=0\)

Giải biệt thức sẽ ra 2 nghiệm \(5+\sqrt{7}\)và \(5-\sqrt{7}\)

Gửi Cool: Lần sau đừng quên tìm điều kiện nhé. Câu c. ĐK: x khác 3 và x khác -3

Xét hiệu \(S_1-S_2=\frac{a^2-b^2}{a+b}+\frac{b^2-c^2}{b+c}+\frac{c^2-a^2}{c+a}\)

                         \(=\frac{\left(a-b\right)\left(a+b\right)}{a+b}+\frac{\left(b-c\right)\left(b+c\right)}{b+c}+\frac{\left(c-a\right)\left(c+a\right)}{c+a}\)

                         \(=a-b+b-c+c-a\)

                           \(=0\)

\(\Rightarrow S_1=S_2\)

+) Áp dụng bđt AM-GM ta có:

\(\frac{a^2}{a+b}+\frac{a+b}{4}\ge2\sqrt{\frac{a^2}{a+b}.\frac{a+b}{4}}=a\)

\(\frac{b^2}{b+c}+\frac{b+c}{4}\ge2\sqrt{\frac{b^2}{b+c}.\frac{b+c}{4}}=b\)

\(\frac{c^2}{c+a}+\frac{c+a}{4}\ge2\sqrt{\frac{c^2}{c+a}.\frac{c+a}{4}}=c\)

Cộng theo vế các đẳng thức trên ta được:

\(S_1+\frac{a+b+c}{2}\ge a+b+c\)

\(\Rightarrow S_1\ge\frac{a+b+c}{2}\left(đpcm\right)\)

dit me may