1.\(\left(x+1\right)\left(x+4\right)=\left(2-x\right)\left(2+x\right)\)

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

\(\Leftrightarrow x^2+5x+4=4-x^2\)

\(\Leftrightarrow x^2+5x+4-4+x^2=0\)

\(\Leftrightarrow2x^2+6x=0\)

\(\Leftrightarrow2x\left(x+3\right)=0\)

\(\Rightarrow2x=0\)hoặc \(x+3=0\)

Giải 2 pt:

\(2x=0\Leftrightarrow x=0\)

\(x+3=0\Leftrightarrow x=-3\)

Vậy \(S=\left\{0;-3\right\}\)

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

\(\Leftrightarrow x^2+5x+4=4-x^2\)

\(\Leftrightarrow x^2+5x+4-4+x^2=0\)

\(\Leftrightarrow2x^2+5x=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{5}{2}\end{cases}}}\)

b,\(x^3-x^2=1-x\)

\(\Leftrightarrow x^3-x^2+x-1=0\)

\(\Leftrightarrow x^2\left(x-1\right)+\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-1\\x=1\end{cases}\Leftrightarrow}x=1}\)

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

\(\Leftrightarrow2x\left(x+1\right)-\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x-x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)^2=0\)

\(\Leftrightarrow x=-1\)

4)\(\left(x-2\right)\left(2x+5\right)=\left(2x-4\right)\left(x+1\right)\)

\(\Leftrightarrow\left(x-2\right)\left(3x+5\right)-2\left(x-2\right)\left(x+1\right)\)

\(\Leftrightarrow\left(x-2\right)\left(3x+5-2x-2\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)

\(\frac{\left(x-2\right)\left(x+10\right)}{3}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{\left(x-2\right)\left(x+4\right)}{4}\)

\(\Leftrightarrow\frac{4\left(x-2\right)\left(x+10\right)}{12}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{3\left(x-2\right)\left(x+4\right)}{12}\)

\(\Leftrightarrow4\left(x-2\right)\left(x+10\right)-\left(x+4\right)\left(x+10\right)=3\left(x-2\right)\left(x+4\right)\)

\(\Leftrightarrow4\left(x^2-2x+10x-20\right)-\left(x^2+4x+10x+40\right)=3\left(x^2-2x+4x-8\right)\)

\(\Leftrightarrow4\left(x^2+8x-20\right)-\left(x^2+14x+40\right)=3\left(x^2+2x-8\right)\)

\(\Leftrightarrow4x^2+32x-80-x^2-14x-40=3x^2+6x-24\)

\(\Leftrightarrow3x^2+18x-120=3x^2+6x-24\)

\(\Leftrightarrow12x=96\)\(\Leftrightarrow x=8\)

Vậy tập nghiệm của phương trình là \(S=\left\{8\right\}\)

\(\frac{\left(x-2\right)\left(x+10\right)}{3}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{\left(x-2\right)\left(x+4\right)}{4}\)

\(\Leftrightarrow\frac{4\left(x-2\right)\left(x+10\right)}{12}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{3\left(x-2\right)\left(x+4\right)}{12}\)

\(\Leftrightarrow4\left(x^2+8x-20\right)-\left(x^2+14x+40\right)=3\left(x^2+2x-8\right)\)

\(\Leftrightarrow4x^2+32x-80-x^2-14x-40-3x^2-6x+24=0\)

\(\Leftrightarrow12x-96=0\)

\(\Leftrightarrow x=8\)

\(\frac{1}{a+1}\ge1-\frac{1}{b+1}+1-\frac{1}{c+1}=\frac{b}{b+1}+\frac{c}{c+1}\ge2\sqrt{\frac{bc}{\left(b+1\right)\left(c+1\right)}}\).

Tương tự ta có: \(\frac{1}{b+1}\ge2\sqrt{\frac{ac}{\left(a+1\right)\left(c+1\right)}}\), \(\frac{1}{c+1}\ge2\sqrt{\frac{ab}{\left(a+1\right)\left(b+1\right)}}\).

Nhân 3 bất đẳng thức trên theo vế ta được: 

\(\frac{1}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge\frac{8abc}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\)

\(\Leftrightarrow abc\le\frac{1}{8}\).

Áp dụng bất đẳng thức Cauchy - Schwarz với 2 bộ số \(\left(a,b,c\right)\)và \(\left(1,1,1\right)\)ta có: 

\(\left(a^2+b^2+c^2\right)\left(1^2+1^2+1^2\right)\ge\left(a.1+b.1+c.1\right)^2=1\)

\(\Rightarrow a^2+b^2+c^2\ge\frac{1}{3}\).

Dấu \(=\)xảy ra khi và chỉ khi \(a=b=c=\frac{1}{3}\).

Còn cách khác :3 

Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có ngay :

\(a^2+b^2+c^2=\frac{a^2}{1}+\frac{b^2}{1}+\frac{c^2}{1}\ge\frac{\left(a+b+c\right)^2}{1+1+1}=\frac{1^2}{3}=\frac{1}{3}\)

Đẳng thức xảy ra <=> a = b = c = 1/3

Vậy ta có điều phải chứng minh 

\(A=\frac{1}{1\left(2n-1\right)}+\frac{1}{3\left(2n-3\right)}+...+\frac{1}{\left(2n-1\right).1}\)

\(A=\frac{1}{2n}\left[\frac{2n-1+1}{1\left(2n-1\right)}+\frac{2n-3+3}{3\left(2n-3\right)}+...+\frac{1+2n-1}{\left(2n-1\right).1}\right]\)

\(A=\frac{1}{2n}\left[\frac{1}{1}+\frac{1}{2n-1}+\frac{1}{3}+\frac{1}{2n-3}+...+\frac{1}{2n-1}+\frac{1}{1}\right]\)

\(A=\frac{1}{n}\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2n-3}+\frac{1}{2n-1}\right)\)

\(\Rightarrow\frac{a}{b}=\frac{1}{n}\).

1) Theo Ta-let: \(\frac{x}{8,6}=\frac{2}{2+4}=\frac{1}{3}\Rightarrow x=\frac{8,6}{3}=\frac{43}{15}\left(cm\right)\)

2) Theo Ta-let: \(\frac{3}{x}=\frac{1}{6}\Rightarrow x=6.3=18\left(cm\right)\)

a, Vì MN // BC Suy ra : \(\frac{AM}{MB}=\frac{MN}{BC}\)( theo định lí Ta lét )

\(\Rightarrow\frac{2}{4}=\frac{x}{8,6}\Rightarrow x=\frac{2.8,6}{4}=\frac{17,2}{4}=4,3\)cm 

b, Vì MN // DE Suy ra : \(\frac{NI}{ND}=\frac{MN}{DE}\)( theo hệ quả Ta lét )

mà \(ND=NI+ID=4+6=10\)cm 

\(\Rightarrow\frac{4}{10}=\frac{3}{x}\Rightarrow x=\frac{3.10}{4}=\frac{30}{4}=7,5\)cm 

1 giờ vòi 1 chảy được : 1 : 4 = 1/4 bể

1 giờ vòi 2 chảy được : 1 : 5 = 1/5 bể

1 giờ vòi 3 chảy được  1:6 = 1/6 bể

=> 1 giờ 3 vòi chảy được : 1/4 + 1/5 + 1/6 = 37/60 bể

=> Thời gian để 3 vòi chảy đầy bể là 1 : 37/60 = 60/37 giờ