\(\frac{\left(x+1\right)-x}{x\left(x+1\right)}+\frac{\left(x+2\right)-\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}+...+\frac{\left(x+100\right)-\left(x+99\right)}{\left(x+99\right)\left(x+100\right)}=\frac{100}{101}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+99}-\frac{1}{x+100}=\frac{100}{101}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+100}=\frac{100}{101}\)
Tự giải nha

1/x -1/x+100 = 100/101

=> ĐK:  \(x\ne\left\{0;-1;-2;...;-99;-100\right\}\)

Đây là dạng dãy số đặc biệt, bạn có thể giải như sau:

Ta có:

\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+99\right)\left(x+100\right)}=\frac{100}{101}\)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+99}-\frac{1}{x+100}=\frac{100}{101}\)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+100}=\frac{100}{101}\)

\(\Leftrightarrow\frac{x+100-x}{x.\left(x+100\right)}=\frac{100}{101}\)

\(\Leftrightarrow\frac{100}{x^2+100x}=\frac{100}{101}\)

\(\Leftrightarrow x^2+100x=101\)

\(\Leftrightarrow x^2+100x-101=0\)

\(\Leftrightarrow x^2+101x-x-101=0\)

\(\Leftrightarrow x\left(x+101\right)-\left(x+101\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+101\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+101=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\left(n\right)\\x=-101\left(n\right)\end{cases}}\)

Vậy: S={1;-101)

Để \(2x^3-3x^2+x+a⋮\left(x+2\right)\) thì:

\(15x+a=15x+30\)

\(\Leftrightarrow a=30\)

vì phép chia trên là phép chia hết nên số dư cuối cùng bằng 0. Để dư bằng 0 thì a=30

(áp dụng lược đồ horner)

We have:\(\hept{\begin{cases}a^2+b^2+c^2=\frac{1}{3}\\a,b,c>0\end{cases}\Rightarrow0< a,b,c< \frac{1}{\sqrt{3}}}\)

We prove to:

\(4x+\frac{2}{3x}\ge-3x^2+\frac{11}{3}\)  with  \(0< x< \frac{1}{\sqrt{3}}\)

\(\Leftrightarrow4x+\frac{2}{3x}+3x^2-\frac{11}{3}\ge0\)

\(\Leftrightarrow9x^3+12x^2-11x+2\ge0\)

\(\Leftrightarrow\left(3x+1\right)^2\left(x+2\right)\ge0\)   Always true to all \(0< x< \frac{1}{\sqrt{3}}\) 

\(\Rightarrow VT\ge-3a^2+\frac{11}{3}-3b^2+\frac{11}{3}-3c^2+\frac{11}{3}\)

\(=-3\left(a^2+b^2+c^2\right)+11=-3.\frac{1}{3}+11=10\) \(\left(đpcm\right)\)

Đặt biểu thức trên là \(A\)

Ta có : \(A=\left(4a+\frac{2}{3a}\right)+\left(4b+\frac{2}{3b}\right)+\left(4c+\frac{2}{3c}\right)\)

Cần chứng minh \(4a+\frac{2}{3a}\ge-3a^2+\frac{11}{3}\) (*)

Thật vậy \(BĐT\Leftrightarrow4a+\frac{2}{3a}+3a^2-\frac{11}{3}\ge0\)

\(\Leftrightarrow\frac{12a^2+2+9a^3-11a}{3a}\ge0\Leftrightarrow\frac{\left(a+2\right)\left(3a-1\right)^2}{3a}\ge0\) (luôn đúng)

Tương tự : \(4b+\frac{2}{3b}\ge-3b^2+\frac{11}{3}\)   và \(4c+\frac{2}{3c}\ge-3c^2+\frac{11}{3}\)

Cộng các bất dẳng thức vừa CM đc ta có :

\(A\ge-3\left(a^2+b^2+c^2\right)+\frac{11}{3}.3=-3.\frac{1}{3}+11=10\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)

2x² - 8x + 3x - 12 + x² - 2x - 5x + 10 = 3x² - 12x - 5x + 20

2x² + x² - 3x² - 8x + 3x - 2x - 5x + 12x + 5x = 20 - 10 +12

5x = 22

x = 4,4

a/ \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)

<=> \(\left(2x-3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)-\left(3x-5\right)\left(x-4\right)=0\)

<=> \(2x^2-8x-3x+12+x^2-2x-5x+10-\left(3x^2-12x-5x+20\right)=0\)

<=> \(2x^2-11x+12+x^2-7x+10-3x^2+17x-20=0\)

<=> \(2-x=0\)

<=> \(x=2\)