\(100^{12}:\left(-100\right)^5=\left(-100\right)^{12}:\left(-100\right)^5=\left(-100\right)^7\)

=> 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)

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

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

Câu 1:

a. \(\left(x-1\right)\left(x+2\right)-x^2+3=5\)

\(x^2+2x-x-2-x^2+3=5\)

\(x+1=5\)

\(x=4\)

b. \(\left(2x+1\right)\left(x-3\right)-2x\left(x+7\right)=100\)

\(2x^2-6x+x-3-2x^2-14x=100\)

\(-19x-3=100\)

\(x=\frac{103}{-19}\)

\(x=-7\)

c. \(\left(3x-1\right)\left(x+2\right)-\left(2-3x\right)\left(x+3\right)=12\)

\(3x^2+6x-x-2-\left(2x+6-3x^2-9x\right)=12\)

\(3x^2+6x-x-2-2x-6+3x^2+9x=12\)

\(6x^2+12x-8=12\)

\(6x^2+12x=20\)

Câu 2:

\(\left(x-5\right)\left(2x+3\right)-2x\left(x-3\right)+x+7\)

\(=2x^2+3x-10x-15-2x^2+6x+x+7\)

\(=-8\) (không phụ thuộc vào biến)

d: \(\Leftrightarrow x^3+6x^2+12x+8-x^3+6x^2-12x+8=12x^2-12x-8\)

\(\Leftrightarrow12x^2+16=12x^2-12x-8\)

=>-12x=24

hay x=-2

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

\(\Leftrightarrow x^2+7x+10-12x+9=x^2-10x+25\)

=>-5x+19=-10x+25

=>5x=6

hay x=6/5

f: \(\dfrac{x-5}{100}+\dfrac{x-4}{101}+\dfrac{x-3}{102}=\dfrac{x-100}{5}+\dfrac{x-101}{4}+\dfrac{x-102}{3}\)

=>x-105=0

hay x=105

a, Ta có: \(x^2\ge0\)

\(\Leftrightarrow x^2+5\ge5\)

\(\Leftrightarrow\left(x^2+5\right)^2\ge25\)

\(\Leftrightarrow\left(x^2+5\right)^2+4\ge29\)

Dấu " = " khi \(x^2=0\Leftrightarrow x=0\)

Vậy \(MIN_{\left(x^2+5\right)^2+4}=29\) khi x = 0

c, Đặt \(C=x\left(x-6\right)+100\)

\(=x^2-6x+100=x^2-6x+9+91\)

\(=\left(x-3\right)^2+91\)

Ta có: \(\left(x-2\right)^2+91\ge91\)

Dấu " = " khi \(\left(x-2\right)^2=0\Leftrightarrow x=2\)

Vậy \(MIN_C=91\) khi x = 2

b,

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

\(=4\left(x^2+\dfrac{1}{2}x-\dfrac{5}{4}\right)\)

\(=4\left(x^2+2.x.\dfrac{1}{4}+\dfrac{1}{16}\right)-5-\dfrac{1}{4}\)

\(=4\left(x+\dfrac{1}{4}\right)^2-\dfrac{21}{4}\)

Mà \(4\left(x+\dfrac{1}{4}\right)^2\ge0=>4\left(x+\dfrac{1}{4}\right)^2-\dfrac{21}{4}\ge-\dfrac{21}{4}\)

Vậy \(Min_Q=-\dfrac{21}{4}\Leftrightarrow x=-\dfrac{1}{4}\)

Vì tú ko lm câu b nên mk chỉ làm câu b thoy .

a) \(\dfrac{2}{3x+9}-\dfrac{x-3}{3x^2+9x}\)

\(=\dfrac{2}{3\left(x+3\right)}-\dfrac{x-3}{3x\left(x+3\right)}\)

\(=\dfrac{2x}{3x\left(x+3\right)}-\dfrac{x-3}{3x\left(x+3\right)}\)

\(=\dfrac{2x-x+3}{3x\left(x+3\right)}\)

\(=\dfrac{x+3}{3x\left(x+3\right)}\)

\(=\dfrac{1}{3x}\)

b) \(\dfrac{x^2+x}{5x^2-10x+5}:\dfrac{3x+3}{5x-5}\)

\(=\dfrac{x\left(x+1\right)}{5\left(x^2-2x+1\right)}:\dfrac{3\left(x+1\right)}{5\left(x-1\right)}\)

\(=\dfrac{x\left(x+1\right)}{5\left(x-1\right)^2}:\dfrac{3\left(x+1\right)}{5\left(x-1\right)}\)

\(=\dfrac{x\left(x+1\right)}{5\left(x-1\right)^2}.\dfrac{5\left(x-1\right)}{3\left(x+1\right)}\)

\(=\dfrac{x}{\left(x-1\right).3}\)

\(=\dfrac{x}{3x-3}\)

c) \(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+...+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)

\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+99}-\dfrac{1}{x+100}\)

\(=\dfrac{1}{x}-\dfrac{1}{x+100}\)

\(=\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)

\(=\dfrac{x+100-x}{x\left(x+100\right)}\)

\(=\dfrac{100}{x\left(x+100\right)}\)

1)

\(\dfrac{x-5}{100}+\dfrac{x-4}{101}+\dfrac{x-3}{102}=\dfrac{x-100}{5}+\dfrac{x-101}{4}+\dfrac{x-102}{3}\)

\(\Leftrightarrow\dfrac{x-5}{100}+1+\dfrac{x-4}{101}+1+\dfrac{x-3}{102}+1=\dfrac{x-100}{5}+1+\dfrac{x-101}{4}+1+\dfrac{x-102}{3}+1\)

\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}=\dfrac{x-105}{5}+\dfrac{x-105}{4}+\dfrac{x-105}{3}+\dfrac{x-105}{2}\)

\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}-\dfrac{x-105}{5}-\dfrac{x-105}{4}-\dfrac{x-105}{3}-\dfrac{x-105}{2}=0\)

\(\Leftrightarrow\left(x-105\right)\left(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}-\dfrac{1}{5}-\dfrac{1}{4}-\dfrac{1}{3}-\dfrac{1}{2}\right)=0\)\(\Leftrightarrow105-x=0\)

\(\Leftrightarrow x=105\)

b)

\(\dfrac{29-x}{21}+\dfrac{27-x}{23}+\dfrac{25-x}{25}+\dfrac{23-x}{27}+\dfrac{21-x}{29}=0\)

\(\Leftrightarrow\dfrac{29-x}{21}+1+\dfrac{27-x}{23}+1+\dfrac{25-x}{25}+1+\dfrac{23-x}{27}+1+\dfrac{21-x}{29}+1=0\)

\(\Leftrightarrow\dfrac{50-x}{21}+\dfrac{50-x}{23}+\dfrac{50-x}{25}+\dfrac{20-x}{27}+\dfrac{50-x}{29}=0\)

\(\Leftrightarrow\left(50-x\right)\left(\dfrac{1}{21}+\dfrac{1}{23}+\dfrac{1}{25}+\dfrac{1}{27}+\dfrac{1}{29}\right)=0\)

\(\Leftrightarrow50-x=0\)

\(\Leftrightarrow x=50\)

2)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}5x+1=3x-2\\5x+1=-3x+2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=\dfrac{1}{8}\end{matrix}\right.\)

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

\(\Leftrightarrow x^3+6x^2+12x+8=8x^3+12x^2+6x+1\)

\(\Leftrightarrow-7x^3-6x^2+6x+7=0\)

\(\Leftrightarrow-7x^3+7x^2-13x^2+13x-7x+7=0\)

\(\Leftrightarrow-7x^2\left(x-1\right)-13x\left(x-1\right)-7\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-7x^2-13x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\-7x^2-13x-7=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-7\left(x^2+\dfrac{13}{7}x+1\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-7\left(x+\dfrac{13}{14}\right)^2-\dfrac{169}{196}=0\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow x=1\)