\(\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{2.4}\right).\left(1+\dfrac{1}{3.5}\right).........\left[1+\dfrac{1}{x.\left(x+2\right)}\right]=\dfrac{31}{16}\)

\(\Rightarrow\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}........\dfrac{\left(x+1\right)^2}{x.\left(x+2\right)}=\dfrac{31}{16}\)

\(\Rightarrow\dfrac{\left[2.3.4.............\left(x+1\right)\right].\left[2.3.4.............\left(x+1\right)\right]}{\left(1.2.3...................x\right).\left(3.4.5..........................\left(x+2\right)\right)}=\dfrac{31}{16}\)

\(\Rightarrow\dfrac{\left(x+1\right).2}{1.\left(x+2\right)}=\dfrac{31}{16}\)

\(\Leftrightarrow16.2\left(x+1\right)=31.\left(x+2\right)\)

\(\Rightarrow32x+32=31x+62\)

\(\Rightarrow x=30\)

Vậy x=30

Chúc bn học tốt

ĐKXĐ: \(x\notin\left\{0;-2\right\}\)

Ta có: \(\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\left(1+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(1+\dfrac{1}{x\left(x+2\right)}\right)=\dfrac{31}{16}\)

\(\Leftrightarrow\dfrac{1\cdot3+1}{1\cdot3}+\dfrac{1+2\cdot4}{2\cdot4}+\dfrac{1+3\cdot5}{3\cdot5}\cdot...\cdot\dfrac{1+x\left(x+2\right)}{x\left(x+2\right)}=\dfrac{31}{16}\)

\(\Leftrightarrow\dfrac{2\cdot2}{1\cdot3}+\dfrac{3\cdot3}{2\cdot4}+\dfrac{4\cdot4}{3\cdot5}+...+\dfrac{\left(x+1\right)\left(x+1\right)}{x\left(x+2\right)}=\dfrac{31}{16}\)

\(\Leftrightarrow\dfrac{1\cdot2\cdot3\cdot...\cdot\left(x+1\right)}{1\cdot2\cdot3\cdot...\cdot x}\cdot\dfrac{2\cdot3\cdot4\cdot...\cdot\left(x+1\right)}{3\cdot4\cdot5\cdot...\cdot\left(x+2\right)}=\dfrac{31}{16}\)

\(\Leftrightarrow\left(x+1\right)\cdot\dfrac{2}{x+2}=\dfrac{31}{16}\)

\(\Leftrightarrow\dfrac{2x+2}{x+2}=\dfrac{31}{16}\)

\(\Leftrightarrow\dfrac{32x+32}{16\left(x+2\right)}=\dfrac{31\left(x+2\right)}{16\left(x+2\right)}\)

Suy ra: \(32x+32=31x+62\)

\(\Leftrightarrow x=30\)(thỏa ĐK)

Vậy: S={30}

Ta có: \(\left|x+\frac{1}{2}\right|\ge0;\left|x+\frac{1}{6}\right|\ge0;\left|x+\frac{1}{12}\right|\ge0;...;\left|x+\frac{1}{110}\right|\ge0\)

=> VT \(\ge\)0  

=>VP  \(\ge\)0  => 11x \(\ge\)0 => x \(\ge\)0.

=> \(\left|x+\frac{1}{2}\right|=x+\frac{1}{2};\left|x+\frac{1}{6}\right|=x+\frac{1}{6};\left|x+\frac{1}{12}\right|=x+\frac{1}{12};...;\left|x+\frac{1}{110}\right|=x+\frac{1}{110}\)

Phương trình <=> \(x+\frac{1}{2}+x+\frac{1}{6}+x+\frac{1}{12}+...+x+\frac{1}{110}=11x\)

<=> \(\left(x+x+x+...+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)=11x\)

<=> \(10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)=11x\)

<=> \(1-\frac{1}{11}=11x-10x\)

<=> \(\frac{10}{11}=x\)

<=> \(x=\frac{10}{11}\left(tm\right)\)

Bởi vì 

\(\frac{1}{2}=\frac{1}{1.2};\frac{1}{6}=\frac{1}{2.3};...;\frac{1}{110}=\frac{1}{10.11}\)

nên từ \(\frac{1}{2}\)đến \(\frac{1}{110}\)chỉ có 10 số

nên chỉ có 10 x

Ta có : \(|x+\frac{1}{2}|\ge0;|x+\frac{1}{6}|\ge0;....;|x+\frac{1}{110}|\ge0\)

\(\Rightarrow|x+\frac{1}{2}|+|x+\frac{1}{6}|+...+|x+\frac{1}{110}|\ge0\)

\(\Rightarrow11x\ge0\)

\(\Rightarrow x\ge0\)

\(\Rightarrow x+\frac{1}{2}>0;x+\frac{1}{6}>0;....;x+\frac{1}{110}>0\)

\(\Rightarrow|x+\frac{1}{2}|=x+\frac{1}{2};|x+\frac{1}{6}|=x+\frac{1}{6};....;|x+\frac{1}{110}|=x+\frac{1}{110}\)

\(\Rightarrow\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+...+\left(x+\frac{1}{110}\right)=11x\)

\(\Rightarrow10x+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)=11x\)

\(\Rightarrow10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)=11x\)

\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{10}-\frac{1}{11}\right)=11x\)

\(\Rightarrow10x+\frac{1}{10}=11x\)

\(\Rightarrow x=\frac{10}{11}\)

Vậy.....

\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+\left|x+\frac{1}{12}\right|+...+\left|x+\frac{1}{110}\right|=11x\)

Với mọi x ta có:

+) \(\left\{{}\begin{matrix}\left|x+\frac{1}{2}\right|\ge0\\\left|x+\frac{1}{6}\right|\ge0\\.........\\\left|x+\frac{1}{110}\right|\ge0\end{matrix}\right.\)

\(\Rightarrow\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{110}\right|\ge0\) \(\forall x.\)

Mà \(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{110}\right|=11x\)

\(\Rightarrow11x\ge0\)

\(\Rightarrow x\ge0.\)

Với \(x\ge0\) thì:

\(\left\{{}\begin{matrix}\left|x+\frac{1}{2}\right|=x+\frac{1}{2}\\\left|x+\frac{1}{6}\right|=x+\frac{1}{6}\\..........\\\left|x+\frac{1}{110}\right|=x+\frac{1}{110}\end{matrix}\right.\)

\(\Rightarrow x+\frac{1}{2}+x+\frac{1}{6}+...+x+\frac{1}{110}=11x\)

\(\Rightarrow11x+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)=11x\)

\(\Rightarrow\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)=11x-11x\)

\(\Rightarrow\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}=0x\) (vô lí).

\(\Rightarrow x\in\varnothing.\)

Vậy không tồn tại giá trị của x thỏa mãn yêu cầu đề bài.

Chúc bạn học tốt!

mình sửa đề chút nhé + \(\left|x+\frac{1}{110}\right|=11x\)

(x-1)³ + 1 + 3x (x - 4) = 0

=> (x³-3.x².1+3.x.1²-1³) +1 + 3x² -12x = 0

=> ( x³-3x²+3x) +1 +3x² -12x=0

=> x³-3x² + 3x +1 +3x² -12x = 0

=> x3 + (-3x²+3x²) + (3x-12x) =0

=> x³ - 9x = 0

=> x = 0

Nhân hết ra,giải phương trình bậc cao đi

( x + 1 ) + ( x + 3 ) + ( x + 5 ) + ( x + 7 ) + ( x + 9 ) + ( x + 11 ) + (x + 13 ) = 119

x X 7 + ( ! + 3 + 5 + 7 + 9 + 11 + 13 )  = 119

x X 7 + 49  = 119

x X 7 = 119 - 49

x X 7 = 70

x = 70 : 7 

x = 10

tk mk và gửi kết bạn cho mk nha huyền trang

các bạn cũng tk và gửi kết bạn cho mk nha

(x+1)+(x+3)+..+(x+13)=119

=>7x+(1+3+..+13)     =119

=>7x+49                  =119

=>7x                       =119-49

=>7x                       =70

=> x                        =10

Vậy x=10

Chúc bn hok giỏi!

7 x X + (1 + 3 + 5 + ...+ 13) = 119

7 x X + 49 = 119

7 x X = 119 - 49 = 70

X = 70 : 7

X = 10

\(1,\left(2x+1\right)\left(x-1\right)-x\left(2x-3\right)+3=0\)

\(\Rightarrow2x^2-2x+x-1-\left(2x^2-3x\right)+3=0\)

\(\Rightarrow2x^2-2x+x-1-2x^2+3x+3=0\)

\(\Rightarrow2x=-2\Rightarrow x=-1\)

\(2,\left(x^2+x-2\right)\left(x^2-x-2\right)-x^2\left(x^2-2\right)+8=0\)

\(\Rightarrow[\left(x^2\right)^2-\left(x-2\right)^2]-x^2\left(x^2-2\right)+8=0\)

\(\Rightarrow x^4-\left(x^2-4x+4\right)-x^4+2x^2+8=0\)

\(\Rightarrow x^4-x^2+4x-4-x^4+2x^2+8=0\)

\(\Rightarrow x^2+4x+4=0\)

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

a) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=46\)

\(\Leftrightarrow16x^2-9-16x^2+40x-25=46\)

\(\Leftrightarrow40x=46+9+25=80\)

\(\Leftrightarrow x=2\)

b) \(\left(x+1\right)^3+2x-\left(x-1\right)^3-3\left[\left(x+1\right)^2+\left(x-1\right)^2\right]+5=0\)

\(=x^3+3x^2+3x+1+2x-x^3+3x^2-3x+1-3\left(x^2+2x+1+x^2-2x+1\right)+5=0\)

\(=6x^2+2x+2-3\left(2x^2+2\right)+5=0\)

\(\Leftrightarrow6x^2+2x+2-6x^2-6+5=0\)

\(\Leftrightarrow2x=-2+6-5=-1\)

\(\Leftrightarrow x=\frac{1}{2}\)