\(\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{3}=\dfrac{23}{12}\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{23}{12}+\dfrac{1}{3}=\dfrac{9}{4}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{3}{2}\\x-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}+\dfrac{1}{2}=2\\x=-\dfrac{3}{2}+\dfrac{1}{2}=-1\end{matrix}\right.\)

\(\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{3}=\dfrac{23}{12}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{23}{12}+\dfrac{1}{3}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{9}{4}=\left(\dfrac{3}{2}\right)^2\)

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

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

x = 2 hoặc x = -1

Sửa đề

\(\dfrac{2}{1^2}\cdot\dfrac{6}{2^2}\cdot\dfrac{12}{3^3}\cdot.......\cdot\dfrac{110}{10^2}\cdot x=-20\)

\(\dfrac{2}{1\cdot1}\cdot\dfrac{2\cdot3}{2\cdot2}\cdot\cdot\cdot\cdot\dfrac{11\cdot10}{10\cdot10}\cdot x=-20\)

\(\dfrac{\left(2\cdot3\cdot4\cdot....\cdot11\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot10\right)}\cdot\dfrac{\left(1\cdot2\cdot3\cdot4\cdot5\cdot...\cdot10\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot10\right)}\cdot x=-20\)

\(11\cdot x=-20\\ x=-\dfrac{20}{11}\)

Ko đề cho thêm \(\dfrac{20}{4²}\) mà 

mấy bạn giỏi toán ơi giúp mk vs

Nguyễn Thanh Hằng Nhã Doanh ngonhuminh nguyen thi vang mấy ban giup mk voi

a/Vì \(\sqrt{x^6+1}=x^2\Rightarrow\left(\sqrt{1+x^6}\right)^2=x^4\Rightarrow1+x^6=x^4\)

\(\Rightarrow1=x^4-x^6\)

\(\Rightarrow1=x^4-x^6\)(Vô lí)

Vì \(1>0\) và \(x^4-x^6< 0\) (với mọi x)

Vậy ko tìm đc giá trị x thỏa mãn

b/\(\left(2x+1\right)^5=\left(2x+1\right)^{2010}\)

\(\Rightarrow\left(2x+1\right)^{2005+5}-\left(2x+1\right)^5=0\)

\(\Rightarrow\left(2x+1\right)^5\cdot\left[\left(2x+1\right)^{2005}-1\right]=0\)

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

\(\Rightarrow\left\{{}\begin{matrix}\left(2x+1\right)^5=0\\2x+1-1=0\\2x+1+1=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2x+1=0\Rightarrow2x=-1\Rightarrow x=-\dfrac{1}{2}\\2x=0\Rightarrow x=0\\2x+2=0\Rightarrow2x=-2\Rightarrow x=-1\end{matrix}\right.\)

Vậy...

c/Nếu \(x-3\ge0\) hay \(x\ge3\) ta sẽ có : \(\left|x-3\right|=x-3\)

Khi đó phương trình ( hay đề bài,đẳng thức) có dạng:

\(12-\left(x-3\right)=5x+8\)

\(\Rightarrow12-x+3=5x+8\)

\(\Rightarrow15-8=5x+x\)

\(\Rightarrow7=6x\Rightarrow x=\dfrac{7}{6}\) (không thỏa mãn \(x\ge3\))

Nếu \(x-3< 0\) hay \(x< 3\) ta sẽ có: \(\left|x-3\right|=3-x\)

Khi đó phương trình có dạng:

\(12-\left(3-x\right)=5x+8\)

\(\Rightarrow9-8=5x-x\)

\(\Rightarrow1=4x\Rightarrow x=\dfrac{1}{4}\) (thỏa mãn)

Vậy....

\(\dfrac{4}{3.5}+\dfrac{8}{5.9}+\dfrac{12}{9.15}+...+\dfrac{32}{x\left(x+16\right)}=\dfrac{16}{15}\)

\(2.\left(\dfrac{2}{3.5}+\dfrac{4}{5.9}+\dfrac{6}{9.15}+..+\dfrac{16}{X.\left(X+16\right)}\right)=\dfrac{16}{15}\)

\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{15}+...+\dfrac{1}{X}-\dfrac{1}{X+16}=\dfrac{8}{15}\)

\(\dfrac{1}{X+16}=\dfrac{1}{3}-\dfrac{8}{15}\)

\(\dfrac{1}{X+16}=\dfrac{-1}{5}\)

\(X+16=-5\)

\(X=-21\)

\(A=\dfrac{x+3}{\left(x-3\right)^2}:\dfrac{12-x^2+x+x^2-9}{x\left(x-3\right)}\)

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

\(=\dfrac{2}{\left(x-2\right)\left(x-3\right)}-\dfrac{2}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x+6-2x+4}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\dfrac{10}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}\)

= 1/48

a) 1/8 x 1/6 = 1/48

1/48

ib fb t gửi cho

\(=\left[\dfrac{x^3}{x^2\left(x^2-4\right)}+\dfrac{12}{6\left(2-x\right)}+\dfrac{1}{x+2}\right]:\left(\dfrac{10-x^2}{x+2}+x-2\right)\)

\(=\left[\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right]:\left(\dfrac{10-x^2}{x+2}+x-2\right)\)

\(=\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\left(\dfrac{10-x^2}{x+2}+x-2\right)\)

\(=-\dfrac{6}{\left(x-2\right)\left(x+2\right)}:\dfrac{10-x^2+x^2+2x-2x-4}{x+2}\)

\(=-\dfrac{6}{\left(x-2\right)\left(x+2\right)}.\dfrac{x+2}{6}\)

\(=\dfrac{-1}{x-2}\)