ây da nhầm rồi để mk làm lại cho kết quả là -4025 mới phải

Có : \(\frac{x+2011}{2013}+\frac{x+2012}{2012}=\frac{x+2010}{2014}+\frac{x+2013}{2011}\)

\(\Leftrightarrow\) \(\left(\frac{x+2011}{2013}+1\right)\)+ \(\left(\frac{x+2012}{2012}+1\right)\) - \(\left(\frac{x+2010}{2014}+1\right)\) - \(\left(\frac{x+2013}{2011}+1\right)=0\)

\(\Leftrightarrow\) \(\frac{x+2025}{2013}+\frac{x+2025}{2012}-\frac{x+2025}{2014}-\frac{x+2025}{2011}\)

\(\Leftrightarrow\) \(\left(x+2025\right)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2014}-\frac{1}{2015}\right)\)=0

\(\Leftrightarrow\)\(x+2025=0\) ( vì \(\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2014}-\frac{1}{2011}\right)\ne0\))

\(\Rightarrow\)\(x=-2025\)

Vậy \(x=-2025\)

Có : \(\frac{2-x}{2005}-1\) = \(\frac{1-x}{2006}-\frac{x}{2007}\)

\(\Leftrightarrow\) \(\frac{2-x}{2005}+1=\left(\frac{1-x}{2006}+1\right)-\left(\frac{x}{2007}+1\right)\)

\(\Leftrightarrow\) \(\frac{-x+2007}{2005}=\frac{-x+2007}{2006}+\frac{-x+2007}{2007}\)

\(\Leftrightarrow\) \(\left(-x+2007\right)\left(\frac{1}{2005}-\frac{1}{2006}-\frac{1}{2007}\right)=0\)

\(\Leftrightarrow\) \(-x+2007=0\) ( vì \(\left(\frac{1}{2005}-\frac{1}{2006}-\frac{1}{2007}\right)\ne0\)

\(\Leftrightarrow\) \(x=2007\)

Vậy \(x=2007\)

Cho dãy các kim loại: Li, Na, Al, Ca, Sr. Số kim loại kiềm trong dãy là

A. 1

B. 3

C. 2

D. 4

cái này là thi trên mạng (violympic) hay là thầy cô cho bài tập vậy bạn thi violympic taons cấp huyện rồi à toán lớp toán đúng hk?

Có : \(2x^2+3x+2\)

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

\(\Leftrightarrow\) \(\left(x^2+2.x.1+1^2\right)\) + \(\left(x^2+2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}\right)\)

\(\Leftrightarrow\) \(\left(x+1\right)^2+\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)

Vì \(\left(x+1\right)^2\ge0và\left(x+\frac{1}{2}\right)^2\ge0\)

\(\Rightarrow\) \(\left(x+1\right)^2+\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

Vậy \(2x^2+3x+2>0\left(\forall_x\right)\)

Có : \(\frac{x+1}{96}+\frac{x+2}{95}=\frac{x+3}{94}+\frac{x+4}{93}\)

\(\Leftrightarrow\)\(\left(\frac{x+1}{96}+1\right)+\left(\frac{x+2}{95}+1\right)\)= \(\left(\frac{x+3}{94}+1\right)+\left(\frac{x+4}{93}+1\right)\)

\(\Leftrightarrow\) \(\frac{x+97}{96}+\frac{x+97}{95}=\frac{x+97}{94}+\frac{x+97}{93}\)

\(\Leftrightarrow\) \(\frac{x+97}{96}+\frac{x+97}{95}-\frac{x+97}{94}-\frac{x+97}{93}=0\)

\(\Leftrightarrow\) \(\left(x+97\right)\left(\frac{1}{96}+\frac{1}{95}-\frac{1}{94}-\frac{1}{93}\right)=0\)

\(\Leftrightarrow\) \(\left(x+97\right)=0\) ( \(\frac{1}{96}+\frac{1}{95}-\frac{1}{94}-\frac{1}{93}\)) \(\ne0\)

\(\Leftrightarrow\)\(x=-97\)

Vậy \(x=-97\)

b. Có : \(\frac{\left(x+2\right)^2}{2x-3}-1\)= \(\frac{x^2+10}{2x-3}\)

ĐKXĐ: \(x\ne\frac{3}{2}\)

\(\Leftrightarrow\) \(\left(x+2\right)^2-1\left(2x-3\right)=x^2+10\)

\(\Leftrightarrow\) \(x^2+4x+4-2x+3=x^2+10\)

\(\Leftrightarrow\)\(x^2+2x+7=x^2+10\)

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

\(\Leftrightarrow\) \(2x=3\)

\(\Leftrightarrow\) \(x=\frac{3}{2}\) ( loại)

Vậy phương trình vô nghiệm \(S=\varnothing\)

đúng rồi đó