Câu này tớ giải hơn 10 lần rồi cậu ( ko xàm :) 

\(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)

\(\Leftrightarrow\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)=\left(\frac{x+6}{94}+1\right)+\left(\frac{x+8}{92}+1\right)\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)

Vì \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\)

Do đó \(x+100=0\Leftrightarrow x=-100\)

Vậy pt có nghiệm : x=-100

Ta có : \(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)

=> \(\frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+8}{92}+1\)

=> \(\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)

=> \(\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)

=> \(\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)

Vì \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\)

=> x + 100 = 0

=> x = - 100

Vậy x = - 100

\(\Leftrightarrow\frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+8}{98}+1\)

\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}\right)=\left(x+100\right)\left(\frac{1}{94}+\frac{1}{92}\right)\)

Do \(\frac{1}{98}+\frac{1}{96}\ne\frac{1}{94}+\frac{1}{92}\)\(\Rightarrow x+100=0\)

                                                    \(\Leftrightarrow x=-100\)

           Vậy \(x=-100\)

~Hok Tốt~

Điều kiện: \(x\ne2\)

Pt: \(\Leftrightarrow2^{\dfrac{3x}{x+2}}=2^2.3^{4-x}\Leftrightarrow3^{\dfrac{x-4}{x+2}}=3^{4-x}\)

\(\Leftrightarrow\dfrac{x-4}{x+2}\log_32=4-x\)

\(\Leftrightarrow\left(x-4\right)\left(x+2+\log_32\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2-\log_32\end{matrix}\right.\)

\(\sqrt{x^2-4x+4}=2-x\)

\(\Leftrightarrow\sqrt{\left(x-2\right)^2}=2-x\)

=>\(\left|x-2\right|=2-x\)

=>x-2<=0

=>x<=2

\(\sqrt{x^2-4x+4}=2-x\left(x\le2\right)\)

\(\Leftrightarrow\sqrt{x^2-2\cdot x\cdot2+2^2}=2-x\)

\(\Leftrightarrow\sqrt{\left(x-2\right)^2}=2-x\)

\(\Leftrightarrow\left|x-2\right|=2-x\)

+) \(x-2=2-x\)

\(\Leftrightarrow x+x=2+2\)

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=2\left(tm\right)\)

+) \(x-2=-\left(2-x\right)\)

\(\Leftrightarrow x-2=x-2\)

\(\Leftrightarrow0=0\) (luôn đúng)

Vậy phương trình thỏa mãn với mọi \(x\le2\)

a: Khi m=1 thì phương trình sẽ là x^2-2x-3=0

=>x=3 hoặc x=-1

b: Δ=(m+1)^2-4(m-4)

=m^2+2m+1-4m+16

=m^2-2m+17

=(m-1)^2+16>=16>0

=>Phương trình luôn có hai nghiệm phân biệt

x1+x2=m+1;x2x1=m-4

(x1^2-mx1+m)(x2^2-mx2+m)=2

=>(x1*x2)^2-m*x2*x1^2+m*x1^2-m*x1*x2^2+m*x1*x2-m^2*x1+m*x2^2-m^2*x2+m^2=2

=>(x1*x2)^2-m*x1*x2(x1+x2)+mx1^2+m*(m-4)-m^2*x1+m*x2^2-m^2*x2+m^2=2

=>(m-4)^2-m*(m-4)(m+1)+m(m-4)-m^2(x1+x2)+m*(x1^2+x2^2)+m^2=2

=>(m-4)^2-m(m^2-3m-4)+m^2-4m-m^2(m+1)+m*[(m+1)^2-2(m-4)]+m^2=2

=>m^2-8m+16-m^3+3m^2+4m+m^2-4m-m^3-m^2+m^2+m[m^2+2m+1-2m+8]=2

=>-2m^3+3m^2-8m+16+m^3+9m-2=0

=>-m^3+3m^2+m+14=0

=>\(m\simeq4,08\)