\(\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)

\(\Leftrightarrow\dfrac{x+2+98}{98}+\dfrac{x+3+97}{97}-\dfrac{x+4+96}{96}-\dfrac{x+5+95}{95}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\right)=0\)

\(\Rightarrow x+100=0\)

\(\Leftrightarrow x=-100\)

Vậy...

a: 78x(x-97)-x+97=0

=>(x-97)(78x-1)=0

=>\(\left[{}\begin{matrix}x=97\\x=\dfrac{1}{78}\end{matrix}\right.\)

b: \(\dfrac{2}{3}x\left(x^2-4\right)=0\)

=>\(x\left(x-2\right)\left(x+2\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)

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

=>\(x^2+4x+4-x^2+4=0\)

=>4x+8=0

=>x+2=0

=>x=-2

\(a,78x\left(x-97\right)-x+97=0\)

\(\Leftrightarrow78x\left(x-97\right)-\left(x-97\right)=0\)

\(\Leftrightarrow\left(x-97\right)\left(78x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-97=0\\78x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=97\\x=\dfrac{1}{78}\end{matrix}\right.\)

\(b,\dfrac{2}{3}x\left(x^2-4\right)=0\)

\(\Leftrightarrow\dfrac{2}{3}x\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=0\\x-2=0\\x+2=0\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left(x+2\right)\left[\left(x+2\right)-\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(x+2\right)\left(x+2-x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\cdot4=0\)

\(\Leftrightarrow x+2=0\)

\(\Leftrightarrow x=-2\)

\(\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{\left(x+3\right)}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)

\(\Rightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}=\dfrac{x+100}{96}+\dfrac{x+100}{95}\)

\(\Rightarrow\left(x+100\right).\left(\dfrac{1}{98}+\dfrac{1}{97}\right)=\left(x+100\right).\left(\dfrac{1}{96}+\dfrac{1}{95}\right)\)

\(\Rightarrow\dfrac{1}{98}+\dfrac{1}{97}=\dfrac{1}{96}+\dfrac{1}{95}\) (Vô lí)

Vậy PTVN

a) \(\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)

\(\Rightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}=\dfrac{x+100}{96}+\dfrac{x+100}{95}\)

\(\Rightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}-\dfrac{x+100}{96}-\dfrac{x+100}{95}=0\)

\(\Rightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\right)=0\)

Vì \(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\ne0\) nên \(x+100=0\Leftrightarrow x=-100\)

b) \(\dfrac{x+1}{1998}+\dfrac{x+2}{1997}=\dfrac{x+3}{1996}+\dfrac{x+4}{1995}\)

\(\Rightarrow\dfrac{x+1}{1998}+1+\dfrac{x+2}{1997}+1=\dfrac{x+3}{1996}+1+\dfrac{x+4}{1995}+1\)

\(\Rightarrow\dfrac{x+1999}{1998}+\dfrac{x+1999}{1997}=\dfrac{x+1999}{1996}+\dfrac{x+1999}{1995}\)

\(\Rightarrow\dfrac{x+1999}{1998}+\dfrac{x+1999}{1997}-\dfrac{x+1999}{1996}-\dfrac{x+1999}{1995}=0\)

\(\Rightarrow\left(x+1999\right)\left(\dfrac{1}{1998}+\dfrac{1}{1997}-\dfrac{1}{1996}-\dfrac{1}{1995}\right)=0\)

Vì \(\dfrac{1}{1998}+\dfrac{1}{1997}-\dfrac{1}{1996}-\dfrac{1}{1995}\ne0\) nên \(x+1999=0\Leftrightarrow x=-1999\)

c) \(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\)

\(\Rightarrow\dfrac{201-x}{99}+1+\dfrac{203-x}{97}+1+\dfrac{205-x}{95}+1=0\)

\(\Rightarrow\dfrac{300-x}{99}+\dfrac{300-x}{97}+\dfrac{300-x}{95}=0\)

\(\Rightarrow\left(300-x\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)=0\)

Vì \(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\ne0\) nên \(300-x=0\Leftrightarrow x=300\)

a/Viết đề mà cx sai đc nữa: \(\left(\frac{x+2}{98}+1\right)\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)\left(\frac{x+5}{95}+1\right)\)

\(\Leftrightarrow\frac{x+100}{98}.\frac{x+100}{97}-\frac{x+100}{96}.\frac{x+100}{95}=0\)

\(\Leftrightarrow\left(x+100\right)^2\left(\frac{1}{98.97}-\frac{1}{96.95}\right)=0\)

\(\Rightarrow x=-100\)

b/\(\Leftrightarrow\left(\frac{x+1}{1998}+1\right)+\left(\frac{x+2}{1997}+1\right)=\left(\frac{x+3}{1996}+1\right)+\left(\frac{x+4}{1995}+1\right)\)

\(\Leftrightarrow\frac{x+1999}{1998}+\frac{x+1999}{1997}-\frac{x+1999}{1996}-\frac{x+1999}{1995}=0\)

\(\Leftrightarrow\left(x+1999\right)\left(...\right)=0\Rightarrow x=-1999\)

b,\(\frac{x+1}{1998}+\frac{x+2}{1997}=\frac{x+3}{1996}+\frac{x+4}{1995}\)

=>\(\frac{x+1}{1998}+1\frac{x+2}{1997}+1=\frac{x+3}{1996}+1+\frac{x+4}{1995}+1\)

\(\Leftrightarrow\)\(\frac{x+1999}{1998}+\frac{x+1999}{1997}=\frac{x+1999}{1996}+\frac{x+1999}{1995}\)

\(\Leftrightarrow\)\(\frac{x+1999}{1998}+\frac{x+1999}{1997}-\frac{x+1999}{1996}-\frac{x+1999}{1995}\)=0

\(\Leftrightarrow\)\(\left(x+1999\right)\left(\frac{1}{1998}+\frac{1}{1997}-\frac{1}{1996}-\frac{1}{1995}\right)\)=0

\(\Leftrightarrow\)x+1999=0(Vì \(\frac{1}{1998}+\frac{1}{1997}-\frac{1}{1996}-\frac{1}{1995}\ne0\))

\(\Leftrightarrow\)x=-1999

Vậy x=-1999

a: \(A=\dfrac{1}{x-1}\cdot5\sqrt{3}\cdot\left|x-1\right|\cdot\sqrt{x-1}\)

\(=\dfrac{5\sqrt{3}}{x-1}\cdot\left(x-1\right)\cdot\sqrt{x-1}=5\sqrt{3}\cdot\sqrt{x-1}\)

b: \(B=10\sqrt{x}-3\cdot\dfrac{10\sqrt{x}}{3}-\dfrac{4}{x}\cdot\dfrac{x\sqrt{x}}{2}\)

\(=10\sqrt{x}-10\sqrt{x}-\dfrac{4\sqrt{x}}{2}=-2\sqrt{x}\)

c: \(C=x-4+\left|x-4\right|\)

=x-4+x-4

=2x-8

\(\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)

\(\Rightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}=\dfrac{x+100}{96}+\dfrac{x+100}{95}\)

\(\Rightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}-\dfrac{x+100}{96}-\dfrac{x+100}{95}=0\)

\(\Rightarrow\left(x+100\right).\left(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\right)=0\)

Mà \(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\ne0\Rightarrow x+100=0\Rightarrow x=-100\)

Vậy x=-100

\(\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)

\(\Rightarrow\left[\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)\right]-\left[\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\right]=0\)

\(\Rightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}-\dfrac{x+100}{96}-\dfrac{x+100}{95}=0\)

\(\Rightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\right)=0\)

\(\Rightarrow x+100=0\Rightarrow x=-100\)

a/ \(\left\{{}\begin{matrix}xy+1+x+y=10\\\left(x+y\right)\left(xy+1\right)=1\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+y=a\\xy+1=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=10\\ab=1\end{matrix}\right.\)

Theo Viet đảo, a và b là nghiệm:

\(t^2-10t+1=0\) \(\Rightarrow\left[{}\begin{matrix}t=5+2\sqrt{6}\\t=5-2\sqrt{6}\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x+y=5+2\sqrt{6}\\xy=4-2\sqrt{6}\end{matrix}\right.\)

Theo Viet đảo, x và y là nghiệm:

\(t^2-\left(5+2\sqrt{6}\right)t+4-2\sqrt{6}=0\) (bấm máy, số xấu quá)

TH2: \(\left\{{}\begin{matrix}x+y=5-2\sqrt{6}\\xy=4+2\sqrt{6}\end{matrix}\right.\)

Ta có \(\left(5-2\sqrt{6}\right)^2-4\left(4+2\sqrt{6}\right)=33-28\sqrt{6}< 0\) nên vô nghiệm

b/ \(\left\{{}\begin{matrix}x^4+y^4=97\\xy\left(x^2+y^2\right)=78\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+y^2\right)^2-2x^2y^2=97\\xy\left(x^2+y^2\right)=78\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x^2+y^2=a>0\\xy=b\end{matrix}\right.\) với \(a\ge2b\) hệ trở thành:

\(\left\{{}\begin{matrix}a^2-2b^2=97\\ab=78\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a^2-2b^2=97\\b=\frac{78}{a}\end{matrix}\right.\)

\(\Rightarrow a^2-2\left(\frac{78}{a}\right)^2=97\)

\(\Leftrightarrow a^4-97a^2-12168=0\Rightarrow\left[{}\begin{matrix}a^2=169\\a^2=-72\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}a=13\Rightarrow b=6\\a=-13< 0\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x^2+y^2=13\\xy=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^2+y^2=13\\y=\frac{6}{x}\end{matrix}\right.\)

\(\Rightarrow x^2+\frac{36}{x^2}=13\Leftrightarrow x^4-13x^2+36=0\) \(\Rightarrow\left[{}\begin{matrix}x^2=9\\x^2=4\end{matrix}\right.\)

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