Bài 2:
a: ||x-3|+7|=10
=>\(\left[{}\begin{matrix}\left|x-3\right|+7=10\\\left|x-3\right|+7=-10\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}\left|x-3\right|=3\\\left|x-3\right|=-17\left(vôlý\right)\end{matrix}\right.\)
=>\(\left|x-3\right|=3\)
=>\(\left[{}\begin{matrix}x-3=3\\x-3=-3\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=6\\x=0\end{matrix}\right.\)
c: ĐKXĐ: x<>1
\(\dfrac{x-1}{-4}=\dfrac{-9}{x-1}\)
=>\(\left(x-1\right)\cdot\left(x-1\right)=\left(-9\right)\cdot\left(-4\right)\)
=>\(\left(x-1\right)^2=36\)
=>\(\left[{}\begin{matrix}x-1=6\\x-1=-6\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-5\left(nhận\right)\end{matrix}\right.\)
d: \(\dfrac{x+2}{98}+\dfrac{x+4}{96}=\dfrac{x+3}{97}+\dfrac{x+5}{95}\)
=>\(\dfrac{x+2}{98}+1+\dfrac{x+4}{96}+1=\dfrac{x+3}{97}+1+\dfrac{x+5}{95}+1\)
=>\(\dfrac{x+2+98}{98}+\dfrac{x+4+96}{96}=\dfrac{x+3+97}{97}+\dfrac{x+5+95}{95}\)
=>\(\dfrac{x+100}{98}+\dfrac{x+100}{96}-\dfrac{x+100}{97}-\dfrac{x+100}{95}=0\)
=>\(\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{97}-\dfrac{1}{95}\right)=0\)
=>x+100=0
=>x=-100
bài 3:
a: \(A=\dfrac{x^2+4}{x^2+1}\)
\(=\dfrac{x^2+1+3}{x^2+1}\)
\(=1+\dfrac{3}{x^2+1}\)
\(x^2>=0\forall x\)
=>\(x^2+1>=1\forall x\)
=>\(\dfrac{3}{x^2+1}< =\dfrac{3}{1}=3\forall x\)
=>\(\dfrac{3}{x^2+1}+1< =3+1=4\forall x\)
=>\(A< =4\forall x\)
Dấu '=' xảy ra khi x=0
b: \(B=\dfrac{2\left(x-1\right)^2+10}{\left(x-1\right)^2+3}\)
\(=\dfrac{2\left(x-1\right)^2+6+4}{\left(x-1\right)^2+3}\)
\(=\dfrac{2\left[\left(x-1\right)^2+3\right]}{\left(x-1\right)^2+3}+\dfrac{4}{\left(x-1\right)^2+3}\)
\(=2+\dfrac{4}{\left(x-1\right)^2+3}\)
\(\left(x-1\right)^2\ge0\forall x\)
=>\(\left(x-1\right)^2+3>=3\forall x\)
=>\(\dfrac{4}{\left(x-1\right)^2+3}< =\dfrac{4}{3}\forall x\)
=>\(\dfrac{4}{\left(x-1\right)^2+3}+2< =\dfrac{4}{3}+2=\dfrac{10}{3}\forall x\)
=>\(B< =\dfrac{10}{3}\forall x\)
Dấu '=' xảy ra khi x=1
c: \(\left|2x-1\right|>=0\forall x\)
\(\left(x-y\right)^4>=0\forall x\),y
=>\(\left|2x-1\right|+\left(x-y\right)^4>=0\forall x,y\)
=>\(-\left|2x-1\right|-\left(x-y\right)^4< =0\forall x,y\)
=>\(-\left|2x-1\right|-\left(x-y\right)^4+7< =7\forall x,y\)
=>\(C< =7\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}2x-1=0\\x-y=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x=1\\x=y\end{matrix}\right.\Leftrightarrow x=y=\dfrac{1}{2}\)
