Câu 4:

giả sử: \(\left\{{}\begin{matrix}x\ne y\\x>0,y>0\end{matrix}\right.\) ,tồn tại đẳng thức:

\(\dfrac{1}{x}=\dfrac{1}{x-y}+\dfrac{1}{y}\)

\(\Leftrightarrow\)\(\dfrac{1}{x-y}+\dfrac{1}{y}-\dfrac{1}{x}=0\)

\(\Leftrightarrow\)\(\dfrac{x.y}{x.y\left(x-y\right)}+\dfrac{x.\left(x-y\right)}{x.y.\left(x-y\right)}-\dfrac{y\left(x-y\right)}{x.y.\left(x-y\right)}=0\)

\(\Leftrightarrow\)\(\dfrac{xy+x^2-xy-xy+y^2}{xy\left(x-y\right)}=0\)

\(\Leftrightarrow x^2-xy+y^2=0\)(vô lý)

ta có \(x^2-xy+y^2=x^2-\dfrac{1}{2}xy-\dfrac{1}{2}xy+\dfrac{1}{4}y^2+\dfrac{3}{4}y^2\\ =x\left(x-\dfrac{1}{2}y\right)-\dfrac{1}{2}y\left(x-\dfrac{1}{2}y\right)+\dfrac{3}{4}y^2\\ =\left(x-\dfrac{1}{2}y\right)\left(x-\dfrac{1}{2}y\right)+\dfrac{3}{4}y^2\\ =\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2\)

ta có\(\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2=0\\ \Leftrightarrow\left(x-\dfrac{1}{2}y\right)^2=\dfrac{3}{4}y^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}y\\y=0\end{matrix}\right.\Leftrightarrow x=y=0\)

x=y=0(trái với GT ban đầu\(\left\{{}\begin{matrix}x\ne y\\x>0,y>0\end{matrix}\right.\))=> không tồn tại đẳng thức

\(\dfrac{1}{x}=\dfrac{1}{x-y}+\dfrac{1}{y}\)

câu 2:đặt S=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{98^2}+\dfrac{1}{99^2}\)

ta có với a>1 thì:

\(a^2=a.a>\left(a-1\right).a\\ \Rightarrow\dfrac{1}{a^2}=\dfrac{1}{a.a}< \dfrac{1}{\left(a-1\right).a}\\ =\dfrac{1}{a-1}-\dfrac{1}{a}\)

nên

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}=\dfrac{1}{1}-\dfrac{1}{2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\)

\(\dfrac{1}{4^2}< \dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4}\)

....

\(\dfrac{1}{98^2}< \dfrac{1}{97.98}=\dfrac{1}{97}-\dfrac{1}{98}\)

\(\dfrac{1}{99^2}< \dfrac{1}{98.99}=\dfrac{1}{98}-\dfrac{1}{99}\)

cộng vế theo vế ta được:

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{98^2}+\dfrac{1}{99^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}\)

\(\Rightarrow S< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}\\ \Rightarrow S< \dfrac{1}{1}-\dfrac{1}{99}< 1\left(đ.p.cm\right)\)

2b.

\(H\left(x\right)=x^2+2x+\dfrac{19}{3}\\ =x^2+x+x+1-1+\dfrac{19}{3}\\ =x\left(x+1\right)+\left(x+1\right)+\dfrac{16}{3}\\ =\left(x+1\right)\left(x+1\right)+\dfrac{16}{3}=\left(x+1\right)^2+\dfrac{16}{3}\)

ta có \(\left(x+1\right)^2\ge0\forall x\Leftrightarrow\left(x+1\right)^2+\dfrac{16}{3}>0\forall x\)

\(\Rightarrow H\left(x\right)\) >0\(\forall x\)

hay \(H\left(x\right)\)không có nghiệm

\(2a.\\ A\left(x\right)=x^5-3x^2+2x^4-x^2+19x-\dfrac{2}{3}\\ =x^5+2x^4-3x^3-x^2+19x-\dfrac{2}{3}\)\(B\left(x\right)=2x^4+x^5-3x^3-2x^2+17x-7\\ =x^5+2x^4-3x^3-2x^2+17x-7\)

\(H\left(x\right)=A\left(x\right)-B\left(x\right)\\ =\left(x^5+2x^4-3x^3-x^2+19x-\dfrac{2}{3}\right)-\left(x^5+2x^4-3x^3-2x^2+17x-7\right)\\ =x^5+2x^4-3x^3-x^2+19x-\dfrac{2}{3}-x^5-2x^4+3x^3+2x^2-17x+7\\ =x^2+2x+\dfrac{19}{3}\)

câu 1b.

trước khi làm bài này có chú ý này:\(0^n=0\)với n\(\ne0\) và \(a^0=1\)với a\(\ne0\)

đặt: \(t=\left(x-5\right)\Rightarrow\left\{{}\begin{matrix}\left(x-5\right)^{x+1}=\left(x-5\right)^{x-5+6}=t^{t+6}\\\left(x-5\right)^{x+2015}=\left(x-5\right)^{x-5+2020}=t^{t+2020}\end{matrix}\right.\)

=> \(\left(x-5\right)^{x+1}-\left(x-5\right)^{x+2015}=0\)

\(\Leftrightarrow\)\(t^{t+6}-t^{t+2020}=0\Leftrightarrow t^{t+6}\left(1-t^{2014}\right)=0\Leftrightarrow\left[{}\begin{matrix}t^{t+6}=0^{t+6}\\1-t^{2014}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=0\\t^{2014}=1=1^{2014}\Rightarrow t=1\end{matrix}\right.\)với t=0 => x-5=0=> x=5

với t=1=> x-5=1=>x=6

câu 1.

đặt A=\(\dfrac{15}{11.14}+\dfrac{15}{14.17}+...+\dfrac{15}{65.68}+\dfrac{15}{68.71}\)

xét \(\dfrac{A}{3}\)=\(\dfrac{15}{3.11.14}+\dfrac{15}{3.14.17}+...+\dfrac{15}{3.65.68}+\dfrac{15}{3.68.71}\)

ta có:+ \(\dfrac{15}{3.11.14}=\dfrac{15}{3}\left(\dfrac{1}{11}-\dfrac{1}{14}\right)=\dfrac{15}{3.11}-\dfrac{15}{3.14}\)

tương tự ta có:

+\(\dfrac{15}{3.11.14}=\dfrac{15}{3.11}-\dfrac{15}{3.14}\)

+\(\dfrac{15}{3.14.17}=\dfrac{15}{3.14}-\dfrac{15}{3.17}\)

....

+\(\dfrac{15}{3.65.68}=\dfrac{15}{3.65}-\dfrac{15}{3.68}\)

+\(\dfrac{15}{3.68.71}=\dfrac{15}{3.68}-\dfrac{15}{3.71}\)

cộng vế theo vế ta đc:

\(\dfrac{15}{3.11.14}+\dfrac{15}{3.14.17}+...+\dfrac{15}{3.65.68}+\dfrac{15}{3.68.71}\)

=\(\dfrac{15}{3.11}-\dfrac{15}{3.14}+\dfrac{15}{3.14}-\dfrac{15}{3.17}+...+\dfrac{15}{3.65}-\dfrac{15}{3.68}+\dfrac{15}{3.68}-\dfrac{15}{3.71}=\dfrac{15}{3.11}-\dfrac{15}{3.71}\)

=> \(\dfrac{A}{3}\)=\(\dfrac{15}{3.11}-\dfrac{15}{3.71}\)

=> A= \(\dfrac{15}{11}-\dfrac{15}{17}=\dfrac{90}{187}\)

\(\left(x-2\right)\left(x-4\right)< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2< 0\\x-4>0\end{matrix}\right.=>4< x< 2\left(1\right)\\\left\{{}\begin{matrix}x-2>0\\x-4< 0\end{matrix}\right.=>2< x< 4\left(2\right)}\end{matrix}\right.\)(1 ) vô lý=> loại

=> (x-2).(x-4)<0 <=> 2<x<4

b. ta có\(x^2+1>0\forall x\)

=>(x² -1).(x²+1)<0 <=> (x² -1)<0 <=> x²<1

<=> -1<x<1

câu c bạn làm tương tự

cho minh xin yeu cau de bai