Ta có: \(C=\dfrac{2x+1}{x^2+x-2}=\dfrac{2x+1}{\left(x-1\right)\left(x+2\right)}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne1\\x\ne-2\end{matrix}\right.\) 

\(\left|2x+5\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+5=7\left(x\ge-\dfrac{5}{2}\right)\\2x+5=-7\left(x< -\dfrac{5}{2}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-12\end{matrix}\right.\) 

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(ktm\right)\\x=-6\left(tm\right)\end{matrix}\right.\)

Thay x=-6 vào C ta có:

\(C=\dfrac{2\cdot-6+1}{\left(-6\right)^2+\left(-6\right)-2}=\dfrac{-12+1}{36-6-2}=\dfrac{-11}{28}\)

(2x^2-3x+1) (x^2-5)-(x^2-x) (2x^2-x-10)=5

<=>2x⁴-3x³+x²-10x²+15x-5-(2x⁴-x³-10x²-2x³+x²+10x)=5

<=>2x⁴-3x³+x²-10x²+15x-5-2x⁴+x³+10x²+2x³-x²-10x=5

<=>5x-5=5

<=>5x=10

<=>x=2

Ta có: \(\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-3}{5}\)

nên \(\dfrac{2x-2}{4}=\dfrac{y+1}{3}=\dfrac{z-3}{5}\)

mà 2x+y-z=0

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{2x-2}{4}=\dfrac{y+1}{3}=\dfrac{z-3}{5}=\dfrac{2x+y-z-2+1+3}{4+3-5}=\dfrac{2}{2}=1\)

Do đó: x=3; y=2; z=8

`Answer:`

`1/5+2/7-1<x<\frac{13}{3}+6/5+\frac{4}{15}`

`VT =1/5+2/7-1=\frac{17}{35}-1=\frac{-18}{35}`

`VP=\frac{13}{3}+6/5+\frac{4}{15}=\frac{83}{15}+\frac{4}{15}=\frac{203}{35}`

`=>\frac{-18}{35}<x<\frac{203}{35}`

`=>-18<x<203`

Vậy `-18<x<203` với `x\inZZ`

\(\dfrac{1}{5}+\dfrac{2}{7}-1< x< \dfrac{13}{3}+\dfrac{6}{5}+\dfrac{4}{15}\)

\(\Leftrightarrow\dfrac{7}{35}+\dfrac{10}{35}-\dfrac{35}{35}< x< \dfrac{65}{15}+\dfrac{18}{15}+\dfrac{4}{15}\)

\(\Leftrightarrow\dfrac{-18}{35}< x< \dfrac{29}{5}\)

\(\Leftrightarrow\dfrac{-18}{35}< \dfrac{35x}{35}< \dfrac{203}{35}\)

\(\Leftrightarrow-18< 35x< 203\)

\(\Leftrightarrow x\in\left\{0;1;2;3;4;5\right\}\)

\(x^2+\frac{1}{x^2}=7\Leftrightarrow\left(x+\frac{1}{x}\right)^2-2=7\Leftrightarrow\left(x+\frac{1}{x}\right)^2=9\Leftrightarrow x+\frac{1}{x}=3\)(vì x>0)

<=>\(\left(x+\frac{1}{x}\right)^3=27\Leftrightarrow x^3+3\left(x+\frac{1}{x}\right)+\frac{1}{x^3}=27\Leftrightarrow x^3+\frac{1}{x^3}+3.3=27\Leftrightarrow x^3+\frac{1}{x^3}=18\)

Xét \(\left(x+\frac{1}{x}\right)\left(x^4+\frac{1}{x^4}\right)=x^5+x^3+\frac{1}{x^3}+\frac{1}{x^5}=x^5+\frac{1}{x^5}+18\)

Mặt khác: 

\(\left(x+\frac{1}{x}\right)\left(x^4+\frac{1}{x^4}\right)=\left(x+\frac{1}{x}\right)\left[\left(x^2+\frac{1}{x^2}\right)^2-2\right]=\left(x+\frac{1}{x}\right)\left(7^2-2\right)=3.47=141\)

=>\(x^5+\frac{1}{x^5}+18=141\Leftrightarrow x^5+\frac{1}{x^5}=123\)

Cảm ơn bạn rất nhiều kết bạn nha!

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