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a, \(\frac{x+9}{10}+\frac{x+10}{9}=\frac{9}{x+10}+\frac{10}{x+9}\)(1)
ĐKXĐ: \(\hept{\begin{cases}x+9\ne0\\x+10\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne-9\\x\ne-10\end{cases}}}\)
(1)\(\Leftrightarrow\frac{9.\left(x+9\right)}{90}+\frac{10.\left(x+10\right)}{90}=\frac{9.\left(x+9\right)}{\left(x+9\right)\left(x+10\right)}+\frac{10.\left(x+10\right)}{\left(x+9\right)\left(x+10\right)}\)
\(\Leftrightarrow9.\left(x+9\right)+10.\left(x+10\right)=9.\left(x+9\right)+10.\left(x+10\right)\)
\(\Leftrightarrow9x+81+10x+100=9x+81+10x+100\)
\(\Leftrightarrow9x+10x-9x-10x=81+100-81-100\)
\(\Leftrightarrow0x=0\)
\(\Rightarrow x\in R\)trừ -9 và -10
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1) Ta có: \(4x+8=3x-1\)
\(\Leftrightarrow4x-3x=-1-8\)
\(\Leftrightarrow x=-9\)
2) Ta có: \(10-5\left(x+3\right)>3\left(x-1\right)\)
\(\Leftrightarrow10-5x-15-3x+3>0\)
\(\Leftrightarrow-8x>2\)
hay \(x< \dfrac{-1}{4}\)
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\(\dfrac{2x-1}{5}-\dfrac{4x}{3}=2x-\dfrac{x}{10}\\ \Leftrightarrow\dfrac{6\left(2x-1\right)}{30}-\dfrac{40x}{30}=\dfrac{60x}{30}-\dfrac{3x}{30}\\ \Leftrightarrow12x-6-40x=60x-3x\\ \Leftrightarrow-28x-6=57x\\ \Leftrightarrow57x+28x+6=0\\ \Leftrightarrow85x=-6\\ \Leftrightarrow x=-\dfrac{6}{85}\)
Làm nhầm làm lại nhé
Ta có:
|x + 10|5 + |x - 10|5
= |x + 10|5 + |10 - x|5
= |x5 + 50x4 + 1000x3 + 10000x2 + 50000x + 100000| + |-x5 + 50x4 - 1000x3 + 10000x2 - 50000x + 100000|
\(\ge\)|100x4 + 20000x2 + 200000|
\(\ge\)|200000| = 200000
Dấu = xảy ra khi x = 0
Ta có:
|x + 10|5 + |x - 10|5
= |x + 10|5 + |10 - x|5
= |x5 + 50x4 + 1000x3 + 10000x2 + 50000x + 1000000| + |-x5 + 50x4 - 1000x3 + 10000x2 - 50000x + 1000000|
\(\ge\)|100x4 + 20000x2 + 2000000|
\(\ge\)|2000000| = 2000000
Dấu = xảy ra khi x = 0