Vì \(\left|x+\frac{25}{47}\right|\ge0\forall x\inℝ\); \(\left|y-\frac{9}{17}\right|\ge0\forall y\inℝ\)
\(\Rightarrow\left|x+\frac{25}{47}\right|+\left|y-\frac{9}{17}\right|\ge0\forall x;y\inℝ\)
Dấu " = " xảy ra <=> \(\hept{\begin{cases}x+\frac{25}{47}=0\\y-\frac{9}{17}=0\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{-25}{47}\\y=\frac{9}{17}\end{cases}}\)
\(\frac{59-x}{41}+\frac{57-x}{43}+\frac{55-x}{45}+\frac{53-x}{47}+\frac{51-x}{49}=-5\) tìm x
\(\frac{x-5}{1990}+\frac{x-15}{1980}+\frac{x-25}{1970}=\frac{x-1990}{5}+\frac{x-1980}{15}+\frac{x-1970}{25}\) tìm x
\(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+17}{83}+\frac{x+116}{4}=0\) tìm x
\(\frac{x}{50}+\frac{x-1}{49}+\frac{x-2}{48}+\frac{x-3}{47}+\frac{x-150}{25}=0\)
Tìm x
Tìm x, y, z biết: \(\frac{3x+y}{47}=\frac{x+y}{-17}=\frac{-2}{x^2}=\frac{-xz^2-yz^2}{z^2+1}\left(x\ne0\right)\)
cho các số x,y,z khác 0 va thoả mãn :\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0.t\text{ính}gi\text{á}tr\text{ị}bi\text{ểu}th\text{ức}P=\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)
tìm x , y\(\left|\frac{1}{2}-\frac{1}{3}+x\right|=-\frac{1}{4}-\left|y\right|\)
\(\left|x-y\right|+\left|y+\frac{9}{25}\right|=0\)
|x-y|+|y+\(\frac{9}{25}\)|=0
( Bài hơi dài , mong mọi người giúp ạ ! )
\(\text{Tính P :}\)
\(P=\frac{0,8:\left(\frac{4}{5}\times1,25\right)}{0,64-\frac{1}{25}}+\frac{\left(1,08-\frac{2}{25}\right):\frac{4}{7}}{\left(6\frac{5}{9}-3\frac{1}{4}\right)\times2\frac{2}{17}}+\left(1,2\times0,5\right):\frac{4}{5}\)
Cho \(\frac{x+16}{9}=\frac{y-25}{16}=\frac{z+9}{25}\)và \(\frac{9-x}{7}+\frac{11-x}{9}=2\),Tính x + y + z
\(|x-y|+|y+\frac{9}{25}|=0\)
Tìm x
