x(x-1)+1/x(1/x-1)=0. Tìm x giúp mik vs😭😭😭😭 Đang cần gấp
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\(\dfrac{x}{6}=\dfrac{7}{4}\Rightarrow x=\dfrac{6\cdot7}{4}=\dfrac{21}{2}\\ \dfrac{3}{x}=\dfrac{21}{17}\Rightarrow x=\dfrac{3\cdot17}{21}=\dfrac{17}{7}\)
\(12.\left(x-1\right)=0\)
\(x-1=0:12\)
\(x-1=0\)
\(x=0+1\)
\(x=1\)
Vì | x+5 | >=0 với mọi x
| y - 4 | >=0 với mọi y
=> |x +5| +|y-4|>=0
Mà |x+5|+|y-4|<=0
=> \(\hept{\begin{cases}x+5=0\\y-4=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-5\\y=4\end{cases}}\)
vậy ...........
hok tốt
Ta có: \(\hept{\begin{cases}\left|x+5\right|\ge0\forall x\\\left|y-4\right|\ge0\forall y\end{cases}}\)
\(\Rightarrow\left|x+5\right|+\left|y-4\right|\ge0\)
\(\Rightarrow\left|x+5\right|+\left|y+4\right|=0\)
\(\Leftrightarrow\hept{\begin{cases}\left|x+5\right|=0\\\left|y-4\right|=0\end{cases}\Rightarrow\hept{\begin{cases}x+5=0\\y-4=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-5\\y=4\end{cases}}}\)
vậy x =-5; y = 4
hok tốt!!
\(\left(2.x+\frac{1}{3}\right)^2=\frac{16}{25}\)
\(\Leftrightarrow2.x+\frac{1}{3}=\pm\sqrt{\frac{16}{25}}\)
\(\Leftrightarrow2.x+\frac{1}{3}=\pm\frac{4}{5}\)
\(\Leftrightarrow\orbr{\begin{cases}2.x+\frac{1}{3}=\frac{4}{5}\\2.x+\frac{1}{3}=-\frac{4}{5}\end{cases}}\Leftrightarrow\orbr{\begin{cases}2.x=\frac{7}{15}\\2.x=-\frac{17}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{30}\\x=-\frac{17}{30}\end{cases}}\)
\(\left(2.x+\frac{1}{3}\right)^2=\frac{16}{25}\)
\(\left(2.x+\frac{1}{3}\right)^2=\left(\frac{4}{5}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}2.x+\frac{1}{3}=\frac{4}{5}\\2.x+\frac{1}{3}=\frac{-4}{5}\end{cases}\Rightarrow\orbr{\begin{cases}2.x=\frac{4}{5}-\frac{1}{3}\\2.x=\frac{-4}{5}-\frac{1}{3}\end{cases}\Rightarrow}\orbr{\begin{cases}2.x=\frac{12}{15}-\frac{5}{15}\\2.x=\frac{-12}{15}-\frac{5}{15}\end{cases}\Rightarrow}\orbr{\begin{cases}2.x=\frac{7}{15}\\2.x=\frac{-17}{15}\end{cases}}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{15}:2\\x=\frac{-17}{15}:2\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{15}.\frac{1}{2}\\x=\frac{-17}{15}.\frac{1}{2}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{7}{30}\\x=\frac{-17}{30}\end{cases}}}\)
Vậy \(x=\frac{7}{30}\)hoặc \(x=\frac{-17}{30}\)
a) \(x^3+2x^2-4x+1\)
\(=\left(x^3+3x^2-x\right)-\left(x^2+3x-1\right)\)
\(=x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\)
\(=\left(x-1\right)\left(x^2+3x-1\right)\)
c) cho da thuc P(x) =2x^4-7x^3 -2x^2 +13x +6? | Yahoo Hỏi & Đáp
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