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\(5\left(x-3\right)+\left(x-2\right)\left(5x-1\right)=5x^2\)
\(\Leftrightarrow5x-15-\left(5x^2-11x+2\right)=5x^2\)
\(\Leftrightarrow5x-15-5x^2+11x-2=5x^2\)
\(\Leftrightarrow-10x^2+16x-17=0\)
\(\cdot\Delta=16^2-4.\left(-10\right).\left(-17\right)=-304< 0\)
Vậy pt vô nghiệm
a) \(A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2x+10}{\left(x+5\right)\left(x-5\right)}\)
\(A=\dfrac{x-5+2x+10-2x-10}{\left(x+5\right)\left(x-5\right)}=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}=\dfrac{1}{x+5}\)
b) \(A=-3\Rightarrow\dfrac{1}{x+5}=-3\)
\(\Leftrightarrow x+5=-\dfrac{1}{3}\Leftrightarrow x=-\dfrac{1}{3}-5=\dfrac{-16}{3}\)
\(9x^2-42x+49=\left(3x-7\right)^2=\left(3.\dfrac{-16}{3}-7\right)^2=\left(-23\right)^2=529\) \(\left(x=\dfrac{-16}{3}\right)\)
a) \(\left(7-14x\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}7-14x=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}14x=7-0\\x=2\end{cases}\Leftrightarrow}}\orbr{\begin{cases}14x=7\\x=2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=2\end{cases}}}\)
Vậy \(x\in\left\{\frac{1}{2};2\right\}\)
b) \(\left(2x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=-1\\x=3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}}\)
Vậy \(x\in\left\{-\frac{1}{2};3\right\}\)
= x3 + 33 -x(x2 -1) -27 =0 ( tổng các lập phuong)
x =0
CX100%
\(x^5+x^4+x^3+x^2+x+1=0\)
\(\Rightarrow x^4\left(x+1\right)+x^2\left(x+1\right)+\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x^4+x^2+1\right)=0\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=-1\)
\(x^5-x^3+x^2-1=x^3\left(x^2-1\right)+\left(x^2-1\right)=\left(x^2-1\right)\left(x^3+1\right)=\left(x-1\right)\left(x+1\right)^2\left(x^2-x+1\right)\)