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\(91-3.\left(7+x\right)=64\)
\(3.\left(7+x\right)=91-64\)
\(3.\left(7+x\right)=27\)
\(7+x=27:3\)
\(7+x=9\)
\(x=9-7\)
\(x=2\)
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C=\(^{5x^2+20x+2010}\)
Vì C \(\ge\)2010
Nên GTNN của C là 2010
Khi \(5x^2+20x=0\)
x=0
A=XÉT \(X\le201Ó\)
TA ĐC X-2010+X-2011=2010-X+2011-X
<=>4021-2X
=>CÓ X\(\le\)2010 =>-X\(\le\) 2010 =>-2X\(\ge\)-4021
DẤU '' ='' XẢY RA KHI X=2010
B.,
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\(3-\frac{x}{5}-x=\frac{x}{x-1}\)
\(\Rightarrow\frac{15\left(x-1\right)}{5\left(x-1\right)}-\frac{x\left(x-1\right)}{5\left(x-1\right)}-\frac{5x\left(x-1\right)}{5\left(x-1\right)}=\frac{5x}{5\left(x-1\right)}\)
\(\Rightarrow15\left(x+1\right)-x\left(x-1\right)-5x\left(x-1\right)=5x\)
\(\Rightarrow15x+15-x^2+x-5x^2+5x=5x\)
Bạn tự làm tiếp theo ha
\(\frac{3-x}{5-x}=\frac{x}{x+1}\)
\(\left(3-x\right)\left(x+1\right)=\left(5-x\right)x\)
\(3\left(x+1\right)-x\left(x+1\right)=5x-x^2\)
\(3x+3-x^2-x=5x-x^2\)
\(2x+3-x^2=5x-x^2\)
\(2x+3=5x\)
\(3=5x-2x\)
\(3x=3\)
\(x=1\)
Vậy x = 1
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|x + 1| + |(x - 1)(x + 1)| = 0
|x + 1| + |x2 - 1| = 0
Vì |x + 1| ≥ 0 ; |x2 - 1| ≥ 0 với mọi x
=> |x + 1| + |x2 - 1| ≥ 0
Mà |x + 1| + |x2 - 1| = 0 => |x + 1| = 0 ; |x2 - 1| = 0
=> x + 1 = 0; x2 = 1 => x = - 1
Vậy x = - 1
TH1 : \(91-3x< 7+x\Rightarrow3x+x>91-7\Rightarrow4x>84\Rightarrow x>21\left(1\right)\)
TH2 : \(7+x\ge64\Rightarrow x\ge57\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow x\ge57\)
91 - 3\(x\) < 7 + \(x\) ≥ 64
⇒ \(\left\{{}\begin{matrix}91-3x< 7+x\\7+x\ge64\end{matrix}\right.\)
\(\left\{{}\begin{matrix}7+x+3x>91\\x\ge64-7\end{matrix}\right.\)
\(\left\{{}\begin{matrix}4x>91-7\\x\ge64-7\end{matrix}\right.\)
\(\left\{{}\begin{matrix}4x>84\\x\ge57\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x>84:4\\x\ge57\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x>21\\x\ge57\end{matrix}\right.\)
\(x\ge\) 57