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a/ Ta có :
\(\left\{{}\begin{matrix}\left(x-1\right)^4\ge0\\\left(y-3\right)^4\ge0\end{matrix}\right.\)
Mà \(\left(x-1\right)^4+\left(y-3\right)^4=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^4=0\\\left(y-3\right)^4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)
Vậy ................
b/ Ta thấy :
\(\left\{{}\begin{matrix}\left(x+y\right)^{2006}\ge0\\2000\left|y-1\right|\ge0\end{matrix}\right.\)
Mà \(\left(x+y\right)^{2006}+2000\left|y-1\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^{2006}=0\\2000\left|y-1\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\\left|y-1\right|=0\end{matrix}\right.\)
+) \(\left|y-1\right|=0\)
\(\Leftrightarrow y-1=0\)
\(\Leftrightarrow y=1\)
Mà \(x+y=0\)
\(\Leftrightarrow x=-1\)
Vậy ........
c/ Tương tự như b
a: \(\left(x-1\right)^4+\left(y-3\right)^4=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)
b: \(\left(x+y\right)^{2006}+2000\left|y-1\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
c: \(\left|x-y-5\right|+\left(y+3\right)^{2000}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=5\\y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+5=-3+5=2\\y=-3\end{matrix}\right.\)
Bài 1 :
\(\frac{x-1}{x-5}=\frac{6}{7}\Leftrightarrow7x-7=6x-30\)
\(\Leftrightarrow x=-23\)
\(\frac{x-2}{x-1}=\frac{x+4}{x+7}\)ĐK : \(x\ne1;-7\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=\left(x+4\right)\left(x-1\right)\)
\(\Leftrightarrow x^2+5x-14=x^2+3x-4\)
\(\Leftrightarrow2x-10=0\Leftrightarrow x=5\)
B1: Đk: 5x ≥ 0 => x ≥ 0
Vì |x + 1| ≥ 0 => |x + 1| = x + 1
|x + 2| ≥ 0 => |x + 2| = x + 2
|x + 3| ≥ 0 => |x + 3| = x + 3
|x + 4| ≥ 0 => |x + 4| = x + 4
=> |x + 1| + |x + 2| + |x + 3| + |x + 4| = 5x
=> x + 1 + x + 2 + x + 3 + x + 4 = 5x
=> 4x + 10 = 5x
=> x = 10
B2: Ta có: |x - 2018| = |2018 - x|
=> A=|x + 2000| + |2018 - x| ≥ |x + 2000 + 2018 - x| = |4018| = 4018
Dấu " = " xảy ra <=> (x + 2000)(x - 2018) ≥ 0
Th1: \(\hept{\begin{cases}x+2000\ge0\\x-2018\ge0\end{cases}\Rightarrow}\hept{\begin{cases}x\ge-2018\\x\le2018\end{cases}}\Rightarrow-2018\le x\le2018\)
Th2: \(\hept{\begin{cases}x+2000\le0\\x-2018\le0\end{cases}\Rightarrow}\hept{\begin{cases}x\le-2018\\x\ge2018\end{cases}}\)(vô lý)
Vậy GTNN của A = 4018 khi -2018 ≤ x ≤ 2018
B3:
a, Vì |x + 1| ≥ 0 ; |2y - 4| ≥ 0
=> |x + 1| + |2y - 4| ≥ 0
Dấu " = " xảy ra <=> \(\hept{\begin{cases}x+1=0\\2y-4=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-1\\y=2\end{cases}}\)
Vậy...
b, Vì |x - y + 1| ≥ 0 ; (y - 3)2 ≥ 0
=> |x - y + 1| + (y - 3)2 ≥ 0
Dấu " = " xảy ra <=> \(\hept{\begin{cases}x-y+1=0\\y-3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x-y=-1\\y=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x-3=-1\\y=3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=3\end{cases}}\)
Vậy...
c, Vì |x + y| ≥ 0 ; |x - z| ≥ 0 ; |2x - 1| ≥ 0
=> |x + y| + |x - z| + |2x - 1| ≥ 0
Dấu " = " xảy ra <=> \(\hept{\begin{cases}x+y=0\\x-z=0\\2x-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+y=0\\x=z\\x=\frac{1}{2}\end{cases}\Leftrightarrow}}\hept{\begin{cases}\frac{1}{2}+y=0\\x=z=\frac{1}{2}\end{cases}\Leftrightarrow}\hept{\begin{cases}y=\frac{-1}{2}\\x=z=\frac{1}{2}\end{cases}}\)
NX:\(\left(x-1\right)^4\ge0\forall x\)
\(\left(y-3\right)^4\ge0\forall y\)
\(\Rightarrow\left(x-1\right)^4+\left(y-3\right)^4\ge0\forall x,y\)
\(\Rightarrow\left(x-1\right)^4+\left(y-3\right)^4=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)
b)làm tương tự phần a:
NX :|y-1| \(\ge\)0 với mọi y
=> 2000|y-1|\(\ge\)0 với mọi y
(x+y)^2006\(\ge\)0 với mọi x
=> 2000|y-1|+ (x+y)^2006\(\ge\)0 với mọi x,y
=> 2000|y-1|+ (x+y)^2006=0
<=> \(\left\{{}\begin{matrix}x+y=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-y\\y=1\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
c) nhận xét |x-y-5| lớn hơn hoặc bằng 0 rồi làm tương tự