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a, \(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)
=>\(2^{x-1}+\frac{5}{2}.2^{x-1}=\frac{7}{32}\)
=>\(2^{x-1}\left(1+\frac{5}{2}\right)=\frac{7}{32}\)
=>\(2^{x-1}\cdot\frac{7}{2}=\frac{7}{32}\)
=>\(2^{x-1}=\frac{1}{16}=\frac{1}{2^4}=2^{-4}\)
=>x-1=-4
=>x=-5
b, |x - 4| + |x - 10| + |x + 101| + |x + 990| + |x + 1000| = |4-x|+|10-x|+|x+101|+|x+990|+|x+1000|
Ta có: \(\left|4-x\right|\ge4-x;\left|10-x\right|\ge10-x;\left|x+990\right|\ge x+990;\left|x+1000\right|\ge x+1000\)
\(\Rightarrow\left|4-x\right|+\left|10-x\right|+\left|x+990\right|+\left|x+1000\right|\ge4-x+10-x+x+990+x+1000\)
\(\Rightarrow\left|4-x\right|+\left|10-x\right|+\left|x+101\right|+\left|x+990\right|+\left|x+1000\right|\ge2004+\left|x+101\right|\)
\(\Rightarrow2005\ge2004+\left|x+101\right|\)
\(\Rightarrow\left|x+1\right|\le1\)
\(\Rightarrow-1\le x+101\le1\)
\(\Rightarrow-102\le x\le-100\)
Vì \(x\in Z\)
\(\Rightarrow x\in\left\{-102;-101;-100\right\}\)
a)\(\left|4-x\right|+\left|x+1\right|=5\)
\(\left|4-x\right|+\left|x+1\right|\ge\left|4-x+x+1\right|=5\)
dấu = xảy ra khi (4-x).(x+1)=0
=> \(-1\le x\le4\)
b) \(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|=4\)
\(\left|x+1\right|+\left|x+4\right|=\left|x+1\right|+\left|-x-4\right|\ge\left|x+1-x-4\right|=\left|-3\right|=3\)
dấu = xảy ra khi \(\left(x+1\right).\left(-x-4\right)\ge0\)
\(-4\le x\le-1\)
\(\left|x+2\right|+\left|x+3\right|=\left|x+2\right|+\left|-x-3\right|\ge\left|x+2-x-3\right|=\left|-1\right|=1\)
dấu = xảy ra khi \(\left(x+2\right).\left(-x-3\right)\ge0\)
\(-3\le x\le-2\)
eiiiiiii sorry nha thiếu, làm tiếp nè =))
\(để\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|=4\)
=> dấu = xảy ra khi đồng thời \(\hept{\begin{cases}\left|x+1\right|+\left|-x-4\right|=3\\\left|x+2\right|+\left|-x-3\right|=1\end{cases}}\)
Vậy \(-3\le x\le-2\)
Bài 1: (1/2x - 5)20 + (y2 - 1/4)10 < 0 (1)
Ta có: (1/2x - 5)20 \(\ge\)0 \(\forall\)x
(y2 - 1/4)10 \(\ge\)0 \(\forall\)y
=> (1/2x - 5)20 + (y2 - 1/4)10 \(\ge\)0 \(\forall\)x;y
Theo (1) => ko có giá trị x;y t/m
Bài 2. (x - 7)x + 1 - (x - 7)x + 11 = 0
=> (x - 7)x + 1.[1 - (x - 7)10] = 0
=> \(\orbr{\begin{cases}\left(x-7\right)^{x+1}=0\\1-\left(x-7\right)^{10}=0\end{cases}}\)
=> \(\orbr{\begin{cases}x-7=0\\\left(x-7\right)^{10}=1\end{cases}}\)
=> x = 7
hoặc : \(\orbr{\begin{cases}x-7=1\\x-7=-1\end{cases}}\)
=> x = 7
hoặc : \(\orbr{\begin{cases}x=8\\x=6\end{cases}}\)
Bài 3a) Ta có: (2x + 1/3)4 \(\ge\)0 \(\forall\)x
=> (2x +1/3)4 - 1 \(\ge\)-1 \(\forall\)x
=> A \(\ge\)-1 \(\forall\)x
Dấu "=" xảy ra <=> 2x + 1/3 = 0 <=> 2x = -1/3 <=> x = -1/6
Vậy Min A = -1 tại x = -1/6
b) Ta có: -(4/9x - 2/5)6 \(\le\)0 \(\forall\)x
=> -(4/9x - 2/15)6 + 3 \(\le\)3 \(\forall\)x
=> B \(\le\)3 \(\forall\)x
Dấu "=" xảy ra <=> 4/9x - 2/15 = 0 <=> 4/9x = 2/15 <=> x = 3/10
vậy Max B = 3 tại x = 3/10