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1. a) Ta có: M = |x + 15/19| \(\ge\)0 \(\forall\)x
Dấu "=" xảy ra <=> x + 15/19 = 0 <=> x = -15/19
Vậy MinM = 0 <=> x = -15/19
b) Ta có: N = |x - 4/7| - 1/2 \(\ge\)-1/2 \(\forall\)x
Dấu "=" xảy ra <=> x - 4/7 = 0 <=> x = 4/7
Vậy MinN = -1/2 <=> x = 4/7
2a) Ta có: P = -|5/3 - x| \(\le\)0 \(\forall\)x
Dấu "=" xảy ra <=> 5/3 - x = 0 <=> x = 5/3
Vậy MaxP = 0 <=> x = 5/3
b) Ta có: Q = 9 - |x - 1/10| \(\le\)9 \(\forall\)x
Dấu "=" xảy ra <=> x - 1/10 = 0 <=> x = 1/10
Vậy MaxQ = 9 <=> x = 1/10
1.
b) \(B=\left|x+8\right|+\left|x+18\right|+\left|x+50\right|\)
Ta có:
\(B=\left|x+8\right|+\left|x+18\right|+\left|x+50\right|\ge\left(\left|x+8\right|+\left|-50-x\right|\right)+\left|x+18\right|\)
\(\Rightarrow B=\left(\left|x+8-50-x\right|\right)+\left|x+18\right|\)
\(\Rightarrow B=\left|-42\right|+\left|x+18\right|\)
\(\Rightarrow B=42+\left|x+18\right|\ge42\)
\(\Rightarrow MIN_B=42\) khi và chỉ khi:
\(\left\{{}\begin{matrix}x+8\ge0\\x+18=0\\x+50\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge-8\\x=-18\\x\ge-50\end{matrix}\right.\Rightarrow x=-18.\)
Vậy \(MIN_B=42\) khi \(x=-18.\)
3.
b) \(\left|x-3\right|-\left|2x+1\right|=0\)
\(\Rightarrow\left|x-3\right|=\left|2x+1\right|\)
\(\Rightarrow\left[{}\begin{matrix}x-3=2x+1\\x-3=-2x-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-2x=1+3\\x+2x=\left(-1\right)+3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}-1x=4\\3x=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4:\left(-1\right)\\x=2:3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=\frac{2}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{-4;\frac{2}{3}\right\}.\)
Chúc bạn học tốt!
Bài 2
\(a,\left(x-3\right)^2=9\Leftrightarrow\left(x-3\right)^2=3^2\Leftrightarrow x-3=3\Leftrightarrow x=6\)
\(b,\left(\frac{1}{2}+x\right)^2=16\Leftrightarrow\left(\frac{1}{2}+x\right)^2=4^2\Leftrightarrow\frac{1}{2}+x=4\Leftrightarrow x=\frac{7}{2}\)
Nhận xét : Lũy thừa bậc chẵn hay giá trị tuyệt đối của 1 số hữu tỉ luôn lớn hơn hoặc bằng 0(bằng 0 khi số hữu tỉ đó là 0)
1)\(\left(2x+\frac{1}{3}\right)^4\ge0\Rightarrow\left(2x+\frac{1}{3}\right)^4-10\ge-10\).Vậy GTNN của A là -10 khi :
\(\left(2x+\frac{1}{3}\right)^4=0\Rightarrow2x+\frac{1}{3}=0\Rightarrow2x=\frac{-1}{3}\Rightarrow x=\frac{-1}{6}\)
\(|2x-\frac{2}{3}|\ge0;\left(y+\frac{1}{4}\right)^4\ge0\Rightarrow|2x-\frac{2}{3}|+\left(y+\frac{1}{4}\right)^4-1\ge-1\).Vậy GTNN của B là -1 khi :
\(\hept{\begin{cases}|2x-\frac{2}{3}|=0\Rightarrow2x-\frac{2}{3}=0\Rightarrow2x=\frac{2}{3}\Rightarrow x=\frac{1}{3}\\\left(y+\frac{1}{4}\right)^4=0\Rightarrow y+\frac{1}{4}=0\Rightarrow y=\frac{-1}{4}\end{cases}}\)
2)\(\left(\frac{3}{7}x-\frac{4}{15}\right)^6\ge0\Rightarrow-\left(\frac{3}{7}x-\frac{4}{15}\right)^6\le0\Rightarrow-\left(\frac{3}{7}x-\frac{4}{15}\right)+3\le3\).Vậy GTLN của C là 3 khi :
\(\left(\frac{3}{7}x-\frac{4}{15}\right)^6=0\Rightarrow\frac{3}{7}x-\frac{4}{15}=0\Rightarrow\frac{3}{7}x=\frac{4}{15}\Rightarrow x=\frac{4}{15}:\frac{3}{7}=\frac{28}{45}\)
\(|x-3|\ge0;|2y+1|\ge0\Rightarrow-|x-3|\le0;-|2y+1|\le0\Rightarrow-|x-3|-|2y+1|+15\le15\)
Vậy GTLN của D là 15 khi :\(\hept{\begin{cases}|x-3|=0\Rightarrow x-3=0\Rightarrow x=3\\|2y+1|=0\Rightarrow2y+1=0\Rightarrow2y=-1\Rightarrow y=\frac{-1}{2}\end{cases}}\)
Bài 1 :
a) Ta thấy : \(\left(x^2-9\right)^2\ge0\)
\(\left|y-2\right|\ge0\)
\(\Leftrightarrow A=\left(x^2-9\right)^2+\left|y-2\right|-1\ge-1\)
Dấu " = " xảy ra :
\(\Leftrightarrow\hept{\begin{cases}x^2-9=0\\y-2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\in\left\{3;-3\right\}\\y=2\end{cases}}\)
Vậy \(Min_A=-1\Leftrightarrow\left(x;y\right)\in\left\{\left(3;2\right);\left(-3;2\right)\right\}\)
b) Ta thấy : \(B=x^2+4x-100\)
\(=\left(x+4\right)^2-104\ge-104\)
Dấu " = " xảy ra :
\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Vậy \(Min_B=-104\Leftrightarrow x=-4\)
c) Ta thấy : \(C=\frac{4-x}{x-3}\)
\(=\frac{3-x+1}{x-3}\)
\(=-1+\frac{1}{x-3}\)
Để C min \(\Leftrightarrow\frac{1}{x-3}\)min
\(\Leftrightarrow x-3\)max
\(\Leftrightarrow x\)max
Vậy để C min \(\Leftrightarrow\)\(x\)max
p/s : riêng câu c mình không tìm được C min :( Mong bạn nào giỏi tìm hộ mình
Bài 2 :
a) Ta thấy : \(x^2\ge0\)
\(\left|y+1\right|\ge0\)
\(\Leftrightarrow3x^2+5\left|y+1\right|-5\ge-5\)
\(\Leftrightarrow C=-3x^2-5\left|y+1\right|+5\le-5\)
Dấu " = " xảy ra :
\(\Leftrightarrow\hept{\begin{cases}x=0\\y+1=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\y=-1\end{cases}}\)
Vậy \(Max_A=-5\Leftrightarrow\left(x;y\right)=\left(0;-1\right)\)
b) Để B max
\(\Leftrightarrow\left(x+3\right)^2+2\)min
Ta thấy : \(\left(x+3\right)^2\ge0\)
\(\Leftrightarrow\left(x+3\right)^2+2\ge2\)
Dấu " = " xảy ra :
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)
Vậy \(Max_B=\frac{1}{2}\Leftrightarrow x=-3\)
c) Ta thấy : \(\left(x+1\right)^2\ge0\)
\(\Leftrightarrow x^2+2x+1\ge0\)
\(\Leftrightarrow-x^2-2x-1\le0\)
\(\Leftrightarrow C=-x^2-2x+7\le8\)
Dấu " = " xảy ra :
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy \(Max_C=8\Leftrightarrow x=-1\)
\(a^2+2ab+b^2=\left(a+b\right)^2\ge0\forall a,b\)
\(a^2-2ab+b^2=\left(a-b\right)^2\ge0\forall a,b\)
\(A^{2n}\ge0\forall A\)
\(-A^{2n}\le0\forall A\)
\(\left|A\right|\ge0\forall A\)
\(-\left|A\right|\le0\forall A\)
\(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)
\(\left|A\right|-\left|B\right|\le\left|A-B\right|\)
\(C=\dfrac{x^2+8}{x^2-2}\)
\(x^2\ge0\Leftrightarrow\left\{{}\begin{matrix}x^2+8\ge8\\x^2-2\ge-2\end{matrix}\right.\)
Mà \(C\) nhỏ nhất nên: \(x^2-2\) phải lớn nhất
\(x^2+8>0\Leftrightarrow x^2-2< 0\) ( để C nhỏ nhất)
\(\Leftrightarrow x^2-2=-1\Rightarrow x^2=1\)
\(min_C=\dfrac{1+8}{1-2}=-9\) Xảy ra khi
\(x^2=1\Leftrightarrow x=\pm1\)
2)
\(A=x^4+3x^2\)
\(A=x^4+3x^2+\dfrac{9}{4}-\dfrac{9}{4}\)
\(A=\left(x^2+\dfrac{3}{2}\right)^2-\dfrac{9}{4}\)
\(x^2\ge0\Rightarrow x^2+\dfrac{3}{2}\ge\dfrac{3}{2}\Leftrightarrow\left(x^2+\dfrac{3}{2}\right)^2\ge\dfrac{9}{4}\)
\(A=\left(x^2+\dfrac{3}{2}\right)^2-\dfrac{9}{4}\ge0\)
Dấu "=" xảy ra khi:
\(x^2=0\Leftrightarrow x=0\)
\(B=\left(x^4+5\right)^2+2\)
\(x^4\ge0\Leftrightarrow x^4+5\ge5\Leftrightarrow\left(x^4+5\right)^2\ge25\)
\(B=\left(x^4+5\right)^2+2\ge27\)
Dấu "=" xảy ra khi:
\(x^4=0\Leftrightarrow x=0\)
a) Ta thấy: \(\left|\dfrac{2}{5}-x\right|\ge0\forall x\)
\(\Rightarrow Q=\dfrac{9}{2}+\left|\dfrac{2}{5}-x\right|\ge\dfrac{9}{2}\forall x\)
Dấu \("="\) xảy ra khi: \(\left|\dfrac{2}{5}-x\right|=0\Leftrightarrow\dfrac{2}{5}-x=0\Leftrightarrow x=\dfrac{2}{5}\)
Vậy \(Min_Q=\dfrac{9}{2}\) khi \(x=\dfrac{2}{5}\).
\(---\)
b) Ta thấy: \(\left|x+\dfrac{2}{3}\right|\ge0\forall x\)
\(\Rightarrow M=\left|x+\dfrac{2}{3}\right|-\dfrac{3}{5}\ge-\dfrac{3}{5}\forall x\)
Dấu \("="\) xảy ra khi: \(\left|x+\dfrac{2}{3}\right|=0\Leftrightarrow x+\dfrac{2}{3}=0\Leftrightarrow x=-\dfrac{2}{3}\)
Vậy \(Min_M=-\dfrac{3}{5}\) khi \(x=-\dfrac{2}{3}\).
\(---\)
c) Ta thấy: \(\left|\dfrac{7}{4}-x\right|\ge0\forall x\)
\(\Rightarrow-\left|\dfrac{7}{4}-x\right|\le0\forall x\)
\(\Rightarrow N=-\left|\dfrac{7}{4}-x\right|-8\le-8\forall x\)
Dấu \("="\) xảy ra khi: \(\left|\dfrac{7}{4}-x\right|=0\Leftrightarrow\dfrac{7}{4}-x=0\Leftrightarrow x=\dfrac{7}{4}\)
Vậy \(Max_N=-8\) khi \(x=\dfrac{7}{4}\).
a) Ta có: \(\left|\dfrac{2}{5}-x\right|\ge0\forall x\)
\(\Rightarrow Q=\dfrac{9}{2}+\left|\dfrac{2}{5}-x\right|\ge\dfrac{9}{2}\forall x\)
Dấu "=" xảy ra khi:
\(\dfrac{2}{5}-x=0\)
\(\Rightarrow x=\dfrac{2}{5}\)
Vậy: ...
b) Ta có: \(\left|x+\dfrac{2}{3}\right|\ge0\forall x\)
\(\Rightarrow M=\left|x+\dfrac{2}{3}\right|-\dfrac{3}{5}\ge-\dfrac{3}{5}\)
Dấu "=" xảy ra:
\(x+\dfrac{2}{3}=0\)
\(\Rightarrow x=-\dfrac{2}{3}\)
Vậy: ...
c) Ta có: \(-\left|\dfrac{7}{4}-x\right|\le0\forall x\)
\(\Rightarrow N=-\left|\dfrac{7}{4}-x\right|-8\le-8\)
Dấu "=" xảy ra:
\(\dfrac{7}{4}-x=0\)
\(\Rightarrow x=\dfrac{7}{4}\)
Vậy: ...