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1/a) Ta có: \(A=x^4+\left(y-2\right)^2-8\ge-8\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x=0\\y-2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\y=2\end{cases}}\)
Vậy GTNN của A = -8 khi x=0, y=2.
b) Ta có: \(B=|x-3|+|x-7|\)
\(=|x-3|+|7-x|\ge|x-3+7-x|=4\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x\ge3\\x\le7\end{cases}}\Rightarrow3\le x\le7\)
Vậy GTNN của B = 4 khi \(3\le x\le7\)
2/ a) Ta có: \(xy+3x-7y=21\Rightarrow xy+3x-7y-21=0\)
\(\Rightarrow x\left(y+3\right)-7\left(y+3\right)=0\Rightarrow\left(x-7\right)\left(y+3\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=7\\y=-3\end{cases}}\)
b) Ta có: \(\frac{x+3}{y+5}=\frac{3}{5}\)và \(x+y=16\)
Áp dụng tính chất bằng nhau của dãy tỉ số, ta có:
\(\frac{x+3}{y+5}=\frac{3}{5}\Rightarrow\frac{x+3}{3}=\frac{y+5}{5}=\frac{x+y+8}{8}=\frac{16+8}{8}=\frac{24}{8}=3\)
\(\Rightarrow\hept{\begin{cases}\frac{x+3}{3}=3\Rightarrow x+3=9\Rightarrow x=6\\\frac{y+5}{5}=3\Rightarrow y+5=15\Rightarrow y=10\end{cases}}\)
Bài 3: đề không rõ.
Bài 1:\(a,A=x^4+\left(y-2\right)^2-8\)
Có \(x^4\ge0;\left(y-2\right)^2\ge0\)
\(\Rightarrow A\ge0+0-8=-8\)
Dấu "=" xảy ra khi \(MinA=-8\Leftrightarrow x=0;y=2\)
\(b,B=\left|x-3\right|+\left|x-7\right|\)
\(\Rightarrow B=\left|x-3\right|+\left|7-x\right|\)
\(\Rightarrow B\ge\left|x-3+7-x\right|\)
\(\Rightarrow B\ge\left|-10\right|=10\)
Dấu "=" xảy ra khi \(MinB=10\Leftrightarrow3\le x\le7\Rightarrow x\in\left(3;4;5;6;7\right)\)
a) A = (x - 1)2 + 12
Do (x - 1)2 \(\ge\)0 \(\forall\)x
=> (x - 1)2 + 12 \(\ge\)12 \(\forall\)x
Dấu "="xảy ra <=> x - 1 = 0 <=> x = 1
Vậy MinA = 12 khi x = 1
b) B = |x + 3| + 2020
Do |x + 3| \(\ge\)0 \(\forall\)x
=> |x + 3| + 2020 \(\ge\)2020 \(\forall\)x
Dấu "=" xảy ra <=> x + 3 = 0 <=> x = -3
Vậy MinB = 2020 khi x = -3
(c;d max hay min ?)
a) \(A=\left(x-1\right)^2+12\ge12\left(\forall x\right)\)
\("="\Leftrightarrow x=1\)
b) \(B=\left|x+3\right|+2020\ge2020\left(\forall x\right)\)
\("="\Leftrightarrow x=-3\)
c) \(C=\frac{5}{x-2}\ge\frac{5}{-1}=-5\left(\forall x\right)\)
\("="\Leftrightarrow x=1\)
d) \(D=\frac{x+5}{x-4}=1+\frac{9}{x-4}\ge1+\frac{9}{-1}=-8\left(\forall x\right)\)
\("="\Leftrightarrow x=3\)
\(A=\left(5-x\right)^{2016}+|2y+6|-2015\)
Vì \(\left(5-x\right)^{2016}=[\left(5-x\right)^{1008}]^2\ge0,\forall x\)
\(|2y+6|\ge0,\forall y\)
nên \(A=\left(5-x\right)^{2016}+|2y+6|-2015\)\(\ge0+0-2015=2015,\forall x,y\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(5-x\right)^{2016}=0\\|2y+6|=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}5-x=0\\2y+6=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=5\\y=-3\end{cases}}\)
Vậy GTNN của A bằng -2015 \(\Leftrightarrow\hept{\begin{cases}x=5\\y=-3\end{cases}}\)
\(B=\frac{-144}{\left(2x+1\right)^4+12}\)
Vì \(\left(2x+1\right)^4=[\left(2x+1\right)^2]^2\ge0,\forall x\)
nên \(\left(2x+1\right)^4+12\ge0+12=12,\forall x\)
\(\Rightarrow B=\frac{-144}{\left(2x+1\right)^4+12}\ge\frac{-144}{12}=-12,\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\left(2x+1\right)^4=0\)
\(\Leftrightarrow2x+1=0\)
\(\Leftrightarrow x=-\frac{1}{2}\)
Vậy GTNN của B bằng -12\(\Leftrightarrow x=-\frac{1}{2}\)
Chúc bạn học tốt ! Nguyen thi ngoc yen
a, Ta có: \(\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-1\right)^2-3\ge-3\forall x\)
Hay: \(A\ge3\forall x\)
Vậy: Min A = 3 tại \(\left(x-1\right)^2=0\Rightarrow x=1\)
b,Ta có: \(\left(x-2\right)^2\ge0\forall x\)
=> \(4+\left(x-2\right)^2\ge4\forall x\)
Hay: \(B\ge4\forall x\)
Vậy: Min B = 4 tại \(\left(x-2\right)^2=0\Rightarrow x=2\)
=.= hk tốt!!
\(\text{a) }\left(x-1\right)^2-3\)
\(\text{Vì }\left(x-1\right)^2\ge0\text{ }\forall x\)
\(\Rightarrow A=\left(x-1\right)^2-3\ge-3\)
\(\text{Dấu ''='' xảy ra khi :}\)
\(\left(x-1\right)^2=0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
\(\text{Vậy Min}_A=-3\Leftrightarrow x=1\)
\(\text{b) }B=4+\left(x-2\right)^2\)
\(\text{Vì }\left(x-2\right)^2\ge0\forall x\)
\(\Rightarrow B=4+\left(x-2\right)^2\ge4\)
\(\text{Dấu ''='' xảy ra khi :}\)
\(\left(x-2\right)^2=0\)
\(\Rightarrow x-2=0\)
\(\Rightarrow x=2\)
\(\text{Vậy Min}_B=4\Leftrightarrow x=2\)
c,\(43+x=2.5^2-\left(x-57\right)\)
\(< =>43+x=50-x+57\)
\(< =>2x=50+57-43\)
\(< =>x=\frac{107-43}{2}=32\)
d,\(-3.2^2\left(x-5\right)+7\left(3-x\right)=5\)
\(< =>-12.\left(x-5\right)+7.\left(3-x\right)=5\)
\(< =>-12x+60+21-7x=5\)
\(< =>-19x=5-81=-76\)
\(< =>x=-\frac{76}{-19}=4\)
Bài 2:
a) \(A=\left|x-3\right|+10\)
Vì \(\left|x-3\right|\ge0\forall x\)\(\Rightarrow\left|x-3\right|+10\ge10\forall x\)
hay \(A\ge10\)
Dấu " = " xảy ra \(\Leftrightarrow x-3=0\)\(\Leftrightarrow x=3\)
Vậy \(minA=10\Leftrightarrow x=3\)
b) \(B=-7+\left(x-1\right)^2\)
Vì \(\left(x-1\right)^2\ge0\forall x\)\(\Rightarrow-7+\left(x-1\right)^2\ge-7\forall x\)
hay \(B\ge-7\)
Dấu " = " xảy ra \(\Leftrightarrow x-1=0\)\(\Leftrightarrow x=1\)
Vậy \(minB=-7\Leftrightarrow x=1\)
Bài làm:
a) \(A=\left|x-0,25\right|\ge0\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left|x-0,25\right|=0\Rightarrow x=0,25\)
Vậy GTNN A là 0 khi x = 0,25
b) \(B=\left|x+0,25\right|+1,75\ge1,75\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left|x+0,25\right|=0\Rightarrow x=-0,25\)
Vậy GTNN B là 1,75 khi x = -0,25
a) Có: \(|\left(x-0,25\right)|\ge0\forall x\)
=> \(A\ge0\)
DẤU "=" XẢY RA <=> \(|\left(x-0,25\right)|=0\)
<=> \(x=0,25\)
b)
CÓ: \(|\left(x+0,25\right)|\ge0\forall x\)
=> \(|\left(x-0,25\right)|+1,75\ge1,75\forall x\)
=> \(B\ge1,75\)
DẤU "=" XẢY RA <=> \(x=0,25\)