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a, Với mọi giá trị của x;y ta có:
\(\left(x+1\right)^2+\left(y-\dfrac{1}{3}\right)^2\ge0\)
\(\Rightarrow\left(x+1\right)^2+\left(y-\dfrac{1}{3}\right)^2-10\ge-10\)
Hay \(C\ge-10\)với mọi giá trị của x;y
Để \(C=-10\) thì \(\left(x+1\right)^2+\left(y-\dfrac{1}{3}\right)^2-10=-10\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x+1\right)^2=0\\\left(y-\dfrac{1}{3}\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy................
b, Với mọi giá trị của x ta có:
\(\left(2x-1\right)^2+3\ge3\Rightarrow\dfrac{5}{\left(2x-1\right)^2+3}\ge\dfrac{5}{3}\)
Hay \(D\ge\dfrac{5}{3}\) với mọi giá trị của x.
Để \(D=\dfrac{5}{3}\) thì \(\dfrac{5}{\left(2x-1\right)^2+3}=\dfrac{5}{3}\)
\(\Rightarrow\left(2x-1\right)^2=0\Rightarrow x=\dfrac{1}{2}\)
Vậy..................
Chúc bạn học tốt!!!
\(C=\left(x+1\right)^2+\left(y-\dfrac{1}{3}\right)^2-10\)
\(\left(x+1\right)^2\ge0;\left(y-\dfrac{1}{3}\right)^2\ge0\)
\(C_{MIN}\Rightarrow\left(x+1\right)^2_{MIN};\left(y-\dfrac{1}{3}\right)^2_{MIN}\)
\(\left(x+1\right)^2_{MIN}=0;\left(y-\dfrac{1}{3}\right)^2_{MIN}=0\)
\(\Rightarrow C_{MIN}=0+0-10=-10\)
\(D=\dfrac{5}{\left(2x-1\right)^2+3}\)
\(D_{MAX}\Rightarrow\left(2x-1\right)^2+3_{MIN}\)
\(\left(2x-1\right)^2\ge0\)
\(\left(2x-1\right)^2+3_{MIN}\Rightarrow\left(2x-1\right)^2_{MIN}=0\)
\(\Rightarrow\left(2x-1\right)^2+3_{MIN}=0+3=3\)
\(\Rightarrow D_{MAX}=\dfrac{5}{3}\)
\(a,A=5-3\left(2x-1\right)^2\le5\left(vì3\left(2x-1\right)^2\ge0\forall xnên-3\left(2x-1\right)^2\le0\right)\\ Dấu"="xảyrakhi:\\ 3\left(2x-1\right)^2=0\\ \Leftrightarrow x=\frac{1}{2}\\ Vậy.....\)
b,
\(B=\frac{1}{2\left(x-1\right)^2+3}\le\frac{1}{0+3}=\frac{1}{3}\left(vì2\left(x-1\right)^2\ge0\forall x\right)\\ Dấu"="xảyrakhi:\\ 2\left(x-1\right)^2=0\\ \Leftrightarrow x=1\\ Vậy...\)
c,
\(C=\frac{x^2+8}{x^2+2}=1+\frac{6}{x^2+2}\le1+\frac{6}{0+2}=4\left(vìx^2\ge0\forall x\right)\\ Dấu"="xảyrakhi:\\ x^2=0\Leftrightarrow x=0\\ Vậy......\)
a) \(A=5-3.\left(3x-1\right)^2=-\left[3\left(3x-1\right)^2-5\right]\)
Ta có: \(\left(3x-1\right)^2\ge0\forall x\)
\(\Rightarrow3.\left(3x-1\right)^2\ge0\)
\(\Rightarrow3\left(3x-1\right)^2-5\ge-5\forall x\)
\(\Rightarrow-\left[3\left(3x-1\right)^2-5\right]\ge5\forall x\)
Vậy \(MinA=5\Leftrightarrow x=\dfrac{1}{3}\)
a) C = 20013 - |5−2x|
do \(-\left|5-2x\right|\le0\forall x\)
=> 20013-\(\left|5-2x\right|\le20013\)
=>A≤20013
=> GTLN C =20013 khi 5-2x=0
=> 2x=5
=> x=\(\dfrac{5}{2}\)
vậy GTLN C = 20013 khi x=\(\dfrac{5}{2}\)
b) D = 7 - \(\left|\dfrac{2}{3}+\dfrac{1}{4}x\right|\)
do \(-\left|\dfrac{2}{3}+\dfrac{1}{4}x\right|\le0\forall x\)
=> 7-\(\left|\dfrac{2}{3}+\dfrac{1}{4}x\right|\le7\)
=> D≤7
=> GTLN D =7 khi \(\dfrac{2}{3}+\dfrac{1}{4}x=0\)
=> x=-\(\dfrac{8}{3}\)
Bài 1:
\(S=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
\(=\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{c+a}+1\right)+\left(\dfrac{c}{a+b}+1\right)-3\)
\(=\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}+\dfrac{a+b+c}{a+b}-3\)
\(=\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)-3\)
\(=2007.\dfrac{1}{90}-3\)
\(=19,3\)
Vậy S = 19,3
5b)\(S=1+3+3^2+...+3^{2013}\)
\(\Rightarrow3S=3+3^2+3^3+...+3^{2014}\)
\(\Rightarrow3S-S=3^{2014}-1\)
\(\Rightarrow S=\dfrac{3^{2014}-1}{2}\)
a)
\(3(2x-\frac{1}{2})+2(\frac{3}{8}-x)=2,75\)
\(\Leftrightarrow 6x-\frac{3}{2}+\frac{3}{4}-2x=2,75\)
\(\Leftrightarrow 4x=\frac{7}{2}\Rightarrow x=\frac{7}{8}\)
b)
\(x-\frac{1}{3}(5-3x)=1\frac{1}{2}x+5\frac{1}{2}\)
\(\Leftrightarrow x-\frac{5}{3}+x=x+\frac{1}{2}x+\frac{11}{2}\)
\(\Leftrightarrow \frac{1}{2}x=\frac{43}{6}\) \(\Rightarrow x=\frac{43}{3}\)
c) \(\sqrt{x-1}=4\Rightarrow x-1=4^2\Rightarrow x=4^2+1=17\)
d)
\(|x|-5\frac{3}{7}|-x|-\frac{3}{4}=2|x|-1\frac{1}{7}\)
\(\Leftrightarrow |x|-\frac{38}{7}|x|-\frac{3}{4}=2|x|-\frac{8}{7}\)
\(\Leftrightarrow |x|(1-\frac{38}{7}-2)=\frac{3}{4}-\frac{8}{7}\)
\(\Leftrightarrow |x|.\frac{-45}{7}=\frac{-11}{28}\)
\(\Leftrightarrow |x|=\frac{11}{180}\Rightarrow \left[\begin{matrix} x=\frac{11}{180}\\ x=-\frac{11}{180}\end{matrix}\right.\)
a.\(3^{x-1}=243\)
\(3^x:3^1=243\)
\(3^x=729\)
\(\Leftrightarrow3^6=729\)
\(\Leftrightarrow x=6\)
b.\(\left(\dfrac{2}{3}\right)^{x+1}=\dfrac{8}{4}\)
\(\left(\dfrac{2}{3}\right)^x.\left(\dfrac{2}{3}\right)=\dfrac{8}{4}\)
\(\left(\dfrac{2}{3}\right)^x=3\)
Câu b tính đến đây rồi không mò đc x nữa.
a) (x+2)2+\(\left(y-\dfrac{1}{5}\right)^2-10\ge-10\)
Dau = xay ra khi : \(\left\{{}\begin{matrix}x+2=0\\y-\dfrac{1}{5}=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=-2\\y=\dfrac{1}{5}\end{matrix}\right.\)
Vay GTNN cua A=-10 khi : x=-2 , y=1/5
b) ta co : (2x-3)2+5≥5
=> B=\(\dfrac{4}{\left(2x-3\right)^2+5}\le\dfrac{4}{5}\)
Dau = xay ra khi : 2x-3=0
=> x=3/2
Vậy GTLN của B=4/5 khí x=3/2
mk giúp bn bài này lun
Giải :
1, Ta có: (x + 2)2 ≥ 0 ∀ x, \(\left(y-\dfrac{1}{5}\right)^2\) ≥ 0 ∀ y
=> (x + 2)2 + \(\left(y-\dfrac{1}{5}\right)^2\) ≥ 0
=> (x + 2)2 + \(\left(y-\dfrac{1}{5}\right)^2\) - 10 ≥ 0 + (-10) = -10
=> A ≥ -10
Dấu "=" xảy ra khi (x + 2)2 = 0 và \(\left(y-\dfrac{1}{5}\right)^2\)= 0
=> x + 2 = 0 và \(y-\dfrac{1}{5}\) = 0
=> x = -2 và y = \(\dfrac{1}{5}\)
Vậy min A =10 khi x = -2 ; y = \(\dfrac{1}{5}\)
b, Ta có: ( 2x - 3)2 ≥ 0 ∀ x
=> ( 2x - 3)2 +5 ≥ 0 + 5 = 5
=> B ≤ \(\dfrac{4}{5}\)
Dấu " = " xảy ra khi (2x - 3)2 = 0 => 2x- 3 = 0
=> x = \(\dfrac{3}{2}\)
Vậy max A = \(\dfrac{4}{5}\) tại x = \(\dfrac{3}{2}\)
c)C=\(\dfrac{x^2+8}{x^2+2}=\dfrac{\left(x^2+2\right)+6}{x^2+2}=1+\dfrac{6}{x^2+2}\)
Để C đạt GTLN thì \(\dfrac{6}{x^2+2}\) đạt GTNN
\(x^2\ge0\Rightarrow x^2+2\ge2\)
Max C=4 khi x=0
a)A= 5-3.\(\left(2x-1\right)^2\)
\(\left(2x-1\right)^2\)\(\ge0\) nên 3.\(\left(2x-1\right)^2\)\(\ge0\)
Max A=5 khi x=\(\dfrac{1}{2}\)
b) Để B=\(\dfrac{1}{2.\left(x-1\right)^2+3}\)đạt GTLN thì \(2.\left(x-1\right)^2+3\) đạt GTNN
\(\left(x-1\right)^2\ge0\Rightarrow2.\left(x-1\right)^2\ge0\Rightarrow2.\left(x-1\right)^2+3\ge3\)
Max B=\(\dfrac{1}{3}\)khi x=1
câu c thiếu đề phải ko bạn