b. Tìm giá trị lớn nhất của biểu thức B =\(\dfrac{x-1}{\left|x-2\right|}\)với x là số nguyên.
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5 tháng 10 2021
a: Ta có: \(x^2=3-2\sqrt{2}\)
nên \(x=\sqrt{2}-1\)
Thay \(x=\sqrt{2}-1\) vào A, ta được:
\(A=\dfrac{\left(\sqrt{2}+1\right)^2}{\sqrt{2}-1}=\dfrac{3+2\sqrt{2}}{\sqrt{2}-1}=7+5\sqrt{2}\)
TD
31 tháng 3 2017
2.
a/\(A=5-I2x-1I\)
Ta thấy: \(I2x-1I\ge0,\forall x\)
nên\(5-I2x-1I\le5\)
\(A=5\)
\(\Leftrightarrow5-I2x-1I=5\)
\(\Leftrightarrow I2x-1I=0\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy GTLN của \(A=5\Leftrightarrow x=\frac{1}{2}\)
b/\(B=\frac{1}{Ix-2I+3}\)
Ta thấy : \(Ix-2I\ge0,\forall x\)
nên \(Ix-2I+3\ge3,\forall x\)
\(\Rightarrow B=\frac{1}{Ix-2I+3}\le\frac{1}{3}\)
\(B=\frac{1}{3}\)
\(\Leftrightarrow B=\frac{1}{Ix-2I+3}=\frac{1}{3}\)
\(\Leftrightarrow Ix-2I+3=3\)
\(\Leftrightarrow Ix-2I=0\)
\(\Leftrightarrow x=2\)
Vậy GTLN của\(A=\frac{1}{3}\Leftrightarrow x=2\)
H9
HT.Phong (9A5)
CTVHS
17 tháng 9 2023
a) Ta có:
\(Q=\sqrt{\left(1-3x\right)\left(x+\dfrac{1}{2}\right)}\) Q có nghĩa khi:
\(\left(1-3x\right)\left(x+\dfrac{1}{2}\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}1-3x\ge0\\x+\dfrac{1}{2}\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}1-3x\le0\\x+\dfrac{1}{2}\le\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3x\le1\\x\ge-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}3x\ge1\\x\le-\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le\dfrac{1}{3}\\x\ge-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge\dfrac{1}{3}\\x\le-\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-\dfrac{1}{2}\le x\le\dfrac{1}{3}\\x\in\varnothing\end{matrix}\right.\)
\(\Leftrightarrow-\dfrac{1}{2}\le x\le\dfrac{1}{3}\)
b) Ta có: \(Q=\sqrt{\left(1-3x\right)\left(x+\dfrac{1}{2}\right)}\)
\(Q=\sqrt{x+\dfrac{1}{2}-3x^2-\dfrac{3}{2}x}\)
\(Q=\sqrt{-\left(3x^2+\dfrac{1}{2}x-\dfrac{1}{2}\right)}\)
\(Q=\sqrt{-3\left(x^2+\dfrac{1}{6}x-\dfrac{1}{6}\right)}\)
\(Q=\sqrt{-3\left(x^2+2\cdot\dfrac{1}{12}\cdot x+\dfrac{1}{144}-\dfrac{25}{144}\right)}\)
\(Q=\sqrt{-3\left(x+\dfrac{1}{12}\right)^2+\dfrac{25}{144}}\)
Mà: \(Q=\sqrt{-3\left(x+\dfrac{1}{12}\right)^2+\dfrac{25}{144}}\le\sqrt{\dfrac{25}{144}}=\dfrac{5}{12}\)
Dấu "=" xảy ra khi:
\(\Leftrightarrow-3\left(x+\dfrac{1}{12}\right)^2=0\)
\(\Leftrightarrow x+\dfrac{1}{12}=0\)
\(\Leftrightarrow x=-\dfrac{1}{12}\)
Vậy: \(Q_{max}=\dfrac{5}{12}.khi.x=-\dfrac{1}{12}\)
20 tháng 6 2021
a) đK: \(x\ne0;2\)
B = \(\dfrac{3x-4}{x\left(x-2\right)}.\dfrac{x\left(x-2\right)}{x^2-4-x^2}=\dfrac{3x-4}{-4}=\dfrac{4-3x}{4}\) \(\dfrac{x-4+2x}{x\left(x-2\right)}:\dfrac{\left(x-2\right)\left(x+2\right)-x^2}{x\left(x-2\right)}\)
= \(\dfrac{3x-4}{x\left(x-2\right)}.\dfrac{x\left(x-2\right)}{x^2-4-x^2}=\dfrac{4-3x}{4}\)
b) Thay x = -2 (TMDK) vào B, ta có:
\(B=\dfrac{4-3.\left(-2\right)}{4}=\dfrac{4+6}{4}=\dfrac{5}{2}\)
c) Để \(\left|B\right|-2x=5\)
<=> \(\left|\dfrac{4-3x}{4}\right|-2x=5\)
TH1: \(x\le\dfrac{4}{3}\)
<=> \(\left|\dfrac{4-3x}{4}\right|=\dfrac{4-3x}{4}\)
PT <=> \(\dfrac{4-3x}{4}-2x=5\)
<=> \(\dfrac{4-3x-8x}{4}=5\)
<=> \(4-11x=20\)
<=> x = \(\dfrac{-16}{11}\) (Tm)
TH2: \(x>\dfrac{4}{3}\)
<=> \(\left|\dfrac{4-3x}{4}\right|=\dfrac{3x-4}{4}\)
PT <=> \(\dfrac{3x-4}{4}-2x=5\)
<=> \(\dfrac{3x-4-8x}{4}=5\)
<=> \(-5x-4=20\)
<=> \(x=\dfrac{-24}{5}\left(l\right)\)
d) Xét (2-x)B = \(\dfrac{\left(2-x\right)\left(4-3x\right)}{4}\) = \(\dfrac{3x^2-10x+8}{4}\)
= \(\dfrac{3\left(x-\dfrac{5}{3}\right)^2-\dfrac{1}{3}}{4}\)
Mà \(3\left(x-\dfrac{5}{3}\right)^2\ge\) 0
=> (2-x)B \(\ge\dfrac{\dfrac{-1}{3}}{4}=\dfrac{-1}{12}\)
Dấu "=" <=> x = \(\dfrac{5}{3}\left(tm\right)\)
e) Số nguyên âm lớn nhất là -1
Để B = -1
<=> \(\dfrac{4-3x}{4}=-1\)
<=> 4 - 3x = -4
<=> \(x=\dfrac{8}{3}\left(tm\right)\)
g)
TH1: \(x\le\dfrac{4}{3}\)
<=> \(\left|\dfrac{4-3x}{4}\right|=\dfrac{4-3x}{4}\)
BDT <=> \(\dfrac{4-3x}{4}< 2x-4\)
<=> \(4-3x< 8x-16\)
<=> \(x>\dfrac{20}{11}\left(l\right)\)
TH2: \(x>\dfrac{4}{3}\)
<=> \(\left|\dfrac{4-3x}{4}\right|=\dfrac{3x-4}{4}\)
BDT <=> \(\dfrac{3x-4}{4}< 2x-4\)
<=> \(3x-4< 8x-16\)
<=> x > \(\dfrac{12}{5}\)
KHDK: \(x>\dfrac{12}{5}\)
31 tháng 10 2021
\(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\ge\dfrac{1}{9}\\ A_{min}=\dfrac{1}{9}\Leftrightarrow x=\dfrac{3}{5}\\ B=\dfrac{2009}{2008}-\left|x-\dfrac{3}{5}\right|\le\dfrac{2009}{2008}\\ B_{max}=\dfrac{2009}{2008}\Leftrightarrow x=\dfrac{3}{5}\\ C=-2\left|\dfrac{1}{3}x+4\right|+1\dfrac{2}{3}\le1\dfrac{2}{3}\\ C_{max}=1\dfrac{2}{3}\Leftrightarrow\dfrac{1}{3}x=-4\Leftrightarrow x=-12\)
Biểu thức không có max. Bạn coi lại đề.
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