Tìm GTLN của: 1-3x-5x2
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CB
2
27 tháng 9 2021
\(-5x^2-2xy-2y^2+14x+10y-1\\ =-\left(x^2+2xy+y^2\right)-\left(4x^2-2\cdot2\cdot\dfrac{7}{2}x+\dfrac{49}{4}\right)-\left(y^2-10y+25\right)+\dfrac{55}{4}\\ =-\left(x+y\right)^2-\left(2x-\dfrac{7}{2}\right)^2-\left(y-5\right)^2+\dfrac{55}{4}\le\dfrac{55}{4}\\ Max\Leftrightarrow\left\{{}\begin{matrix}x=-y\\2x=\dfrac{7}{2}\\y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=\dfrac{7}{4}\\y=5\end{matrix}\right.\Leftrightarrow x,y\in\varnothing\)
Vậy dấu \("="\) ko xảy ra
28 tháng 9 2021
a: Ta có: \(-x^2+3x\)
\(=-\left(x^2-3x+\dfrac{9}{4}-\dfrac{9}{4}\right)\)
\(=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{9}{4}\le\dfrac{9}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)
NM
2
21 tháng 10 2021
a: Ta có: \(A=2x^2-8x+1\)
\(=2\left(x^2-4x+\dfrac{1}{2}\right)\)
\(=2\left(x^2-4x+4-\dfrac{7}{2}\right)\)
\(=2\left(x-2\right)^2-7\ge-7\forall x\)
Dấu '=' xảy ra khi x=2
AH
Akai Haruma
Giáo viên
28 tháng 8 2021
Lời giải:
$K=-5x^2+20x-2021=-2001-5(x^2-4x+4)=-2001-5(x-2)^2$
Vì $(x-2)^2\geq 0, \forall x\in\mathbb{R}$
$\Rightarrow K=-2001-5(x-2)^2\leq -2001$
Vậy $K_{\max}=-2001$ khi $(x-2)^2=0\Leftrightarrow x=2$
28 tháng 8 2021
Ta có: \(K=-5x^2+20x-2021\)
\(=-5\left(x^2-4x+\dfrac{2021}{5}\right)\)
\(=-5\left(x^2-4x+4+\dfrac{2001}{5}\right)\)
\(=-5\left(x-2\right)^2-2001\le-2001\forall x\)
Dấu '=' xảy ra khi x=2
MH
26 tháng 9 2023
\(D=2023-8x+2y+4xy-y^2-5x^2\)
\(=-\left(y^2+5x^2-4xy-2y+8x-2023\right)\)
\(=-\left(y^2-2.y.\left(2x+1\right)+\left(2x+1\right)^2-\left(2x+1\right)^2+5x^2+8x-2023\right)\)
\(=-\left[\left(y-2x-1\right)^2-4x^2-4x-1+5x^2+8x-2023\right]\)
\(=-\left[\left(y-2x-1\right)^2+x^2+4x-2024\right]\)
\(=-\left[\left(y-2x-1\right)^2+\left(x+2\right)^2\right]+2028\)
Vì \(-\left[\left(y-2x-1\right)^2+\left(x+2\right)^2\right]\le0\forall x,y\)
\(MaxD=2028\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-3\end{matrix}\right.\)
K
1
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
25 tháng 9 2023
\(C=-\left(x^2+4x+4\right)-\left(y^2-8y+16\right)+22\\ =-\left(x^2+2x.2+2^2\right)-\left(y^2-2.y.4+4^2\right)+22\\ =-\left(x+2\right)^2-\left(y-4\right)^2+22\\ Vậy:max_C=22.khi.x=-2.và.y=4\)
4 tháng 1 2023
(x−1)(5x2−3x+2)=x(5x2−3x+2)−1(5x2−3x+2)
=x.5x2+x.(−3x)+x.2+(−1).5x2+(−1)(−3x)+(−1).2=x.5x^2+x.\left(-3x\right)+x.2+\left(-1\right).5x^2+\left(-1\right)\left(-3x\right)+\left(-1\right).2=x.5x2+x.(−3x)+x.2+(−1).5x2+(−1)(−3x)+(−1).2
=5x3−3x2+2x−5x2+3x−2=5x^3-3x^2+2x-5x^2+3x-2=5x3−3x2+2x−5x2+3x−2
=5x3−8x2+5x−2=5x^3-8x^2+5x-2=5x3−8x2+5x−2.
4 tháng 1 2023
(x−1)(5x2−3x+2)=x(5x2−3x+2)−1(5x2−3x+2)
=x.5x2+x.(−3x)+x.2+(−1).5x2+(−1)(−3x)
=5x3−3x2+2x−5x2+3x−2=5x^3-3x^2+2x-5x^2+3x-2=5x3−3x2+2x−5x2+3x−2
=5x3−8x2+5x−2=5x^3-8x^2+5x-2=5x3−8x2+5x−2.
HH
2
H9
HT.Phong (9A5)
CTVHS
25 tháng 10 2023
A) \(A=-3x^2+x+1\)
\(A=-3\left(x^2-\dfrac{1}{3}x-\dfrac{1}{3}\right)\)
\(A=-3\left(x^2-2\cdot\dfrac{1}{6}\cdot x+\dfrac{1}{36}-\dfrac{13}{36}\right)\)
\(A=-3\left(x-\dfrac{1}{6}\right)^2+\dfrac{13}{12}\)
Mà: \(-3\left(x-\dfrac{1}{6}\right)^2\le0\forall x\)
\(\Rightarrow A=-3\left(x-\dfrac{1}{6}\right)^2+\dfrac{13}{12}\le\dfrac{13}{12}\forall x\)
Dấu "=" xảy ra khi:
\(x-\dfrac{1}{6}=0\Rightarrow x=\dfrac{1}{6}\)
Vậy: \(A_{max}=\dfrac{13}{12}.khi.x=\dfrac{1}{6}\)
B) \(B=2x^2-8x+1\)
\(B=2\left(x^2-4x+\dfrac{1}{2}\right)\)
\(B=2\left(x^2-4x+4-\dfrac{7}{2}\right)\)
\(B=2\left(x-2\right)^2-7\)
Mà: \(2\left(x-2\right)^2\ge0\forall x\)
\(\Rightarrow B=2\left(x-2\right)^2-7\ge-7\forall x\)
Dấu "=" xảy ra khi:
\(x-2=0\Rightarrow x=2\)
Vậy: \(B_{min}=2.khi.x=2\)
3 tháng 10 2021
\(A=\left(4x^2-4xy+y^2\right)+\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{21}{4}\\ A=\left(2x-y\right)^2+\left(x+\dfrac{3}{2}\right)^2-\dfrac{21}{4}\ge-\dfrac{21}{4}\\ A_{min}=-\dfrac{21}{4}\Leftrightarrow\left\{{}\begin{matrix}2x=y\\x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-3\end{matrix}\right.\)
\(B=\left[\left(x-1\right)\left(x+2\right)\right]\left[x\left(x+1\right)\right]=\left(x^2+x-2\right)\left(x^2+x\right)\\ B=\left(x^2+x\right)^2-2\left(x^2+x\right)\\ B=\left(x^2+x\right)^2-2\left(x^2+x\right)+1-1=\left(x^2+x-1\right)^2-1\ge-1\\ B_{min}=-1\Leftrightarrow x^2+x-1=0\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2-\dfrac{5}{4}=0\\ \Leftrightarrow\left(x+\dfrac{1-\sqrt{5}}{2}\right)\left(x+\dfrac{1+\sqrt{5}}{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{5}}{2}\\x=\dfrac{1+\sqrt{5}}{2}\end{matrix}\right.\)
Ta có
\(1-3x-5x^2=\left(-5x^2-\frac{2.\sqrt{5}x.3}{2\sqrt{5}}-\frac{9}{20}\right)+1+\frac{9}{20}\)
\(=\frac{29}{20}-\left(\sqrt{5}x+\frac{3}{2\sqrt{5}}\right)^2\le\frac{29}{20}\)
Vậy GTLN là \(\frac{29}{20}\) dạt được khi \(\sqrt{5}x+\frac{3}{2\sqrt{5}}=0\Leftrightarrow x=\frac{-3}{10}\)