Tìm gtnn của (x^2+x+1)^2-4(x+2)^2+15
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16 tháng 10 2023
1:
ĐKXĐ: x>=0; x<>4
\(P=\dfrac{\sqrt{x}+\sqrt{x}-2}{x-4}\cdot\dfrac{\sqrt{x}-2}{2}\)
\(=\dfrac{2\sqrt{x}-2}{2}\cdot\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
3: \(P-1=\dfrac{\sqrt{x}-1-\sqrt{x}-2}{\sqrt{x}+2}=\dfrac{-3}{\sqrt{x}+2}< 0\)
=>P<1
NT
1
TN
30 tháng 10 2016
(x-1)(x-2)(x-3)(x-4)+15
=(x2-5x+4)(x2-5x+6)+15
Đặt t=x2-5x+4 ta có:
t(t+2)+15=t2+2t+15
=t2+2t+1+14=(t+1)2+14\(\ge\)14
Dấu = khi t=-1 => x2-5x+4=-1 =>x=\(\frac{5\pm\sqrt{5}}{2}\)
Vậy....
AH
Akai Haruma
Giáo viên
30 tháng 7 2021
1.
$x(x+2)(x+4)(x+6)+8$
$=x(x+6)(x+2)(x+4)+8=(x^2+6x)(x^2+6x+8)+8$
$=a(a+8)+8$ (đặt $x^2+6x=a$)
$=a^2+8a+8=(a+4)^2-8=(x^2+6x+4)^2-8\geq -8$
Vậy $A_{\min}=-8$ khi $x^2+6x+4=0\Leftrightarrow x=-3\pm \sqrt{5}$
AH
Akai Haruma
Giáo viên
30 tháng 7 2021
2.
$B=5+(1-x)(x+2)(x+3)(x+6)=5-(x-1)(x+6)(x+2)(x+3)$
$=5-(x^2+5x-6)(x^2+5x+6)$
$=5-[(x^2+5x)^2-6^2]$
$=41-(x^2+5x)^2\leq 41$
Vậy $B_{\max}=41$. Giá trị này đạt tại $x^2+5x=0\Leftrightarrow x=0$ hoặc $x=-5$
CV
10 tháng 7 2018
1.(√x -2)^2 ≥ 0 --> x -4√x +4 ≥ 0 --> x+16 ≥ 12 +4√x --> (x+16)/(3+√x) ≥4
--> Pmin=4 khi x=4
4 tháng 5 2021
2. Đặt \(\sqrt{x^2-4x+5}=t\ge1\)1
=> M=2x2-8x+\(\sqrt{x^2-4x+5}\)+6=2(t2-5)+t+6
<=> M=2t2+t-4\(\ge\)2.12+1-4=-1
Mmin=-1 khi t=1 hay x=2
BS
2
1 tháng 8 2017
a) vì \(\left|x+\frac{15}{19}\right|\ge0\text{ }\forall\text{ }x\)
\(\Rightarrow\)Mmin \(\Leftrightarrow\)M = 0 \(\Rightarrow\)x = \(\frac{-15}{19}\)
b) vì \(\left|x-\frac{4}{7}\right|\ge0\text{ }\forall\text{ }x\)
\(\Rightarrow\)\(\left|x-\frac{4}{7}\right|-\frac{1}{2}\ge\frac{-1}{2}\)
\(\Rightarrow\)Nmin \(\Leftrightarrow\)N = \(\frac{-1}{2}\)\(\Rightarrow\)\(x=\frac{4}{7}\)
D
1
3 tháng 8 2017
a) vì | x + 15/19 | \(\ge\)0 \(\forall\)x
\(\Rightarrow\)Mmin \(\Leftrightarrow\)M = 0 \(\Rightarrow\)x = -15/19
b) vì | x - 4/7 | \(\ge\)0 \(\forall\)x
\(\Rightarrow\)|x - 4/7 | - 1/2 \(\ge\)-1/2
\(\Rightarrow\)Nmin \(\Leftrightarrow\)N = -1/2 \(\Rightarrow\)x = 4/7
\(A=\left(x^2+x+1\right)^2-4\left(x+2\right)^2+15\)
\(\Rightarrow A=\left(x^2+x+1\right)^2-\left[2\left(x+2\right)\right]^2+15\)
\(\Rightarrow A=\left(x^2+x+1+2x+2\right)\left(x^2+x+1-2x-2\right)+15\)
\(\Rightarrow A=\left(x^2+3x+3\right)\left(x^2-x-1\right)+15\)
\(\Rightarrow A=\left(x^2+3x+\dfrac{9}{4}-\dfrac{9}{4}+3\right)\left(x^2-x+\dfrac{1}{4}-\dfrac{1}{4}-1\right)+15\)
\(\Rightarrow A=\left[\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\right]\left[\left(x-\dfrac{1}{2}\right)^2-\dfrac{5}{4}\right]+15\left(1\right)\)
Ta có : \(\left\{{}\begin{matrix}\left(x+\dfrac{3}{2}\right)^2\ge0,\forall x\\\left(x-\dfrac{1}{2}\right)^2\ge0,\forall x\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4},\forall x\\\left(x-\dfrac{1}{2}\right)^2-\dfrac{5}{4}\ge-\dfrac{5}{4},\forall x\end{matrix}\right.\)
\(\left(1\right)\Rightarrow\left[{}\begin{matrix}A\ge\dfrac{3}{4}.\left[\left(-\dfrac{3}{2}-\dfrac{1}{2}\right)^2-\dfrac{5}{4}\right]+15\left(x=-\dfrac{3}{2}\right)\\A\ge\left[\left(\dfrac{1}{2}+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\right].\left(-\dfrac{5}{4}\right)+15\left(x=\dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}A\ge\dfrac{3}{4}.\left[4-\dfrac{5}{4}\right]+15\left(x=-\dfrac{3}{2}\right)\\A\ge\left[4+\dfrac{3}{4}\right].\left(-\dfrac{5}{4}\right)+15\left(x=\dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}A\ge\dfrac{3}{4}.\dfrac{9}{4}+15\left(x=-\dfrac{3}{2}\right)\\A\ge\dfrac{19}{4}.\left(-\dfrac{5}{4}\right)+15\left(x=\dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}A\ge\dfrac{27}{16}+15\left(x=-\dfrac{3}{2}\right)\\A\ge-\dfrac{95}{16}+15\left(x=\dfrac{1}{2}\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}A\ge\dfrac{267}{16}\left(x=-\dfrac{3}{2}\right)\\A\ge\dfrac{145}{16}\left(x=\dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Rightarrow A\ge\dfrac{145}{16}\left(x=\dfrac{1}{2}\right)\)
\(\Rightarrow GTNN\left(A\right)=\dfrac{145}{16}\left(x=\dfrac{1}{2}\right)\)