Tìm GTNN của: \(B=\dfrac{4x^2-6x+1}{\left(2x-1\right)^2}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
BC
1
KK
7 tháng 3 2018
A= \(\dfrac{x^2-4x+1}{x^2}\)
ĐKXĐ x≠0
A= \(\dfrac{x^2}{x^2}-\dfrac{4x}{x^2}+\dfrac{1}{x^2}\)
=\(1-\dfrac{4}{x}+\dfrac{1}{x^2}\)
đặt \(\dfrac{1}{x}=y\) ta có
1-4y+y2
= y2-4y+1
=(y2-4y+4)-3
= (y-2)2 -3
do (y-2)2 ≥ 0 ∀x
=> (y-2)2 -3 ≥ -3
=> A ≥ -3
=> Amin =-3dấu '=' xảy ra khi
y-2=0
=> y=2
=> \(\dfrac{1}{x}=2\)
=> x=\(\dfrac{1}{2}\)
vậy GTNN A =-3 khi x=\(\dfrac{1}{2}\)
ND
8 tháng 3 2018
a.
\(A=\dfrac{x^2-4x+1}{x^2}\)
\(\Rightarrow A=\dfrac{x^2-4x+4-3}{x^2}\)
\(\Rightarrow A=\dfrac{\left(x-2\right)^2-3}{x^2}\)
Ta có: \(\left(x-2\right)^2-3\ge-3\)
\(\Rightarrow x=2\)
Khi đó ta được Min A = \(\dfrac{\left(2-2\right)-3}{2^2}\ge\dfrac{-3}{4}\)
Vậy Min A = \(\dfrac{-3}{4}\)
15 tháng 3 2021
a) Ta có: \(P=\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}-\dfrac{2x^2}{4\left(2-x\right)+x^2\left(2-x\right)}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}-\dfrac{2x^2}{\left(2-x\right)\left(x^2+4\right)}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\left(\dfrac{\left(x^2-2x\right)\left(x-2\right)}{2\left(x-2\right)\left(x^2+4\right)}+\dfrac{4x^2}{2\left(x-2\right)\left(x^2+4\right)}\right)\cdot\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\dfrac{x^3-x^2-2x^2+4x+4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\dfrac{x^3+x^2+4x}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\dfrac{x\left(x^2+x+4\right)}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{\left(x^2+x+4\right)\left(x+1\right)}{2x\left(x^2+4\right)}\)
H
1
KB
19 tháng 4 2022
P/s : Mik nghĩ là \(\left(2x+1\right)^2\)
\(C=x+\dfrac{1}{4x}+\dfrac{x}{\left(2x+1\right)^2}=\left[\dfrac{x}{\left(2x+1\right)^2}+\dfrac{2x+1}{16}+\dfrac{2x+1}{16}+\dfrac{1}{16x}\right]+\dfrac{3}{4}\left(x+\dfrac{1}{4x}\right)-\dfrac{1}{8}\)
AD BĐT AM - GM ta được : \(\dfrac{x}{\left(2x+1\right)^2}+\dfrac{2x+1}{16}+\dfrac{2x+1}{16}+\dfrac{1}{16x}\ge4\sqrt[4]{\dfrac{1}{16^3}}=\dfrac{1}{2}\)
\(x+\dfrac{1}{4x}\ge2\sqrt{\dfrac{1}{4}}=1\)
Suy ra : \(C\ge\dfrac{1}{2}+\dfrac{3}{4}.1-\dfrac{1}{8}=\dfrac{9}{8}\)
" = " \(\Leftrightarrow x=\dfrac{1}{2}\)
BC
0
NH
2
ND
Nguyễn Đức Trí
VIP
19 tháng 8 2023
\(C=\dfrac{5}{3-\left(4x+1\right)^2}\)
Điều kiện xác định khi
\(3-\left(4x+1\right)^2\ne0\Leftrightarrow\left[{}\begin{matrix}4x+1\ne\sqrt[]{3}\\4x+1\ne-\sqrt[]{3}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ne\dfrac{\sqrt[]{3}-1}{4}\\x\ne\dfrac{-\sqrt[]{3}-1}{4}\end{matrix}\right.\)
Ta có :
\(\left(4x+1\right)^2\ge0,\forall x\)
\(\Leftrightarrow3-\left(4x+1\right)^2\le3\)
\(\Leftrightarrow C=\dfrac{5}{3-\left(4x+1\right)^2}\ge\dfrac{5}{3}\)
Vậy \(GTNN\left(C\right)=\dfrac{5}{3}\left(tạix=-\dfrac{1}{4}\right)\)
ND
Nguyễn Đức Trí
VIP
19 tháng 8 2023
\(B=\left(2x\right)^2+2\left(y-1\right)^2-5\)
vì \(\left\{{}\begin{matrix}\left(2x\right)^2\ge0,\forall x\\2\left(y-1\right)^2\ge0,\forall y\end{matrix}\right.\)
\(\Rightarrow B=\left(2x\right)^2+2\left(y-1\right)^2-5\ge-5\)
Dấu "=" xảy tại khi
\(\left\{{}\begin{matrix}2x=0\\2\left(y-1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\)
Vậy \(GTNN\left(B\right)=-5\left(tạix=0;y=1\right)\)
18 tháng 3 2018
\(B=\dfrac{3x^2-2x+3}{x^2+1}=\dfrac{2x^2+x^2-2x+1+2}{x^2+1}\\ =\dfrac{\left(2x^2+2\right)+\left(x^2-2x+1\right)}{x^2+1}\\ =\dfrac{2\left(x^2+1\right)}{x^2+1}+\dfrac{x^2-2x+1}{x^2+1}\\ =2+\dfrac{\left(x-1\right)^2}{x^2+1}\)
Do \(\dfrac{\left(x-1\right)^2}{x^2+1}\ge0\forall x\)
\(\Rightarrow B=\dfrac{\left(x-1\right)^2}{x^2+1}+2\ge2\forall x\)
Dấu "=" xảy ra khi :
\(\dfrac{\left(x-1\right)^2}{x^2+1}=0\\ \Leftrightarrow\left(x-1\right)^2=0\\ \Leftrightarrow x-1=0\\ \Leftrightarrow x=1\)
Vậy \(B_{\left(Min\right)}=2\) khi \(x=1\)
18 tháng 3 2018
\(A=\dfrac{4x^2-6x+1}{\left(2x-1\right)^2}=\dfrac{4x^2-4x-2x+1+1-1}{\left(2x-1\right)^2}\\ =\dfrac{\left(4x^2-4x+1\right)-\left(2x-1\right)-1}{\left(2x-1\right)^2}\\ =\dfrac{\left(2x-1\right)^2}{\left(2x-1\right)^2}-\dfrac{2x-1}{\left(2x-1\right)^2}-\dfrac{1}{\left(2x-1\right)^2}\\ =1-\dfrac{1}{2x-1}-\dfrac{1}{\left(2x-1\right)^2}\)
Đặt \(-\dfrac{1}{2x-1}=y\)
\(\Rightarrow A=1+y+y^2\\ =y^2+y+\dfrac{1}{4}+\dfrac{3}{4}\\ =\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}\\ =\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Do \(\left(y+\dfrac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Dấu "=" xảy ra khi:
\(\left(y+\dfrac{1}{2}\right)^2=0\\ \Leftrightarrow y+\dfrac{1}{2}=0\\ \Leftrightarrow y=-\dfrac{1}{2}\\ \Leftrightarrow-\dfrac{1}{2x-1}=-\dfrac{1}{2}\\ \Leftrightarrow2x-1=2\\ \Leftrightarrow2x=3\\ \Leftrightarrow x=\dfrac{3}{2}\)
Vậy \(A_{\left(Min\right)}=\dfrac{3}{4}\) khi \(x=\dfrac{3}{2}\)
Ta có: \(\left(2x-1\right)^2\ge0\)
\(\Rightarrow\) B nhỏ nhất khi \(4x^2-6x+1\)có giá trị nhỏ nhất
Mà: \(4x^2-6x+1=4\left(x^2-2.\dfrac{3}{4}x+\dfrac{9}{16}\right)-\dfrac{5}{4}=4\left(x-\dfrac{3}{4}\right)^2-\dfrac{5}{4}\ge\dfrac{-5}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{3}{4}\)
\(\Rightarrow\min\limits_{\left(4x^2-6x+1\right)}=\dfrac{-5}{4}.\) khi \(x=\dfrac{3}{4}\)
\(\Rightarrow\left(2x-1\right)^2=\dfrac{1}{4}\)
\(\Rightarrow\min\limits_B=\dfrac{-5}{4}:\dfrac{1}{4}=\dfrac{-5}{4}.4=-5\) Khi \(x=\dfrac{3}{4}\)
Ta có: (2x−1)2≥0(2x−1)2≥0
⇒⇒ B nhỏ nhất khi 4x2−6x+14x2−6x+1có giá trị nhỏ nhất
Mà: 4x2−6x+1=4(x2−2.34x+916)−54=4(x−34)2−54≥−544x2−6x+1=4(x2−2.34x+916)−54=4(x−34)2−54≥−54
Dấu "=" xảy ra ⇔x=34⇔x=34
⇒min(4x2−6x+1)=−54.⇒min(4x2−6x+1)=−54. khi x=34x=34
⇒(2x−1)2=14⇒(2x−1)2=14
⇒minB=−54:14=−54.4=−5⇒minB=−54:14=−54.4=−5 Khi x=34