NL
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
ND
8 tháng 3 2018
a. \(A+1=\dfrac{27-12x+x^2+9}{x^2+9}\)
\(\Rightarrow A+1=\dfrac{x^2-12x+36}{x^2+9}\)
\(\Rightarrow A+1=\dfrac{\left(x-6\right)^2}{x^2+9}\ge0\)
Min A+1 = 0
=> Min A = -1
Dấu = xảy ra khi và chỉ khi x = 6
\(4-A=\dfrac{4x^2+36-27+12x}{x^2+9}\)
\(4-A=\dfrac{4x^2+12x+9}{x^2+9}\)
\(4-A=\dfrac{\left(2x+3\right)^2}{x^2+9}\)
\(A=4-\dfrac{\left(2x+3\right)^2}{x^2+9}\le4\)
=> Max A= 4
Dấu = xảy ra khi và chỉ khi \(x=\dfrac{-3}{2}\)
8 tháng 3 2018
B=\(\dfrac{8x+3}{4x^2+1}=\dfrac{4x^2+8x+4-4x^2-1}{4x^2+1}\)
=\(\dfrac{\left(4x^2+8x+4\right)-\left(4x^2+1\right)}{4x^2+1}=\dfrac{4\left(x^2+2x+1\right)}{4x^2+1}-1\)
=\(\dfrac{4\left(x+1\right)^2}{4x^2+1}-1\)
=> Min B=-1 dấu = xảy ra khi x=-1
B=\(\dfrac{8x+3}{4x^2+1}=\dfrac{16x^2+4-16x^2+8x-1}{4x^2+1}\)
=\(\dfrac{\left(16x^2+4\right)-\left(16x^2-8x+1\right)}{4x^2+1}=\dfrac{4\left(4x^2+1\right)-\left(4x-1\right)^2}{4x^2+1}\)
=\(\dfrac{4\left(4x^2+1\right)}{4x^2+1}-\dfrac{\left(4x-1\right)^2}{4x^2+1}\)=\(4-\dfrac{\left(4x-1\right)^2}{4x^2+1}\)
=> Max B=4 dấu = xảy ra khi x=\(\dfrac{1}{4}\)
12 tháng 6 2018
\(A=\dfrac{2x+1}{x^2+2}\)
*Min A:
Ta có: \(A=\dfrac{2x+1}{x^2+2}\)
\(=\dfrac{4x+2}{2\left(x^2+2\right)}=\dfrac{\left(x^2+4x+4\right)-\left(x^2+2\right)}{2\left(x^2+2\right)}\)
\(=\dfrac{\left(x+2\right)^2}{2\left(x^2+1\right)}+\dfrac{1}{2}\ge\dfrac{1}{2},\forall x\in R\)
Vậy \(Min_A=\dfrac{1}{2}khi\left(x+2\right)^2=0\)
\(\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
*Max A:
Ta có: \(A=\dfrac{2x+1}{x^2+2}\)
\(=\dfrac{x^2+2-x^2+2x-1}{x^2+2}\)
\(=\dfrac{(x^2+2)-(x^2-2x+1)}{x^2+2}\)
\(=\dfrac{x^2+2}{x^2+2}-\dfrac{\left(x-1\right)^2}{x^2+2}\)
\(=1-\dfrac{\left(x-1\right)^2}{x^2+2}\le0,\forall x\in R\)
Vậy \(Max_A=1khi\left(x-1\right)^2=0\)
\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)
20 tháng 11 2017
\(\text{a) }\dfrac{x^2+x+1}{x^2+2x+1}\\ =\dfrac{x^2+2x-x+1+1-1}{x^2+2x+1}\\ =\dfrac{\left(x^2+2x+1\right)-\left(x+1\right)+1}{x^2+2x+1}\\ =\dfrac{x^2+2x+1}{x^2+2x+1}-\dfrac{x+1}{\left(x+1\right)^2}+\dfrac{1}{\left(x+1\right)^2}\\ =1-\dfrac{1}{x+1}+\dfrac{1}{\left(x+1\right)^2}\left(1\right)\\ Đặt\text{ }\dfrac{1}{x+1}=y\\ \Rightarrow\left(1\right)=1-y+y^2\\ =y^2-y+\dfrac{1}{4}+\dfrac{3}{4}\\ =\left(y^2-y+\dfrac{1}{4}\right)+\dfrac{3}{4}\\ =\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\\ Do\text{ }\left(y-\dfrac{1}{2}\right)^2\ge0\forall x\\ \Rightarrow\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\\ Dấu\text{ }"="\text{ }xảy\text{ }ra\text{ }khi:\\ \left(y-\dfrac{1}{2}\right)^2=0\\ \Leftrightarrow y-\dfrac{1}{2}=0\\ \Leftrightarrow y=\dfrac{1}{2}\\ \Leftrightarrow\dfrac{ 1}{x+1}=\dfrac{1}{2}\\ \Leftrightarrow x+1=2\\ \Leftrightarrow x=1\\ Vậy\text{ }GTNN\text{ }của\text{ }phân\text{ }thức\text{ }là\text{ }\dfrac{3}{4}\text{ }khi\text{ }x=1\)
\(\text{b) }\dfrac{4x^2-6x+1}{\left(2x-1\right)^2}\\ =\dfrac{4x^2-4x-2x+1+1-1}{\left(2x-1\right)^2}\\ =\dfrac{\left(4x^2-4x+1\right)-\left(2x-1\right)-1}{\left(2x-1\right)^2}\\ =\dfrac{\left(2x-1\right)^2}{\left(2x-1\right)^2}-\dfrac{2x-1}{\left(2x-1\right)^2}-\dfrac{1}{\left(2x-1\right)^2}\\ =1-\dfrac{1}{2x-1}-\dfrac{1}{\left(2x-1\right)^2}\left(1\right)\\ Đặt\text{ }-\dfrac{1}{2x-1}=y\\ \Rightarrow\left(1\right)=1+y+y^2\\ =y^2+y+\dfrac{1}{4}+\dfrac{3}{4}\\ =\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}\\ =\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\\ Do\text{ }\left(y+\dfrac{1}{2}\right)^2\ge0\forall x\\ \Rightarrow\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\\ Dấu\text{ }"="\text{ }xảy\text{ }ra\text{ }khi:\\ \left(y+\dfrac{1}{2}\right)^2=0\\ \Leftrightarrow y+\dfrac{1}{2}=0\\ \Leftrightarrow y=-\dfrac{1}{2}\\ \Leftrightarrow-\dfrac{1}{2x-1}=-\dfrac{1}{2}\\ \Leftrightarrow2x-1=2\\ \Leftrightarrow2x=3\\ \Leftrightarrow x=\dfrac{3}{2}\\ Vậy\text{ }GTNN\text{ }của\text{ }biểu\text{ }thức\text{ }là\text{ }\dfrac{3}{4}\text{ }khi\text{ }x=\dfrac{3}{2}\)
S
10 tháng 2 2019
a. \(\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}=\dfrac{9}{x^2-9}\) (ĐKXĐ: \(x\ne\pm3\))
\(\Leftrightarrow\left(x+3\right)^2-\left(x-3\right)^2=9\)
\(\Leftrightarrow x^2+6x+9-x^2+6x-9=9\)
\(\Leftrightarrow12x=9\Leftrightarrow x=\dfrac{3}{4}\left(tm\right)\)
\(\Rightarrow S=\left\{\dfrac{3}{4}\right\}\)
b. \(\dfrac{x+2}{4}-x+3=\dfrac{1-x}{8}\)
\(\Leftrightarrow2\left(x+2\right)-8\left(x-3\right)=1-x\)
\(\Leftrightarrow2x+4-8x+24=1-x\)
\(\Leftrightarrow2x-8x+x=1-4-24\)
\(\Leftrightarrow-3x=-27\Leftrightarrow x=9\)
\(\Rightarrow S=\left\{9\right\}\)
-Mệt -.-
ND
29 tháng 6 2018
3.
\(A=\dfrac{2x+1}{x^2+2}=\dfrac{x^2+2-x^2+2x-1}{x^2+2}=\dfrac{\left(x^2+2\right)-\left(x^2-2x+1\right)}{x^2+2}=1-\dfrac{\left(x-1\right)^2}{x^2+2}\)
Ta có: \(\dfrac{\left(x-1\right)^2}{x^2+2}\ge0\forall x\in R\)
⇒ \(A=1-\dfrac{\left(x-1\right)^2}{x^2+2}\le1\)
Vậy: \(Max_A=1\Leftrightarrow x=1\)
* \(A=\dfrac{2x+1}{x^2+2}=\dfrac{2\left(2x+1\right)}{2\left(x^2+2\right)}=\dfrac{4x+2}{2\left(x^2+2\right)}=\dfrac{-x^2-2+x^2+4x+4}{2\left(x^2+2\right)}\)
\(=-\dfrac{1}{2}+\dfrac{x^2+4x+4}{x^2+2}=-\dfrac{1}{2}+\dfrac{\left(x+2\right)^2}{x^2+2}\ge-\dfrac{1}{2}\)
Vậy: \(Min_A=-\dfrac{1}{2}\Leftrightarrow x=-2\)
* \(B=\dfrac{4x+3}{x^2+1}\) ( 1 cách khác)
\(\Rightarrow B\left(x^2+1\right)=4x+3\)
\(\Rightarrow Bx^2-4x+B-3=0\) (1) \(\left(a=B;b=-4,c=B-3\right)\)
* Với B = 0, pt (1) có nghiệm x = \(-\dfrac{3}{4}\)
* Với B ≠ 0, pt (1) có nghiệm khi và chỉ khi:
\(\Delta=b^2-4ac\ge0\)
\(\Rightarrow\left(-4\right)^2-4.B.\left(B-3\right)\ge0\)
\(\Rightarrow16-4B^2+12B\ge0\)
\(\Rightarrow\left(B-4\right)\left(B+1\right)\ge0\)
\(\Rightarrow-1\le B\le4\)
Suy ra: \(Min_B=-1\Leftrightarrow x=\dfrac{-b}{2a}=\dfrac{4}{2.\left(-1\right)}=-2\)
\(Max_B=4\Leftrightarrow x=\dfrac{-b}{2a}=\dfrac{4}{2.4}=\dfrac{1}{2}\)
29 tháng 6 2018
\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=4\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{ac}+\dfrac{2}{bc}=4\)
<=>\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\) +\(2\left(\dfrac{c}{abc}+\dfrac{b}{abc}+\dfrac{a}{abc}\right)=4\)
<=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{a+b+c}{abc}\right)=4\)
<=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{abc}{abc}\right)=4\) (vì a+b+c =abc)
<=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2=4\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\left(đpcm\right)\)
\(A=\dfrac{1}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x+1}=\dfrac{1}{\left(x-3\right)\left(x+1\right)}\)
\(=\dfrac{1}{x^2-2x-3}=\dfrac{x^2-2x-3+4}{x^2-2x-3}\)
Để A nhỏ nhất thì x^2-2x-3 lớn nhất
=>KHông có giá trị nhỏ nhất của A nha bạn