tìm GTLN,NN (nếu có) của: a) \(A=\frac{8x+3}{4x^2+1}\) b)\(B=\frac{x}{\left(x+2010\right)^2}\)
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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
D
27 tháng 6 2018
\(a,\)
\(A=\left(\frac{4x}{x+2}-\frac{x^3-8}{x^3+8}.\frac{4x^2-4x+16}{x^2-4}\right):\frac{16}{x+2}.\frac{x^2+3x+2}{x^2+x+1}\)\(ĐKXĐ:x\ne\pm2\)
\(A=[\frac{4x}{x+2}-\frac{\left(x-2\right)\left(x^2+2x+4\right).4\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)\left(x-2\right)\left(x+2\right)}]:\frac{16}{x+2}.\frac{\left(x+1\right)\left(x+2\right)}{x^2+x+1}\)
\(A=[\frac{4x}{x+2}-\frac{4\left(x^2+2x+4\right)}{\left(x+2\right)^2}].\frac{x+2}{16}.\frac{\left(x+1\right)\left(x+2\right)}{x^2+x+1}\)
\(A=\frac{4x^2+8x-4x^2-8x-16}{\left(x+2\right)^2}.\frac{x+2}{16}.\frac{\left(x+1\right)\left(x+2\right)}{x^2+x+1}\)
\(A=\frac{16\left(x+2\right)}{\left(x+2\right)^2.16}.\frac{\left(x+1\right)\left(x+2\right)}{x^2+x+1}\)
\(A=\frac{-\left(x+1\right)}{x^2+x+1}\)
\(B=\frac{x^2+x-2}{x^3-1}\)\(ĐKXĐ:x\ne1\)
\(B=\frac{\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(B=\frac{x+2}{x^2+x+1}\)
\(b,\)
Ta có:
\(A+B=\frac{-\left(x+1\right)}{x^2+x+1}+\frac{x+2}{x^2+x+1}\)
\(=\frac{-x-1+x+2}{x^2+x+1}\)
\(=\frac{1}{x^2+x+1}\)
\(\Rightarrow A+B=\frac{1}{x^2+x+1}=\frac{1}{x^2+2.x.\left(\frac{1}{2}\right)^2+\frac{3}{4}}=\frac{1}{\left(x+\frac{1}{2}\right)^2}+\frac{3}{4}\)
Vì:\(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)
\(\Rightarrow\frac{1}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}\le\frac{1}{\frac{3}{4}}\)
\(\Rightarrow A+B\le\frac{4}{3}\)
\(\Rightarrow GTLN\)của \(A+B=\frac{4}{3}\Leftrightarrow x+\frac{1}{2}=0\)
\(\Leftrightarrow x=\frac{-1}{2}\left(TMĐK\right)\)
Vậy........
\(A=\dfrac{8x+3}{4x^2+1}=\dfrac{4\left(4x^2+1\right)-\left(4x-1\right)^2}{4x^2+1}=4-\dfrac{\left(4x-1\right)^2}{4x^2+1}\le4\)
Vậy GTLN của A là 4 . Dấu " = " xảy ra khi \(\left(4x-1\right)^2=0\Leftrightarrow x=\dfrac{1}{4}\)
\(\text{a)* }A=\dfrac{8x+3}{4x^2+1}=\dfrac{\left(4x^2+8x+4\right)-\left(4x^2+1\right)}{4x^2+1}\\ =\dfrac{4x^2+8x+4}{4x^2+1}-\dfrac{4x^2+1}{4x^2+1}=\dfrac{4\left(x+1\right)^2}{4x^2+1}-1\ge-1\)
Dấu \("="\) xảy ra khi \(\left(x+1\right)^2=0\)
\(\Leftrightarrow x=-1\)
\(\text{* }A=\dfrac{8x+3}{4x^2+1}=\dfrac{-\left(16x^2-8x+1\right)+\left(16x^2+4\right)}{4x^2+1}\\ =\dfrac{-\left(16x^2-8x+1\right)}{4x^2+1}+\dfrac{16x^2+4}{4x^2+1}\\ =\dfrac{-\left(16x^2-8x+1\right)}{4x^2+1}+\dfrac{4\left(4x^2+1\right)}{4x^2+1}\\ =\dfrac{-\left(4x-1\right)^2}{4x^2+1}+4\)
Dấu \("="\) xảy ra khi \(4x-1=0\)
\(\Leftrightarrow x=\dfrac{1}{4}\)
Vậy \(A_{Min}=-1\Leftrightarrow x=-1\)
\(A_{Max}=4\Leftrightarrow x=\dfrac{1}{4}\)