tìm GTNN
\(9x+\dfrac{4x+3}{2x+1}\)
Tìm GTNN của các biểu thức sau
A=\(\dfrac{2}{6x-5-9x^2}\)
B=\(\dfrac{4x^2-6x+3}{2x^2-3x+2}\)
C=\(\dfrac{3x^2-8x+6}{x^2-2x+1}\)
GIÚP MÌNH 3 CÂU NÀY VỚI MÌNH CẢM ƠN!!!
Mình nghĩ ra câu C rồi bạn nào giúp mình nghĩ nốt câu A,B hộ mình nhé mình cảm ơn!
a:6x-5-9x^2
=-(9x^2-6x+5)
=-(9x^2-6x+1+4)
=-(3x-1)^2-4<=-4
=>A>=2/-4=-1/2
Dấu = xảy ra khi x=1/3
b: \(B=\dfrac{4x^2-6x+4-1}{2x^2-3x+2}=2-\dfrac{1}{2x^2-3x+2}\)
2x^2-3x+2=2(x^2-3/2x+1)
=2(x^2-2*x*3/4+9/16+7/16)
=2(x-3/4)^2+7/8>=7/8
=>-1/2x^2-3x+2<=-1:7/8=-8/7
=>B<=-8/7+2=6/7
Dâu = xảy ra khi x=3/4
Tìm GTNN của phân thức: \(\dfrac{3+\left|2x-1\right|}{14}\)
Tìm GTLN của phân thức: \(\dfrac{-4x^2+4x}{15}\)
\(\left|2x-1\right|+3\ge3\Leftrightarrow\dfrac{3+\left|2x-1\right|}{14}\ge\dfrac{3}{14}\)
Dấu \("="\Leftrightarrow2x-1=0\Leftrightarrow x=\dfrac{1}{2}\)
\(\dfrac{-4x^2+4x}{15}=\dfrac{-4x^2+4x-1+1}{15}=\dfrac{-\left(2x-1\right)^2+1}{15}\)
Ta có \(-\left(2x-1\right)^2+1\le1\Leftrightarrow\dfrac{-\left(2x-1\right)^2+1}{15}\le\dfrac{1}{15}\)
Dấu \("="\Leftrightarrow2x-1=0\Leftrightarrow x=\dfrac{1}{2}\)
2.tìm x
a)\(\sqrt{x^2-6x+9}\)
b)\(\sqrt{x^2-2x+1}\)
c)\(\sqrt{4x+12}-3\sqrt{x+3}+7\sqrt{9x+27}=20\)
d)\(\sqrt{4x+20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=6\)
a) \(\sqrt{x^2-6x+9}\)
\(=\sqrt{\left(x^2-2.x.3+3^2\right)}\)
\(=\sqrt{\left(x-3\right)^2}\) ≥0,∀x
⇒x∈\(R\)
b) \(\sqrt{x^2-2x+1}\)
\(=\sqrt{\left(x^2-2.x.1+1^2\right)}\)
\(=\sqrt{\left(x-1\right)^2}\) ≥0,∀x
⇒x∈\(R\)
a, Tìm GTNN: A = \(\dfrac{x^2-2x+2013}{x^2}\) ; x>0
b, Tìm GTLN và GTNN của: B = \(\dfrac{4x+1}{4x^2+2}\)
a.
\(A=\dfrac{2013}{x^2}-\dfrac{2}{x}+1=2013\left(\dfrac{1}{x}-\dfrac{1}{2013}\right)^2+\dfrac{2012}{2013}\ge\dfrac{2012}{2013}\)
Dấu "=" xảy ra khi \(x=2013\)
b.
\(B=\dfrac{4x^2+2-4x^2+4x-1}{4x^2+2}=1-\dfrac{\left(2x-1\right)^2}{4x^2+2}\le1\)
\(B_{max}=1\) khi \(x=\dfrac{1}{2}\)
\(B=\dfrac{-2x^2-1+2x^2+4x+2}{4x^2+2}=-\dfrac{1}{2}+\dfrac{\left(x+1\right)^2}{2x^2+1}\ge-\dfrac{1}{2}\)
\(B_{max}=-\dfrac{1}{2}\) khi \(x=-1\)
Tìm GTNN
\(A=x^2-2x+5\)
\(B=4x^2+4x+3\)
\(C=9x^2-6x+7\)
D\(=5x^2+3x+8\)
`A=x^2-2x+5`
`=x^2-2x+1+4`
`=(x-1)^2+4>=4`
Dấu "=" `<=>x=1`
`B=4x^2+4x+3`
`=4x^2+4x+1+2`
`=(2x+1)^2+2>=2`
Dấu "=" xảy ra khi `x=-1/2`
`C=9x^2-6x+7`
`=9x^2-6x+1+6`
`=(3x-1)^2+6>=6`
Dấu '=' xảy ra khi `x=1/3`
`D=5x^2+3x+8`
`=5(x^2+3/5x)+8`
`=5(x^2+3/5x+9/100-9/100)+8`
`=5(x+3/10)^2+151/20>=151/20`
Dấu "=" xảy ra khi `x=-3/10`
\(A=x^2-2x+5=x^2-2x+1+4=\left(x-1\right)^2+4\)
Ta có: \(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2+4\ge4\Rightarrow A_{min}=4\) khi \(x=1\)
\(B=4x^2+4x+3=4x^2+4x+1+2=\left(2x+1\right)^2+2\)
Ta có: \(\left(2x+1\right)^2\ge0\Rightarrow\left(2x+1\right)^2+2\ge2\Rightarrow B_{min}=2\) khi \(x=-\dfrac{1}{2}\)
\(C=9x^2-6x+7=9x^2-6x+1+6=\left(3x-1\right)^2+6\)
Ta có: \(\left(3x-1\right)^2\ge0\Rightarrow\left(3x-1\right)^2+6\ge6\Rightarrow C_{min}=6\) khi \(x=\dfrac{1}{3}\)
\(D=5x^2+3x+8\Rightarrow5\left(x^2+2.x.\dfrac{3}{10}+\dfrac{9}{100}\right)+\dfrac{151}{20}=5\left(x+\dfrac{3}{10}\right)^2+\dfrac{151}{20}\)
Ta có: \(5\left(x+\dfrac{3}{10}\right)^2\ge0\Rightarrow5\left(x+\dfrac{3}{10}\right)^2+\dfrac{151}{20}\ge\dfrac{151}{20}\)
\(\Rightarrow D_{min}=\dfrac{151}{20}\) khi \(x=-\dfrac{3}{10}\)
- A = (x-1)2 + 4 \(\ge4\)
Dấu "=" <=> x = 1
- B = (2x+1)2 +2 \(\ge2\)
Dấu "=" xảy ra <=> x = \(\dfrac{-1}{2}\)
- C = (3x - 1)2 + 6 \(\ge6\)
Dấu "=" <=> x = \(\dfrac{1}{3}\)
- D = \(5\left(x^2+\dfrac{3}{5}x+\dfrac{9}{100}\right)+\dfrac{151}{20}=5\left(x+\dfrac{3}{10}\right)^2+\dfrac{151}{20}\ge\dfrac{151}{20}\)
Dấu "=" <=> x = \(\dfrac{-3}{10}\)
Tìm giá trị lớn nhất (GTNN) của các biểu thức sau:
A= \(\dfrac{4+5\left|1-2x\right|}{7}\)
B= \(\dfrac{x^2+4x-6}{3}\)
C= \(\dfrac{5}{x^2-2x+3}\)
Tìm GTNN
A = 9x^2+5-6x
B = 1+x^2-x
C = 4x^2+2x-1
D = 2x^2-x-3
E = 3x^2-1+4x
\(a,A=9x^2+5-6x=9x^2-6x+1+4\)
\(=\left(3x-1\right)^2+4\)
Vì: \(\left(3x-1\right)^2+4\ge4\forall x\)
\(\Rightarrow\)GTNN của A là 4 tại \(\left(3x-1\right)^2=0\Rightarrow x=\frac{1}{3}\)
b,\(B=1+x^2-x=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì: \(\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)
\(\Rightarrow\)GTNN của B là 3/4 tại \(\left(x-\frac{1}{2}\right)^2=0\Rightarrow x=\frac{1}{2}\)
Các phần cn lại lm tg tự nha bn
tìm GTNN
\(x+\dfrac{4x+2}{2x-1}\)
với \(x\ge\dfrac{1}{2}\)(điều kiện chắc vậy)
A=\(x+\dfrac{4x+2}{2x-1}=\dfrac{x\left(2x-1\right)}{2x-1}+\dfrac{4x+2}{2x-1}\)
\(=\dfrac{2x^2-x+4x+2}{2x-1}=\dfrac{2x^2+3x+2}{2x-1}\)
\(=>2A=\)\(\dfrac{4x^2+6x+4}{2x-1}\)
\(=\dfrac{4x^2-4x+1+10x+3}{2x-1}\)
\(=\dfrac{\left(2x-1\right)^2+5\left(2x-1\right)+8}{2x-1}=2x-1+\dfrac{8}{2x-1}+5\)
\(\ge2\sqrt{8}+5\)
=>\(A\ge\dfrac{2\sqrt{8}+5}{2}=\sqrt{8}+\dfrac{5}{2}\)
Dấu"=" xảy ra<=>\(x=\dfrac{1}{2}\left(1+2\sqrt{2}\right)\)(TM)
Vậy min A=\(\sqrt{8}+\dfrac{5}{2}\)
rút gọn rồi tính giá trị biểu thức
a,\(\dfrac{9x^2-6x+1}{9x^2+1}\) tại x =-3
b, \(\dfrac{x^2-6x+9}{-9x+3x^2}\) tại x=-\(\dfrac{1}{3}\)
c, \(\dfrac{x^2-4x+4}{2x^2-4x}\) tại x=-\(\dfrac{1}{2}\)
a) \(\dfrac{9x^2-6x+1}{9x^2-1}\)
\(=\dfrac{\left(3x-1\right)^2}{\left(3x-1\right)\left(3x+1\right)}\)
\(=\dfrac{3x-1}{3x+1}\)
\(=\dfrac{3\cdot\left(-3\right)-1}{3\cdot\left(-3\right)+1}=\dfrac{-9-1}{-9+1}=\dfrac{-10}{-8}=\dfrac{5}{4}\)
b) Ta có: \(\dfrac{x^2-6x+9}{3x^2-9x}\)
\(=\dfrac{\left(x-3\right)^2}{3x\left(x-3\right)}\)
\(=\dfrac{x-3}{3x}\)
\(=\dfrac{-\dfrac{1}{3}-3}{3\cdot\dfrac{-1}{3}}=\dfrac{-\dfrac{10}{3}}{-1}=\dfrac{10}{3}\)
c) Ta có: \(\dfrac{x^2-4x+4}{2x^2-4x}\)
\(=\dfrac{\left(x-2\right)^2}{2x\left(x-2\right)}\)
\(=\dfrac{x-2}{2x}\)
\(=\dfrac{\dfrac{-1}{2}-2}{2\cdot\dfrac{-1}{2}}=\dfrac{-\dfrac{5}{2}}{-1}=\dfrac{5}{2}\)