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a: Để A<0 thì 2*căn x-4<0
=>căn x<2
=>0<=x<4
=>\(x\in\left\{0;1;2;3\right\}\)
b: \(A-2=\dfrac{2\sqrt{x}-4-2\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{-6}{\sqrt{x}+1}< 0\)
=>A<2
c: A<1
=>A-1<0
=>\(\dfrac{2\sqrt{x}-4-\sqrt{x}-1}{\sqrt{x}+1}< 0\)
=>căn x-5<0
=>0<=x<25
d: A>-1
=>A+1>0
=>\(\dfrac{2\sqrt{x}-4+\sqrt{x}+1}{\sqrt{x}+1}>0\)
=>3*căn x-3>0
=>x>1
e: A<=(-x+6căn x-8)/(căn x+1)
=>2*căn x-4<=-x+6căn x-8
=>x-4căn x+4<=0
=>x=4
\(P=\frac{\sqrt{x}-1}{\sqrt{x}}-9\sqrt{x}=1-\left(\frac{1}{\sqrt{x}}+9\sqrt{x}\right)\)
\(\frac{1}{\sqrt{x}}+9\sqrt{x}\ge2\sqrt{\frac{1}{\sqrt{x}}\cdot9\sqrt{x}}=6\)
\(\Rightarrow P\le1-6=-5\)
Dấu "=" xảy ra khi \(\frac{1}{\sqrt{x}}=9\sqrt{x}\Leftrightarrow x=\frac{1}{9}\)
Vậy MaxP =-5 đạt được khi \(x=\frac{1}{9}\)
P=\(\frac{\sqrt{x}-1-9x}{\sqrt{x}}=\frac{-5\sqrt{x}-\left(9x-6\sqrt{x}+1\right)}{\sqrt{x}}=-5-\frac{\left(3\sqrt{x}-1\right)^2}{\sqrt{x}}\le-5\)
Dấu "=" xảy ra\(\Leftrightarrow3\sqrt{x}-1=0\Leftrightarrow x=\frac{1}{9}\)
Vậy: Pmax = -5 \(\Leftrightarrow x=\frac{1}{9}\)
Ta có:
\(A=\sqrt{1-x}+\sqrt{1+x}\) \(\left(-1\le x\le1\right)\)
\(=1.\sqrt{1-x}+1.\sqrt{1+x}\)
Áp dụng BĐT Bunhiacopxki, ta có:
\(A=1.\sqrt{1-x}+1.\sqrt{1+x}\)
\(\le\sqrt{\left(1^2+1^2\right).\left(1-x+1+x\right)}=\sqrt{2.2}=2\)
Vậy \(A_{max}=2\), đạt được khi và chỉ khi \(\dfrac{1}{\sqrt{1-x}}=\dfrac{1}{\sqrt{1+x}}\Leftrightarrow1-x=1+x\Leftrightarrow x=0\)
Lời giải:
$\frac{\sqrt{x}+1}{\sqrt{x}+4}=\frac{\sqrt{x}+4-3}{\sqrt{x}+4}=1-\frac{3}{\sqrt{x}+4}$
Vì $\sqrt{x}\geq 0$ nên $\sqrt{x}+4\geq 4$
$\Rightarrow \frac{3}{\sqrt{x}+4}\leq \frac{3}{4}$
$\Rightarrow \frac{\sqrt{x}+1}{\sqrt{x}+4}=1-\frac{3}{\sqrt{x}+4}\geq 1-\frac{3}{4}=\frac{1}{4}$
Vậy $M=\frac{1}{4}$
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$N=\frac{\sqrt{x}+5}{\sqrt{x}+2}=1+\frac{3}{\sqrt{x}+2}$
Do $\sqrt{x}\geq 0$ nên $\sqrt{x}+2\geq 2$
$\Rightarrow \frac{3}{\sqrt{x}+2}\leq \frac{3}{2}$
$\Rightarrow \frac{\sqrt{x}+5}{\sqrt{x}+2}\leq 1+\frac{3}{2}=\frac{5}{2}$
Vậy $N=\frac{5}{2}$
$\Rightarrow 2M+N =2.\frac{1}{4}+\frac{5}{2}=3$
Đáp án C.
a: \(A=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{x-9}=\dfrac{-3\sqrt{x}-9}{x-9}\)
\(=\dfrac{-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{-3}{\sqrt{x}-3}\)
b: A=1/3
=>\(\dfrac{-3}{\sqrt{x}-3}=\dfrac{1}{3}\)
=>căn x-3=-9
=>căn x=-6(loại)
c: căn x-3>=-3
=>3/căn x-3<=-1
=>-3/căn x-3>=1
Dấu = xảy ra khi x=0
\(P\le\sqrt{2\left(3x-5+7-3x\right)}=2\)
\(P_{max}=2\) khi \(3x-5=7-3x\Rightarrow x=2\)
\(A=2\left(x-1\right)+\dfrac{9}{x-1}+2\ge2\sqrt{\dfrac{18\left(x-1\right)}{x-1}}+2=6\sqrt{2}+2\)
\(A_{min}=6\sqrt{2}+2\) khi \(x=\dfrac{2+3\sqrt{2}}{2}\)
\(3=x+y+xy\le\sqrt{2\left(x^2+y^2\right)}+\dfrac{x^2+y^2}{2}\)
\(\Rightarrow\left(\sqrt{x^2+y^2}-\sqrt{2}\right)\left(\sqrt{x^2+y^2}+3\sqrt{2}\right)\ge0\)
\(\Rightarrow x^2+y^2\ge2\)
\(\Rightarrow-\left(x^2+y^2\right)\le-2\)
\(P=\sqrt{9-x^2}+\sqrt{9-y^2}+\dfrac{x+y}{4}\le\sqrt{2\left(9-x^2+9-y^2\right)}+\dfrac{\sqrt{2\left(x^2+y^2\right)}}{4}\)
\(P\le\sqrt{2\left(18-x^2-y^2\right)}+\dfrac{1}{4}.\sqrt{2\left(x^2+y^2\right)}\)
\(P\le\left(\sqrt{2}-1\right)\sqrt{18-x^2-y^2}+\sqrt[]{2}\sqrt{\dfrac{\left(18-x^2-y^2\right)}{2}}+\dfrac{1}{2}\sqrt{\dfrac{x^2+y^2}{2}}\)
\(P\le\left(\sqrt{2}-1\right).\sqrt{18-2}+\sqrt{\left(2+\dfrac{1}{4}\right)\left(\dfrac{18-x^2-y^2+x^2+y^2}{2}\right)}=\dfrac{1+8\sqrt{2}}{2}\)
Dấu "=" xảy ra khi \(x=y=1\)
\(ĐKXĐ:1\le x\le9\)
Đặt \(A=\sqrt{x-1}+\sqrt{9-x}\)
\(A^2=\left(\sqrt{x-1}+\sqrt{9-x}\right)^2\)
\(A^2=x-1+2\sqrt{\left(x-1\right)\left(9-x\right)}+9-x\)
\(A^2\ge8+\left(x-1\right)+\left(9-x\right)=8+8=16\)(1)
Dấu " = " xảy ra \(\Leftrightarrow x-1=9-x\Leftrightarrow x=5\)
\(\left(1\right)\Leftrightarrow Max_{A^2}=16\Leftrightarrow Max_A=4\Leftrightarrow x=5\)
Sửa cho mình dấu \(\ge\)thành \(\le\)nhé !