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`A=(9(x-2)+18)/(2-x)+2/x`
`=-9+18/(2-x)+2/x`
`=-9+2(9/(2-x)+1/x)`
Áp dụng bđt cosi-schwarts ta có:
`9/(2-x)+1/x>=(3+1)^2/(2-x+x)=8`
`=>A>=16-9=7`
Dấu "=" xảy ra khi `3/(2-x)=1/x`
`<=>3x=2-x`
`<=>4x=2<=>x=1/2(tm)`
b
`y=x/(1-x)+5/x`
`=(x-1+1)/(1-x)+5/x`
`=1/(1-x)+5/x-1`
Áp dụng cosi-schwarts ta có:
`1/(1-x)+5/x>=(1+sqrt5)^2/(1-x+x)=(1+sqrt5)^2=6+2sqrt5`
`=>y>=5+2sqrt5`
Dấu "=" xảy ra khi `1/(1-x)=sqrt5/x`
`<=>x=sqrt5-sqrt5x`
`<=>x(1+sqrt5)=sqrt5`
`<=>x=sqrt5/(sqrt5+1)=(sqrt5(sqrt5-1))/(5-1)=(5-sqrt5)/4`
`c)C=2/(1-x)+1/x`
Áp dụng bđt cosi schwarts ta có:
`C>=(sqrt2+1)^2/(1-x+x)=3+2sqrt2`
Dấu "=" xảy ra khi `sqrt2/(1-x)=1/x`
`<=>sqrt2x=1-x`
`<=>x(sqrt2+1)=1`
`<=>x=1/(sqrt2+1)=(sqrt2-1)/(2-1)=sqrt2-1`
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
a: Ta có: \(x=\sqrt{28-16\sqrt{3}}+2\sqrt{3}\)
\(=4-2\sqrt{3}+2\sqrt{3}\)
=4
Thay x=4 vào B, ta được:
\(B=\dfrac{2-4}{2}=-1\)
1. \(x=\frac{1}{9}\) thỏa mãn đk: \(x\ge0;x\ne9\)
Thay \(x=\frac{1}{9}\) vào A ta có:
\(A=\frac{\sqrt{\frac{1}{9}}+1}{\sqrt{\frac{1}{9}}-3}=-\frac{1}{2}\)
2. \(B=...\)
\(B=\frac{3\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{4x+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{3x-9\sqrt{x}+x+3\sqrt{x}-4x-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
3. \(P=A:B=\frac{\sqrt{x}+1}{\sqrt{x}-3}:\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{\sqrt{x}+3}{-6}\)
Vì \(\sqrt{x}+3\ge3\forall x\)\(\Rightarrow\frac{\sqrt{x}+3}{-6}\le\frac{3}{-6}=-\frac{1}{2}\)
hay \(P\le-\frac{1}{2}\)
Dấu "=" xảy ra <=> x=0
1, Với \(x\ge0,x\ne1\) ta có :
\(P=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-1\right)\)
\(=\dfrac{\sqrt{x}+1+\sqrt{x}}{x-1}:\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\)
2, Ta có \(P=\dfrac{7}{4}\)
\(\Rightarrow\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}=\dfrac{7}{4}\)
\(\Leftrightarrow4\left(2\sqrt{x}+1\right)=7\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow8\sqrt{x}+4=7\sqrt{x}=7\)
\(\Leftrightarrow\sqrt{x}=3\)
\(\Leftrightarrow x=9\left(tm\right)\)
1) Ta có: \(P=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-1\right)\)
\(=\left(\dfrac{\sqrt{x}+1+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}-1}\right)\)
\(=\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}-\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\)
2) Để \(P=\dfrac{7}{4}\) thì \(\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}=\dfrac{7}{4}\)
\(\Leftrightarrow4\cdot\left(2\sqrt{x}+1\right)=7\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow8\sqrt{x}+4=7\sqrt{x}+7\)
\(\Leftrightarrow8\sqrt{x}-7\sqrt{x}=7-4\)
\(\Leftrightarrow\sqrt{x}=3\)
hay x=9(nhận)
Vậy: Để \(P=\dfrac{7}{4}\) thì x=9
a: \(A=\dfrac{2\sqrt{x}+6+\sqrt{x}-3}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{3\left(\sqrt{x}+1\right)}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{3}{\sqrt{x}+3}\)
b: \(\sqrt{x}+3>=3\)
=>A<=1
Dấu = xảy ra khi x=0
c: \(P=A:\left(B-1\right)=\dfrac{3}{\sqrt{x}+3}:\dfrac{2\sqrt{x}+1-\sqrt{x}-3}{\sqrt{x}+3}=\dfrac{3}{\sqrt{x}-2}\)
Để P nguyên thì căn x-2\(\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{1;25\right\}\)
\(a,\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\left(l\right)\\x=-2\left(l\right)\end{matrix}\right.\Leftrightarrow x\in\varnothing\Leftrightarrow A\in\varnothing\\ b,\text{ý bạn là rút gọn A hả?}\\ A=\dfrac{x-2+2x+3x+6}{\left(x-2\right)\left(x+2\right)}=\dfrac{6x+4}{\left(x-2\right)\left(x+2\right)}\)
a.
\(A=x^2+\dfrac{2021}{x}=x^2+\dfrac{2021}{2x}+\dfrac{2021}{2x}\ge3\sqrt[3]{\dfrac{2021^2}{4x^2}}=3\sqrt[3]{\dfrac{2021^2}{4}}\)
Dấu "=" xảy ra khi \(x=\sqrt[3]{\dfrac{2021}{3}}\)
b.
\(B=4\left(x-1\right)+\dfrac{25}{x-1}+4\ge2\sqrt{\dfrac{100\left(x-1\right)}{x-1}}+4=24\)
Dấu "=" xảy ra khi \(x=\dfrac{7}{2}\)
c.
\(C=3x+\dfrac{16}{x^3}=x+x+x+\dfrac{16}{x^3}\ge4\sqrt[4]{\dfrac{16x^3}{x^3}}=8\)
\(A_{min}=8\) khi \(x=2\)
d.
\(D=x+\dfrac{1}{x}=\left(\dfrac{x}{4}+\dfrac{1}{x}\right)+\dfrac{3}{4}.x\ge2\sqrt{\dfrac{x}{4x}}+\dfrac{3}{4}.2=\dfrac{5}{2}\)
Dấu "=" xảy ra khi \(x=2\)
e.
\(E=\dfrac{9\left(x-2\right)+18}{2-x}+\dfrac{2}{x}=2\left(\dfrac{1}{x}+\dfrac{9}{2-x}\right)-9\ge\dfrac{2.\left(1+3\right)^2}{x+2-x}-9=7\)
\(E_{min}=7\) khi \(x=\dfrac{1}{5}\)
f.
\(F=\dfrac{3}{1-x}+\dfrac{4}{x}\ge\dfrac{\left(\sqrt{3}+2\right)^2}{1-x+x}=7+4\sqrt{3}\)
Dấu "=" xảy ra khi \(x=4-2\sqrt{3}\)