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\(a,x=16\Rightarrow A=\dfrac{\sqrt{16}+2}{\sqrt{16}-3}=\dfrac{4+2}{4-3}=6\)
\(b,B=\dfrac{\sqrt{x}+5}{\sqrt{x}+1}+\dfrac{\sqrt{x}-7}{1-x}\left(dk:x\ge0,x\ne1,x\ne9\right)\\ =\dfrac{\sqrt{x}+5}{\sqrt{x}+1}-\dfrac{\sqrt{x}-7}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-7\right)}{x-1}\\ =\dfrac{x+4\sqrt{x}-5-\sqrt{x}+7}{x-1}\\ =\dfrac{x+3\sqrt{x}+2}{x-1}\\ =\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\left(dpcm\right)\)
\(c,\dfrac{4A}{A}\le\dfrac{x}{\sqrt{x}-3}\Leftrightarrow\dfrac{4\left(\sqrt{x}+2\right)}{\sqrt{x}-3}:\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\le\dfrac{x}{\sqrt{x}-3}\)
\(\Leftrightarrow\dfrac{4\left(\sqrt{x}+2\right)}{\sqrt{x}-3}.\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\le\dfrac{x}{\sqrt{x}-3}\)
\(\Leftrightarrow4-\dfrac{x}{\sqrt{x}-3}\le0\)
\(\Leftrightarrow\dfrac{4\sqrt{x}-12-x}{\sqrt{x}-3}\le0\)
\(\Leftrightarrow\) Pt vô nghiệm
Vậy không có giá trị x thỏa yêu cầu đề bài.
a, Ta có : \(x=9\Rightarrow\sqrt{x}=3\)
Thay vào biểu thức A ta được : \(A=\frac{2}{3-2}=2\)
b, Với \(x\ge0;x\ne4\)
\(B=\frac{\sqrt{x}}{\sqrt{x}+2}+\frac{4\sqrt{x}}{x-4}=\frac{\sqrt{x}\left(\sqrt{x}-2\right)+4\sqrt{x}}{x-4}\)
\(=\frac{x+2\sqrt{x}}{x-4}=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{x-4}=\frac{\sqrt{x}}{\sqrt{x}-2}\)( đpcm )
c, Ta có : \(A+B=\frac{3x}{\sqrt{x}-2}\)hay
\(\frac{2}{\sqrt{x}-2}+\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{2+\sqrt{x}}{\sqrt{x}-2}=\frac{3x}{\sqrt{x}-2}\)
\(\Rightarrow2+\sqrt{x}=3x\Leftrightarrow3x-2-\sqrt{x}=0\)
\(\Leftrightarrow\sqrt{x}=3x-2\Leftrightarrow x=9x^2-12x+4\)
\(\Leftrightarrow\left(9x-4\right)\left(x-1\right)=0\Leftrightarrow x=\frac{4}{9}\left(ktm\right);x=1\)( đk : \(x\ge\frac{2}{3}\))
a, Ta có : \(x=4\Rightarrow\sqrt{x}=2\)
Thay vào biểu thức A ta được : \(\frac{1}{2-1}=1\)
b, Với \(x\ge0;x\ne1\)
\(Q=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{x-1}-1=\frac{\sqrt{x}\left(\sqrt{x}+1\right)-2-x+1}{x-1}\)
\(=\frac{x+\sqrt{x}-2-x+1}{x-1}=\frac{\sqrt{x}-1}{x-1}=\frac{1}{\sqrt{x}+1}\)
c, Ta có : \(\frac{1}{Q}+P\le4\)hay\(1:\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}\le4\)ĐK : \(x\ne1\)
\(\Leftrightarrow\frac{x-1+1}{\sqrt{x}-1}-4\le0\Leftrightarrow\frac{x-4\sqrt{x}+4}{\sqrt{x}-1}\le0\)
\(\Leftrightarrow\frac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-1}\le0\Rightarrow\sqrt{x}-1\le0\Leftrightarrow\sqrt{x}\le1\Leftrightarrow x\le1\)do \(\left(\sqrt{x}-2\right)^2\ge0\)
Kết hợp với đk, vậy \(x< 1\)
1, thay x=4 (TMĐKXĐ) vào P ta được:
P=\(\dfrac{1}{\sqrt{4}-1}\)=1
vậy khi x=4 thì P =1
2,với x≥0,x≠1:
Q=\(\dfrac{\sqrt{x}}{\sqrt{x}-1}\)-\(\dfrac{2}{\sqrt{x}-1}-1\)=\(\dfrac{\sqrt{x}-2-\sqrt{x}+1}{\sqrt{x}-1}\)=\(\dfrac{-1}{\sqrt{x}-1}\)
vậy Q=\(\dfrac{-1}{\sqrt{x}-1}\)
3,\(\dfrac{1}{Q}+P\le4\)
⇒1/\(\dfrac{-1}{\sqrt{x}-1}\)+\(\dfrac{1}{\sqrt{x}-1}\)≤4⇔\(\dfrac{-\sqrt{x}-1}{1}+\dfrac{1}{\sqrt{x}-1}\le4\)⇔\(\dfrac{-x+1+1}{\sqrt{x}-1}-4\le0\)⇔\(\dfrac{-x+2-4\sqrt{x}+4}{\sqrt{x}-1}\le0\)⇔\(\dfrac{-x-4\sqrt{x}+6}{\sqrt{x}-1}\le0\)⇔\(\dfrac{x+4\sqrt{x}-6}{\sqrt{x}-1}\le0\)⇔\(\dfrac{x+4\sqrt{x}+4-10}{\sqrt{x}-1}\le0\)
\(\dfrac{ \left(\sqrt{x}+2\right)^2-10}{\sqrt{x}-1}\le0\)⇒\(\sqrt{x}-1\le0\) (vì (\(\sqrt{x}+2\))\(^2\)≥0 ∀ x hay (\(\sqrt{x}+2\))\(^2\)-10>0 ∀ x)
⇔x≤1 (KTM)
vậy không có giá trị nào của x TM để \(\dfrac{1}{Q}+P\le4\)
`A)đk:x>=0,x ne 25`
`A=9=>A=(3+2)/(3-5)=-5/2`
`B)B=(3sqrtx-15+20-2sqrtx)/(x-25)`
`=(sqrtx+5)/(x-25)`
`=1/(sqrtx-5)`
`A=B.|x-4|`
`<=>A/B=|x-4|`
`<=>\sqrtx+2=|x-4|`
`<=>\sqrtx+2=(sqrtx+2)|sqrtx-2|`
`<=>|sqrtx-2|=1`
`+)sqrtx-2=1<=>x=9(tm)`
`+)sqrtx-2=-1<=>x=1(tm)`
Vậy `S={1,9}`
a, Thay x=9 vào biểu thức A ta có
\(A=\dfrac{\sqrt{9}+2}{\sqrt{9}-5}\)
\(A=\dfrac{3+2}{3-5}=\dfrac{5}{-2}=-2,5\)
Vậy A =-2,5 khi x=9
a: Khi x=25 thì \(A=\dfrac{7\cdot5-2}{5-2}=\dfrac{33}{3}=11\)
b: P=A*B
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{2}{\sqrt{x}-1}-\dfrac{4\sqrt{x}}{x-1}\right)\cdot\dfrac{7\sqrt{x}-2}{\sqrt{x}-2}\)
\(=\dfrac{x-\sqrt{x}+2\sqrt{x}+2-4\sqrt{x}}{x-1}\cdot\dfrac{7\sqrt{x}-2}{\sqrt{x}-2}\)
\(=\dfrac{x-3\sqrt{x}+2}{x-1}\cdot\dfrac{7\sqrt{x}-2}{\sqrt{x}-2}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\cdot\left(7\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{7\sqrt{x}-2}{\sqrt{x}+1}\)