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Bài 1:
a: \(B=\dfrac{\sqrt{x}+x+\sqrt{x}-x}{1-x}\cdot\dfrac{x-1}{3-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-3}\)
b: Để B=-1 thì \(2\sqrt{x}=-\sqrt{x}+3\)
=>3 căn x=3
=>căn x=1
hay x=1(loại)
\(1.a.A=\left(1-\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}+1}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\left(x\ge0;x\ne4;x\ne9\right)\)
\(b.A< 0\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\)
\(\Leftrightarrow\sqrt{x}-2< 0\)
\(\Leftrightarrow x< 4\)
Kết hợp với ĐKXĐ , ta có : \(0\le x< 4\)
KL............
\(2.\) Tương tự bài 1.
\(3a.A=\dfrac{1}{x-\sqrt{x}+1}=\dfrac{1}{x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)
\(\Rightarrow A_{Max}=\dfrac{4}{3}."="\Leftrightarrow x=\dfrac{1}{4}\)
\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\frac{1}{x+1}\right).\frac{x+1}{\sqrt{x}-1}\)ĐK x>=0 x khác -1
=\(\frac{\sqrt{x}+1}{x+1}.\frac{x+1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b/ x =\(\frac{2+\sqrt{3}}{2}=\frac{4+2\sqrt{3}}{4}=\frac{3+2\sqrt{3}+1}{4}=\frac{\left(\sqrt{3}+1\right)^2}{4}\)
\(\Rightarrow\sqrt{x}=\frac{\sqrt{3}+1}{2}\)
Em thay vào tính nhé!
c) với x>1
A=\(\frac{\sqrt{x}+1}{\sqrt{x}-1}.\sqrt{x}=\frac{x+\sqrt{x}}{\sqrt{x}-1}=\sqrt{x}+2+\frac{2}{\sqrt{x}-1}=\sqrt{x}-1+\frac{2}{\sqrt{x}-1}+3\)
Áp dụng bất đẳng thức Cosi
A\(\ge2\sqrt{2}+3\)
Xét dấu bằng xảy ra ....
\(a)P=\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}+\dfrac{2x}{x-1}\\ P=\dfrac{\sqrt{x}+1+\sqrt{x}-1+2x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ P=\dfrac{2\sqrt{x}+2x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ P=\dfrac{2\sqrt{x}\left(1+\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ P=\dfrac{2\sqrt{x}}{\sqrt{x}-1}\)
a: \(P=\dfrac{\sqrt{x}+1+\sqrt{x}-1+2x}{x-1}=\dfrac{2x+2\sqrt{x}}{x-1}=\dfrac{2\sqrt{x}}{\sqrt{x}-1}\)
b: Để P>-1/2 thì P+1/2>0
\(\Leftrightarrow\dfrac{2\sqrt{x}}{\sqrt{x}-1}+\dfrac{1}{2}>0\)
=>\(\dfrac{5\sqrt{x}-1}{2\left(\sqrt{x}-1\right)}>0\)
=>1/5<căn x<1
=>1/25<x<1
có phải/....
1) \(A=\dfrac{x+3}{\sqrt{x}-2}\)
\(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\sqrt{x}-2}{x-4}\) hay \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\left(\sqrt{x}-2\right)}{x-4}\)
2) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)
`a)`\(P=A:B\)
\(P=\left(\dfrac{x+1}{x-1}+\dfrac{x-1}{x+1}\right):\left(\dfrac{2}{x^2-1}-\dfrac{x}{x-1}+\dfrac{1}{x+1}\right)\)
\(P=\dfrac{\left(x+1\right)^2+\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}:\dfrac{2-x\left(x+1\right)+\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(P=\dfrac{2x^2+2}{\left(x-1\right)\left(x+1\right)}:\dfrac{1-x^2}{\left(x-1\right)\left(x+1\right)}\)
\(P=-\dfrac{2x^2+2}{\left(x-1\right)\left(x+1\right)}\)
`b)`\(P=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{2x^2+2}{\left(x-1\right)\left(x+1\right)}=\dfrac{1}{2}\)
\(\Leftrightarrow2\left(2x^2+2\right)=\left(x-1\right)\left(x+1\right)\)
\(\Leftrightarrow4x^2+4=x^2-1\)
\(\Leftrightarrow3x^2=-5\) ( vô lý )
Vậy không có giá trị `x` thỏa mãn `P=1/2`
help^^