Question

giúp mình vs từ 17-20

Answer 1

Bài 20:

a: \(A=1-\dfrac{\sqrt{x}}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}+1}\)

Thay \(x=6-2\sqrt{5}\) vào A, ta được:

\(A=\dfrac{1}{\sqrt{6-2\sqrt{5}}+1}=\dfrac{1}{\sqrt{\left(\sqrt{5}-1\right)^2}+1}\)

\(=\dfrac{1}{\sqrt{5}-1+1}=\dfrac{1}{\sqrt{5}}=\dfrac{\sqrt{5}}{5}\)

b: P=A:B

\(=\dfrac{1}{\sqrt{x}+1}:\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}-\dfrac{10-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)

\(=\dfrac{1}{\sqrt{x}+1}:\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-10+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)

\(=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-4\sqrt{x}+3-x+4-10+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt[]{x}-3\right)}\)

\(=\dfrac{1}{\sqrt{x}+1}:\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)

c: \(P=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1-3}{\sqrt{x}+1}=1-\dfrac{3}{\sqrt{x}+1}\)

\(\sqrt{x}+1>=1\forall x\) thỏa mãn ĐKXĐ

=>\(\dfrac{3}{\sqrt{x}+1}< =3\forall x\) thỏa mãn ĐKXĐ

=>\(-\dfrac{3}{\sqrt{x}+1}>=-3\forall x\) thỏa mãn ĐKXĐ

=>\(P=-\dfrac{3}{\sqrt{x}+1}+1>=-3+1=-2\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi x=0

Bài 19:

a: \(P=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{1-x}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{1-\sqrt{x}}{\sqrt{x}+1}\right)\)

\(=\dfrac{\left(\sqrt{x}+1\right)^2+\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{\left(\sqrt{x}+1\right)^2+\left(1-\sqrt{x}\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x+2\sqrt{x}+1+x-\sqrt{x}-\sqrt{x}}{x-1}\cdot\dfrac{x-1}{x+2\sqrt{x}+1-\left(\sqrt{x}-1\right)^2}\)

\(=\dfrac{2x+1}{x+2\sqrt{x}+1-\left(x-2\sqrt{x}+1\right)}=\dfrac{2x+1}{4\sqrt{x}}\)

b: \(x=\dfrac{2-\sqrt{3}}{2}=\dfrac{4-2\sqrt{3}}{4}=\dfrac{\left(\sqrt{3}-1\right)^2}{4}\)

Thay \(x=\dfrac{\left(\sqrt{3}-1\right)^2}{4}\) vào P, ta được:

\(P=\dfrac{2\cdot\dfrac{2-\sqrt{3}}{2}+1}{4\cdot\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{4}}}\)

\(=\dfrac{2-\sqrt{3}+1}{4\cdot\dfrac{\left(\sqrt{3}-1\right)}{2}}=\dfrac{3-\sqrt{3}}{2\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}}{2}\)

c: \(P-\dfrac{1}{2}=\dfrac{2x+1}{4\sqrt{x}}-\dfrac{1}{2}\)

\(=\dfrac{2x+1-2\sqrt{x}}{4\sqrt{x}}=\dfrac{2\left(x-\sqrt{x}+\dfrac{1}{2}\right)}{4\sqrt{x}}\)

\(=\dfrac{2\left(x-\sqrt{x}+\dfrac{1}{4}+\dfrac{1}{4}\right)}{4\sqrt{x}}=\dfrac{2\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{1}{2}}{4\sqrt{x}}>0\)

=>\(P>\dfrac{1}{2}\)