1×9999÷98767×0299999=30371,37911

nhớ tít nha

ĐK: \(x\ge0,x\ne9\).

\(P=\frac{x+2\sqrt{x}-10}{x-\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}+2}+\frac{\sqrt{x}-2}{3-\sqrt{x}}\)

\(=\frac{x+2\sqrt{x}-10}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}+\frac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{x+2\sqrt{x}-10+x-6\sqrt{x}+9-x+4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{x-4\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)

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

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

\(P=\frac{\sqrt{3}-1-2}{\sqrt{3}-1+4}=\frac{\sqrt{3}-3}{\sqrt{3}+3}=-2+\sqrt{3}\).

\(P=\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{\sqrt{x}+2-3}{\sqrt{x}+2}=1-\frac{3}{\sqrt{x}+2}\)

Dễ thấy \(P< 1\).

\(\sqrt{x}+2\ge2\Rightarrow\frac{3}{\sqrt{x}+2}\le\frac{3}{2}\Rightarrow P\ge1-\frac{3}{2}=-\frac{1}{2}\)

Suy ra \(-\frac{1}{2}\le P< 1\)do đó \(P\)chỉ có thể nhận \(1\)giá trị nguyên duy nhất là \(P=0\).

Với \(P=0\Rightarrow x=1\). 

Do đó ta có đpcm. 

5-1000000=

em còn chx lên lớp 9 luôn á

trông hack não z =)

\(N=\frac{2\sin30^o-2\sin30^o\cos30^o}{\cos^230^o-\left(\cos^230^o-\sin^230^o\right)}=\frac{2\sin30^o\left(1-\cos30^o\right)}{\sin^230^o}=\)

\(=\frac{2\left(1-\cos30^o\right)}{\sin30^o}=\frac{2\left(1-\frac{\sqrt{3}}{2}\right)}{\frac{1}{2}}=2-\sqrt{3}\)

x=100/9

ĐK: x không âm

\(\sqrt{9x}+\sqrt{16x}=14\)

\(\Leftrightarrow\left(3+4\right)\sqrt{x}=14\Leftrightarrow\sqrt{x}=\frac{7}{6}\)

\(\Leftrightarrow x=\frac{49}{36}\)