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dễ mà bạn quy đồng biến đỗi là ra chứ làm đánh mấy bài này ra tốn tg lắm
\(Q=\frac{\sqrt{x}\cdot\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\cdot\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(Q=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(Q=x+1\)
Không thể tìm được GTLN hay GTNN của Q.
b)
\(\frac{3x+3}{\sqrt{x}}=3\sqrt{x}+\frac{3}{\sqrt{x}}\)
Để \(\frac{3Q}{\sqrt{x}}\) nguyên thì \(\frac{3}{\sqrt{x}}\)nguyên hay \(\sqrt{x}\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Vì \(\sqrt{x}\)dương nên \(\sqrt{x}\in\left\{1;3\right\}\)
Vậy x=1, x=9 là các giá trị cần tìm
\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\frac{1}{x+1}\right).\frac{x+1}{\sqrt{x}-1}\)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 ....
\(A=\left(\dfrac{2x+\sqrt{x}-1}{1-x}+\dfrac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\right):\dfrac{2\sqrt{x}-1}{\sqrt{x}-x}=\left[\dfrac{\left(2x+\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}{\left(1-x\right)\left(x-\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}-x\right)\left(2x+\sqrt{x}-1\right)}{\left(1-x\right)\left(x-\sqrt{x}+1\right)}\right]:\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}=\dfrac{\left(2x+\sqrt{x}-1\right)\left(x-\sqrt{x}+1+\sqrt{x}-x\right)}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}\left(1-\sqrt{x}\right)}{2\sqrt{x}-1}=\dfrac{\sqrt{x}\left(2x+2\sqrt{x}-\sqrt{x}-1\right)}{\left(1+\sqrt{x}\right)\left(2\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\)
\(M=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}-\dfrac{3\sqrt{x}-1}{\sqrt{x}+1}\right)\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x^3}-1}-\dfrac{2x+1}{\sqrt{x^3}-1}\right)\)
\(M=\left(\dfrac{x+\sqrt{x}-3\sqrt{x}+1}{\sqrt{x}+1}\right)\left(\dfrac{x-\sqrt{x}-2x-1}{\sqrt{x^3}-1}\right)\)
\(M=\left(\dfrac{x-2\sqrt{x}+1}{\sqrt{x}+1}\right)\left(\dfrac{-x-\sqrt{x}-1}{\sqrt{x^3}-1}\right)\)
\(M=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\dfrac{-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(M=\dfrac{1-\sqrt{x}}{\sqrt{x}+1}\)
\(P=\dfrac{2x+\sqrt{x}}{\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x-\sqrt{x}+1}+1\)
\(=\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+1\)
\(=2\sqrt{x}+1-\sqrt{x}-1+1\)
\(=\sqrt{x}+3\)
\(P=\dfrac{2x+\sqrt{x}}{\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x-\sqrt{x}+1}+1\)
\(P=\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+1\)
\(P=2\sqrt{x}+1-\sqrt{x}-1+1\)
\(P=\sqrt{x}+1\)