Câu 32:

$P = \dfrac{x^2 +\sqrt{x}}{x - \sqrt{x} + 1} + 1 - \dfrac{2x + \sqrt{x}}{\sqrt{x}}\\=\dfrac{\sqrt{x}(\sqrt{x^3} + 1)}{x - \sqrt{x} + 1} + 1 - \dfrac{\sqrt{x}(2\sqrt{x} + 1)}{\sqrt{x}}\\=\dfrac{\sqrt{x}(\sqrt{x} + 1)(x - \sqrt{x} + 1)}{x - \sqrt{x} + 1} + 1 - (2\sqrt{x} + 1)\\=\sqrt{x^2} + \sqrt{x} + 1 - 2\sqrt{x} - 1\\=x - \sqrt{x}$

Câu 33:

$\sqrt{48} - 2\sqrt{75} + \sqrt{108}\\=\sqrt{16.3} - 2\sqrt{25.3} + \sqrt{36.3}\\=4\sqrt{3} - 10\sqrt{3}+6\sqrt{3}\\=0$

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

Đáp án:$32a,x-\sqrt{x}$

$33a, 0$ ; $33b , P=\frac{-\sqrt{x}}{1+\sqrt{x}}$

Giải thích các bước giải:

Câu 32:

$P=\frac{x^2+\sqrt{2}}{x-\sqrt{x}+1}+1-\frac{2x+\sqrt{x}}{\sqrt{x}}$

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

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

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

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

$P=x+\sqrt{x}-2\sqrt{x}=x-\sqrt{x}$

Câu 33:

a,$\sqrt{48}-2\sqrt{75}+\sqrt{108}$

$=4.\sqrt{3}-2.5.\sqrt{3}+6\sqrt{3}$

$=4\sqrt{3}-10\sqrt{3}+6\sqrt{3}=0$

b,

$P=(\frac{1}{1-\sqrt{x}}-\frac{1}{1+\sqrt{x}}).(1-\frac{1}{\sqrt{x}})$

$P=(\frac{1+\sqrt{x}}{1-x}-\frac{1-\sqrt{x}}{1-x}).(\frac{\sqrt{x}-1}{\sqrt{x}}$

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

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

$P=\frac{-\sqrt{x}}{(\sqrt{x}-1)(1+\sqrt{x})}.({x}-1)$

$P=\frac{-\sqrt{x}}{1+\sqrt{x}}$

Vậy $P=\frac{-\sqrt{x}}{1+\sqrt{x}}$

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