Bạn tham khảo:
75/
a/$\frac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}$
=$\frac{\sqrt{x^3}-\sqrt{y^3}}{\sqrt{x}-\sqrt{y}}$
=$\frac{(\sqrt{x}-\sqrt{y})(x+\sqrt{x}\sqrt{y}+y)}{\sqrt{x}-\sqrt{y}}$
=$x+$$\sqrt{x}$ $\sqrt{y}$ $+y$
b/
$\frac{x-\sqrt{3x}+3}{x\sqrt{x}+3\sqrt{3}}$
=$\frac{x-\sqrt{x}.\sqrt{3}-3}{\sqrt{x^3}+\sqrt{9}}$
=$\frac{x-3\sqrt{x}+3}{(\sqrt{x}+\sqrt{3})(x-3\sqrt{x}+3)}$
=$\frac{1}{\sqrt{x}-\sqrt{3}}$
76/
a/ $\frac{1}{\sqrt{3}+\sqrt{2}+1}$
=$\frac{\sqrt{3}+1-\sqrt{2}}{(\sqrt{3}+1+\sqrt{2})(\sqrt{3}+1-\sqrt{2})}$
=$\frac{\sqrt{3}+1-\sqrt{2}}{(\sqrt{3}+1)^2-2}$
=$\frac{\sqrt{3}+1-\sqrt{2}}{2(\sqrt{3}+1)(\sqrt{3}-1)}$
Làm tương 1 lần nữa ta được
=$\frac{2-\sqrt{6}+\sqrt{2}}{4}$
b/Làm tương tự nhé:
77/
a/$\sqrt{2x-3}=1+$$\sqrt{2}$
⇔2x+3=1+2$\sqrt{2}$+2
⇔2x=2$\sqrt{2}$
⇔x=$\sqrt{2}$
b/Cậu viết lại đề nhé:
10+$\sqrt{3}x$=$(2+\sqrt{6})^{2}$
⇔10+$\sqrt{x}3=10$+4$\sqrt{6}$
⇔x=$4\sqrt{2}$
c.$\sqrt{3x-2}=2-$$\sqrt{3}$
⇔$3x-2=4-$$4\sqrt{3}+3$
⇔$3x-2=7-$$4\sqrt{3}$
⇔x=$3-\frac{4\sqrt{3}}{3}$
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