Có ai bt lm bài này k giúp mk vs mk đg rất rất cần mong các bạn giúp cho (làm bài 1, bài 2 vs 3 thôi nhé)
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\(3,\\ a,\dfrac{\left(1+\sqrt{x}\right)^2-4\sqrt{x}}{1-\sqrt{x}}\\ =\dfrac{\sqrt{x}-2\sqrt{x}+1}{1-\sqrt{x}}=\dfrac{\left(1-\sqrt{x}\right)^2}{1-\sqrt{x}}=1-\sqrt{x}=1-\sqrt{2}\)
\(b,\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+4\sqrt{xy}}{1+\sqrt{xy}}\\ =\dfrac{x+2\sqrt{xy}+y}{1+\sqrt{xy}}=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2}{1+\sqrt{xy}}\\ =\dfrac{\left(\sqrt{2}+\sqrt{3}\right)^2}{1+\sqrt{6}}=\dfrac{5+2\sqrt{6}}{1+\sqrt{6}}\\ =\dfrac{\left(5+2\sqrt{6}\right)\left(\sqrt{6}-1\right)}{5}\\ =\dfrac{3\sqrt{6}+7}{5}\)
sin 650=cos 350
\(cos70^0=sin30^0\)
\(tan80^0=cot20^0\)
\(cot68^0=tan32^0\)
\(1,\\ a,=\dfrac{\left(3+2\sqrt{3}\right)\sqrt{3}}{3}+\dfrac{\left(2+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{1}\\ =\dfrac{3\sqrt{3}+6}{3}+\sqrt{2}=\sqrt{3}+1+\sqrt{2}\\ b,=\left(\dfrac{\sqrt{5}+\sqrt{2}}{3}-\dfrac{\sqrt{5}-\sqrt{2}}{3}+1\right)\cdot\dfrac{1}{\left(\sqrt{2}+1\right)^2}\\ =\dfrac{\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}+3}{3}\cdot\dfrac{1}{\left(\sqrt{2}+1\right)^2}\\ =\dfrac{2\sqrt{2}+3}{3\left(3+2\sqrt{2}\right)}=\dfrac{1}{3}\)
\(2,\\ A=2x+\sqrt{\left(x-3\right)^2}=2x+\left|x-3\right|\\ =2\left(-5\right)+\left|-5-3\right|=-10+8=-2\\ B=\dfrac{\sqrt{\left(2x+1\right)^2}}{\left(x-4\right)\left(x+4\right)}\left(x-4\right)^2=\dfrac{\left|2x+1\right|\left(x-4\right)}{x+4}\\ B=\dfrac{17\cdot4}{12}=\dfrac{17}{3}\)
Bài 3:
\(a,=\sqrt[3]{\left(x-1\right)^3}-\sqrt[3]{\left(5x+1\right)^3}=x-1-5x-1=-4x-2\\ b,=6a-6a+20a=20a\)
Bài 2:
\(a,=2\sqrt[3]{6}+3\sqrt[3]{5}-4\sqrt[3]{6}-2\sqrt[3]{5}=\sqrt[3]{5}-2\sqrt[3]{6}\\ b,=\sqrt[3]{8}-4\sqrt[3]{27}+2\sqrt[3]{64}=2-12+16=6\\ c,=\sqrt[3]{64}+\sqrt[3]{48}+\sqrt[3]{36}-\sqrt[3]{48}-\sqrt[3]{36}-\sqrt[3]{27}=4-3=1\\ d,=\sqrt[3]{162\left(-2\right)\cdot\dfrac{2}{3}}=\sqrt[3]{-216}=-6\)