Câu hỏi của James Pham

Question

Bài 1: Rút gọn$3\sqrt{9a^6}-6a^3$ (với mọi a)$\sqrt{\left(x-1\right)^2}+\sqrt{\left(1-3x\right)^2}$ (Với $\dfrac{1}{3}$ < x ≤ 1 )$\sqrt{2-\sqrt{3}}.\left(\sqrt{6}+\sqrt{2}\right)$\(\left(\sqrt...

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$3\sqrt{9a^6}-6a^3=3\left|3a^3\right|-6a^3$Xét $a\ge0\Rightarrow$ biểu thức $=9a^3-6a^3=3a^3$Xét $a< 0\Rightarrow$ biểu thức $=-9a^3-6a^3=-15a^3$$\sqrt{\left(x-1\right)^2}+\sqrt{\left(1-3x\right)^2}=\left|x-1\right|+\left|1-3x\right|$$=1-x+3x-1\left(\dfrac{1}{3}< x\le1\right)=2x$$\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)=\sqrt{2-\sqrt{3}}.\sqrt{2}\left(\sqrt{3}+1\right)=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}+1\right)$$=\sqrt{\left(\sqrt{3}-1\right)^2}\left(\sqrt{3}+1\right)=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=2$$\left(\sqrt{10}+\sqrt{2}\right)\left(6-2\sqrt{5}\right)\sqrt{3+\sqrt{5}}=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{2}.\sqrt{3+\sqrt{5}}$$=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{6+2\sqrt{5}}=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{\left(\sqrt{5}+1\right)^2}$$=\left(\sqrt{5}+1\right)^2\left(\sqrt{5}-1\right)^2=4^2=16$$\sqrt{23-8\sqrt{7}}+\sqrt{8-2\sqrt{7}}=\sqrt{\left(2\sqrt{7}-4\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}$$=2\sqrt{7}-4+\sqrt{7}-1=3\sqrt{7}-5$$\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}$$=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}$$=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|$$=\sqrt{x-1}+1+1-\sqrt{x-1}=2$$\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}$$=\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}$$=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}=\left|\sqrt{x-4}+2\right|+\left|\sqrt{x-4}-2\right|$Xét $x\ge8\Rightarrow\sqrt{x-4}\ge2\Rightarrow$biểu thức $=\sqrt{x-4}+2+\sqrt{x-4}-2$$=2\sqrt{x-4}$Xét $x< 8\Rightarrow\sqrt{x-4}< 2\Rightarrow$ biểu thức $=\sqrt{x-4}+2+2-\sqrt{x-4}=4$ Câu trả lời: $3\sqrt{9a^6}-6a^3=3\left|3a^3\right|-6a^3$Xét $a\ge0\Rightarrow$ biểu thức $=9a^3-6a^3=3a^3$Xét $a< 0\Rightarrow$ biểu thức $=-9a^3-6a^3=-15a^3$$\sqrt{\left(x-1\right)^2}+\sqrt{\left(1-3x\right)^2}=\left|x-1\right|+\left|1-3x\right|$$=1-x+3x-1\left(\dfrac{1}{3}< x\le1\right)=2x$$\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)=\sqrt{2-\sqrt{3}}.\sqrt{2}\left(\sqrt{3}+1\right)=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}+1\right)$$=\sqrt{\left(\sqrt{3}-1\right)^2}\left(\sqrt{3}+1\right)=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=2$$\left(\sqrt{10}+\sqrt{2}\right)\left(6-2\sqrt{5}\right)\sqrt{3+\sqrt{5}}=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{2}.\sqrt{3+\sqrt{5}}$$=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{6+2\sqrt{5}}=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{\left(\sqrt{5}+1\right)^2}$$=\left(\sqrt{5}+1\right)^2\left(\sqrt{5}-1\right)^2=4^2=16$$\sqrt{23-8\sqrt{7}}+\sqrt{8-2\sqrt{7}}=\sqrt{\left(2\sqrt{7}-4\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}$$=2\sqrt{7}-4+\sqrt{7}-1=3\sqrt{7}-5$$\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}$$=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}$$=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|$$=\sqrt{x-1}+1+1-\sqrt{x-1}=2$$\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}$$=\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}$$=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}=\left|\sqrt{x-4}+2\right|+\left|\sqrt{x-4}-2\right|$Xét $x\ge8\Rightarrow\sqrt{x-4}\ge2\Rightarrow$biểu thức $=\sqrt{x-4}+2+\sqrt{x-4}-2$$=2\sqrt{x-4}$Xét $x< 8\Rightarrow\sqrt{x-4}< 2\Rightarrow$ biểu thức $=\sqrt{x-4}+2+2-\sqrt{x-4}=4$

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