`@` `\text {Ans}`
`\downarrow`
`4 \sqrt {(-3)^6} + 5 \sqrt {(-2)^4`
`= 4* \sqrt {729} + 5 \sqrt {16}`
`= 4*27 + 5*4`
`= 108 + 20`
`= 128`
Đúng 0
Bình luận (0)
Các câu hỏi tương tự
1> Rut gon
a)\(\sqrt{6-2\sqrt{2}+2\sqrt{3}-2\sqrt{6}}\)
b) \(\left(\sqrt{2}+1\right)\left(\left(\sqrt{2}\right)^2+1\right)\left(\left(\sqrt{2}\right)^4+1\right)\left(\left(\sqrt{2}\right)^8+1\right)\left(\left(\sqrt{2}\right)^{16}+1\right)\)
c)\(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
d) \(\sqrt{3-\frac{4\sqrt{5}}{3}}+\sqrt{3+\frac{4\sqrt{5}}{3}}\)
RÚT GỌN BIỂU THỨC
A= \(\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right)\)\(\left(2+\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}\right)\)
B= \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\)\(\left(\sqrt{6}+11\right)\)
\(\left(6\right)\dfrac{3\sqrt{x}}{5\sqrt{x}-1}\le-3\)
\(\left(7\right)\dfrac{8\sqrt{x}+8}{6\sqrt{x}+9}>\dfrac{8}{3}\)
\(\left(8\right)\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}< -4\)
\(\left(9\right)\dfrac{4\sqrt{x}+6}{5\sqrt{x}+7}\le-\dfrac{2}{3}\)
\(\left(10\right)\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}>-6\)
Gấp lắm . Giúp mình cảm ơn ạBài 1 2sqrt{left(1+sqrt{3}right)^{ }3}-sqrt{left(2sqrt{3}-3right)^2}left(1+sqrt{3}-sqrt{5}right).left(1+sqrt{3}+sqrt{5}right)left(sqrt[]{dfrac{8}{3}}-sqrt{5}right)xsqrt{6}left(5+4sqrt{2}right).left(3+2sqrt{1}+sqrt{2}right).left(3-2sqrt{1}+2right)sqrt{3+2sqrt{2}}-sqrt{3-2sqrt{2}}
Đọc tiếp
Gấp lắm . Giúp mình cảm ơn ạ
Bài 1
\(2\sqrt{\left(1+\sqrt{3}\right)^{ }3}-\sqrt{\left(2\sqrt{3}-3\right)^2}\)
\(\left(1+\sqrt{3}-\sqrt{5}\right).\left(1+\sqrt{3}+\sqrt{5}\right)\)
\(\left(\sqrt[]{\dfrac{8}{3}}-\sqrt{5}\right)x\sqrt{6}\)
\(\left(5+4\sqrt{2}\right).\left(3+2\sqrt{1}+\sqrt{2}\right).\left(3-2\sqrt{1}+2\right)\)
\(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)
\(\left(5\right)\sqrt{x+3-4\sqrt{x-1}}\sqrt{x+8+6\sqrt{x-1}}=5\)
\(\left(6\right)2x^2+3x+\sqrt{2x^2+3x+9}=33\)
\(\left(7\right)\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+30}=8\)
\(\left(8\right)x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\)
a)\(\left(\sqrt{5}+2\right).\left(17-4\sqrt{9+4\sqrt{5}}\right)?\)
b)\(\left(\sqrt{3-1}\right).\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
c) \(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}}\)
d) \(\left(\sqrt{2}+1\right)\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\left(5-2\sqrt{2}-\sqrt{3}\right)\)
\(\left(\sqrt{6}+\sqrt{10}\right).\sqrt{4-\sqrt{15}}\)
\(\left(3+\sqrt{15}\right).\left(\sqrt{10}-2\right).\sqrt{3-\sqrt{5}}\)
\(\left(4+\sqrt{15}\right).\left(\sqrt{10}-\sqrt{6}\right).\sqrt{4-\sqrt{15}}\)
\(\sqrt{\left(11-6\sqrt{2}\right)^2}+\sqrt{\left(11+6\sqrt{2}\right)^2}\)
\(\sqrt{\left(10-4\sqrt{6}\right)^2}-\sqrt{\left(10+4\sqrt{6}\right)^2}\)
\(\sqrt{\left(4-\sqrt{5}\right)^2}+\sqrt{\left(1-\sqrt{5}\right)^2}\)
\(\sqrt{\left(7+\sqrt{2}\right)^2}-\sqrt{\left(1-\sqrt{2}\right)^2}\)
1)\(\sqrt{6+2\sqrt{5-\sqrt{13+48}}}\)
2)\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
3)\(\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)
4)\(C=\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}\)
5)\(D=\left(3\sqrt{2}+\sqrt{6}\right)\sqrt{6-\sqrt{3}}\)
so sánh\(\sqrt[3]{\left(1-\sqrt{3}\right)\left(4-2\sqrt{3}\right)}v\text{à}\sqrt[3]{\left(1-\sqrt{5}\right)\left(6-2\sqrt{5}\right)}\)