Rút Gọn
\(a,\sqrt{3}-\frac{1}{3}\sqrt{27}+2\sqrt{507}\) b, \(5\sqrt{12a}+\sqrt{49a}-\sqrt{64a}vớia>0\)
c,\(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right):\sqrt{3}\)
Giúp mk nha mk khẩn lắm
Rút gọn bt
\(a,A=\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\left(\sqrt{5-2\sqrt{6}}\right)}{9\sqrt{3}-11\sqrt{2}}\)
\(b,C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(c,\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(d,\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
d/ \(x=\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
\(\Leftrightarrow x^3=3+\sqrt{9+\frac{125}{27}}+3-\sqrt{9+\frac{125}{27}}-3\left(\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\right)\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}.\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
\(\Leftrightarrow x^3=6-3x\sqrt[3]{9-9-\frac{125}{27}}\)
\(\Leftrightarrow x^3=6-5x\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow x=1\)
c/
\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{12}+4}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{4+2\sqrt{3}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
\(=3-1=2\)
b/ \(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{25}}\)
\(=\sqrt{4+5}=3\)
Rút gọn biểu thức sau :
1) \(\sqrt{\frac{4}{3}}-\frac{\sqrt{24}}{2\sqrt{2}}+\frac{3}{\sqrt{27}}\)
2) \(4\sqrt{20}-3\sqrt{125}+5\sqrt{45}-15\sqrt{\frac{1}{5}}\)
3) \(\sqrt{343a}+\sqrt{63a}-\sqrt{28a}\left(a\ge0\right)\)
4) \(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right):\sqrt{3}\)
Giúp mik vs
a,=0
b,\(5\sqrt{5}\)
c=\(8\sqrt{7a}\)
d,=\(11\sqrt{3}\)
câu 2 rút gọn A= \(\sqrt{12}+2\sqrt{27}+3\sqrt{45}-9\sqrt{48}\)
B=\(\left(\sqrt{48}-2\sqrt{75}+\sqrt{108}-\sqrt{147}\right):\sqrt{3}\)
\(A=\sqrt{12}+2\sqrt{27}+3\sqrt{45}-9\sqrt{48}\)
\(=2\sqrt{3}+6\sqrt{3}+9\sqrt{5}-36\sqrt{3}\)
\(=9\sqrt{5}-28\sqrt{3}\)
\(B=\left(\sqrt{48}-2\sqrt{75}+\sqrt{108}-\sqrt{147}\right):\sqrt{3}\)
\(=4-2\cdot5+6-7\)
\(=4-10+6-7\)
=-7
A=\(\sqrt{12}\)+2\(\sqrt{27}\)+3\(\sqrt{45}\) -9\(\sqrt{48}\)
=\(\sqrt{4.3}\) +2\(\sqrt{9.3}\)+3\(\sqrt{9.5}\) -9\(\sqrt{16.3}\)
=2\(\sqrt{3}\) +6\(\sqrt{3}\)+9\(\sqrt{5}\) -36\(\sqrt{3}\)
=\(\sqrt{3}\)(2+6-36) + 9\(\sqrt{5}\)
=9\(\sqrt{5}\)- 28\(\sqrt{3}\)
Rút Gọn
a,\(\sqrt{75}-\sqrt{5\frac{1}{3}}+\frac{9}{2}\sqrt{2\frac{2}{3}}+2\sqrt{27}\)
b,\(\sqrt{48}+\sqrt{5\frac{1}{3}}+2\sqrt{75}-5\sqrt{1\frac{1}{3}}\)
c,\(\left(\sqrt{12}+2\sqrt{27}\right)\frac{\sqrt{3}}{2}-\sqrt{150}\)
d,\(\left(\sqrt{18}+\sqrt{0,5}-3\sqrt{\frac{1}{3}}\right)-\left(\sqrt{\frac{1}{8}-\sqrt{75}}\right)\)
e,\(6\sqrt{\frac{8}{9}}-5\sqrt{\frac{32}{25}}+14\sqrt{\frac{18}{49}}\)
f,\(2\sqrt{\frac{16}{3}}-3\sqrt{\frac{1}{27}}-6\sqrt{\frac{4}{75}}\)
g,\(\left(2\sqrt{\frac{16}{3}}-3\sqrt{\frac{1}{27}}-6\sqrt{\frac{4}{75}}\right)\sqrt{3}\)
h,\(\left(6\sqrt{\frac{8}{9}}-5\sqrt{\frac{32}{25}}+14\sqrt{\frac{18}{49}}\right)\sqrt{\frac{1}{2}}\)
i,\(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\)
j,\(\left(\sqrt{\frac{1}{7}}-\sqrt{\frac{16}{7}}+7\right):\sqrt{7}\)
Rút gọn biểu thức:
\(\sqrt{\frac{2a}{3}}.\sqrt{\frac{3a}{8}}vớia\ge0\)\(\sqrt{5a}.\sqrt{45a}-3avớia\ge0\)\(4\sqrt{16a^6}-6a^3\rightarrow kq2TH\)\(\left(3-a\right)^2-\sqrt{0,2}.\sqrt{180a^4}\)\(\sqrt{\frac{27.\left(a-3\right)^2}{48}}vớia< 3\)\(\frac{\sqrt{63y^3}}{\sqrt{7y}}vớiy>0\)\(\frac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^2}}vớia< 0,b\ne0\)\(\frac{a-b}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{a^3}+\sqrt{b^3}}{a-b}\left(a\ge0;b\ge0;a\ne b\right)\)\(\frac{2a+\sqrt{ab}-3b}{2a-5\sqrt{ab}+3b}\left(a,b\ge0;4a\ne9b\right)\)Bài 1: Rút gọn biểu thức
1) \(\sqrt{12}-\sqrt{27}+\sqrt{48}\) 2) \(\left(\sqrt{25}+\sqrt{20}-\sqrt{80}\right):\sqrt{5}\)
3) \(2\sqrt{27}-\sqrt{\frac{16}{3}}-\sqrt{48}-\sqrt{8\frac{1}{3}}\) 4) \(\frac{1}{\sqrt{5}-\sqrt{3}}-\frac{1}{\sqrt{5}+\sqrt{3}}\)
5) \(\left(\sqrt{125}-\sqrt{12}-2\sqrt{5}\right)\left(3\sqrt{5}-\sqrt{3}+\sqrt{27}\right)\) 6) \(\left(3\sqrt{20}-\sqrt{125}-15\sqrt{\frac{1}{5}}\right).\sqrt{5}\)
7) \(\left(6\sqrt{128}-\frac{3}{5}\sqrt{50}+7\sqrt{8}\right):3\sqrt{2}\) 8) \(\left(2\sqrt{48}-\frac{3}{2}\sqrt{\frac{4}{3}}+\sqrt{27}\right).2\sqrt{3}\)
9) \(\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{8}-4\right)^2}\) 10) \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}-3\right)^2}\)
11) \(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}\) 12) \(\left(1-\frac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\frac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
13) \(\sqrt{15-6\sqrt{6}}\) 14) \(\sqrt{8-2\sqrt{15}}\) 15) \(\sqrt[3]{-2}.\sqrt[3]{32}+\sqrt{2}.\sqrt{32}\)
a)\(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right):\sqrt{3}\)
b)\(\sqrt{27\left(1-\sqrt{3}\right)^4}:3\sqrt{15}\)
c) \(\frac{\sqrt{6+2\sqrt{5}}}{\sqrt{5}+1}\)
Bài 1: Rút gọn biểu thức:
a) \(\left(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{2}+\sqrt{20}-\dfrac{5}{4}\sqrt{\dfrac{4}{5}+\sqrt{5}}\right)\)
b) \(\dfrac{1}{3}\sqrt{48}+3\sqrt{75}-\sqrt{27}-10\sqrt{1\dfrac{1}{3}}\)
c) \(\dfrac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)
d) \(\sqrt{\dfrac{3}{4}}+\sqrt{\dfrac{1}{3}}+\sqrt{\dfrac{1}{12}}\)
Bài 2: Giải các phương trình sau:
a) \(x^2+4x+5=2\sqrt{2x+3}\)
b) \(x^2+9x+20=2\sqrt{3x+10}\)
c) \(x^2+7x+14=2\sqrt{x+4}\)
d) \(4\sqrt{x+1}=x^2-5x+14\)
e) \(\sqrt{6-x}=3x-4\)
f) \(\sqrt{5x-9}=9-2x\)
Mọi người làm ơn giúp mình với. Mình đang cần gấp ạ. Cảm ơn mọi người rất nhiều
Bài 1:
a) Ta có: \(\left(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{2}\sqrt{20}-\dfrac{5}{4}\sqrt{\dfrac{4}{5}}+\sqrt{5}\right)\)
\(=\left(\sqrt{5}+\sqrt{5}-\dfrac{5}{4}\cdot\dfrac{2}{\sqrt{5}}+\sqrt{5}\right)\)
\(=3\sqrt{5}-\dfrac{1}{2}\sqrt{5}\)
\(=\dfrac{5}{2}\sqrt{5}\)
c) Ta có: \(\dfrac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)
\(=\dfrac{\sqrt{35}\left(\sqrt{5}-\sqrt{7}+2\sqrt{2}\right)}{\sqrt{35}}\)
\(=2\sqrt{2}+\sqrt{5}-\sqrt{7}\)
Bài 2:
e) ĐKXĐ: \(\dfrac{4}{3}\le x\le6\)
Ta có: \(\sqrt{6-x}=3x-4\)
\(\Leftrightarrow6-x=\left(3x-4\right)^2\)
\(\Leftrightarrow9x^2-24x+16+6-x=0\)
\(\Leftrightarrow9x^2-25x+22=0\)
\(\Delta=\left(-25\right)^2-4\cdot9\cdot22=625-792< 0\)
Vậy: Phương trình vô nghiệm
Rút gọn
1,\(2\sqrt{\frac{16}{3}}-3\sqrt{\frac{1}{27}}-6\sqrt{\frac{4}{75}}\)
2,\(\left(2\sqrt{\frac{16}{3}}-3\sqrt{\frac{1}{27}}-6\sqrt{\frac{4}{75}}\right)\sqrt{3}\)
3,\(\left(6\sqrt{\frac{8}{9}}-5\sqrt{\frac{32}{25}}+14\sqrt{\frac{18}{49}}\right)\sqrt{\frac{1}{2}}\)
4,\(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\)
5,\(\left(\sqrt{\frac{1}{7}}-\sqrt{\frac{16}{7}}+\sqrt{7}\right):\sqrt{7}\)