a: \(\Leftrightarrow\sqrt{6}\left(x+1\right)=5\sqrt{6}\)
=>x+1=5
=>x=4
b: =>x^2/10=1,1
=>x^2=11
=>x=căn 11 hoặc x=-căn 11
c: =>(4x+3)/(x+1)=9 và (4x+3)/(x+1)>=0
=>4x+3=9x+9
=>-5x=6
=>x=-6/5
d: =>(2x-3)/(x-1)=4 và x-1>0 và 2x-3>=0
=>2x-3=4x-4 và x>=3/2
=->-2x=-1 và x>=3/2
=>x=1/2 và x>=3/2
=>Ko có x thỏa mãn
e: Đặt căn x=a(a>=0)
PT sẽ là a^2-a-5=0
=>\(\left[{}\begin{matrix}a=\dfrac{1+\sqrt{21}}{2}\left(nhận\right)\\a=\dfrac{1-\sqrt{21}}{2}\left(loại\right)\end{matrix}\right.\)
=>x=(1+căn 21)^2/4=(11+căn 21)/2
a) \(2\sqrt{24}-5\sqrt{54}+\sqrt{10+4\sqrt{6}}\)
b) \(\dfrac{\sqrt{18}-\sqrt{12}}{\sqrt{6}-2}+\dfrac{4}{\sqrt{3}+1}+\sqrt{\left(3\sqrt{3}-12\right)^2}\)
\(a,=4\sqrt{6}-15\sqrt{6}+\sqrt{\left(2+\sqrt{6}\right)^2}=-11\sqrt{6}+2+\sqrt{6}=2-10\sqrt{6}\\ b,=\dfrac{\sqrt{3}\left(\sqrt{6}-2\right)}{\sqrt{6}-2}+\dfrac{4\left(\sqrt{3}-1\right)}{2}+\left|3\sqrt{3}-12\right|=\sqrt{3}+2\sqrt{3}-2+12-3\sqrt{3}=10\)
rút gọn biểu thức sau
a,\(5\sqrt{48}-4\sqrt{27}-2\sqrt{57}+\sqrt{108}\)
b,\(2\sqrt{24}-2\sqrt{54}+3\sqrt{6}-\sqrt{150}\)
a) \(5\sqrt{48}-4\sqrt{27}-2\sqrt{57}+\sqrt{108}\)
\(=20\sqrt{3}-12\sqrt{3}-2\sqrt{57}+6\sqrt{3}\)
\(=\left(20-12+6\right)\sqrt{3}-2\sqrt{57}\)
\(=14\sqrt{3}-2\sqrt{57}\)
b) \(2\sqrt{24}-2\sqrt{54}+3\sqrt{6}-\sqrt{150}\)
\(=4\sqrt{6}-6\sqrt{6}+3\sqrt{6}-5\sqrt{6}\)
\(=\left(4-6+3-5\right)\sqrt{6}\)
\(=-4\sqrt{6}\)
\(x^3=\left(\sqrt[3]{5+2\sqrt{6}}+\sqrt[3]{5-2\sqrt{6}}\right)^3=\sqrt[3]{5+2\sqrt{6}}^3\)
\(+3\sqrt[3]{\left(5+2\sqrt{6}\right)^2}.\sqrt[3]{5-2\sqrt{6}}+3\sqrt[3]{5+2\sqrt{6}}.\sqrt[3]{\left(5-2\sqrt{6}\right)^2}+\sqrt[3]{5-2\sqrt{6}}^3\)
\(=5+2\sqrt{6}+3\sqrt[3]{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)}.\sqrt[3]{5+2\sqrt{6}}\)
\(+3\sqrt[3]{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)}.\sqrt[3]{5-2\sqrt{6}}+5-2\sqrt{6}\)
\(=5+5+3\sqrt[3]{\left(25-4.6\right)}.\sqrt[3]{5+2\sqrt{6}}+3\sqrt[3]{\left(25-4.6\right)}.\sqrt[3]{5-2\sqrt{6}}\)
\(=10+ 3\sqrt[3]{5+2\sqrt{6}}+3\sqrt[3]{5-2\sqrt{6}}\)
p/s : có bạn hỏi nên mình đăng , các bạn đừng report nhé
\(\dfrac{2\sqrt{30}}{\sqrt{5}+\sqrt{6}+\sqrt{7}} \)
\(\sqrt{24}+6\sqrt{\dfrac{2}{3}+\dfrac{10}{\sqrt{6}-1}}\)
\(\dfrac{2\sqrt{15}+\sqrt{16}}{\sqrt{84}+\sqrt{6}}\)
\(2\sqrt{40\sqrt{2}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)
\(\dfrac{\left(2+\sqrt{3}\right)^2-1}{\left(\sqrt{3}+1\right)^2}:\dfrac{\left(3+\sqrt{5}\right)^2-4}{\left(\sqrt{5}+1\right)^2}\)
giúp em với ạ
\(2\sqrt{40\sqrt{3}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)
\(=2\cdot\sqrt{40\sqrt{3}}-2\cdot\sqrt{5\sqrt{3}}-3\cdot\sqrt{20\sqrt{3}}\)
\(=2\cdot2\sqrt{10}\cdot\sqrt{\sqrt{3}}-2\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}-6\sqrt{5}\cdot\sqrt{\sqrt{3}}\)
\(=4\sqrt{10}\sqrt{\sqrt{3}}-4\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}\)
bài 1 rút gọn biểu thức sau:
a)\(\sqrt{16+6\sqrt{7}}\)- \(\sqrt{8-2\sqrt{7}}\) b)K=\(\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}\)
c)\(\sqrt{60-24\sqrt{6}}\)+\(\sqrt{40-16\sqrt{6}}\) d)B=(3+\(\sqrt{3}\))\(\sqrt{12-6\sqrt{13}}\)
e)\(\sqrt{6-4\sqrt{2}}\)-\(\sqrt{\left(\sqrt{2}-\sqrt{6}\right)^2}\)
bài 2 cho biểu thức A=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{3}{\sqrt{x}-3}\right).\dfrac{\sqrt{x}+3}{x+9}\)( với x≥0 và x≠ 9)
a) rút gọn biểu thức A
b) tính giá trị biểu thức\(x=4+2\sqrt{3}\)
\(1,\\ a,=\sqrt{\left(3+\sqrt{7}\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}=3+\sqrt{7}-\sqrt{7}+1=4\\ b,K=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}-1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\ c,=\sqrt{\left(6-2\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-4\right)^2}=6-2\sqrt{6}+2\sqrt{6}-4=2\\ e,=\sqrt{\left(2-\sqrt{2}\right)^2}-\left(\sqrt{6}-\sqrt{2}\right)=2-\sqrt{2}-\sqrt{6}+\sqrt{2}=2-\sqrt{6}\)
\(2,\\ a,A=\dfrac{x-3\sqrt{x}+3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+3}{x+9}\\ A=\dfrac{x+9}{\left(\sqrt{x}-3\right)\left(x+9\right)}=\dfrac{1}{\sqrt{x}-3}\\ b,x=4+2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{3}+1\\ \Leftrightarrow A=\dfrac{1}{\sqrt{3}+1-3}=\dfrac{1}{\sqrt{3}+2}=2-\sqrt{3}\)
\(\sqrt{7+\sqrt{24}}\)
\(\sqrt{2-\sqrt{3}}\)
\(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt[]{6}}\)
\(\sqrt{7+\sqrt{24}}=\sqrt{7+2\sqrt{6}}=\sqrt{\left(\sqrt{6}\right)^2+2.\sqrt{6}.1+1^2}\)
\(=\sqrt{\left(\sqrt{6}+1\right)^2}=\left|\sqrt{6}+1\right|=\sqrt{6}+1\)
\(\sqrt{2-\sqrt{3}}=\sqrt{\dfrac{4-2\sqrt{3}}{2}}=\sqrt{\dfrac{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1^2}{2}}\)
\(=\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{2}}=\dfrac{\left|\sqrt{3}-1\right|}{\sqrt{2}}=\dfrac{\sqrt{3}-1}{\sqrt{2}}\)
\(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}\)
\(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\left|\sqrt{3}+\sqrt{2}\right|-\left|\sqrt{3}-\sqrt{2}\right|\)
\(=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)
Tính:
a/ \(\sqrt{8+2\sqrt{15}}+\frac{2}{\sqrt{5}+\sqrt{3}}\)
b/ \(\sqrt{7+2\sqrt{6}}+\frac{6-2\sqrt{6}}{\sqrt{6}}-\sqrt{54}\)
Lời giải:
a)
\(\sqrt{8+2\sqrt{15}}+\frac{2}{\sqrt{5}+\sqrt{3}}=\sqrt{3+5+2\sqrt{3}.\sqrt{5}}+\frac{2}{\sqrt{5}+\sqrt{3}}\)
\(=\sqrt{(\sqrt{3}+\sqrt{5})^2}+\frac{2(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}=\sqrt{3}+\sqrt{5}+\frac{2(\sqrt{5}-\sqrt{3})}{5-3}\)
\(=\sqrt{3}+\sqrt{5}+\sqrt{5}-\sqrt{3}=2\sqrt{5}\)
b)
\(\sqrt{7+2\sqrt{6}}+\frac{6-2\sqrt{6}}{\sqrt{6}}-\sqrt{54}=\sqrt{6+1+2\sqrt{6}.\sqrt{1}}+\sqrt{6}-2-3\sqrt{6}\)
\(=\sqrt{(\sqrt{6}+1)^2}+\sqrt{6}-2-3\sqrt{6}\)
\(=\sqrt{6}+1+\sqrt{6}-2-3\sqrt{6}=-(\sqrt{6}+1)\)
\(a.\sqrt{8+2\sqrt{15}}+\frac{2}{\sqrt{5}+\sqrt{3}}\\ =\sqrt{5+2\cdot\sqrt{5}\cdot\sqrt{3}+3}+\frac{2}{\sqrt{5}+\sqrt{3}}\\ =\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}+\frac{2}{\sqrt{5}+\sqrt{3}}\\ =\sqrt{5}+\sqrt{3}+\frac{2}{\sqrt{5}+\sqrt{3}}\\ =\frac{\left(\sqrt{5}+\sqrt{3}\right)^2+2}{\sqrt{5}+\sqrt{3}}\\ =\frac{8+2\sqrt{15}+2}{\sqrt{5}+\sqrt{3}}\\ =\frac{10+2\sqrt{15}}{\sqrt{5}+\sqrt{3}}=\frac{2\sqrt{5}\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{5}+\sqrt{3}}=2\sqrt{5}\)
Tính
a) \(\left(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}+4\sqrt{\dfrac{3}{2}}\right)\times\left(2\sqrt{\dfrac{2}{3}}-\sqrt{12}-\sqrt{6}\right)\)
b) \(\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
c) \(\left(\sqrt{11+2\sqrt{24}}-\sqrt{11-2\sqrt{24}}\right):2\sqrt{3}\)
d) \(\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}\)
e) \(\sqrt{5+6\sqrt{2}}-\sqrt{9-6\sqrt{2}}-\sqrt{21-12\sqrt{3}}\)
f) \(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
b: \(=\sqrt{5}-1-\sqrt{5}-1=-2\)
c: \(=\dfrac{\left(2\sqrt{2}+\sqrt{3}-2\sqrt{2}+\sqrt{3}\right)}{2\sqrt{3}}=1\)
d: \(=\dfrac{\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}-1-\sqrt{5}-1}{\sqrt{2}}=-\sqrt{2}\)
Bài 1: Chứng minh:
a) \(\sqrt{9+\sqrt{17}}-\sqrt{9-\sqrt{17}}-\sqrt{2}=0\)
b) \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}=\sqrt{5}+1\)
c) \(\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}< 24\)
d) \(\sqrt{6+\sqrt{6+\sqrt{6+,,,+\sqrt{6}}}}=3\)