bài 1: tính
\(\sqrt{4+\sqrt{5\sqrt{3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}}\)
Bài 2: cho biểu thức: \(\left(x+\sqrt{x^2+2006}\right)\left(y+\sqrt{y^2+2006}\right)=2006\)
Hãy tính tổng: S= x+y
Chứng tỏ rằng
\(\frac{2}{3\left(1+\sqrt{2}\right)}+\frac{2}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{2}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{2}{4011\left(\sqrt{2005}+\sqrt{2006}\right)}<1-\frac{1}{\sqrt{2006}}\)
Chứng tỏ rằng
\(\frac{2}{3\left(1+\sqrt{2}\right)}+\frac{2}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{2}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{2}{4011\left(\sqrt{2005}+\sqrt{2006}\right)}<1-\frac{1}{\sqrt{2006}}\)
chứng tỏ rằng :
\(\frac{2}{3\left(1+\sqrt{2}\right)}+\frac{2}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{2}{7\left(\sqrt{3}+\sqrt{4}\right)}+...\frac{2}{4011\left(\sqrt{2005}+\sqrt{2006}\right)}
\(VT=\frac{2\left(\sqrt{2}-\sqrt{1}\right)}{3.\left(2-1\right)}+\frac{2\left(\sqrt{3}-\sqrt{2}\right)}{5\left(3-2\right)}+...+\frac{2\left(\sqrt{2006}-\sqrt{2005}\right)}{4011\left(2006-2005\right)}\)
\(VT=\frac{2\left(\sqrt{2}-\sqrt{1}\right)}{3}+\frac{2\left(\sqrt{3}-\sqrt{2}\right)}{5}+...+\frac{2\left(\sqrt{2006}-\sqrt{2005}\right)}{4011}\)
Nhận xét: (a-b)2 \(\ge\) 0 => a2 + b2 \(\ge\) 2ab
Áp dụng ta có: \(3=\left(\sqrt{2}\right)^2+\left(\sqrt{1}\right)^2\ge2.\sqrt{2}.\sqrt{1}\)
\(5=\left(\sqrt{3}\right)^2+\left(\sqrt{2}\right)^2\ge2.\sqrt{3}.\sqrt{2}\)
...
\(4011=\left(\sqrt{2006}\right)^2+\left(\sqrt{2005}\right)^2\ge2.\sqrt{2006}.\sqrt{2005}\)
=> \(VT
Bài 1: Tính
a) \(\left(\sqrt{x}-\sqrt{y}\right).\left(\sqrt{x}+\sqrt{y}\right)\)
b) \(\left(\sqrt{x}-3\right).\left(\sqrt{x}+2\right)\)
c) \(\sqrt{\left(2-\sqrt{5}\right)^2.\left(2+\sqrt{5}\right)^2}\)
d) \(\sqrt{\left(-5\right)^2.3^2}\)
e) \(\sqrt{\frac{5}{27}.\frac{8}{20}}\)
Bài 2: Tính
a) \(\left(\sqrt{14}+\sqrt{6}\right).\sqrt{5-\sqrt{27}}\)
b) \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
c) \(\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
1)
a)\(\left(\sqrt{x}-\sqrt{y}\right).\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}\right)^2-\left(\sqrt{y}\right)^2=x-y\)
b)\(\left(\sqrt{x}-3\right).\left(\sqrt{x}+2\right)=\left(\sqrt{x}\right)^2+2\sqrt{x}-3\sqrt{x}-6=x-\sqrt{x}-6\)
c)\(\sqrt{\left(2-\sqrt{5}\right)^2.\left(2+\sqrt{5}\right)^2}=\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)=4-\left(\sqrt{5}\right)^2\)
=\(4-5=1\)
d)\(\sqrt{\left(-5\right)^2.3^2}=\left(5\right).3=15\)
e)\(\sqrt{\frac{5}{27}.\frac{8}{20}}=\sqrt{\frac{2}{27}}\)
ĐÂy này nhớ **** vài câu nha
a, \(\left(\sqrt{14}+\sqrt{6}\right)\sqrt{5-\sqrt{21}}=\left(\sqrt{7}+\sqrt{3}\right)\sqrt{2}\sqrt{5-\sqrt{21}}\)
\(=\left(\sqrt{7}+\sqrt{3}\right)\sqrt{10-2\sqrt{21}}=\left(\sqrt{7}+\sqrt{3}\right)\sqrt{7-2.\sqrt{3}.\sqrt{7}+3}\)
\(=\left(\sqrt{7}+\sqrt{3}\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)=7-3=4\)
1/Cho \(x+y+z+\sqrt{xyz}=4\)
Tính giá trị biểu thức \(T=\sqrt{x\left(4-y\right)\left(4-z\right)}+\sqrt{y\left(4-x\right)\left(4-z\right)}+\sqrt{z\left(4-x\right)\left(4-y\right)}-\sqrt{xyz}\)
2/Cho \(x=\sqrt[3]{4+2\sqrt{2}}+\sqrt[3]{4-2\sqrt{2}}\)
Tính giá trị biểu thức \(F=\left(x^3-6x-10\right)^{2019}\)
3/Cho \(x=\sqrt{\frac{1}{2\sqrt{3}-2}-\frac{3}{2\sqrt{3}+2}}\)
Tính giá trị biểu thức \(P=x^2+\frac{x-1}{2}\)
4/Cho \(x=\sqrt{28-10\sqrt{3}}\)
Tính giá trị biểu thức \(F=\frac{2x^4-21x^3+55x^2-32x-4012}{x^2-10x+20}\)
Bài 1: Tính
a) \(\sqrt{27}+\dfrac{1}{2}\sqrt{48}-\sqrt{108}\)
b) \(\left(\sqrt{14}-\sqrt{10}\right)\sqrt{6+\sqrt{35}}\)
c) \(\dfrac{\sqrt{15}+\sqrt{3}}{1+\sqrt{5}}-\dfrac{2}{\sqrt{3}-1}\)
Bài 2: Cho biểu thức
A = \(\dfrac{x-5}{x+2\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}+\dfrac{2}{\sqrt{x}-1}\)
a) Rút gọn A
b) Tìm x để A = 2
c) Tìm các số nguyên của x để A ∈ Z
Bài 1:
a: \(\sqrt{27}+\dfrac{1}{2}\sqrt{48}-\sqrt{108}\)
\(=3\sqrt{3}+\dfrac{1}{2}\cdot4\sqrt{3}-6\sqrt{3}\)
\(=-3\sqrt{3}+2\sqrt{3}=-\sqrt{3}\)
b: \(\left(\sqrt{14}-\sqrt{10}\right)\cdot\sqrt{6+\sqrt{35}}\)
\(=\left(\sqrt{7}-\sqrt{5}\right)\cdot\sqrt{2}\cdot\sqrt{6+\sqrt{35}}\)
\(=\left(\sqrt{7}-\sqrt{5}\right)\cdot\sqrt{12+2\sqrt{35}}\)
\(=\left(\sqrt{7}-\sqrt{5}\right)\cdot\sqrt{\left(\sqrt{7}+\sqrt{5}\right)^2}\)
\(=\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)=7-5=2\)
c: \(\dfrac{\sqrt{15}+\sqrt{3}}{1+\sqrt{5}}-\dfrac{2}{\sqrt{3}-1}\)
\(=\dfrac{\sqrt{3}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}-\dfrac{2\left(\sqrt{3}+1\right)}{3-1}\)
\(=\sqrt{3}-\sqrt{3}-1=-1\)
Bài 2:
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\end{matrix}\right.\)
\(A=\dfrac{x-5}{x+2\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}+\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-5}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}+3}+\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-5+\sqrt{x}-1+2\sqrt{x}+6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
b: A=2
=>\(\sqrt{x}=2\left(\sqrt{x}-1\right)\)
=>\(2\sqrt{x}-2=\sqrt{x}\)
=>\(\sqrt{x}=2\)
=>x=4(nhận)
c: Để A là số nguyên thì \(\sqrt{x}⋮\sqrt{x}-1\)
=>\(\sqrt{x}-1+1⋮\sqrt{x}-1\)
=>\(\sqrt{x}-1\inƯ\left(1\right)\)
=>\(\sqrt{x}-1\in\left\{1;-1\right\}\)
=>\(\sqrt{x}\in\left\{2;0\right\}\)
=>\(x\in\left\{4;0\right\}\)
Cho biểu thức \(P=x^3+y^3-3\left(x+y\right)+2021\). Tính giá trị biểu thức P với :
\(x=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)
và \(y=\sqrt[3]{3-2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\)
Cho biểu thức P=\(x^3+y^3-3\left(x+y\right)+1993\) . Tính giá trị biểu thức P với : \(x=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\) và \(y=\sqrt[3]{3-2\sqrt{2}}+\sqrt[3]{3+2\sqrt{2}}\)
\(x=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\\ \Leftrightarrow x^3=9+4\sqrt{5}+9-4\sqrt{5}+3\sqrt[3]{\left(9-4\sqrt{5}\right)\left(9+4\sqrt{5}\right)}\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\\ \Leftrightarrow x^3=18+3x\sqrt[3]{81-80}=18-3x\\ \Leftrightarrow x^3-3x=18\\ y=\sqrt[3]{3-2\sqrt{2}}+\sqrt[3]{3+2\sqrt{2}}\\ \Leftrightarrow y^3=6+3\sqrt[3]{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}\left(\sqrt[3]{3-2\sqrt{2}}+\sqrt[3]{3+2\sqrt{2}}\right)\\ \Leftrightarrow y^3=6+3y\sqrt[3]{9-8}=6+3y\\ \Leftrightarrow y^3-3y=6\\ \Leftrightarrow P=x^3+y^3-3\left(x+y\right)+1993\\ P=x^3+y^3-3x-3y+1993=18+6+1993=2017\)
Áp dụng: \(\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3=a^3+b^3+3ab\left(a+b\right)\)
\(x=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)
\(\Rightarrow x^3=9+4\sqrt{5}+9-4\sqrt{5}+3\sqrt[3]{\left(9+4\sqrt{5}\right)\left(9-4\sqrt{5}\right)}\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\)
\(=18+3\sqrt[3]{81-80}.x=18+3x\)
\(y=\sqrt[3]{3-2\sqrt{2}}+\sqrt[3]{3+2\sqrt{2}}\)
\(\Rightarrow y^3=3-2\sqrt{2}+3+2\sqrt{2}+3\sqrt[3]{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}\left(\sqrt[3]{3-2\sqrt{2}}+\sqrt[3]{3+2\sqrt{2}}\right)\)
\(=6+3\sqrt[3]{9-8}y=6+3y\)
\(P=x^3+y^3-3\left(x+y\right)+1993\)
\(=18+3x+6+3y-3x-3y+1993=2017\)
Tính giá trị các biểu thức:
a.\(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right)\sqrt{3}\)
b.\(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)
c.\(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)3\sqrt{6}\)
d.\(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)
a) Ta có: \(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right)\cdot\sqrt{3}\)
\(=\left(7\cdot4\sqrt{3}+3\cdot3\sqrt{3}-2\cdot2\sqrt{3}\right)\cdot\sqrt{3}\)
\(=33\sqrt{3}\cdot\sqrt{3}\)
=99
b) Ta có: \(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)
\(=\left(12\cdot5\sqrt{2}-8\cdot10\sqrt{2}+7\cdot15\sqrt{2}\right):\sqrt{10}\)
\(=\dfrac{85\sqrt{2}}{\sqrt{10}}=\dfrac{85}{\sqrt{5}}=17\sqrt{5}\)
c) Ta có: \(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\cdot2\sqrt{2}\right)\cdot3\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+3\sqrt{2}\right)\cdot3\sqrt{6}\)
\(=36-36\sqrt{2}+18\sqrt{3}\)
d) Ta có: \(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)
\(=3\cdot\sqrt{75\sqrt{2}}+5\cdot\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)
\(=3\cdot5\sqrt{2}\cdot\sqrt{\sqrt{2}}+4\sqrt{3}\sqrt{\sqrt{2}}\)
\(=15\sqrt{\sqrt{8}}+4\sqrt{\sqrt{18}}\)
a,=\(\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right).\sqrt{3}\)
\(=28.3+9.3-4.3=99\)
b,\(=\left(60\sqrt{2}-80\sqrt{2}+175\sqrt{2}\right):\sqrt{10}\)
\(=155\sqrt{2}:\sqrt{10}=\dfrac{155}{\sqrt{5}}\)
d,Ta có:\(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)
\(=3\sqrt{75\sqrt{2}}+5\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)
\(=15\sqrt{3\sqrt{2}}+20\sqrt{3\sqrt{2}}-16\sqrt{3\sqrt{2}}\)
\(=19\sqrt{3\sqrt{2}}\)