Tính giá trị biểu thức:\(\frac{\sqrt{5+\sqrt{3}}+\sqrt{5-\sqrt{3}}}{\sqrt{5\sqrt{22}}}\)+\(\sqrt{11-6\sqrt{2}}\)
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tinh gia trị biểu thức P=\(\dfrac{\sqrt{5+\sqrt{3}}+\sqrt{5-\sqrt{3}}}{\sqrt{5+\sqrt{22}}}=\sqrt{11-6\sqrt{2}}\)
Tính giá trị ccacs biểu thức:
\(\frac{2}{\sqrt{6}-2}+\frac{2}{\sqrt{6}+2}+\frac{5}{\sqrt{6}}\)
\(\frac{1}{\sqrt{3}+\sqrt{2}-\sqrt{5}}-\frac{1}{\sqrt{3}+\sqrt{2}+\sqrt{5}}\)
\(\frac{2}{\sqrt{6}-2}+\frac{2}{\sqrt{6}+2}+\frac{5}{\sqrt{6}}.\)
\(=\frac{\sqrt{2}.\sqrt{2}}{\sqrt{2}\left(\sqrt{3}-1\right)}+\frac{\sqrt{2}.\sqrt{2}}{\sqrt{2}\left(\sqrt{3}+1\right)}+\frac{5}{\sqrt{6}}\)
\(=\frac{\sqrt{2}\left(\sqrt{3}+1\right)}{3-1}+\frac{\sqrt{2}\left(\sqrt{3}-1\right)}{3+1}+\frac{5}{\sqrt{6}}\)
\(=\frac{\left(\sqrt{3}+1\right)}{\sqrt{2}}+\frac{\sqrt{3}-1}{\sqrt{8}}+\frac{5}{\sqrt{6}}\)
\(=...\)
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\(a,\frac{2}{\sqrt{6}-2}+\frac{2}{\sqrt{6}+2}+\frac{5}{\sqrt{6}}\)
\(=\frac{2.\left(\sqrt{6}+2+\sqrt{6}-2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}+\frac{5\sqrt{6}}{6}\)
\(=\frac{4\sqrt{6}}{6-2^2}+\frac{5\sqrt{6}}{6}=2\sqrt{6}+\frac{5\sqrt{6}}{6}\)
\(=\frac{17\sqrt{6}}{6}\)
\(b,\frac{1}{\sqrt{3}+\sqrt{2}-\sqrt{5}}-\frac{1}{\sqrt{3}+\sqrt{2}+\sqrt{5}}\)
\(=\frac{\sqrt{3}+\sqrt{2}+\sqrt{5}-\left(\sqrt{3}+\sqrt{2}-\sqrt{5}\right)}{\left(\sqrt{3}+\sqrt{2}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{2}+\sqrt{5}\right)}\)
\(=\frac{2\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}\right)^2-5}\)
\(=\frac{2\sqrt{5}}{5+2\sqrt{6}-5}=\sqrt{\frac{5}{6}}\)
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a) Rút gọn biểu thức:\(\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{\sqrt{5}-5}{1-\sqrt{5}}\right):\frac{1}{\sqrt{2}-\sqrt{5}}\)
b) Tìm giá trị nhỏ nhất của biểu thức B=\(x^2-x\sqrt{3}+1\)
a) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\)
\(=\left[-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\cdot\left(\sqrt{2}-\sqrt{5}\right)\)
\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(2-5\right)\)
\(=-\left(-3\right)\)
\(=3\)
b) Ta có:
\(x^2-x\sqrt{3}+1\)
\(=x^2-2\cdot\dfrac{\sqrt{3}}{2}\cdot x+\left(\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
\(=\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
Mà: \(\left(x-\dfrac{\sqrt{3}}{2}\right)^2\ge0\forall x\) nên
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\)
Dấu "=" xảy ra:
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{\sqrt{3}}{2}\)
Vậy: GTNN của biểu thức là \(\dfrac{1}{4}\) tại \(x=\dfrac{\sqrt{3}}{2}\)
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a)
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\\ =\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}^2-\sqrt{5}^2\right)\\ =-\left(2-5\right)\\ =-\left(-3\right)\\ =3\)
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a) Cho x = \(\frac{\sqrt[3]{10+6\sqrt{3}}\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)Tính giá trị biểu thức: A = \(\left(x^3-4x+1\right)^{2018}\)
b) Cho x = \(\sqrt[3]{7+5\sqrt{2}}-\frac{1}{\sqrt[3]{7+5\sqrt{2}}}\)Tính giá trị biểu thức: B = \(\left(x^3+3x-14\right)^{2018}\)
ai nay dung kinh nghiem la chinh
cau a)
ta thay \(10+6\sqrt{3}=\left(1+\sqrt{3}\right)^3\)
\(6+2\sqrt{5}=\left(1+\sqrt{5}\right)^2\)
khi do \(x=\frac{\sqrt[3]{\left(\sqrt{3}+1\right)^3}\left(\sqrt{3}-1\right)}{\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{5}}\)
\(x=\frac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{1+\sqrt{5}-\sqrt{5}}\)
\(x=\frac{3-1}{1}=2\)
suy ra
x^3-4x+1=1
A=1^2018
A=1
b)
ta thay
\(7+5\sqrt{2}=\left(1+\sqrt{2}\right)^3\)
khi do
\(x=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\frac{1}{\sqrt[3]{\left(1+\sqrt{2}\right)^3}}\)
\(x=1+\sqrt{2}-\frac{1}{1+\sqrt{2}}=\frac{\left(1+\sqrt{2}\right)^2-1}{1+\sqrt{2}}=\frac{2+2\sqrt{2}}{1+\sqrt{2}}\)
x=2
thay vao
x^3+3x-14=0
B=0^2018
B=0
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Tính giá trị của biểu thức:
\(\sqrt{2+\sqrt{3+\sqrt[3]{4+5\sqrt[5]{6+7\sqrt[7]{8+9\sqrt[9]{10+11\sqrt[11]{12}}}}}}}\)
Tính giá trị biểu thức:
\(\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{6}}\)
Tính giá trị biểu thức
\(A=\frac{3+\sqrt{5}}{\sqrt{10}+\sqrt{3}+\sqrt{5}}-\frac{3-\sqrt{5}}{\sqrt{10}+\sqrt{3-\sqrt{5}}}\\ \\ \)
Đề bài : Tính giá trị biểu thức :
\(B=\frac{_{\left(3+\sqrt{5}\right).}\left(3-\sqrt{5}\right)}{\left(3+\sqrt{5}\right)}\)
\(C=\frac{1}{\sqrt{5}+\sqrt{3}}-\frac{1}{\sqrt{5}-\sqrt{3}_{ }}\)
\(D=\frac{2}{\sqrt{3}+1}+\frac{1}{\sqrt{3}-2}+\frac{6}{\sqrt{3}+3}\)
\(B=\frac{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}{3+\sqrt{5}}=3-\sqrt{5}\)
\(C=\frac{1}{\sqrt{5}+\sqrt{3}}-\frac{1}{\sqrt{5}-\sqrt{3}}\)
\(=\frac{\sqrt{5}-\sqrt{3}}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\frac{\sqrt{5}+\sqrt{3}}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}\)
\(=\frac{\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}}{2}\)
\(=\frac{-2\sqrt{3}}{2}=-\sqrt{3}\)
\(D=\frac{2}{\sqrt{3}+1}+\frac{1}{\sqrt{3}-2}+\frac{6}{\sqrt{3}+3}\)
\(=\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}+\frac{\sqrt{3}+2}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}+\frac{6\left(3-\sqrt{3}\right)}{\left(\sqrt{3}+3\right)\left(3-\sqrt{3}\right)}\)
\(=\sqrt{3}-1-\left(\sqrt{3}+2\right)-\left(3-\sqrt{3}\right)\)
\(=\sqrt{3}-1-\sqrt{3}-2-3+\sqrt{3}=\sqrt{3}-6\)
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Tính giá trị của biểu thức
\(H=5\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}-\sqrt{\frac{5}{2}}\right)^2+\)\(\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}-\sqrt{\frac{5}{2}}\right)^2\)
Bạn kiểm tra lại đề. Theo mình
\(H=5\left(\sqrt{2+\sqrt{3}}-\sqrt{3-\sqrt{5}}-\sqrt{\frac{5}{2}}\right)^2+\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}-\sqrt{\frac{3}{2}}\right)^2\)