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DM
0
22 tháng 3 2017
\(a^3=16-8\sqrt{5}+16+8\sqrt{5}+3.\sqrt[3]{16^2-8^2.5}a\)
\(a^3=32+3.\sqrt[3]{4^3\left(4-5\right)}a=32-12a\)
\(f\left(x\right)=\left[\left(32-12a\right)+12a-31\right]^{2016}=1^{2016}=1\)
NH
22 tháng 3 2017
a=\(\sqrt[3]{16-8\sqrt{5}}\)+\(\sqrt[3]{16+8\sqrt{5}}\)
=\(\sqrt[3]{1-3\sqrt{5}+15-5\sqrt{5}}+\sqrt[3]{1+3\sqrt{5}+15+5\sqrt{5}}\)=\(\sqrt[3]{\left(1-\sqrt{5}\right)^3}+\sqrt[3]{\left(1+\sqrt{5}\right)^3}\)
=1-\(\sqrt{5}+1+\sqrt{5}\)=2
thay vào ta được f(a)=(8+24-31)2016=(-1)2016=1
20 tháng 1 2019
\(a^3=16-8\sqrt{5}+16+8\sqrt{5}+96\sqrt[3]{\left(16-8\sqrt{5}\right)\left(16+8\sqrt{5}\right)}\)
\(a^3=32+96\sqrt[3]{-64}=32+96.\left(-4\right)=-352\)
đến đây dễ r
20 tháng 1 2019
\(a^3=32+3\sqrt[3]{\left(16-8\sqrt{5}\right)\left(16+8\sqrt{5}\right)}\left(\sqrt[3]{16+8\sqrt{5}}+\sqrt[3]{16-8\sqrt{5}}\right)\)
HA
2
E
3 tháng 11 2018
\(x^3=16-8\sqrt{5}+16+8\sqrt{5}+3\sqrt[3]{\left(16-8\sqrt{5}\right)\left(16+8\sqrt{5}\right)}\left(\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+5\sqrt{5}}\right)=32+3\sqrt[3]{256-320}.x=32-12x\)
<=> x3 +12x - 32 = 0
<=> x = 2
30 tháng 10 2020
a) Ta có: \(\frac{7\sqrt{2}+2\sqrt{7}}{\sqrt{14}}-\frac{5}{\sqrt{7}+\sqrt{5}}\)
\(=\frac{\sqrt{14}\left(\sqrt{7}+\sqrt{2}\right)}{\sqrt{14}}-\frac{5\left(\sqrt{7}-\sqrt{5}\right)}{\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}\)
\(=\frac{2\left(\sqrt{7}+\sqrt{2}\right)-5\left(\sqrt{7}-\sqrt{5}\right)}{2}\)
\(=\frac{2\sqrt{7}+2\sqrt{2}-5\sqrt{7}+5\sqrt{5}}{2}\)
\(=\frac{2\sqrt{2}-3\sqrt{7}+5\sqrt{5}}{2}\)
b) Ta có: \(\frac{\sqrt{2}\left(3+\sqrt{5}\right)}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{\sqrt{2}\left(3-\sqrt{5}\right)}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(=\frac{\sqrt{2}\left(6+2\sqrt{5}\right)}{4\sqrt{2}+\sqrt{2}\cdot\sqrt{6+2\sqrt{5}}}+\frac{\sqrt{2}\left(6-2\sqrt{5}\right)}{4\sqrt{2}-\sqrt{2}\cdot\sqrt{6-2\sqrt{5}}}\)
\(=\frac{6\sqrt{2}+2\sqrt{10}}{4\sqrt{2}+\sqrt{2}\cdot\sqrt{\left(\sqrt{5}+1\right)^2}}+\frac{6\sqrt{2}-2\sqrt{10}}{4\sqrt{2}-\sqrt{2}\cdot\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\frac{6\sqrt{2}+2\sqrt{10}}{4\sqrt{2}+\sqrt{2}\cdot\left|\sqrt{5}+1\right|}+\frac{6\sqrt{2}-2\sqrt{10}}{4\sqrt{2}-\sqrt{2}\cdot\left|\sqrt{5}-1\right|}\)
\(=\frac{6\sqrt{2}+2\sqrt{10}}{4\sqrt{2}+\sqrt{2}\left(\sqrt{5}+1\right)}+\frac{6\sqrt{2}-2\sqrt{10}}{4\sqrt{2}-\sqrt{2}\cdot\left(\sqrt{5}-1\right)}\)(Vì \(\sqrt{5}>1>0\))
\(=\frac{6\sqrt{2}+2\sqrt{10}}{4\sqrt{2}+\sqrt{10}+\sqrt{2}}+\frac{6\sqrt{2}-2\sqrt{10}}{4\sqrt{2}-\sqrt{10}+\sqrt{2}}\)
\(=\frac{6\sqrt{2}+2\sqrt{10}}{5\sqrt{2}+\sqrt{10}}+\frac{6\sqrt{2}-2\sqrt{10}}{5\sqrt{2}-\sqrt{10}}\)
\(=\frac{6+2\sqrt{5}}{5+\sqrt{5}}+\frac{6-2\sqrt{5}}{5-\sqrt{5}}\)
\(=\frac{\left(\sqrt{5}+1\right)^2}{\sqrt{5}\left(\sqrt{5}+1\right)}+\frac{\left(\sqrt{5}-1\right)^2}{\sqrt{5}\left(\sqrt{5}-1\right)}\)
\(=\frac{\sqrt{5}+1+\sqrt{5}-1}{\sqrt{5}}\)
\(=\frac{2\sqrt{5}}{\sqrt{5}}=2\)
c) Đặt \(A=\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}\)
Ta có: \(A=\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}\)
\(\Leftrightarrow A^3=32-12\cdot\left(\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}\right)\)
\(=32-12A\)
\(\Leftrightarrow A^3+12A-32=0\)
\(\Leftrightarrow A^3-2A^2+2A^2-4A+16A-32=0\)
\(\Leftrightarrow A^2\left(A-2\right)+2A\left(A-2\right)+16\left(A-2\right)=0\)
\(\Leftrightarrow\left(A-2\right)\left(A^2+2A+16\right)=0\)
mà \(A^2+2A+16>0\)
nên A-2=0
hay A=2
Vậy: \(\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}=2\)
AD
2
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 9 2019
\(x=\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16-8\sqrt{5}}\)
\(\Rightarrow x^3=32+3\sqrt[3]{16^2-8^2.5}\left(\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}\right)\)
\(\Rightarrow x^3=32-12x\)
\(\Rightarrow x^3+12x-32=0\)
\(\Rightarrow\left(x-2\right)\left(x^2+2x+16\right)=0\)
\(\Rightarrow x=2\)
Vậy \(\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}=2\)
25 tháng 9 2019
x=3√16−8√5+3√16−8√5x=16−853+16−853
⇒x3=32+33√162−82.5(3√16−8√5+3√16+8√5)⇒x3=32+3162−82.53(16−853+16+853)
⇒x3=32−12x⇒x3=32−12x
⇒x3+12x−32=0⇒x3+12x−32=0
⇒(x−2)(x2+2x+16)=0⇒(x−2)(x2+2x+16)=0
⇒x=2⇒x=2
Vậy 3√16−8√5+3√16+8√5=2
24 tháng 7 2016
2) \(A=\sqrt{15a^2-8a\sqrt{15}+16}\\ =\sqrt{\left(a\sqrt{15}-4\right)^2}\)
b) Khi a=\(\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}\) thì
\(A=\sqrt{\left[\left(\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}\right)\sqrt{15}-4\right]^2}\)
\(=\sqrt{\left[\left(3+5\right)-4\right]^2}\)
\(=\sqrt{4^2}\)
\(=4\)
18 tháng 6 2019
a, \(\sqrt{3-\sqrt{5}}+\sqrt{7-3\sqrt{5}}\)\(=\sqrt{\frac{1}{2}.\left(6-2\sqrt{5}\right)}\)\(+\sqrt{\frac{1}{2}.\left(14-2.3\sqrt{5}\right)}\)
\(=\sqrt{\frac{1}{2}.\left(\sqrt{5}-1\right)^2}\)\(+\sqrt{\frac{1}{2}.\left(3-\sqrt{5}\right)^2}\)\(=\frac{\sqrt{2}}{2}.\left(\sqrt{5}-1\right)+\frac{\sqrt{2}}{2}.\left(3-\sqrt{5}\right)\)
\(=\frac{\sqrt{2}}{2}.2=\sqrt{2}\)
Câu b đề đúng ko bn
11 tháng 8 2017
\(\sqrt[3]{16+8\sqrt{5}}+\sqrt[3]{16-8\sqrt{5}}=\sqrt[3]{\left(1+\sqrt{5}\right)^3}+\sqrt[3]{\left(1-\sqrt{5}\right)^3}=1+\sqrt{5}+1-\sqrt{5}=2\)
giải chi tiết hộ mk nhá
\(\sqrt[3]{16-8\sqrt{5}}\)=\(\sqrt[3]{1-3\sqrt{5}+15-5\sqrt{5}}\)=\(\sqrt[3]{1-3\sqrt{5}+3\left(\sqrt{5}\right)^2-\left(\sqrt{5}\right)^3}\)=\(\sqrt[3]{\left(1-\sqrt{5}\right)^3}\)=\(1-\sqrt{5}\)
làm tương tự: \(\sqrt[3]{16+8\sqrt{5}}\)=\(1+\sqrt{5}\)
suy ra: a = 2