rút gọn \(B=\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\)
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7 tháng 7 2021
a) Ta có: \(A^3=\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)^3\)
\(=2+\sqrt{5}+2-\sqrt{5}+3\cdot\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)
\(=4-3\cdot A\)
\(\Leftrightarrow A^3+3A-4=0\)
\(\Leftrightarrow A^3-A+4A-4=0\)
\(\Leftrightarrow A\left(A-1\right)\left(A+1\right)+4\left(A-1\right)=0\)
\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)
\(\Leftrightarrow A=1\)
2
24 tháng 9 2023
`a)(\sqrt{14}-3\sqrt{2})^2+6\sqrt{28}`
`=14-12\sqrt{7}+18+12\sqrt{7}=32`
`b)2\sqrt{20}-3\sqrt{20}+\sqrt{125}`
`=4\sqrt{5}-6\sqrt{5}+5\sqrt{5}`
`=3\sqrt{5}`.
H9
HT.Phong (9A5)
CTVHS
24 tháng 9 2023
a) \(\left(\sqrt{14}-3\sqrt{2}\right)^2-6\sqrt{28}\)
\(=\left(\sqrt{14}\right)^2-2\cdot\sqrt{14}\cdot3\sqrt{2}+\left(3\sqrt{2}\right)^2+6\sqrt{28}\)
\(=14-6\sqrt{28}+18+6\sqrt{28}\)
\(=14+18\)
\(=32\)
b) \(2\sqrt{20}-3\sqrt{20}+\sqrt{125}\)
\(=2\cdot2\sqrt{5}-3\cdot2\sqrt{5}+5\sqrt{5}\)
\(=4\sqrt{5}-6\sqrt{5}+5\sqrt{5}\)
\(=3\sqrt{5}\)
TH
1
2 tháng 9 2017
\(A=\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}=\sqrt[3]{2^3+3.2^2.\sqrt{2}+3.2.\left(\sqrt{2}\right)^2+\left(\sqrt{2}\right)^3}+\sqrt[3]{2^3-3.2^2.\sqrt{2}+3.2.\left(\sqrt{2}\right)^2-\left(\sqrt{2}\right)^3}\)\(=\sqrt[3]{\left(2+\sqrt{2}\right)^3}+\sqrt[3]{\left(2-\sqrt{2}\right)^3}=2+\sqrt{2}+2-\sqrt{2}=4.\)
SD
0
ML
16 tháng 7 2015
\(A=\sqrt[3]{\left(\frac{1}{2}+\frac{1}{2}\sqrt{13}\right)^3}+\sqrt[3]{\left(\frac{1}{2}-\frac{1}{2}\sqrt{13}\right)^3}\)
\(=\frac{1}{2}+\frac{\sqrt{13}}{2}+\frac{1}{2}-\frac{\sqrt{13}}{2}=1\)
\(B=\sqrt[3]{\left(2+\sqrt{2}\right)^3}+\sqrt[3]{\left(2-\sqrt{2}\right)^3}=2+\sqrt{2}+2-\sqrt{2}=4\)
11 tháng 8 2017
cau a,b,c thay no co chung 1 dang do la
\(\sqrt[3]{a+m}+\sqrt[3]{a-m}\)
dang nay co 2 cach
C1: nhanh kho nhin de sai
VD: cau B
\(B^3=40+3\sqrt[3]{\left(20+14\sqrt{2}\right)\left(20-14\sqrt{2}\right)}\left(B\right)\)
B^3=40+3(2)(B)
B^3=40+6B
B=4
C2: hoi dai nhung de nhin
dat \(a=\sqrt[3]{20+14\sqrt{2}};b=\sqrt[3]{20-14\sqrt{2}}\)
de thay B=a+b
ab=2
a^3+b^3=40
suy ra B^3=a^3+b^3+3ab(a+b)
B^3=40+6B
B=4
giai tuong tu
con co cach nay nhung it su dung vi kho tim
C3: dua ve tong lap phuong
VD:cau B
\(20+14\sqrt{2}=\left(2+\sqrt{2}\right)^3\)
\(20-14\sqrt{2}=\left(2-\sqrt{2}\right)^3\)
de thay
B=4
cau d)
dung CT nay
\(\sqrt[m]{a}=\sqrt[m\cdot n]{\left(a\right)^n}\)
ap dung vao bai
\(\sqrt[3]{2\sqrt{3}-4\sqrt{2}}=\sqrt[6]{\left(2\sqrt{3}-4\sqrt{2}\right)^2}=\sqrt[6]{44-16\sqrt{6}}\)
nhanh vao
\(\sqrt[6]{\left(44-16\sqrt{6}\right)\left(44+16\sqrt{6}\right)}=\sqrt[6]{400}=\sqrt[3]{20}\)
19 tháng 4 2019
\(A=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\Leftrightarrow A^3=2+\sqrt{5}+2-\sqrt{5}+3\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\Leftrightarrow A^3=4+3\sqrt[3]{-1}.A\Leftrightarrow A^3=4-3A\Leftrightarrow A^3+3A-4=0\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)(1)
Ta có \(A^2+A+4>0\)
Vậy (1)\(\Leftrightarrow A-1=0\Leftrightarrow A=1\)
Vậy A=1
\(B=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\Leftrightarrow B^3=5\sqrt{2}+7-5\sqrt{2}+7-3\sqrt[3]{\left(5\sqrt{2}+7\right)\left(5\sqrt{2}-7\right)}\left(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\right)\Leftrightarrow B^3=14-3\sqrt[3]{1}.B\Leftrightarrow B^3=14-3B\Leftrightarrow B^3+3B-14=0\Leftrightarrow\left(B-2\right)\left(B^2+2B+7\right)=0\left(2\right)\)
Ta lại có \(B^2+2B+7>0\)
Vậy (2)\(\Leftrightarrow B-2=0\Leftrightarrow B=2\)
Vậy B=2
\(C=\sqrt[3]{20+14\sqrt{2}}-\sqrt[3]{14\sqrt{2}-20}=\sqrt[3]{\left(\sqrt{2}\right)^3+3.\left(\sqrt{2}\right)^2.2+3.\sqrt{2}.4+8}-\sqrt[3]{\left(\sqrt{2}\right)^3-3.\left(\sqrt{2}\right)^2.2+3.\sqrt{2}.4-8}=\sqrt[3]{\left(\sqrt{2}+2\right)^2}-\sqrt[3]{\left(\sqrt{2}-2\right)}=\sqrt{2}+2-\sqrt{2}+2=4\)
LL
1
NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 8 2021
Đặt \(x=\sqrt[3]{20+14\sqrt[]{2}}+\sqrt[3]{20-14\sqrt[]{2}}\)
\(\Rightarrow x^3=40+3\sqrt[3]{\left(20+14\sqrt[]{2}\right)\left(20-14\sqrt[]{2}\right)}.\left(\sqrt[3]{20+14\sqrt[]{2}}+\sqrt[3]{20-14\sqrt[]{2}}\right)\)
\(\Rightarrow x^3=40+6x\)
\(\Rightarrow x^3-6x-40=0\)
\(\Rightarrow\left(x-4\right)\left(x^2+4x+10\right)=0\)
\(\Rightarrow x=4\)
Vậy \(\sqrt[3]{20+14\sqrt[]{2}}+\sqrt[3]{20-14\sqrt[]{2}}=4\)
Áp dụng hằng đẳng thức (a+b)3 = a3 + 3a2b + 3ab2 + b3 = a3 + b3 + 3ab.(a +b) ta có:
\(B^3=20+14\sqrt{2}+20-14\sqrt{2}+3\sqrt[3]{20+14\sqrt{2}}.\sqrt[3]{20-14\sqrt{2}}\left(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\right)\)
\(B^3=40+3\sqrt[3]{\left(20+14\sqrt{2}\right)\left(20-14\sqrt{2}\right)}.B\)
\(B^3=40+3.\sqrt[3]{400-392}.B=40+3.\sqrt[3]{8}.B=40+6B\)
=> B3 - 6B - 40 = 0
<=> B3 - 64 - 6B + 24 = 0
<=> (B - 4 ).(B2 + 4B + 16) - 6.(B - 4) = 0
<=> (B - 4).(B2 + 4B + 16 - 6) = 0 <=> B = 4 hoặc B2 + 4B + 10 = 0
B2 + 4B + 10 = 0 Vô nghiêm vì \(\Delta\) = 16 - 40 = -24 < 0
Vậy B = 4