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a) Đặt A=\(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
<=> \(\sqrt{2}\cdot A=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}\)=\(\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}\)
= \(\sqrt{7}+1-\sqrt{7}+1=2\)
=> \(A=\frac{2}{\sqrt{2}}\sqrt{2}\)
b) Ta đặt \(B=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
=> \(B^2=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}\)
= \(8+2\sqrt{6-2\sqrt{5}}=8+2\sqrt{5-2\sqrt{5}+1}\)=\(8+2\sqrt{\left(\sqrt{5}-1\right)^2}=8+2\sqrt{5}-2=6+2\sqrt{5}\)
= \(5+2\sqrt{5}+1=\left(\sqrt{5}+1\right)^2\)
=> B=\(\sqrt{5}+1\)
c) Ta xét \(A=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}\)
=> \(\sqrt{2}\cdot A=\sqrt{8+2\sqrt{3}\cdot\sqrt{5}}+\sqrt{8-2\sqrt{3}\cdot\sqrt{5}}\)
= \(\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
= \(\sqrt{3}+\sqrt{5}+\sqrt{5}-\sqrt{3}\)= \(2\sqrt{5}\)
=> A=\(\sqrt{5}\)
Ta có : \(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
= \(A-\sqrt{6-2\sqrt{5}}\)
= \(\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-\sqrt{5}+1\)=1
\(A=\sqrt{8-2\sqrt{15}}=\sqrt{5-2\sqrt{15}+3}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\left|\sqrt{5}-\sqrt{3}\right|=\sqrt{5}-\sqrt{3}\)
\(B=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
\(\Leftrightarrow\sqrt{4-\sqrt{7}}B=\sqrt{4+\sqrt{7}}\sqrt{4-\sqrt{7}}-\sqrt{\left(4-\sqrt{7}\right)^2}\)
\(\Leftrightarrow\sqrt{4-\sqrt{7}}B=\sqrt{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}-\left|4-\sqrt{7}\right|\)
\(\Leftrightarrow\sqrt{4-\sqrt{7}}B=\sqrt{16-7}-4+\sqrt{7}\)
\(\Leftrightarrow\sqrt{4-\sqrt{7}}B=3-4+\sqrt{7}=-1+\sqrt{7}\)
\(\Leftrightarrow B=\frac{-1+\sqrt{7}}{\sqrt{4-\sqrt{7}}}\)
tíck mình nha bn thanks !!!!!!!!!!
\(A=\sqrt{7-2\sqrt{10}}+\sqrt{7+2\sqrt{10}}\)
\(A^2=\left(7+2\sqrt{10}+7-2\sqrt{10}\right)+2\sqrt{\left(7-2\sqrt{10}\right)\left(7+2\sqrt{10}\right)}\)
\(=14+2\sqrt{49-40}=14+6=20\)
Khi đó:\(A=\sqrt{20}\)
Các câu còn lại bạn làm nốt nhé
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
\(\sqrt[6]{\left(44-16\sqrt{6}\right)\left(44+16\sqrt{6}\right)}=\sqrt[6]{400}=\sqrt[3]{20}\)
\(A=\sqrt{4+\sqrt{7}}-\sqrt{4+\sqrt{7}}\Leftrightarrow\sqrt{2}A=\sqrt{8+2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(\Leftrightarrow\sqrt{2}A=\sqrt{\sqrt{7}^2+2\sqrt{7}+1}-\sqrt{\sqrt{7}^2+2\sqrt{7}+1}\)
\(\Leftrightarrow\sqrt{2}A=\sqrt{7}+1-\sqrt{7}-1=0\)
\(\Leftrightarrow A=0\)