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![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=2.\left|\left(-3\right)\right|^3+2.\left(-2\right)^2-4\left|\left(-2\right)^3\right|\)
\(=54+8-32=30\)
\(B=\left|\sqrt{2}-2\right|+\left|\sqrt{2}-3\right|=2-\sqrt{2}+3-\sqrt{2}\)
\(=5-2\sqrt{2}\)
\(C=\left|3-\sqrt{3}\right|-\left|1+\sqrt{3}\right|=3-\sqrt{3}-1-\sqrt{3}\)
\(=2-2\sqrt{3}\)
\(D=\left|5+\sqrt{6}\right|-\left|\sqrt{6}-5\right|=5+\sqrt{6}-5+\sqrt{6}\)
\(=2\sqrt{6}\)
\(E=\sqrt{15^2}-\sqrt{5^2}=15-5=10\)
`A=2sqrt{(-3)^6}+2sqrt{(-2)^4}-4sqrt{(-2)^6}=2|(-3)^3|+2|(-2)^2|-4|(-2)^3|=54+8-32=30` $\\$ `B=sqrt{(sqrt2-2)^2}+sqrt{(sqrt2-3)^2}=2-sqrt2+3-sqrt2=5-2sqrt2` $\\$ `C=sqrt{(3-sqrt3)^2}-sqrt{(1+sqrt3)^2}=3-sqrt3-sqrt3-1=2-2sqrt3` $\\$ `D=sqrt{(5+sqrt6)^2}-sqrt{(sqrt6-sqrt5)^2}=5+sqrt6-5+sqrt6=2sqrt6` $\\$ `E=sqrt{17^2-8^2}-sqrt{3^2+4^2}=sqrt{289-64}-sqrt{9+16}=sqrt(225)-sqrt{25}=15-5=10`
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=2\sqrt{5}-2\sqrt{5}+\sqrt{5}-1\)
\(=\sqrt{5}-1\)
b) Ta có: \(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(=8\sqrt{2}+10\sqrt{2}-16\sqrt{2}\)
\(=2\sqrt{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left(\sqrt{2}+1\right)\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\left(5-2\sqrt{2}-\sqrt{3}\right)\)
\(=\left(\sqrt{6}+\sqrt{2}+\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\left(5-2\sqrt{2}-\sqrt{3}\right)\)
\(=\left(7+2\sqrt{6}+3\sqrt{3}+4\sqrt{2}\right)\left(5-2\sqrt{2}-\sqrt{3}\right)\)
\(=10\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\sqrt{(3-2\sqrt{2})^2}+\sqrt{(3+2\sqrt{2})^2}=3-2\sqrt{2}+3-2\sqrt{2}=6\)
b\(\sqrt{(5-2\sqrt{6})^2}+\sqrt{(5+2\sqrt{6})^2}=5-2\sqrt{6}+5+2\sqrt{6}=10 \)
các ý còn lại làm tương tự
hình như ở câu a) chỗ sau dấu bằng đầu tiên bạn bị sai dấu trừ cuối cùng
![](https://rs.olm.vn/images/avt/0.png?1311)
i: =-12*căn 3/2căn 3=-6
h: =72căn 2/12căn 2=6
g: =25căn 12/5căn 6=5căn 2
f: =(15:5)*căn 6:3=3căn 2
d: =-1/2*6*căn 10=-3căn 10
\(E=\left(\sqrt{2}+1\right)\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\left(3-2\sqrt{2}+\frac{4-2\sqrt{3}}{2}\right)\)
\(=\left(\sqrt{2}+1\right)\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\left[\left(1-\sqrt{2}\right)^2+\frac{\left(\sqrt{3}-1\right)^2}{2}\right]\)
Ta có : \(\left(\sqrt{2}+1\right)\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\left(\sqrt{2}-1\right)^2\)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\left(\sqrt{2}-1\right)\)
và \(\left(\sqrt{2}+1\right)\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\frac{\left(\sqrt{3}-1\right)^2}{2}\)
\(=\frac{1}{2}.\left(\sqrt{2}+1\right)\left(\sqrt{6}+1\right).\left(3-1\right)\left(\sqrt{3}-1\right)\)
\(=\left(\sqrt{2}+1\right)\left(\sqrt{6}+1\right)\left(\sqrt{3}-1\right)\)
\(\Rightarrow E=\left(\sqrt{6}+1\right)\left[\left(\sqrt{2}+1\right)\left(\sqrt{3}-1\right)+\left(\sqrt{2}-1\right)\left(\sqrt{3}+1\right)\right]\)
\(=\left(\sqrt{6}+1\right)\left(2\sqrt{6}-2\right)=2\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)\)
\(=2\left(6-1\right)=10\)