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\(a,=\sqrt{6+2\sqrt{3-2\sqrt{3}+1}}\)
\(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{6+2\left(\sqrt{3}-1\right)}\)
\(=\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{3+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
\(b,=\sqrt{6-2\sqrt{3+\sqrt{12+2\sqrt{12}+1}}}\)
\(=\sqrt{6-2\sqrt{3+\sqrt{12}+1}}\)
\(=\sqrt{6-2\sqrt{3+2\sqrt{3}+1}}\)
\(=\sqrt{6-2\left(\sqrt{3}+1\right)}=\sqrt{6-2\sqrt{3}-2}=\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{3-2\sqrt{3}+1}=\sqrt{3}-1\)
\(c,=\sqrt{\sqrt{3}+\sqrt{48-10\sqrt{4+2.2\sqrt{3}+3}}}\)
\(=\sqrt{\sqrt{3}+\sqrt{48-10\left(2+\sqrt{3}\right)}}\)
\(=\sqrt{\sqrt{3}+\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{\sqrt{3}+\sqrt{25-2.5\sqrt{3}+3}}\)
\(=\sqrt{\sqrt{3}+5-\sqrt{3}}=\sqrt{5}\)
\(d,=\sqrt{23-6\sqrt{10+4\sqrt{2-2\sqrt{2}+1}}}\)
\(=\sqrt{23-6\sqrt{6+4\sqrt{2}}}\)
\(=\sqrt{23-6\sqrt{4+2.2\sqrt{2}+2}}\)
\(=\sqrt{23-6\sqrt{\left(2+\sqrt{2}\right)^2}}\)
\(=\sqrt{23-12-6\sqrt{2}}=\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{9-2.3\sqrt{2}+2}=3-\sqrt{2}\)
a) Ta có: \(\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{6+2\left(\sqrt{3}-1\right)}\)
\(=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
b) Ta có: \(\sqrt{6-2\sqrt{3+\sqrt{13+4\sqrt{3}}}}\)
\(=\sqrt{6-2\sqrt{4+2\sqrt{3}}}\)
\(=\sqrt{6-2\left(\sqrt{3}+1\right)}\)
\(=\sqrt{4-2\sqrt{3}}=\sqrt{3}-1\)
c) Ta có: \(\sqrt{\sqrt{3}+\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
\(=\sqrt{\sqrt{3}+\sqrt{48-10\left(2+\sqrt{3}\right)}}\)
\(=\sqrt{\sqrt{3}+\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{\sqrt{3}+5-\sqrt{3}}\)
\(=\sqrt{5}\)
d) Ta có: \(\sqrt{23-6\sqrt{10+4\sqrt{3-2\sqrt{2}}}}\)
\(=\sqrt{23-6\sqrt{10+4\left(\sqrt{2}-1\right)}}\)
\(=\sqrt{23-6\sqrt{6-4\sqrt{2}}}\)
\(=\sqrt{23-6\left(2-\sqrt{2}\right)}\)
\(=\sqrt{11+6\sqrt{2}}\)
\(=3+\sqrt{2}\)
\(B=\sqrt{18-4\sqrt{15}-4\sqrt{3}+2\sqrt{5}}-\sqrt{13-4\sqrt{3}}\)
\(=\sqrt{12+5+1-4\sqrt{15}-4\sqrt{3}+2\sqrt{5}}-\sqrt{12+1-4\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{5}+1-2\sqrt{3}\right)^2}-\sqrt{\left(2\sqrt{3}-1\right)^2}\)
\(=2\sqrt{3}-1-\sqrt{5}-2\sqrt{3}+1=-\sqrt{5}\)
Bạn ko nói rõ lớp mấy để đưa ra cách giải phù hợp.
1) Gọi chữ số hàng đơn vị là x (0 < x <9) => chữ số hàng chục là 3x
Số ban đầu có dạng 10.3x + x = 31x
Sau khi đổi chỗ số mới có dạng 10.x + 3x = 13x
Vì số mới nhỏ hơn số đã cho 18 nên có pt 31x - 13x = 18 <=> 18x = 18 => x = 1 (TMĐK)
Suy ra chữ số hàng chục là 3. Vậy số cần tìm là 31.
2) Tóm tắt thôi nhé.
Chữ số hàng chục là a, hàng đơn vị là b. => Số có dạng 10a + b và a+ b = 10
Số mới sau khi đổi chỗ là 10b + a
Giải hệ 2 pt: a + b = 10 và (10a + b) - (10b + a) = 36
được a = 7; b = 3. Vậy số cần tìm là 73.
3) Gọi a là số tự nhiên sau khi đã xóa đi 5. Số ban đầu là 10a + 5
xóa chữ số 5 thì số ấy giảm đi 1787 đơn vị nên ta có pt : 10a + 5 - 1787 = a
=> 9a = 1782 => a = 198 => Số ban đầu là 1985
a)\(A=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)
\(=\sqrt[3]{1+3\sqrt{2}+3\sqrt{2^2}+2\sqrt{2}}-\sqrt[3]{2\sqrt{2}-3\sqrt{2^2}+3\sqrt{2}-1}\)
\(=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\sqrt[.3]{\left(\sqrt{2}-1\right)^3}\)
\(=1+\sqrt{2}-\left(\sqrt{2}-1\right)=2\)
b)\(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)
\(\Leftrightarrow B^3=5+2\sqrt{13}+3\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}\left(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5+2\sqrt{13}}\right)+5-2\sqrt{13}\)
\(\Leftrightarrow B^3=10+3.\sqrt[3]{-27}.B\)
\(\Leftrightarrow B^3+9B-10=0\)
\(\Leftrightarrow\left(B-1\right)\left(B^2+B+10\right)=0\)
\(\Leftrightarrow B=1\) (vì \(B^2+B+10>0\))
c)\(C=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\)
\(\Leftrightarrow2C=\sqrt[3]{8\sqrt{5}+16}-\sqrt[3]{8\sqrt{5}-16}=\sqrt[3]{1+3\sqrt{5}+3\sqrt{5^2}+5\sqrt{5}}-\sqrt[3]{5\sqrt{5}-3\sqrt{5^2}+3\sqrt{5}-1}\)
\(=\sqrt[3]{\left(1+\sqrt{5}\right)^3}-\sqrt[3]{\left(\sqrt{5}-1\right)^3}\)
\(=1+\sqrt{5}-\left(\sqrt{5}-1\right)\)
\(\Rightarrow C=1\)
d) \(D=\dfrac{10}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}\left(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\right)\)
\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{\left(\sqrt[3]{3}+\sqrt[3]{2}\right)\left(\sqrt[3]{9^2}-\sqrt[3]{6}+\sqrt[3]{2^2}\right)}\left(\dfrac{1+\sqrt{2}}{\sqrt{\left(1-\sqrt{3}\right)^2}}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\right)\)
\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{5}.\dfrac{1+\sqrt{2}}{\left|1-\sqrt{3}\right|}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)
\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(\sqrt{2}\right)^2-1}{\left(\sqrt{3}\right)^2-1}\)
\(=\sqrt[3]{3}+\sqrt[3]{2}\)
Vậy...
a: \(=\sqrt{5}-1\)
b: \(=\sqrt{2}-1\)
c: \(=\sqrt{3}+1\)
d: \(=\sqrt{13}+1\)
a) Ta có: \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2\cdot\sqrt{20}\cdot3+9}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5-2\cdot\sqrt{5}\cdot1+1}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}\)
\(=\sqrt{1}=1\)
b) Ta có: \(\sqrt{6+2\sqrt{5}-\sqrt{29-12\sqrt{5}}}\)
\(=\sqrt{6+2\sqrt{5}-\sqrt{20-2\cdot2\sqrt{5}\cdot3+9}}\)
\(=\sqrt{6+2\sqrt{5}-\sqrt{\left(2\sqrt{5}-3\right)^2}}\)
\(=\sqrt{6+2\sqrt{5}-\left(2\sqrt{5}-3\right)}\)
\(=\sqrt{6+3}=3\)
c) Sửa đề: \(\sqrt{2+\sqrt{5+\sqrt{13-\sqrt{48}}}}\)
Ta có: \(\sqrt{2+\sqrt{5+\sqrt{13-\sqrt{48}}}}\)
\(=\sqrt{2+\sqrt{5+\sqrt{12-2\cdot2\sqrt{3}\cdot1+1}}}\)
\(=\sqrt{2+\sqrt{5+\sqrt{\left(2\sqrt{3}-1\right)^2}}}\)
\(=\sqrt{2+\sqrt{5+2\sqrt{3}-1}}\)
\(=\sqrt{2+\sqrt{3+2\sqrt{3}\cdot1+1}}\)
\(=\sqrt{2+\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\sqrt{3+\sqrt{3}}\)
d) Ta có: \(\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)
\(=\dfrac{\left(6-2\sqrt{5}\right)\sqrt{6+2\sqrt{5}}+\left(6+2\sqrt{5}\right)\sqrt{6-2\sqrt{5}}}{2\sqrt{2}}\)
\(=\dfrac{\left(\sqrt{5}-1\right)^2\cdot\left(\sqrt{5}+1\right)+\left(\sqrt{5}+1\right)^2\cdot\left(\sqrt{5}-1\right)}{2\sqrt{2}}\)
\(=\dfrac{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\left(\sqrt{5}-1+\sqrt{5}+1\right)}{2\sqrt{2}}\)
\(=\dfrac{4\cdot2\sqrt{5}}{2\sqrt{2}}\)
\(=\dfrac{8\sqrt{5}}{2\sqrt{2}}=\dfrac{4\sqrt{5}}{\sqrt{2}}=2\sqrt{10}\)
\(1,\\ a,=\sqrt{\left(3+\sqrt{7}\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}=3+\sqrt{7}-\sqrt{7}+1=4\\ b,K=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}-1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\ c,=\sqrt{\left(6-2\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-4\right)^2}=6-2\sqrt{6}+2\sqrt{6}-4=2\\ e,=\sqrt{\left(2-\sqrt{2}\right)^2}-\left(\sqrt{6}-\sqrt{2}\right)=2-\sqrt{2}-\sqrt{6}+\sqrt{2}=2-\sqrt{6}\)
\(2,\\ a,A=\dfrac{x-3\sqrt{x}+3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+3}{x+9}\\ A=\dfrac{x+9}{\left(\sqrt{x}-3\right)\left(x+9\right)}=\dfrac{1}{\sqrt{x}-3}\\ b,x=4+2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{3}+1\\ \Leftrightarrow A=\dfrac{1}{\sqrt{3}+1-3}=\dfrac{1}{\sqrt{3}+2}=2-\sqrt{3}\)
d/ \(x=\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
\(\Leftrightarrow x^3=3+\sqrt{9+\frac{125}{27}}+3-\sqrt{9+\frac{125}{27}}-3\left(\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\right)\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}.\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
\(\Leftrightarrow x^3=6-3x\sqrt[3]{9-9-\frac{125}{27}}\)
\(\Leftrightarrow x^3=6-5x\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow x=1\)
c/
\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{12}+4}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{4+2\sqrt{3}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
\(=3-1=2\)
a, \(\dfrac{2}{5}\sqrt{75}-0,5\sqrt{48}+\sqrt{300}-\dfrac{2}{3}\sqrt{12}=2\sqrt{3}-2\sqrt{3}+10\sqrt{3}-\dfrac{4}{3}\sqrt{3}=\dfrac{26}{3}\sqrt{3}\)
b, \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}=\dfrac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}+\dfrac{3}{\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}\)
\(=\dfrac{\sqrt{6}}{2}+\dfrac{\sqrt{3}}{\sqrt{3}+\sqrt{2}}\)
\(=\dfrac{\sqrt{6}}{2}+\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)\)
\(=\dfrac{\sqrt{6}}{2}+3-\sqrt{6}=\dfrac{6-\sqrt{6}}{2}\)
c, \(3\sqrt{2}-2\sqrt{3}+2\sqrt{3}+3\sqrt{2}=6\sqrt{2}\)
d, \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{\left(\sqrt{6}-3\right)^2}+\sqrt{\left(2\sqrt{6}+3\right)^2}\)
\(=-\sqrt{6}+3+2\sqrt{6}+3=\sqrt{6}+6\)
e, Ghi đúng đề.
\(\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}=\dfrac{a+b-2\sqrt{ab}+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}=2\sqrt{b}\)
Ta có: \(B=\sqrt{13+3\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
\(=\sqrt{13+3\sqrt{2+\sqrt{8+2\cdot2\sqrt{2}\cdot1+1}}}\)
\(=\sqrt{13+3\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\)
\(=\sqrt{13+3\sqrt{2+2\sqrt{2}+1}}\)
\(=\sqrt{13+3\sqrt{\left(\sqrt{2}+1\right)^2}}\)
\(=\sqrt{13+3\cdot\left(\sqrt{2}+1\right)}\)
\(=\sqrt{13+3\sqrt{2}+3}\)
\(=\sqrt{16+3\sqrt{2}}\)
Ta có: \(C=\frac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3-\sqrt{3+\sqrt{12+2\cdot\sqrt{12}\cdot1+1}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3-\sqrt{3+\sqrt{\left(\sqrt{12}+1\right)^2}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3-\sqrt{3+\sqrt{12}+1}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3-\sqrt{3+2\cdot\sqrt{3}\cdot1+1}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\cdot\sqrt{3-\left(\sqrt{3}+1\right)^2}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\cdot\sqrt{3-\sqrt{3}-1}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\cdot\sqrt{2-\sqrt{3}}}{\sqrt{2}\cdot\left(\sqrt{3}-1\right)}=\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{3}-1}\)
\(=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{3}-1}=\frac{\sqrt{3}-1}{\sqrt{3}-1}=1\)
Ta có: \(D=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{18-\sqrt{128}}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{16-2\cdot4\cdot\sqrt{2}+2}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+4-\sqrt{2}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-2}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{1}}\)
\(=\sqrt{6+2\sqrt{2}}\)