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a: \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)
\(=4-\sqrt{15}+\sqrt{15}=4\)
b: \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=2+\sqrt{3}-2+\sqrt{3}\)
\(=2\sqrt{3}\)
c: \(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(=\sqrt{\left(2\sqrt{5}+3\right)^2}-\sqrt{\left(2\sqrt{5}-3\right)^2}\)
\(=2\sqrt{5}+3-2\sqrt{5}+3=6\)
\(a,=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\) \(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2.3\sqrt{20}+9}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\)\(=\sqrt{\sqrt{5}-\sqrt{6-\sqrt{20}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5-2\sqrt{5}+1}}=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}-\sqrt{5}+1}\)
\(=\sqrt{1}=1\)
\(b,=\sqrt{3+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\) \(=\sqrt{3+30\sqrt{2+\sqrt{8+2\sqrt{8}+1}}}\)
\(=\sqrt{3+30\sqrt{2+\sqrt{\left(\sqrt{8}+1\right)^2}}}\)\(=\sqrt{3+30\sqrt{3+\sqrt{8}}}=\sqrt{3+30\sqrt{2+2\sqrt{2}+1}}\)
\(=\sqrt{3+30\sqrt{\left(\sqrt{2}+1\right)^2}}=\sqrt{3+30\sqrt{2}+30}=\sqrt{33+30\sqrt{2}}\)
a) Ta có: \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}\)
=1
b) Ta có: \(\sqrt{3+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
\(=\sqrt{3+30\sqrt{2+2\sqrt{2}+1}}\)
\(=\sqrt{3+30\left(\sqrt{2}+1\right)}\)
\(=\sqrt{33+30\sqrt{2}}\)
Câu 4: a) ĐK: \(x^2\ge9\Leftrightarrow\left[{}\begin{matrix}x\ge3\\x\le-3\end{matrix}\right.\)
b) ĐK: \(x^2-3x+2\ge0\Leftrightarrow\left[{}\begin{matrix}x\le1\\x\ge2\end{matrix}\right.\)
c) Đk: \(-3\le x< 5\)
d) x + 3 và 5 - x đồng dấu. Xét hai trường hợp:
\(\left\{{}\begin{matrix}x+3\ge0\\5-x>0\left(\text{do mẫu phải khác 0}\right)\end{matrix}\right.\Leftrightarrow-3\le x< 5\)
\(\left\{{}\begin{matrix}x+3< 0\\5-x< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< -3\\x>5\end{matrix}\right.\) do x ko thể đồng thời thỏa mãn cả hai nên loại.
Câu 1:
a) Đặt \(A=x+\sqrt{\left(x+2\right)^2}\cdot\left(x-2\right)\)
\(A=x+\left|x+2\right|\cdot\left(x-2\right)\)
+) Với \(x\ge-2\):
\(A=x+\left(x+2\right)\left(x-2\right)=x+x^2-4\)
+) Với \(x< -2\):
\(A=x-\left(x+2\right)\left(x-2\right)=x-x^2+4\)
b) \(B=\sqrt{m^2-6m+9-2m}\)
\(B=\sqrt{m^2-8m+9}\)
Bạn xem lại đề nhé :)
c) \(C=1+\sqrt{\frac{\left(x-1\right)^2}{x-1}}\)
\(C=1+\sqrt{x-1}\)
d) \(D=\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)
\(D=\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}\)
\(D=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}\)
\(D=\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\)
+) Xét \(x\ge8\):
\(D=\sqrt{x-4}+2+\sqrt{x-4}-2=2\sqrt{x-4}\)
+) Xét \(4< x< 8\):
\(D=\sqrt{x-4}+2+2-\sqrt{x-4}=4\)
Vậy....
a: Sửa đề: \(A=\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)
\(=4-\sqrt{15}+\sqrt{15}=4\)
b: \(A=2-\sqrt{3}+\sqrt{3}-1=1\)
c: \(C=3\sqrt{5}-2-3\sqrt{5}-2=-4\)
d: Sửa đề: \(D=\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(=2\sqrt{5}+3-2\sqrt{5}+3\)
=6
a) \(A=\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)
\(A=\left|4-\sqrt{15}\right|+\sqrt{15}\)
\(A=4-\sqrt{15}+\sqrt{15}\)
\(A=4\)
b) \(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1-\sqrt{3}\right)}\)
\(B=\left|2-\sqrt{3}\right|+\left|1-\sqrt{3}\right|\)
\(B=2-\sqrt{3}-1+\sqrt{3}\)
\(B=1\)
c) \(C=\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)
\(C=\sqrt{\left(3\sqrt{5}\right)^2-2\cdot3\sqrt{15}\cdot2+2^2}-\sqrt{\left(3\sqrt{5}\right)^2+2\cdot3\sqrt{5}\cdot2+2^2}\)
\(C=\sqrt{\left(3\sqrt{5}-2\right)^2}-\sqrt{\left(3\sqrt{5}+2\right)^2}\)
\(C=\left|3\sqrt{5}-2\right|-\left|3\sqrt{5}+2\right|\)
\(C=3\sqrt{5}-2-3\sqrt{5}-2\)
\(C=-4\)
d) \(D=\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(D=\sqrt{\left(2\sqrt{5}\right)^2+2\cdot2\sqrt{5}\cdot3+3^2}-\sqrt{\left(2\sqrt{5}\right)^2-2\cdot2\sqrt{5}\cdot3+3^3}\)
\(D=\sqrt{\left(2\sqrt{5}+3\right)^2}-\sqrt{\left(2\sqrt{5}-3\right)^2}\)
\(D=\left|2\sqrt{5}+3\right|-\left|2\sqrt{5}-3\right|\)
\(D=2\sqrt{5}+3-2\sqrt{5}+3\)
\(D=6\)
9: \(A=\dfrac{\sqrt{8+2\sqrt{15}}-\sqrt{14-6\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+\sqrt{3}-3+\sqrt{5}}{\sqrt{2}}=\dfrac{2\sqrt{10}+\sqrt{6}-3\sqrt{2}}{2}\)
10: \(A=\dfrac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
11: \(A=\dfrac{\sqrt{24-6\sqrt{7}}-\sqrt{24+6\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}=-\dfrac{2\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)
12: \(B=\left(3+\sqrt{3}\right)\sqrt{12-6\sqrt{3}}\)
\(=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)\)
=9-3=6
13: \(A=\sqrt{5}-2-\left(3-\sqrt{5}\right)\)
\(=\sqrt{5}-2-3+\sqrt{5}=2\sqrt{5}-5\)
a) Ta có: \(A^3=\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)^3\)
\(=2+\sqrt{5}+2-\sqrt{5}+3\cdot\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)
\(=4-3\cdot A\)
\(\Leftrightarrow A^3+3A-4=0\)
\(\Leftrightarrow A^3-A+4A-4=0\)
\(\Leftrightarrow A\left(A-1\right)\left(A+1\right)+4\left(A-1\right)=0\)
\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)
\(\Leftrightarrow A=1\)
a,
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}=a-2\sqrt{ab}+b=\left(\sqrt{a}-\sqrt{b}\right)^2\)
b,
A=\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+2\sqrt{12}}}}}{\sqrt{6}+\sqrt{2}}=\frac{2\sqrt{3+\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}}{\sqrt{6}+\sqrt{2}}=\frac{2\sqrt{3+\sqrt{5-1-\sqrt{12}}}}{\sqrt{6}+\sqrt{2}}\)\(=\frac{2\sqrt{3+\sqrt{4-2\sqrt{3}}}}{\sqrt{6}+\sqrt{2}}=\frac{2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}}{\sqrt{6}+\sqrt{2}}=\frac{\sqrt{2}\sqrt{4+2\sqrt{3}}}{\sqrt{6}+\sqrt{2}}=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}}=1\)
B=
\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)
a) \(A=\sqrt{a-2-2\sqrt{a-3}}-\sqrt{a+1-4\sqrt{a-3}}=\sqrt{\left(a-3\right)-2\sqrt{a-3}+1}-\sqrt{\left(a-3\right)-4\sqrt{a-3}+4}=\sqrt{\left(\sqrt{a-3}-1\right)^2}-\sqrt{\left(\sqrt{a-3}-2\right)^2}\)Ta có 3≤a≤4\(\Rightarrow\left\{{}\begin{matrix}\sqrt{\left(\sqrt{a-3}-1\right)^2}=1-\sqrt{a-3}\\\sqrt{\left(\sqrt{a-3}-2\right)^2}=2-\sqrt{a-3}\end{matrix}\right.\)
Vậy A=\(1-\sqrt{a-3}-\left(2-\sqrt{a-3}\right)=1-\sqrt{a-3}-2+\sqrt{a-3}=-1\)b) B=\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\times\sqrt{2003-2\sqrt{2005-2\sqrt{2004}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}\right)^2-2.2\sqrt{5}.3+9}}}\times\sqrt{2003-2\sqrt{2004-2\sqrt{2004}+1}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\times\sqrt{2003-2\sqrt{\left(\sqrt{2004}-1\right)^2}}=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\times\sqrt{2003-2\sqrt{2004}+2}=\sqrt{\sqrt{5}-\sqrt{5-2\sqrt{5}+1}}\times\sqrt{2004-2\sqrt{2004}+1}\)
\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\times\sqrt{\left(\sqrt{2004}-1\right)^2}=\sqrt{\sqrt{5}-\sqrt{5}+1}\times\left(\sqrt{2004}-1\right)=\sqrt{1}\times\left(\sqrt{2004}-1\right)=\sqrt{2004}-1\)