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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....
Mình làm luôn nhé :
\(\sqrt{45-2.3\sqrt{5}+1}-\sqrt{20-2.3.2\sqrt{5}+9}\sqrt{8-2.2\sqrt{2}.\sqrt{5}+5-\sqrt{45+2.2.\sqrt{2}.3\sqrt{5}+8}}\left(\sqrt{3}+\sqrt{5}\right).\sqrt{5-2.\sqrt{5}.\sqrt{2}+2}\left(\sqrt{7}-\sqrt{3}\right).\sqrt{7+2.\sqrt{7}.\sqrt{3}+3}\) Tới đây dễ rồi , bạn tự nhóm HĐT là ra ::v
\(b,\) Ta có:
\(\dfrac{1}{n\sqrt{n-1}+\left(n-1\right)\sqrt{n}}\\ =\dfrac{1}{\sqrt{n}.\sqrt{n-1}\left(\sqrt{n}+\sqrt{n-1}\right)}\\ =\dfrac{\sqrt{n}}{\sqrt{n}.\sqrt{n-1}}-\dfrac{\sqrt{n-1}}{\sqrt{n}.\sqrt{n-1}}\\ =\dfrac{1}{\sqrt{n-1}}-\dfrac{1}{\sqrt{n}}\)
Thay:
\(n=2\) \(\Leftrightarrow\dfrac{1}{2\sqrt{1}+1\sqrt{2}}=\dfrac{1}{1}-\dfrac{1}{\sqrt{2}}\)
\(n=3\Leftrightarrow\dfrac{1}{3\sqrt{2}+2\sqrt{3}}=\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}\)
\(...\)
\(n=2007\Leftrightarrow\dfrac{1}{2007\sqrt{2006}+2006\sqrt{2007}}=\dfrac{1}{\sqrt{2006}}-\dfrac{1}{\sqrt{2007}}\\ \)
\(=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}-2\right)\left(2+\sqrt{3}\right)\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}\left(3-4\right)\)
\(=\left(\sqrt{3}-1\right).\left(-1\right)=1-\sqrt{3}\)
b/ \(=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}+1\right)=\sqrt{\left(\sqrt{3}-1\right)^2}\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=3-1=2\)
c/ \(=\sqrt{6+2\sqrt{5}-\sqrt{\left(2\sqrt{5}-3\right)^2}}\)
\(=\sqrt{6+2\sqrt{5}-2\sqrt{5}+3}=\sqrt{9}=3\)
d/ \(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)
à
Bài 1 : \(\sqrt{49-12\sqrt{5}}+\sqrt{49+12\sqrt{5}}\)
\(=\sqrt{45-4\sqrt{45}+4}+\sqrt{45+4\sqrt{45}+4}\)
\(=\sqrt{\left(\sqrt{45}-2\right)^2}+\sqrt{\left(\sqrt{45}+2\right)^2}\)
\(=\sqrt{45}-2+\sqrt{45}+2=2\sqrt{45}\)
Bài 2 : \(\sqrt{29+12\sqrt{5}}+\sqrt{29-12\sqrt{5}}\)
\(=\sqrt{20+6\sqrt{20}+9}+\sqrt{20-6\sqrt{20}+9}\)
\(=\sqrt{\left(\sqrt{20}+3\right)^2}+\sqrt{\left(\sqrt{20}-3\right)^2}\)
\(=\sqrt{20}+3+\sqrt{20}-3=2\sqrt{20}\)
Bài 3 : \(\sqrt{31-12\sqrt{3}}+\sqrt{31+12\sqrt{3}}\)
\(=\sqrt{27-4\sqrt{27}+4}+\sqrt{27+4\sqrt{27}+4}\)
\(=\sqrt{\left(\sqrt{27}-2\right)^2}+\sqrt{\left(\sqrt{27}+2\right)^2}\)
\(=\sqrt{27}-2+\sqrt{27}+2=2\sqrt{27}\)
Chúc bạn học tốt
4 , Ta có :
\(\sqrt{39-12\sqrt{3}}-\sqrt{39+12\sqrt{3}}\)
\(=\sqrt{3-2.6.\sqrt{3}+6^2}-\sqrt{3+2.6.\sqrt{3}+6^2}\)
\(=\sqrt{\left(\sqrt{3}-6\right)^2}-\sqrt{\left(\sqrt{3}+6\right)^2}\)
\(=\left|\sqrt{3}-6\right|-\left|\sqrt{3}+6\right|\)
\(=6-\sqrt{3}-\sqrt{3}-6\)
\(=-2\sqrt{3}\)
\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}×\sqrt{2004-2\sqrt{2006}-2\sqrt{2005}}=\sqrt{2004-2\sqrt{2006-2\sqrt{2005}}}\)