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BT1:
Ta có: \(A\cdot B=\sqrt{4+\sqrt{10+2\sqrt{5}}}\cdot\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(=\sqrt{16-10-2\sqrt{5}}\)
\(=\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)
Từ đó thay vào: \(\left(A-B\right)^2\)
\(=A^2-2AB+B^2\)
\(=4+\sqrt{10+2\sqrt{5}}-2\left(\sqrt{5}-1\right)+4-\sqrt{10+2\sqrt{5}}\)
\(=10-2\sqrt{5}\)
\(\Rightarrow A-B=\sqrt{10-2\sqrt{5}}\)
BT2:
Đặt \(B=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
\(\Leftrightarrow B^2=4+\sqrt{7}-2\sqrt{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}+4-\sqrt{7}\)
\(=8-2\sqrt{16-7}=8-2\cdot3=2\)
\(\Rightarrow B=\sqrt{2}\)
\(\Rightarrow A=B-\sqrt{2}=\sqrt{2}-\sqrt{2}=0\)
BT3:
đk: \(\orbr{\begin{cases}x\ge2\\x< -2\end{cases}}\)
\(C=\frac{x+2+\sqrt{x^2-4}}{x+2-\sqrt{x^2-4}}+\frac{x+2-\sqrt{x^2-4}}{x+2+\sqrt{x^2-4}}\)
\(C=\frac{\left(x+2+\sqrt{x^2-4}\right)^2}{\left(x+2\right)^2-\left(x^2-4\right)}+\frac{\left(x+2-\sqrt{x^2-4}\right)^2}{\left(x+2\right)^2-\left(x^2-4\right)}\)
\(C=\frac{\left(x+2\right)^2+2\left(x+2\right)\sqrt{x^2-4}+x^2-4+\left(x+2\right)^2-2\left(x+2\right)\sqrt{x^2-4}+x^2-4}{x^2+4x+4-x^2+4}\)
\(C=\frac{2x^2+8x+8+2x^2-8}{4x+8}\)
\(C=\frac{4x^2+8x}{4x+8}=x\)
Vậy C = x
a)\(A=\sqrt{2}-\sqrt{12-8\sqrt{2}}\)
\(A=\sqrt{2}-\sqrt{\left(2\sqrt{2}-2\right)^2}\)
\(A=\sqrt{2}-2\sqrt{2}+2\)
\(A=2-\sqrt{2}\)
c)\(C=\dfrac{2\sqrt{3-\sqrt{5}}}{\sqrt{10}-\sqrt{2}}=\dfrac{2\sqrt{3-\sqrt{5}}}{\sqrt{2}\left(\sqrt{5}-1\right)}=\dfrac{\sqrt{2}\sqrt{3-\sqrt{5}}}{\sqrt{5}-1}=\dfrac{\sqrt{6-2\sqrt{5}}}{\sqrt{5}-1}=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{5}-1}=\dfrac{\sqrt{5}-1}{\sqrt{5}-1}=1\)
d)với x,y,x>0 xyz=100 =>\(\sqrt{xyz}=\sqrt{100}=10\)
\(D=\dfrac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+10}+\dfrac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{10\sqrt{z}}{\sqrt{xz}+10\sqrt{z}+10}\)
\(D=\dfrac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+\sqrt{xyz}}+\dfrac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{\sqrt{xyz^2}}{\sqrt{xz}+\sqrt{xyz^2}+\sqrt{xyz}}\)
\(D=\dfrac{1}{\sqrt{y}+1+\sqrt{yz}}+\dfrac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{\sqrt{yz}}{1+\sqrt{yz}+\sqrt{y}}\)
\(D=\dfrac{1+\sqrt{y}+\sqrt{yz}}{\sqrt{yz}+\sqrt{y}+1}=1\)
mình chỉ giải được câu a,c,d còn câu b mình nghĩ sai đề
1.
a, \(\sqrt{7-2\sqrt{10}}+\sqrt{7+2\sqrt{10}}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\)
\(=\sqrt{5}-\sqrt{2}+\sqrt{5}+\sqrt{2}=2\sqrt{5}\)
b, \(\sqrt{8-2\sqrt{15}}+\sqrt{8-2\sqrt{15}}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\)
\(=\sqrt{5}-\sqrt{3}+\sqrt{5}+\sqrt{3}=2\sqrt{5}\)
c, \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}=\frac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{2}}=\frac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\frac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)
\(A=\sqrt{7-2\sqrt{10}}+\sqrt{7+2\sqrt{10}}\)
\(A^2=\left(7+2\sqrt{10}+7-2\sqrt{10}\right)+2\sqrt{\left(7-2\sqrt{10}\right)\left(7+2\sqrt{10}\right)}\)
\(=14+2\sqrt{49-40}=14+6=20\)
Khi đó:\(A=\sqrt{20}\)
Các câu còn lại bạn làm nốt nhé
\(\sqrt{21-3\sqrt{48}}=\sqrt{12+3.2.\sqrt{12}+9}=2\sqrt{3}-3\)
a+b=-1
mình làm mẫu 2 bài nhé 2 bài kia bạn làm tương tự
1)a)\(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{3}=\sqrt{3}+1-\sqrt{3}=1\)
\(\sqrt{10-2\sqrt{21}}+\sqrt{7}=\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}+\sqrt{7}=\sqrt{7}+\sqrt{3}+\sqrt{7}=2\sqrt{7}+\sqrt{3}\)
2)a) \(\sqrt{12-6\sqrt{3}}-\sqrt{3}=\sqrt{\left(3-\sqrt{3}\right)^2}-\sqrt{3}=3-\sqrt{3}-\sqrt{3}=3-2\sqrt{3}\)
b) \(\sqrt{7+2\sqrt{6}}-\sqrt{3}=\sqrt{\left(1+\sqrt{6}\right)^2}-\sqrt{3}=1+\sqrt{6}-\sqrt{3}\)
Bài 1:
a)
\(\sqrt{13-2\sqrt{42}}=\sqrt{6+7-2\sqrt{6.7}}=\sqrt{(\sqrt{7}-\sqrt{6})^2}=|\sqrt{7}-\sqrt{6}|=\sqrt{7}-\sqrt{6}\)
b)
\(\sqrt{46+6\sqrt{5}}=\sqrt{46+2\sqrt{45}}=\sqrt{45+1+2\sqrt{45.1}}=\sqrt{(\sqrt{45}+1)^2}=\sqrt{45}+1\)
\(=3\sqrt{5}+1\)
c)
\(\sqrt{12-3\sqrt{15}}=\sqrt{\frac{24-6\sqrt{15}}{2}}=\sqrt{\frac{24-2\sqrt{135}}{2}}=\sqrt{\frac{15+9-2\sqrt{15.9}}{2}}\)
\(=\sqrt{\frac{(\sqrt{15}-\sqrt{9})^2}{2}}=\frac{\sqrt{15}-\sqrt{9}}{\sqrt{2}}=\frac{\sqrt{15}-3}{\sqrt{2}}\)
d)
\(\sqrt{11+\sqrt{96}}=\sqrt{11+2\sqrt{24}}=\sqrt{8+3+2\sqrt{8.3}}\)
\(=\sqrt{(\sqrt{8}+\sqrt{3})^2}=\sqrt{8}+\sqrt{3}\)
Bài 1:
a)
\(\sqrt{13-2\sqrt{42}}=\sqrt{6+7-2\sqrt{6.7}}=\sqrt{(\sqrt{7}-\sqrt{6})^2}=|\sqrt{7}-\sqrt{6}|=\sqrt{7}-\sqrt{6}\)
b)
\(\sqrt{46+6\sqrt{5}}=\sqrt{46+2\sqrt{45}}=\sqrt{45+1+2\sqrt{45.1}}=\sqrt{(\sqrt{45}+1)^2}=\sqrt{45}+1\)
\(=3\sqrt{5}+1\)
c)
\(\sqrt{12-3\sqrt{15}}=\sqrt{\frac{24-6\sqrt{15}}{2}}=\sqrt{\frac{24-2\sqrt{135}}{2}}=\sqrt{\frac{15+9-2\sqrt{15.9}}{2}}\)
\(=\sqrt{\frac{(\sqrt{15}-\sqrt{9})^2}{2}}=\frac{\sqrt{15}-\sqrt{9}}{\sqrt{2}}=\frac{\sqrt{15}-3}{\sqrt{2}}\)
d)
\(\sqrt{11+\sqrt{96}}=\sqrt{11+2\sqrt{24}}=\sqrt{8+3+2\sqrt{8.3}}\)
\(=\sqrt{(\sqrt{8}+\sqrt{3})^2}=\sqrt{8}+\sqrt{3}\)
1/Em không chắc nha, nhất là câu c ý, nó sai sai hay là em làm sai nhỉ?
a) ĐK \(x\ge0\). Bình phương hai vế:
\(x+5=x+2\sqrt{x}+1\Leftrightarrow2\sqrt{x}=4\Leftrightarrow x=4\) (TMĐK)
b)ĐK \(0\le x\le1\) . Bình phương hai vế:
\(2\sqrt{x\left(1-x\right)}=0\Leftrightarrow x\left(1-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\left(TMĐK\right)\)
c) ĐK: \(\left\{{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\Leftrightarrow5\le x\le3\) (vô lí))
Vậy không tồn tại x thỏa mãn đề bài.