\(\text{1= \sqrt{1} = \sqrt{(−1×−1)} = \sqrt{(−1)}\text{x} \sqrt{(−1)}= i × i = i ^2 = −1 }\)
Suy ra 1 = -1
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Những câu hỏi liên quan
Đưa thừa số ra ngoài dấu căn :
a. \(\sqrt{9\text{6}}.\sqrt{\text{1}2\text{5}}\)
b.\(\sqrt{a^4.\text{6}^{\text{5}}}\)
c.\(\sqrt{a^{\text{6}}.b^{\text{1}\text{1}}}\)
d.\(\:\sqrt{a^{\text{3}}\left(\text{1}-a\right)^4}\)
a) Ta có: \(\sqrt{96}\cdot\sqrt{125}\)
\(=\sqrt{16}\cdot\sqrt{6}\cdot\sqrt{25}\cdot\sqrt{5}\)
\(=20\cdot\sqrt{30}\)
b) Ta có: \(\sqrt{a^4\cdot6^5}\)
\(=a^2\cdot36\cdot\sqrt{6}\)
c) Ta có: \(\sqrt{a^6\cdot b^{11}}\)
\(=\sqrt{a^6}\cdot\sqrt{b^{11}}\)
\(=\left|a^3\right|\cdot\left|b^5\right|\cdot\sqrt{b}\)
\(=a^3b^5\cdot\sqrt{b}\)
d) Ta có: \(\sqrt{a^3\left(1-a\right)^4}\)
\(=\sqrt{a^3}\cdot\sqrt{\left(1-a\right)^4}\)
\(=a\sqrt{a}\cdot\left(1-a\right)^2\)
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Bài 1: Tính
Asqrt{5-2text{√}6}+sqrt{5+2text{√}6}
B left(sqrt{10}+sqrt{6}right)sqrt{8-2text{√}15}
Csqrt{4+text{√}7}+sqrt{4-text{√}7}
Dleft(3+text{√}5right)left(text{√}10-text{√}2right)sqrt{3-text{√}5}
Bài 2: Phân tích thành nhân tử
a, ab+ba+√a+1; a0
b, x-2sqrt{xy}+y left(xge0;yge0right)
c, sqrt{xy}+2text{√}x-3text{√}y-6left(xge0;yge0right)
Bài 3: Rút gọn
M left(frac{1}{text{√}x-1}-frac{1}{text{√}x}right)divleft(frac{text{√}x+1}{text{√}x-2}-frac{text{√}x+2}{text{√}x-1}right)
a, Rút gọn...
Đọc tiếp
Bài 1: Tính
A=\(\sqrt{5-2\text{√}6}+\sqrt{5+2\text{√}6}\)
B= \(\left(\sqrt{10}+\sqrt{6}\right)\sqrt{8-2\text{√}15}\)
C=\(\sqrt{4+\text{√}7}+\sqrt{4-\text{√}7}\)
D=\(\left(3+\text{√}5\right)\left(\text{√}10-\text{√}2\right)\sqrt{3-\text{√}5}\)
Bài 2: Phân tích thành nhân tử
a, ab+ba+√a+1; a>=0
b, x-2\(\sqrt{xy}\)+y \(\left(x\ge0;y\ge0\right)\)
c, \(\sqrt{xy}+2\text{√}x-3\text{√}y-6\)\(\left(x\ge0;y\ge0\right)\)
Bài 3: Rút gọn
M= \(\left(\frac{1}{\text{√}x-1}-\frac{1}{\text{√}x}\right)\div\left(\frac{\text{√}x+1}{\text{√}x-2}-\frac{\text{√}x+2}{\text{√}x-1}\right)\)
a, Rút gọn M
b, Tính giá trị của M khi x=2
c, Tìm x để M>0
Bài 1:
\(A=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}=\sqrt{2+3-2\sqrt{2.3}}+\sqrt{2+3+2\sqrt{2.3}}\)
\(=\sqrt{(\sqrt{2}-\sqrt{3})^2}+\sqrt{\sqrt{2}+\sqrt{3})^2}\)
\(=|\sqrt{2}-\sqrt{3}|+|\sqrt{2}+\sqrt{3}|=\sqrt{3}-\sqrt{2}+\sqrt{2}+\sqrt{3}=2\sqrt{3}\)
\(B=(\sqrt{10}+\sqrt{6})\sqrt{8-2\sqrt{15}}\)
\(=(\sqrt{10}+\sqrt{6}).\sqrt{3+5-2\sqrt{3.5}}\)
\(=(\sqrt{10}+\sqrt{6})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)
\(=\sqrt{2}(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})=\sqrt{2}(5-3)=2\sqrt{2}\)
\(C=\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\)
\(C^2=8+2\sqrt{(4+\sqrt{7})(4-\sqrt{7})}=8+2\sqrt{4^2-7}=8+2.3=14\)
\(\Rightarrow C=\sqrt{14}\)
\(D=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{2}\sqrt{3-\sqrt{5}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{6-2\sqrt{5}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{5+1-2\sqrt{5.1}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{(\sqrt{5}-1)^2}\)
\(=(3+\sqrt{5})(\sqrt{5}-1)^2=(3+\sqrt{5})(6-2\sqrt{5})=2(3+\sqrt{5})(3-\sqrt{5})=2(3^2-5)=8\)
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Bài 2:
a) Bạn xem lại đề.
b) \(x-2\sqrt{xy}+y=(\sqrt{x})^2-2\sqrt{x}.\sqrt{y}+(\sqrt{y})^2=(\sqrt{x}-\sqrt{y})^2\)
c)
\(\sqrt{xy}+2\sqrt{x}-3\sqrt{y}-6=(\sqrt{x}.\sqrt{y}+2\sqrt{x})-(3\sqrt{y}+6)\)
\(=\sqrt{x}(\sqrt{y}+2)-3(\sqrt{y}+2)=(\sqrt{x}-3)(\sqrt{y}+2)\)
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Bài 3:
a) ĐKXĐ:\(x>0; x\neq 1; x\neq 4\)
\(M=\frac{\sqrt{x}-(\sqrt{x}-1)}{(\sqrt{x}-1)\sqrt{x}}:\frac{(\sqrt{x}+1)(\sqrt{x}-1)-(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}-1)}\)
\(=\frac{1}{\sqrt{x}(\sqrt{x}-1)}:\frac{(x-1)-(x-4)}{(\sqrt{x}-2)(\sqrt{x}-1)}=\frac{1}{\sqrt{x}(\sqrt{x}-1)}:\frac{3}{(\sqrt{x}-2)(\sqrt{x}-1)}\)
\(\frac{1}{\sqrt{x}(\sqrt{x}-1)}.\frac{(\sqrt{x}-2)(\sqrt{x}-1)}{3}=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
b)
Khi $x=2$ \(M=\frac{\sqrt{2}-2}{3\sqrt{2}}=\frac{1-\sqrt{2}}{3}\)
c)
Để \(M>0\leftrightarrow \frac{\sqrt{x}-2}{3\sqrt{x}}>0\leftrightarrow \sqrt{x}-2>0\leftrightarrow x>4\)
Kết hợp với ĐKXĐ suy ra $x>4$
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Đưa thừa số ra ngoài dấu căn:
a.\(\sqrt{49.\text{3}\text{6}0}\)
b.\(\sqrt{\text{1}2\text{5}a^2}\) với a<0
c.\(-\sqrt{\text{5}00.\text{1}\text{6}2}\)
d.\(\frac{\text{1}}{\text{3}}\sqrt{22\text{5}a^2}\) với a>0
a) \(\sqrt{49.360}=\sqrt{7^2.6^2.10}=7.6\sqrt{10}=42\sqrt{10}\)
b)\(\sqrt{125a^2}=\sqrt{5^2.5.a^2}=5.\left|a\right|\sqrt{5}=-5a\sqrt{5}\) ( vì a<0)
c)\(-\sqrt{500.162}=-\sqrt{10^2.5.9^2.2}=-10.9\sqrt{5.2}=-90\sqrt{10}\)
d) \(\frac{1}{3}\sqrt{225a^2}=\frac{1}{3}\sqrt{15^2.a^2}=\frac{1}{3}.15.\left|a\right|=\frac{15a}{3}\) ( a>0)
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1) B=\(\left(\frac{\sqrt{x}}{2}\text{+}\frac{1}{2\sqrt{x}}\right)\left(\frac{\sqrt{x}-1}{\sqrt{x}\text{+}1}-\frac{\sqrt{x}\text{+}1}{\sqrt{x}-1}\right)\)
1, P=(\(\dfrac{\text{x-1}}{\text{x+3}\sqrt{\text{x-4}}}+\dfrac{\sqrt{\text{x}}+1}{1-\sqrt{\text{x}}}\)) : \(\dfrac{\text{x}+2\sqrt{\text{x}}+1}{x-1}\)+1
a, Rút gọn P
b, Tìm x để P<0
giải phương trình sau :
\(\sqrt{x}+\sqrt[4]{x\text{(}1-x\text{)}^2}+\sqrt[4]{\text{(}1-x\text{)}^3}=\sqrt{1-x}+\sqrt[4]{x^3}+\sqrt[4]{x^2.\text{(}1-x\text{)}}\)
giải phương trình
\(\text{x}^2-4=3\sqrt{\text{x}^3-4\text{x}}\)
\(9\text{x}+17=6\sqrt{8\text{x}-1}+4\sqrt{\text{x}+3}\)
\(\sqrt{2\text{x}-1}+\text{x}=\sqrt{\text{x}}+\sqrt{\text{x}^2-\text{x}+1}\)
\(2\sqrt{\text{x}^2-\text{x}+1}+\sqrt{\text{x}^2+\text{x}+1}=\sqrt{\text{x}^4+\text{x}^2+1}+2\)
a: Đặt \(x^2-4=a\)
Pt sẽ là \(a=3\sqrt{xa}\)
\(\Rightarrow a^2=9xa\)
\(\Leftrightarrow a\left(a-9x\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(x^2-4-9x\right)=0\)
hay \(x\in\left\{2;-2;\dfrac{9+\sqrt{97}}{2};\dfrac{9-\sqrt{97}}{2}\right\}\)
d: Đặt \(\sqrt{x^2-x+1}=a;\sqrt{x^2+x+1}=b\)
Pt sẽ là 2a+b=ab+2
=>(b-2)(1-a)=0
=>b=2 và 1-a
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+x+1=4\\x^2-x+1=1\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
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Rút gọng:\(\left(\frac{1}{\text{x}-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{\text{x}}+1}{\text{x}-2\sqrt{x}+1}\)
\(choP=\left(\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}}{2}+\frac{1}{2\sqrt{x}}\right)a;R\text{ú}tg\text{ọ}nP....b;T\text{í}nhPkhiX=3-2\sqrt{2}c;t\text{ì}mX\text{đ}\text{ể}P=1\)