Bài 2 Rút gọn
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d) \(\sqrt{17+12\sqrt{2}}=\sqrt{8+12\sqrt{2}+9}=\sqrt{\left(2\sqrt{2}+3\right)^2}=2\sqrt{2}+3\)
Bài 1 :
\(\left(x-2\right)^2-\left(x-3^2\right)=\left(x-2\right)^2-\left(x-9\right)\)
\(=x^2-4x+4-x+9=x^2-5x+13\)
Bài 2 :
a, \(P=\frac{1-4x^2}{4x^2-4x+1}=\frac{\left(1-2x\right)\left(2x+1\right)}{\left(2x-1\right)^2}\)
\(=\frac{-\left(2x-1\right)\left(2x+1\right)}{\left(2x-1\right)^2}=\frac{-\left(2x+1\right)}{2x-1}=\frac{-2x-1}{2x-1}\)
b, Thay x = -4 ta được :
\(\frac{-2.\left(-4\right)-1}{2.\left(-4\right)-1}=\frac{8-1}{-8-1}=-\frac{7}{9}\)
\(\dfrac{7x^2-14x+7}{3x^2-3x}=\dfrac{7\left(x^2-2x+1\right)}{3x\left(x-1\right)}=\dfrac{7\left(x-1\right)^2}{3x\left(x-1\right)}=\dfrac{7\left(x-1\right)}{3x}\)
1:
\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)
2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)
\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)
\(=\dfrac{20-6}{2}=7\)
a, \(\sqrt{23-8\sqrt{7}}\)
\(=\sqrt{16-8\sqrt{7}+7}\)
\(=\sqrt{4^2-2.4.\sqrt{7}+\left(\sqrt{7}\right)^2}\)
\(=\sqrt{\left(4-\sqrt{7}\right)^2}\)
\(=\left|4-\sqrt{7}\right|\)
\(=4-\sqrt{7}\)
b, \(\sqrt{18-8\sqrt{2}}\)
\(=\sqrt{16-8\sqrt{2}+2}\)
\(=\sqrt{4^2-2.4.\sqrt{2}+\left(\sqrt{2}\right)^2}\)
\(=\sqrt{\left(4-\sqrt{2}\right)^2}\)
\(=\left|4-\sqrt{2}\right|\)
\(=4-\sqrt{2}\)
c, \(\sqrt{12+6\sqrt{3}}\)
\(=\sqrt{9+6\sqrt{3}+3}\)
\(=\sqrt{3^2+2.3.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(3+\sqrt{3}\right)^2}\)
\(=\left|3+\sqrt{3}\right|\)
\(=3+\sqrt{3}\)
d, \(\sqrt{17+12\sqrt{2}}\)
\(=\sqrt{9+12\sqrt{2}+8}\)
\(=\sqrt{3^2+3.2.2\sqrt{2}+\left(2\sqrt{2}\right)^2}\)
\(=\sqrt{\left(3+2\sqrt{2}\right)^2}\)
\(=\left|3+2\sqrt{2}\right|\)
\(=3+2\sqrt{2}\)
a) \(\sqrt{23-8\sqrt{7}}=4-\sqrt{7}\)
b) \(\sqrt{18-8\sqrt{2}}=4-\sqrt{2}\)
c) \(\sqrt{12+6\sqrt{3}}=3+\sqrt{3}\)
d) \(\sqrt{17+12\sqrt{2}}=3+2\sqrt{2}\)