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a ) x 2 - 3 = x 2 - ( √ 3 ) 2 = ( x - √ 3 ) ( x + √ 3 ) b ) x 2 - 6 = x 2 - ( √ 6 ) 2 = ( x - √ 6 ) ( x + √ 6 ) c ) x 2 + 2 √ 3 x + 3 = x 2 + 2 √ 3 x + ( √ 3 ) 2 = ( x + √ 3 ) 2 d ) x 2 - 2 √ 5 x + 5 = x 2 - 2 √ 5 x + ( √ 5 ) 2 = ( x - √ 5 ) 2
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a) \(x^2-5=\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\)
b) \(x^2-11=\left(x-\sqrt{11}\right)\left(x+\sqrt{11}\right)\)
c: \(x-2=\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)
d: \(x^2-2\sqrt{5x}+5=\left(x-\sqrt{5}\right)^2\)
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Bài làm:
a) \(x^2-7=\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\)
b) \(4x^2-5=\left(2x-\sqrt{5}\right)\left(2x+\sqrt{5}\right)\)
c) \(3x^2-1=\left(x\sqrt{3}-1\right)\left(x\sqrt{3}+1\right)\)
d) \(x-1=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
e) \(x-4=\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)
f) \(9x-4=\left(3\sqrt{x}-2\right)\left(3\sqrt{x}+2\right)\)
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1, Với x >= 0 ; x khác 1
\(P=\dfrac{\sqrt{x}\left(x-1\right)+2\sqrt{x}\left(\sqrt{x}-1\right)-\left(3x+1\right)\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x\sqrt{x}+2x-3\sqrt{x}-3x\sqrt{x}-3x-\sqrt{x}-1}{\left(x-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-2x\sqrt{x}-x-4\sqrt{x}-1}{\left(x-1\right)\left(\sqrt{x}+1\right)}\)
mình sửa đề câu 2 nhé
a, \(x^2+mx-1=0\)
\(\Delta=m^2-4\left(-1\right)=m^2+4>0\)
Vậy pt luôn có 2 nghiệm pb
b, Theo Vi et : \(\left\{{}\begin{matrix}x_1+x_2=-m\\x_1x_2=-1\end{matrix}\right.\)
Ta có : \(\left(x_1+x_2\right)^2-2x_1x_2=7\)
Thay vào ta được : \(m^2+2=7\Leftrightarrow m^2=5\Leftrightarrow m=\pm\sqrt{5}\)
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\(\sqrt{48\cdot45}=12\sqrt{15}\\ \sqrt{225\cdot17}=15\sqrt{17}\\ \sqrt{a^3b^7}=\left|ab^3\right|\sqrt{ab}=ab^3\sqrt{ab}\\ \sqrt{x^5\left(x-3\right)^2}=\left|x^2\left(x-3\right)\right|\sqrt{x}=x^2\left(x-3\right)\sqrt{x}\)
\(\sqrt{48\cdot45}=4\sqrt{3}\cdot3\sqrt{5}=12\sqrt{15}\)
\(\sqrt{225\cdot17}=15\sqrt{17}\)
Câu a và câu c đề sai
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