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NT
1
29 tháng 8 2019
\(a,x^8+14x^4+1=\left(x^8+14x^4+49\right)-48\)
\(=\left(x^4+7\right)^2-48\)
\(=\left(x^4+7+\sqrt{48}\right)\left(x^4+7-\sqrt{48}\right)\)
\(b,x^8+98x^4+1\)
\(=\left(x^8+98x^4+2401\right)-2400\)
\(=\left(x^4+49\right)^2-2400\)
\(=\left(x^4+49+20\sqrt{6}\right)\left(x^4+49-20\sqrt{6}\right)\)
Mình nghĩ vậy k bt đúng k :)
28 tháng 2 2020
\(1,\hept{\begin{cases}x+2y=5\\3x-y=1\end{cases}\Leftrightarrow}\hept{\begin{cases}3x+6y=15\\3x-y=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=2\end{cases}}\)
\(2,\hept{\begin{cases}9y-2x=10\\4x-2y=12\end{cases}\Leftrightarrow}\hept{\begin{cases}9y-2x=10\\2x-y=6\end{cases}\Leftrightarrow}\hept{\begin{cases}x=4\\y=2\end{cases}}\)
\(3,\hept{\begin{cases}\sqrt{4x-y}=a\\8x-2y=2a^2\end{cases}\Leftrightarrow\hept{\begin{cases}8x-2y=2a^2\\8x-2y=2a^2\end{cases}}\Leftrightarrow khong}cogiatri\)
29 tháng 2 2020
3)\(\hept{\begin{cases}8x-2y=2a^2\\8x-2y=2a^2\end{cases}}\Leftrightarrow8x-2y=2a^2\) có vô số nghiệm em nhé!
1 tháng 8 2018
\(a\text{) }\sqrt{10+\sqrt{9}}=\sqrt{10+3}=\sqrt{13}\)
\(b\text{) }\sqrt{21+6\sqrt{6}}-\sqrt{21-6\sqrt{6}}\\ =\sqrt{18+3+2\sqrt{54}}-\sqrt{18+3-2\sqrt{54}}\\ =\sqrt{\left(\sqrt{18}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{18}-\sqrt{3}\right)^2}\\ =\sqrt{18}+\sqrt{3}-\sqrt{18}+\sqrt{3}\\ =2\sqrt{3}\)
\(d\text{) }\sqrt{x+1+2\sqrt{x}}\left(x\ge0\right)\\ =\sqrt{\left(\sqrt{x}+1\right)^2}=\sqrt{x}+1\)
\(e\text{) }\sqrt{2x+3+2\sqrt{x^2+3x+2}}\left(x\le-2;x\ge-1\right)\\ =\sqrt{\left(x+2\right)+\left(x+1\right)+2\sqrt{\left(x+1\right)\left(x+2\right)}}=\sqrt{\left(\sqrt{x+1}+\sqrt{x+2}\right)^2}=\sqrt{x+1}+\sqrt{x+2}\)
Xem lại đề câu c nha.
1 tháng 8 2018
a)\(\sqrt{10+\sqrt{9}}=\sqrt{10+3}=\sqrt{13}\)
b)\(\sqrt{21+6\sqrt{6}}-\sqrt{21-6\sqrt{6}}\)
=\(\sqrt{\left(3\sqrt{2}\right)^2+2.3\sqrt{2}.\sqrt{3}+\sqrt{3^2}}-\sqrt{\left(3\sqrt{2}\right)^2-2.3.\sqrt{2}.\sqrt{3}+\sqrt{3^2}}\)
=\(\sqrt{\left(3\sqrt{2}+\sqrt{3}\right)^2}-\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\)
=\(3\sqrt{2}+\sqrt{3}-3\sqrt{2}+\sqrt{3}\)
=\(2\sqrt{3}\)
c)\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10-2\sqrt{5}}}\)
ÁP dụng HĐT \(\sqrt{a+b}\pm\sqrt{a-b}=\sqrt{2\left(a.\sqrt{a^2\pm b}\right)}\)ta có:
=\(\sqrt{2\left(4+\sqrt{4^2-10-2\sqrt{5}}\right)}\)
=\(\sqrt{2\left(4+\sqrt{16-10-2\sqrt{5}}\right)}\)
=\(\sqrt{2\left(4+\sqrt{6-2\sqrt{5}}\right)}\)
=\(\sqrt{2\left(4+\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}.1+1^2}\right)}\)
=\(\sqrt{2\left(4+\sqrt{\left(\sqrt{5}-1\right)^2}\right)}\)
=\(\sqrt{2\left(4+\sqrt{5}-1\right)}\)
=\(\sqrt{2\left(3+\sqrt{5}\right)}\)
=\(\sqrt{6+\sqrt{5}}=\sqrt{5}+1\)
d)\(\sqrt{x+1+2\sqrt{x}}=\sqrt{\left(\sqrt{x}\right)^2+2\sqrt{x}.1+1^2}=\sqrt{x}+1\)
2 tháng 1 2019
1.
a) \(\sqrt{3-2\sqrt{2}}+\sqrt{6-4\sqrt{2}}+\sqrt{9-4\sqrt{2}}=\sqrt{2-2\sqrt{2}+1}+\sqrt{4-2.2.\sqrt{2}+2}+\sqrt{8-2.2\sqrt{2}.1+1}=\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.1+1^2}+\sqrt{2^2-2.2.\sqrt{2}+\left(\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}\right)^2-2.2\sqrt{2}.1+1^2}=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}-1\right)^2}=\left|\sqrt{2}-1\right|+\left|2-\sqrt{2}\right|+\left|2\sqrt{2}-1\right|=\sqrt{2}-1+2-\sqrt{2}+2\sqrt{2}-1=2\sqrt{2}\)
b) \(\sqrt{\left(4+\sqrt{10}\right)^2}-\sqrt{\left(4-\sqrt{10}\right)^2}=\left|4+\sqrt{10}\right|-\left|4-\sqrt{10}\right|=4+\sqrt{10}-4+\sqrt{10}=2\sqrt{10}\)
c) \(\dfrac{1}{\sqrt{2013}-\sqrt{2014}}-\dfrac{1}{\sqrt{2014}-\sqrt{2015}}=\dfrac{\sqrt{2013}+\sqrt{2014}}{\left(\sqrt{2013}-\sqrt{2014}\right)\left(\sqrt{2013}+\sqrt{2014}\right)}-\dfrac{\sqrt{2014}+\sqrt{2015}}{\left(\sqrt{2014}-\sqrt{2015}\right)\left(\sqrt{2014}+\sqrt{2015}\right)}=\dfrac{\sqrt{2013}+\sqrt{2014}}{2013-2014}-\dfrac{\sqrt{2014}+\sqrt{2015}}{2014-2015}=-\left(\sqrt{2013}+\sqrt{2014}\right)+\sqrt{2014}+\sqrt{2015}=-\sqrt{2013}-\sqrt{2014}+\sqrt{2014}+\sqrt{2015}=\sqrt{2015}-\sqrt{2013}\)
2.
a) \(x^2-2\sqrt{5}x+5=0\Leftrightarrow x^2-2.x.\sqrt{5}+\left(\sqrt{5}\right)^2=0\Leftrightarrow\left(x-\sqrt{5}\right)^2=0\Leftrightarrow x-\sqrt{5}=0\Leftrightarrow x=\sqrt{5}\)Vậy S={\(\sqrt{5}\)}
b) ĐK:x\(\ge-3\)
\(\sqrt{x+3}=1\Leftrightarrow\left(\sqrt{x+3}\right)^2=1^2\Leftrightarrow x+3=1\Leftrightarrow x=-2\left(tm\right)\)
Vậy S={-2}
3.
a) \(A=\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)=\sqrt{x}\left(\sqrt{x}-1\right)-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1\)
b) Ta có \(A=x-\sqrt{x}+1=x-2\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có \(\left(\sqrt{x}-\dfrac{1}{2}\right)^2\ge0\Leftrightarrow\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\Leftrightarrow A\ge\dfrac{3}{4}\)
Dấu bằng xảy ra khi x=\(\dfrac{1}{4}\)
Vậy GTNN của A=\(\dfrac{3}{4}\)
AH
Akai Haruma
Giáo viên
26 tháng 6 2019
Bài 1:
\(A=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}=\sqrt{2+3-2\sqrt{2.3}}+\sqrt{2+3+2\sqrt{2.3}}\)
\(=\sqrt{(\sqrt{2}-\sqrt{3})^2}+\sqrt{\sqrt{2}+\sqrt{3})^2}\)
\(=|\sqrt{2}-\sqrt{3}|+|\sqrt{2}+\sqrt{3}|=\sqrt{3}-\sqrt{2}+\sqrt{2}+\sqrt{3}=2\sqrt{3}\)
\(B=(\sqrt{10}+\sqrt{6})\sqrt{8-2\sqrt{15}}\)
\(=(\sqrt{10}+\sqrt{6}).\sqrt{3+5-2\sqrt{3.5}}\)
\(=(\sqrt{10}+\sqrt{6})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)
\(=\sqrt{2}(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})=\sqrt{2}(5-3)=2\sqrt{2}\)
\(C=\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\)
\(C^2=8+2\sqrt{(4+\sqrt{7})(4-\sqrt{7})}=8+2\sqrt{4^2-7}=8+2.3=14\)
\(\Rightarrow C=\sqrt{14}\)
\(D=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{2}\sqrt{3-\sqrt{5}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{6-2\sqrt{5}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{5+1-2\sqrt{5.1}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{(\sqrt{5}-1)^2}\)
\(=(3+\sqrt{5})(\sqrt{5}-1)^2=(3+\sqrt{5})(6-2\sqrt{5})=2(3+\sqrt{5})(3-\sqrt{5})=2(3^2-5)=8\)
AH
Akai Haruma
Giáo viên
26 tháng 6 2019
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)
\(\sqrt{xy}+2\sqrt{x}-3\sqrt{y}-6=(\sqrt{x}.\sqrt{y}+2\sqrt{x})-(3\sqrt{y}+6)\)
\(=\sqrt{x}(\sqrt{y}+2)-3(\sqrt{y}+2)=(\sqrt{x}-3)(\sqrt{y}+2)\)
KK
9 tháng 12 2018
a) đkxđ x≥0 , x ≠1
\(K=\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
= \(\dfrac{x-1-4\sqrt{x}+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
= \(\dfrac{x-3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\)b)
\(\dfrac{\sqrt{x}-3}{\sqrt{x}-2}=\dfrac{\sqrt{x}-2-1}{\sqrt{x}-2}=1-\dfrac{1}{\sqrt{x}-2}\)
để K ∈ z thì \(\dfrac{-1}{\sqrt{x}-2}\) nguyên
=> √x -2 ∈ Ư(-1)={-1;1}
=> x ∈ {1; 9}
vậy ...
4 tháng 12 2022
a: \(=\dfrac{x-1-4\sqrt{x}+\sqrt{x}+1}{x-1}\cdot\dfrac{x-1}{x-2\sqrt{x}}\)
\(=\dfrac{x-3\sqrt{x}}{x-2\sqrt{x}}=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\)
b: Để K là số nguyên thì \(\sqrt{x}-2-1⋮\sqrt{x}-2\)
=>\(\sqrt{x}-2\in\left\{1;-1\right\}\)
hay x=9
c: Để K là số âm thì \(\dfrac{\sqrt{x}-3}{\sqrt{x}-2}< 0\)
=>4<x<9
a) = a^10 - a + a^5 - a^2 + a^2 + a + 1
= a(a^9 - 1) + a^2(a^3 - 1) + (a^2 + a + 1)
= a.(a^3-1)(a^6 + a^3 + 1) + a^2(a-1)(a^2+a+1) + (a^2 + a + 1)
= a.(a-1)(a^2 + a + 1)(a^6 + a^3 + 1) + a^2(a-1)(a^2+a+1) + (a^2 + a + 1)
= (a^2 + a + 1)[(a.(a-1)(a^6 + a^3 + 1) + a^2 + 1]
b) x^5 - x^4 - 1 = x^5 - x^4 + x^3 - x^3 + x^2 - x - x^2 + x - 1
= x^3(x^2 - x + 1) - x(x^2 - x + 1) - (x^2 - x + 1)
= (x^2 - x + 1)(x^3 - x - 1)
a) \(a^{10}+a^5+1\)
\(=\left(a^{10}-a^9+a^7-a^6+a^5-a^3+a^2\right)\)
\(+\left(a^9-a^8+a^6-a^5+a^4-a^2+a\right)\)
\(+\left(a^8-a^7+a^5-a^4+a^3-a+1\right)\)
\(=a^2\left(a^8-a^7+a^5-a^4+a^3-a+1\right)\)
\(+a\left(a^8-a^7+a^5-a^4+a^3-a+1\right)\)
\(+\left(a^8-a^7+a^5-a^4+a^3-a+1\right)\)
\(=\left(a^2+a+1\right)\left(a^8-a^7+a^5-a^4+a^3-a+1\right)\)