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23 tháng 5 2022
Bài 4:
a: \(=2-\sqrt{3}+\sqrt{3}-1=1\)
b: \(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
c: \(=\dfrac{\left(15\cdot10\sqrt{2}-3\cdot15\sqrt{2}+2\cdot5\sqrt{2}\right)}{\sqrt{10}}\)
\(=15\cdot\sqrt{20}-3\cdot\sqrt{45}+2\cdot\sqrt{5}\)
\(=30\sqrt{5}-9\sqrt{5}+2\sqrt{5}=33\sqrt{5}\)
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=\(\dfrac{3}{4}\)
LG
9 tháng 10 2018
Điều kiện để biểu thức có nghĩa là:
1) 5x - 10 ≥ 0
⇔ 5x ≥ 10
⇔ x ≥ 2.
2) 1 + x\(^2\) > 1 ∀ x
⇒ Luôn có nghĩa với mọi giá trị x
3) 3 - x ≥ 0 và 2 - x > 0
⇔ x < 3 và x < 2
⇔ x < 2
4) - 1 + x > 0
⇔ x > 1.
28 tháng 10 2022
Bài 2:
a: =>25x=35^2=1225
=>x=49
b: \(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+\dfrac{4}{3}\cdot3\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
=>x+5=4
=>x=-1
30 tháng 6 2017
\(Q=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)
\(Q=\dfrac{2\sqrt{x}-9-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\dfrac{2\sqrt{x}-9-\left(x-9\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(Q=\dfrac{2\sqrt{x}-9-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\dfrac{2\sqrt{x}-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\dfrac{-\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(Q=\dfrac{-\sqrt{x}}{\sqrt{x}-3}\)
4 tháng 9 2022
a: \(P=\left(\dfrac{x-1}{2\sqrt{x}}\right)^2\cdot\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}\)
\(=\dfrac{\left(x-1\right)^2}{4x}\cdot\dfrac{-4\sqrt{x}}{x-1}=\dfrac{-\left(x-1\right)}{\sqrt{x}}\)
b: Khi \(x=4+2\sqrt{3}\) thì \(P=\dfrac{-\left(4+2\sqrt{3}-1\right)}{\sqrt{3}+1}=\dfrac{-\left(3+2\sqrt{3}\right)}{\sqrt{3}+1}=\dfrac{-3-\sqrt{3}}{2}\)
c: Để P<0 thì -(x-1)<0
=>x-1>0
=>x>1
20 tháng 10 2022
a: \(C=\dfrac{3\sqrt{x}-x+x+9}{9-x}:\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\cdot\left(-1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}\)
\(=\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)
b: Để C<-1 thì C+1<0
=>-3 căn x+2 căn x+4<0
=>-căn x<-4
=>x>16
c: \(C+\dfrac{3}{2}=\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}+\dfrac{3}{2}=\dfrac{-3\sqrt{x}+3\sqrt{x}+6}{2\left(\sqrt{x}+2\right)}=\dfrac{3}{\sqrt{x}+2}>0\)
=>C>-3/2
CW
28 tháng 6 2018
a) \(13-\sqrt{\left(8x-1\right)^2}=\sqrt{x^2}\) (*)
\(\Leftrightarrow13-\left|8x-1\right|=\left|x\right|\)
Th1: \(8x-1\ge0\Leftrightarrow x\ge\dfrac{1}{8}\)
(*) \(\Leftrightarrow13-8x+1=x\Leftrightarrow9x=14\Leftrightarrow x=\dfrac{14}{9}\left(N\right)\)
Th2: \(x\le0\)
(*) \(\Leftrightarrow13+8x-1=-x\Leftrightarrow9x=-12\Leftrightarrow x=-\dfrac{4}{3}\left(N\right)\)
Th3: \(\left\{{}\begin{matrix}8x-1\ge0\\x\le0\end{matrix}\right.\Leftrightarrow\dfrac{1}{8}\le x\le0\) (vô lý)
Th4: \(\left\{{}\begin{matrix}8x-1\le0\\x\ge0\end{matrix}\right.\Leftrightarrow0\le x\le\dfrac{1}{8}\)
(*) \(\Leftrightarrow13-8x+1=x\Leftrightarrow9x=14\Leftrightarrow x=\dfrac{14}{9}\left(L\right)\)
Kl: x= 14/9 , x= -4/3
CW
28 tháng 6 2018
b) \(\sqrt{\left(x+1\right)^2}+\sqrt{\left(2x+3\right)^2}=3\Leftrightarrow\left|x+1\right|+\left|2x+3\right|=3\)(*)
Th1: \(x\ge-1\)
(*) \(\Leftrightarrow x+1+2x+3=3\Leftrightarrow3x=-1\Leftrightarrow x=-\dfrac{1}{3}\left(N\right)\)
Th2: \(x\le-\dfrac{3}{2}\)
(*) \(\Leftrightarrow-x-1-2x-3=3\Leftrightarrow-3x=7\Leftrightarrow x=-\dfrac{7}{3}\left(N\right)\)
Th3: \(\left\{{}\begin{matrix}x+1\ge0\\2x+3\le0\end{matrix}\right.\Leftrightarrow-1\le x\le-\dfrac{3}{2}\) (vô lý)
Th4: \(\left\{{}\begin{matrix}x+1\le0\\2x+3\ge0\end{matrix}\right.\Leftrightarrow-\dfrac{3}{2}\le x\le-1\)
(*) \(\Leftrightarrow-x-1-2x-3=3\Leftrightarrow-3x=7\Leftrightarrow x=-\dfrac{7}{3}\left(L\right)\)
Kl: x= -1/3 , x= -7/3
HN
8 tháng 9 2018
Mọi ngươi giúp em với ạ chứ em làm câu a Bài 1 và 2 ra kết quả dài quá :(
31 tháng 8 2022
Bài 1:
a: \(P=\dfrac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)
\(=\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)
b: Để P<1 thì P-1<0
\(\Leftrightarrow\dfrac{\sqrt{a}-4-\sqrt{a}+2}{\sqrt{a}-2}< 0\)
=>căn a-2>0
=>a>4
ĐKXĐ: x>=0
\(M=\dfrac{\sqrt{x}+2+1}{\sqrt{x}+2}=1+\dfrac{1}{\sqrt{x}+2}\)
\(\sqrt{x}+2>=2\)
=>\(\dfrac{1}{\sqrt{x}+2}< =\dfrac{1}{2}\)
=>\(M=1+\dfrac{1}{\sqrt{x}+2}< =\dfrac{1}{2}+1=\dfrac{3}{2}\)
Dấu = xảy ra khi x=0
Dấu = xảy ra khi x=0