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KT
25 tháng 7 2018
\(A=\sqrt{80}+\sqrt{45}+\sqrt{5}=\sqrt{16.5}+\sqrt{9.5}+\sqrt{5}\)
\(=4\sqrt{5}+3\sqrt{5}+\sqrt{5}=8\sqrt{5}\)
\(B=\frac{5}{\sqrt{10}}+3,5\sqrt{40}=\sqrt{\frac{25}{10}}+3,5\sqrt{16.2,5}\)
\(=\sqrt{2,5}+3,5.4\sqrt{2,5}=15\sqrt{2,5}\)
\(C=\frac{1}{\sqrt{3}-2}+\frac{\sqrt{300}}{10}-\sqrt{12}\)
\(=\frac{\sqrt{3}+2}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}+\frac{\sqrt{100.3}}{10}-\sqrt{4.3}\)
\(=-\sqrt{3}-2+\sqrt{3}-2\sqrt{3}=-2\sqrt{3}-2\)
\(D=4\sqrt{x}+2\sqrt{x^2}-\sqrt{16x}=4\sqrt{x}+2x-4\sqrt{x}=2x\) ( do \(x\ge0\))
\(F=\frac{a-2\sqrt{a}}{\sqrt{a}-2}=\frac{\sqrt{a}.\left(\sqrt{a}-2\right)}{\sqrt{a}-2}=\sqrt{a}\)
mk chỉnh đề
\(E=\sqrt{25x+25}-\sqrt{9x+9}+\sqrt{4x+4}\)
\(=\sqrt{25\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}\)
\(=5\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}=4\sqrt{x+1}\)
\(G=\frac{2}{\sqrt{3}+\sqrt{5}}-\frac{2}{\sqrt{5}-\sqrt{7}}=\frac{2\left(\sqrt{3}-\sqrt{5}\right)}{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{3}-\sqrt{5}\right)}-\frac{2\left(\sqrt{5}+\sqrt{7}\right)}{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{5}-\sqrt{7}\right)}\)
\(=\sqrt{3}-\sqrt{5}-\sqrt{5}-\sqrt{7}=\sqrt{3}-\sqrt{7}\)
30 tháng 1 2016
a - b =3
a2 - b2 = 15
=> a+b = 5
=> a =4 ; b = 1
=> 25 - x2 = 16 => x = + -3 thỏa mãn 10 -x2 =1
25 tháng 8 2020
a)ĐKXĐ: \(x\ge0\)
Ta có: \(\sqrt{x}=2\)
\(\Leftrightarrow x=2^2=4\)(nhận)
Vậy: S={4}
b) ĐKXĐ: \(x\ge1\)
Ta có: \(\sqrt{25x-25}-10=0\)
\(\Leftrightarrow\sqrt{25}\cdot\sqrt{x-1}=10\)
\(\Leftrightarrow\sqrt{x-1}=2\)
\(\Leftrightarrow x-1=2^2=4\)
hay x=5(nhận)
Vậy: S={5}
c)ĐKXĐ: \(x\in Z\)
Ta có: \(\sqrt{25-10x+x^2}=7\)
\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=7\)
\(\Leftrightarrow\left|x-5\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=7\\x-5=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\left(nhận\right)\\x=-2\left(nhận\right)\end{matrix}\right.\)
Vậy: S={12;-2}
S
28 tháng 7 2019
\(a,\sqrt{x^4}=7\Leftrightarrow x^2=7\Leftrightarrow x=\pm\sqrt{7}\)
\(Dk:x\ge\frac{2}{3};\sqrt{3x-2}=4\Leftrightarrow3x-2=16\Leftrightarrow3x=18\Leftrightarrow x=6\left(tm\right)\)
\(dk:x\ge\frac{3}{2};\sqrt{2x-3}=\sqrt{x-1}\Leftrightarrow2x-3=x-1\Leftrightarrow x=2\left(tm\right)\)
\(dk:x\ge0;x-10\sqrt{x}+25=0\Leftrightarrow\left(\sqrt{x}-5\right)^2=0\Leftrightarrow\sqrt{x}=5\Leftrightarrow x=25\left(tm\right)\)
\(\sqrt{2x}< 3\Leftrightarrow\sqrt{2}.\sqrt{x}< 3\Leftrightarrow0\le\sqrt{x}< \sqrt{4,5}\Leftrightarrow0\le x< 4,5\)
\(h,dk:x\ge3;\sqrt{\left(x-1\right)^2}=3x-9\Leftrightarrow\left|x-1\right|=3x-9\Leftrightarrow x-1=3x-9\left(x\ge3\right)\Leftrightarrow x=4\left(tm\right)\)
CW
26 tháng 6 2018
a) \(\sqrt{25-x^2}-\sqrt{10-x^2}=3\) (*)
Đk: \(-\sqrt{10}\le x\le\sqrt{10}\)
(*) \(\Leftrightarrow\sqrt{25-x^2}=3+\sqrt{10-x^2}\Leftrightarrow25-x^2=19-x^2+6\sqrt{10-x^2}\)
\(\Leftrightarrow6\sqrt{10-x^2}=6\Leftrightarrow\sqrt{10-x^2}=1\Leftrightarrow\left[{}\begin{matrix}x=-3\left(N\right)\\x=3\left(N\right)\end{matrix}\right.\)
Kl: x = +- 3
b) \(\sqrt{x^2-x-6}+x^2-x-18=0\) (*)
đk: \(\left[{}\begin{matrix}x\le-2\\x\ge3\end{matrix}\right.\)
(*) \(\Leftrightarrow x^2-x-6+\sqrt{x^2-x-6}-12=0\)
Đặt \(t=\sqrt{x^2-x-6}\Rightarrow t^2=x^2-x-6\) (t >/ 0)
phương trình (*) trở thành : \(t^2+t-12=0\Leftrightarrow\left[{}\begin{matrix}t=3\left(N\right)\\t=-4\left(L\right)\end{matrix}\right.\)
Với t=3. ta có: \(\sqrt{x^2-x-6}=3\Leftrightarrow x^2-x-15=0\Leftrightarrow x=\dfrac{1\pm\sqrt{61}}{2}\left(N\right)\)
Kl: \(x=\dfrac{1\pm\sqrt{61}}{2}\)
c) \(\sqrt{x-2009}+\sqrt{y+2008}+\sqrt{z-2}=\dfrac{1}{2}\left(x+y+z\right)\) (*)
Đk: \(\left\{{}\begin{matrix}x\ge2009\\y\ge-2008\\z\ge2\end{matrix}\right.\)
(*) \(\Leftrightarrow2\sqrt{x-2009}+2\sqrt{y+2008}+2\sqrt{z-2}=x+y+z\)
\(\Leftrightarrow\left(x-2009-2\sqrt{x-2009}+1\right)+\left(y+2008-2\sqrt{y+2008}+1\right)+\left(z-2-2\sqrt{z-2}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2009}-1\right)^2+\left(\sqrt{y+2008}-1\right)^2+\left(\sqrt{z-2}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2009}=1\\\sqrt{y+2008}=1\\\sqrt{z-2}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2010\left(N\right)\\y=-2007\left(N\right)\\z=3\left(N\right)\end{matrix}\right.\)
Kl: x= 2010, y= -2007, z=3
c1 \(\sqrt{25-x^2}=3+\sqrt{10-x^2}\) (dk \(-5\le x\le5\) )
bp 2 ve \(25-x^2=9+6\sqrt{10-x^2}+10-x^2\)
\(\Leftrightarrow6\sqrt{10-x^2}=6\)
\(\Leftrightarrow\sqrt{10-x^2}=1\Leftrightarrow10-1=x^2\)
\(\Leftrightarrow x=+-3\)
c2 \(\left(\sqrt{25-x^2}-4\right)-\left(\sqrt{10-x^2}-1\right)=0\)
\(\Leftrightarrow\frac{\left(25-x^2-16\right)}{\left(\sqrt{25-x^2}+4\right)\left(\sqrt{25-x^2}-4\right)}-\frac{10-x^2-1}{\left(\sqrt{10-x^2}-1\right)\left(\sqrt{10-x^2}+1\right)}=0\)
\(\Leftrightarrow\left(9-x^2\right)\left(....\right)=0\)
\(\Leftrightarrow x=+-3\)