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a, Ta có : \(4-2\sqrt{3}=3-2\sqrt{3}+1=\left(\sqrt{3}\right)^2-2\sqrt{3}\times1+1^2=\left(\sqrt{3}-1\right)^2\)
\(\Rightarrow\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\left|\sqrt{3}-1\right|-\sqrt{3}\)
Ta có : \(\sqrt{3}>\sqrt{1}\)(vì 3>1)
\(\Leftrightarrow\sqrt{3}>1\Leftrightarrow\sqrt{3}-1>0\Rightarrow\left|\sqrt{3}-1\right|=\sqrt{3}-1\)
Ta có: \(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\left|\sqrt{3}-1\right|-\sqrt{3}=\sqrt{3}-1-\sqrt{3}=-1\)
a) \(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\)=\(\sqrt{3}-1-\sqrt{3}=-1\)
b) \(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\) = \(\sqrt{\left(3+\sqrt{2}\right)^2}-3+\sqrt{2}\)
= \(3+\sqrt{2}-3+\sqrt{2}\) = \(2\sqrt{2}\)
d) \(x-4+\sqrt{16-8x+x^2}\) \(\left(x>4\right)\) = \(x-4+\sqrt{\left(4-x\right)^2}\)
= \(x-4+\left|4-x\right|\) = \(x-4-4+x\) (vì \(x>4\))
= \(2x-8\)
a) \(A=\sqrt{11+6\sqrt{2}}-3+\sqrt{2}=\sqrt{9+2.3\sqrt{2}+2}-3+\sqrt{2}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}-3+\sqrt{2}=3+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}\)
b) x<0
\(B=\sqrt{9x^2}-2x=\left|3x\right|-2x=-3x-2x=-5x\)
c) x>4
\(C=x-4+\sqrt{16-8x+x^2}=x-4+\sqrt{\left(4-x\right)^2}\)
\(=x-4+\left|4-x\right|=x-4+x-4=2x-8\)
b: \(=\dfrac{\left|x\right|+\left|x-2\right|+1}{2x-1}=\dfrac{x+x-2+1}{2x-1}=\dfrac{2x-1}{2x-1}=1\)
c: \(=\left|x-4\right|+\left|x-6\right|\)
=x-4+6-x=2
a) \(\frac{\sqrt{2x^3}}{\sqrt{8x}}=\sqrt{\frac{2x^3}{8x}}=\frac{1}{2}x\)
b) \(\left(3-\sqrt{5}\right)\left(x+\sqrt{5}\right)=3^2-\left(\sqrt{5}\right)^2=9-5=4\)
c) \(\sqrt{\frac{3x^2y^4}{27}}=0\)
\(y\ne0\)
Thì \(\sqrt{\frac{3x^2y^4}{27}}=\frac{1}{3}xy^2\)
e) \(\frac{y}{x^2}\sqrt{\frac{36x^4}{y^2}}=\frac{y}{x^2}.\frac{6x^2}{\left|y\right|}=\frac{6y}{\left|y\right|}\)
Vì y < 0 nên \(\left|y\right|=-y\)
Vậy \(\frac{6y}{\left|y\right|}=\frac{6y}{-y}=-6\)
f) \(\frac{\sqrt{99999999}}{\sqrt{11111111}}=\sqrt{\frac{99999999}{11111111}}=\sqrt{9}=3\)
điều kiện -4<=x<=4x<=4
\(a,\sqrt{\left(x+4\right)^2}+\sqrt{\left(x-4\right)^2}\)
\(A=\left|x+4\right|+\left|x-4\right|\)
KẾT HỢP ĐIỀU KIỆN
\(A=x+4+4-x\)
\(A=8\)
\(B=\sqrt{\left(3x\right)^2-6x+1}+\sqrt{\left(2x\right)^2-12x+3^2}\)
\(B=\sqrt{\left(3x-1\right)^2}+\sqrt{\left(2x-3\right)^2}\)
\(B=\left|3x-1\right|+\left|2x-3\right|\)
\(TH1:x>=\frac{3}{2}\)
\(B=3x-1+2x-3\)
\(B=5x-4\)
\(TH2:\frac{1}{3}< =x< \frac{3}{2}\)
\(B=3x-1-2x+3\)
\(B=x+2\)
\(TH3:x< \frac{1}{3}\)
\(B=-3x+1-2x+3\)
\(B=4-5x\)
câu c và câu d tương tự
câu c tách ra: \(C=\sqrt{\left(\sqrt{x}-3\right)^2}-\sqrt{\left(2\sqrt{x}+1\right)^2}\)
còn câu d tách ra :\(D=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)
\(D=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
bạn tự làm nốt câu c, d nha
a) Đk: \(\left[{}\begin{matrix}x\le-1\\x\ge1\end{matrix}\right.\)
\(\sqrt{x^2-1}-x^2+1=0\)
\(\Leftrightarrow x^2-1-\sqrt{x^2-1}= 0\)
\(\Leftrightarrow\left(\sqrt{x^2-1}-1\right)\sqrt{x^2-1}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-1}-1=0\\\sqrt{x^2-1}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-1}=1\\x^2-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=2\left(1\right)\\x^2=1\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x=\pm\sqrt{2}\left(N\right)\)
\(\left(2\right)\Leftrightarrow x=\pm1\left(N\right)\)
Kl: \(x=\pm\sqrt{2}\), \(x=\pm1\)
b) Đk: \(\left[{}\begin{matrix}x\le-2\\x\ge2\end{matrix}\right.\)
\(\sqrt{x^2-4}-x+2=0\)
\(\Leftrightarrow\sqrt{x^2-4}=x-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-4=x^2-4x+4\\x\ge2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x=8\\x\ge2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\left(N\right)\\x\ge2\end{matrix}\right.\)
kl: x=2
c) \(\sqrt{x^4-8x^2+16}=2-x\)
\(\Leftrightarrow\sqrt{\left(x^2-4\right)^2}=2-x\)
\(\Leftrightarrow\left|x^2-4\right|=2-x\) (*)
Th1: \(x^2-4< 0\Leftrightarrow-2< x< 2\)
(*) \(\Leftrightarrow x^2-4=x-2\Leftrightarrow x^2-x-2=0\Leftrightarrow\left[{}\begin{matrix}x=2\left(L\right)\\x=-1\left(N\right)\end{matrix}\right.\)
Th2: \(x^2-4\ge0\Leftrightarrow\left[{}\begin{matrix}x\le-2\\x\ge2\end{matrix}\right.\)
(*)\(\Leftrightarrow x^2-4=2-x\Leftrightarrow x^2+x-6=0\Leftrightarrow\left[{}\begin{matrix}x=2\left(N\right)\\x=-3\left(N\right)\end{matrix}\right.\)
Kl: x=-3, x=-1,x=2
d) \(\sqrt{9x^2+6x+1}=\sqrt{11-6\sqrt{2}}\)
\(\Leftrightarrow\sqrt{\left(3x+1\right)^2}=\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(\Leftrightarrow\left|3x+1\right|=3-\sqrt{2}\) (*)
Th1: \(3x+1\ge0\Leftrightarrow x\ge-\dfrac{1}{3}\)
(*) \(\Leftrightarrow3x+1=3-\sqrt{2}\Leftrightarrow x=\dfrac{2-\sqrt{2}}{3}\left(N\right)\)
Th2: \(3x+1< 0\Leftrightarrow x< -\dfrac{1}{3}\)
(*) \(\Leftrightarrow3x+1=-3+\sqrt{2}\Leftrightarrow x=\dfrac{-4+\sqrt{2}}{3}\left(N\right)\)
Kl: \(x=\dfrac{2-\sqrt{2}}{3}\), \(x=\dfrac{-4+\sqrt{2}}{3}\)
e) Đk: \(x\ge-\dfrac{3}{2}\)
\(\sqrt{4^2-9}=2\sqrt{2x+3}\) \(\Leftrightarrow\sqrt{7}=2\sqrt{2x+3}\) \(\Leftrightarrow7=8x+12\)
\(\Leftrightarrow8x=-5\Leftrightarrow x=-\dfrac{5}{8}\left(N\right)\)
kl: \(x=-\dfrac{5}{8}\)
f) Đk: x >/ 5
\(\sqrt{4x-20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\)
\(\Leftrightarrow x=9\left(N\right)\)
kl: x=9
a) \(\sqrt{9x^2}-2x\left(x< 0\right)\)= \(\sqrt{\left(3x\right)^2}-2x\)
= \(\left|3x\right|-2x\) = \(-3x-2x\) (vì \(x< 0\) )
= \(-5x\)
b) \(x-4+\sqrt{16-8x+x^2}\) \(\left(x>4\right)\) = \(x-4+\sqrt{\left(4-x\right)^2}\)
= \(x-4+\left|4-x\right|\) = \(x-4-4+x\) (vì \(x>4\))
= \(2x-8\)
a, \(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\sqrt{3}-1-\sqrt{3}=-1\)
b,\(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}=\sqrt{\left(\sqrt{2}+3\right)^2}-3+\sqrt{2}=\sqrt{2}+3-3+\sqrt{2}=2\sqrt{2}\)
c, \(\sqrt{9x^2}-2x=\sqrt{\left(3x\right)^2}-2x=3x-2x=x\)
d, câu này sai đề rồi , nếu sửa lại phải như này :
\(x-4+\sqrt{16-8x+x^2}=x-4+\sqrt{\left(4-x\right)^2}=x-4+4-x=0\)
a) \(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\)=\(\sqrt{3}-1-\sqrt{3}=-1\)
b) \(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\) = \(\sqrt{\left(3+\sqrt{2}\right)^2}-3+\sqrt{2}\)
= \(3+\sqrt{2}-3+\sqrt{2}\) = \(2\sqrt{2}\)
c) \(\sqrt{9x^2}-2x=\sqrt{\left(3x\right)^2}-2x\) = \(\left|3x\right|-2x=-3x-2x\) (x < 0)
= \(-5x\)
d) \(x-4+\sqrt{16-8x+x^2}\) \(\left(x>4\right)\) = \(x-4+\sqrt{\left(4-x\right)^2}\)
= \(x-4+\left|4-x\right|\) = \(x-4-4+x\) ( \(x>4\))
= \(2x-8\)