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a: \(=\dfrac{\left|x+2\right|}{x-1}\)
b: \(=x-2y-\left|x-2y\right|\)\(=\left[{}\begin{matrix}x-2y-x+2y=0\\x-2y+x-2y=2x-4y\end{matrix}\right.\)
c: \(=\dfrac{\left|x+2\right|}{\left(x+2\right)\left(x-2\right)}=\pm\dfrac{1}{x-2}\)
a) \(x-2y-\sqrt{x^2-4xy+4y^2}\)
\(=x-2y-\sqrt{\left(x-2y\right)^2}\)
\(=x-2y-\left|x-2y\right|\)
TH1: \(x-2y--\left(x-2y\right)\)
\(=x-2y+x-2y\)
\(=2x-4y\)
TH2: \(x-2y-\left(x-2y\right)\)
\(=x-2y-x+2y\)
\(=0\)
b) \(x^2+\sqrt{x^4-8x^2+16}\)
\(=x^2+\sqrt{\left(x^2-4\right)^2}\)
\(=x^2+\left|x^2-4\right|\)
TH1:
\(x^2+-\left(x^2-4\right)\)
\(=x^2-x^2+4\)
\(=4\)
TH2:
\(x^2+\left(x^2-4\right)\)
\(=x^2+x^2-4\)
\(=2x^2-4\)
c) \(2x-1-\sqrt{\dfrac{x^2-10x+25}{x-5}}\) (x>5)
\(=2x-1-\sqrt{\dfrac{\left(x-5\right)^2}{x-5}}\)
\(=2x-1-\sqrt{x-5}\)
d) \(\sqrt{\dfrac{x^4-4x^2+4}{x^2-2}}\) (\(x>\sqrt{2}\))
\(=\sqrt{\dfrac{\left(x^2-2\right)^2}{x^2-2}}\)
\(=\sqrt{x^2-2}\)
e) \(\sqrt{\left(x^2-4\right)^2}+\dfrac{x-4}{\sqrt{x^2-8x+16}}\)
\(=\left|x^2-4\right|+\dfrac{x-4}{\sqrt{\left(x-4\right)^2}}\)
\(=\left|x^2-4\right|+\sqrt{\dfrac{\left(x-4\right)^2}{\left(x-4\right)^2}}\)
\(=\left|x^2-4\right|+1\)
TH1:
\(x^2-4+1\)
\(=x^2-3\)
TH2:
\(-\left(x^2-4\right)+1\)
\(=-x^2+4+1\)
\(=-x^2+5\)
a: \(A=x-2y-\sqrt{x^2-4xy+4y^2}\)
=x-2y-|x-2y|
Khi x>=2y thì A=x-2y-x+2y=0
Khi x<2y thì A=x-2y+x-2y=2x-4y
b: \(B=x^2+\sqrt{x^4-8x^2+16}\)
\(=x^2+\left|x^2-4\right|\)
TH1: x>=2 hoặc x<=-2
B=x^2+x^2-4=2x^2-4
TH2: -2<=x<=2
B=x^2+4-x^2=4
c: \(C=2x-1-\sqrt{\dfrac{x^2-10x+25}{x-5}}\)
\(=2x-1-\sqrt{\dfrac{\left(x-5\right)^2}{x-5}}=2x-1-\sqrt{x-5}\)
d: \(D=\sqrt{\dfrac{x^4-4x^2+4}{x^2-2}}=\sqrt{\dfrac{\left(x^2-2\right)^2}{x^2-2}}=\sqrt{x^2-2}\)
a: \(A=6\sqrt{3}+10\sqrt{3}-12\sqrt{3}=4\sqrt{3}\)
b: \(B=7\sqrt{3}+5\sqrt{3}-12\sqrt{3}=0\)
c: \(=12\sqrt{2}-6+3\left(9-4\sqrt{2}\right)=12\sqrt{2}-6+27-12\sqrt{2}=21\)
d: \(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}=4\sqrt{5}\)
Bạn nên gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề của bạn hơn nhé. Viết thế này khó đọc quá.
\(x+2y-\sqrt{x^2-4xy+4y^2}\)(sửa đề)
\(=x+2y-\sqrt{\left(x-2y\right)^2}\)
\(=x+2y-\left|x-2y\right|\)
\(=x+2y-\left(x-2y\right)\left(vì.x\ge2y\right)\)
\(=x+2y-x+2y\)
\(=4y\)
\(x+2y-\sqrt{x^2-4xy+4y^2}^2\)
\(=x+2y-\sqrt{\left(x-2y\right)^2}^2\)
\(=x+2y-\left(x-2y\right)^2\)
\(=x+2y-x^2+4xy-4y^2\)
\(1,2\sqrt{27}+5\sqrt{12}-3\sqrt{48}\\ =2.3\sqrt{3}+5.2\sqrt{3}-3.4\sqrt{3}\\ =6\sqrt{3}+10\sqrt{3}-12\sqrt{3}\\ =4\sqrt{3}\)
\(2,\sqrt{147}+\sqrt{75}-4\sqrt{27}\\ =7\sqrt{3}+5\sqrt{3}-4.3\sqrt{3}\\ =7\sqrt{3}+5\sqrt{3}-12\sqrt{3}\\ =\sqrt{3}\left(7+5-12\right)\\ =0\)
\(3,3\sqrt{2}\left(4-\sqrt{2}\right)+3\left(1-2\sqrt{2}\right)^2\\ =3\sqrt{2}.\left(4-\sqrt{2}\right)+3\left(1-4\sqrt{2}+8\right)\\ =12\sqrt{2}-6+3-12\sqrt{2}+24\\ =21\)
\(4,2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\\ =2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\\ =\sqrt{5}\left(2-5-4+11\right)\\ =4\sqrt{5}\)
1: =6căn 3+10căn 3-12căn 3=4căn 3
2: =7căn 3+5căn 3-12căn 3=0
3: =12căn 2-6+3(9-4căn 2)
=12căn 2-6+27-12căn 2=21
4: =2căn 5-5căn 5+4căn 5+9 căn 5
=10căn 5
\(a,P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2}{1-x}\right)\left(dkxd:x\ge0,x\ne1\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(=\dfrac{\sqrt{x}.\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\dfrac{x-2}{\sqrt{x}}\)
\(b,x=4+2\sqrt{3}\Rightarrow P=\dfrac{\left(4+2\sqrt{3}\right)-2}{\sqrt{4+2\sqrt{3}}}\)
\(=\dfrac{2\sqrt{3}+4-2}{\sqrt{\sqrt{3}^2+2\sqrt{3}+1}}\)
\(=\dfrac{2\sqrt{3}+2}{\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\dfrac{2\left(\sqrt{3}+1\right)}{\left|\sqrt{3}+1\right|}\)
\(=\dfrac{2\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=2\)
a: \(P=\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{x-1}\)
\(=\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{x-1}{\sqrt{x}+1}=\dfrac{x-2}{\sqrt{x}}\)
b: Khi x=4+2căn 3 thì \(P=\dfrac{2+2\sqrt{3}}{\sqrt{3}+1}=2\)
