a)\(\frac{\sqrt{7\left(-x\right)^2y^4}}{\sqrt{28x^4y^4}}\)với \(x>0;y\ne0\)
b)\(\sqrt{2x+2\sqrt{2x-1}}+\sqrt{2x+2\sqrt{2x-1}}\)
với \(\frac{1}{2}\le x\le1\)
c)\(\frac{1}{3}\sqrt{9x-27}+\sqrt{2x-6}-\sqrt{4x-12}=2-\sqrt{2}\)
K
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NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 9 2019
\(A=\sqrt{\frac{x^2y^2}{x^2+y^2}+\frac{x^2y^2}{\left(x+y\right)^2}+\sqrt{\frac{\left(x^4+x^2y^2\right)^2+\left(y^4+x^2y^2\right)^2+x^4y^4}{\left(x^2+y^2\right)^2}}}\)
\(=\sqrt{\frac{x^2y^2}{x^2+y^2}+\frac{x^2y^2}{\left(x+y\right)^2}+\sqrt{\frac{\left(x^4+x^2y^2\right)^2+2x^4y^4+2x^2y^6+y^8}{\left(x^2+y^2\right)^2}}}\)
\(=\sqrt{\frac{x^2y^2}{x^2+y^2}+\frac{x^2y^2}{\left(x+y\right)^2}+\sqrt{\frac{\left(x^4+x^2y^2\right)^2+2\left(x^4+x^2y^2\right)y^4+y^8}{\left(x^2+y^2\right)^2}}}\)
\(=\sqrt{\frac{x^2y^2}{x^2+y^2}+\frac{x^2y^2}{\left(x+y\right)^2}+\sqrt{\frac{\left(x^4+x^2y^2+y^4\right)^2}{\left(x^2+y^2\right)^2}}}\)
\(=\sqrt{\frac{x^2y^2}{x^2+y^2}+\frac{x^2y^2}{\left(x+y\right)^2}+\frac{x^4+x^2y^2+y^4}{x^2+y^2}}\)
\(=\sqrt{\frac{x^2y^2}{\left(x+y\right)^2}+\frac{x^4+2x^2y^2+y^4}{x^2+y^2}}=\sqrt{\frac{x^2y^2}{\left(x+y\right)^2}+\frac{\left(x^2+y^2\right)^2}{x^2+y^2}}\)
\(=\sqrt{\frac{x^2y^2}{\left(x+y\right)^2}+x^2+y^2}=\sqrt{\frac{\left(x^2+xy\right)^2+\left(y^2+xy\right)^2+x^2y^2}{\left(x+y\right)^2}}\)
\(=\sqrt{\frac{\left(x^2+xy\right)^2+2x^2y^2+2xy^3+y^4}{\left(x+y\right)^2}}=\sqrt{\frac{\left(x^2+xy\right)^2+2\left(x^2+xy\right)y^2+y^4}{\left(x+y\right)^2}}\)
\(=\sqrt{\frac{\left(x^2+xy+y^2\right)^2}{\left(x+y\right)^2}}=\frac{x^2+xy+y^2}{x+y}\)
19 tháng 10 2020
Ta có: \(x^4+y^4+\frac{x^4y^4}{\left(x^2+y^2\right)^2}\)
\(=\left(x^4+2x^2y^2+y^4\right)-2x^2y^2+\frac{x^4y^4}{\left(x^2+y^2\right)}\)
\(=\left(x^2+y^2\right)^2-2x^2y^2+\left(\frac{x^2y^2}{x^2+y^2}\right)^2\)
\(=\left(x^2+y^2-\frac{x^2y^2}{x^2+y^2}\right)^2\)
Thay vào ta tính được:
\(P=\sqrt{\frac{x^2y^2}{x^2+y^2}+\frac{x^2y^2}{\left(x+y\right)^2}+\sqrt{\left(x^2+y^2-\frac{x^2y^2}{x^2+y^2}\right)^2}}\)
Mà \(x^2+y^2-\frac{x^2y^2}{x^2+y^2}=\frac{\left(x^2+y^2\right)^2-x^2y^2}{x^2+y^2}=\frac{x^4+x^2y^2+y^4}{x^2+y^2}>0\left(\forall x,y\right)\)
Khi đó:
\(P=\sqrt{\frac{x^2y^2}{x^2+y^2}+\frac{x^2y^2}{\left(x+y\right)^2}+x^2+y^2-\frac{x^2y^2}{x^2+y^2}}\)
\(P=\sqrt{x^2+y^2+\frac{x^2y^2}{\left(x+y\right)^2}}\)
\(P=\sqrt{\left(x^2+2xy+y^2\right)-2xy+\frac{x^2y^2}{\left(x+y\right)^2}}\)
\(P=\sqrt{\left(x+y\right)^2-2xy+\left(\frac{xy}{x+y}\right)^2}\)
\(P=\sqrt{\left(x+y-\frac{xy}{x+y}\right)^2}\)
\(P=\left|x+y-\frac{xy}{x+y}\right|=\left|\frac{x^2+xy+y^2}{x+y}\right|=\frac{x^2+xy+y^2}{x+y}\)
Vậy \(P=\frac{x^2+xy+y^2}{x+y}\)
6 tháng 3 2022
1, với x > 0 ; x khác 1 ; 4
a, \(P=\left(\dfrac{x+\sqrt{x}-x-2}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}-4}{x-1}\right)\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}:\dfrac{x-4}{x-1}=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
b, Ta có P > 0 => \(\sqrt{x}-1>0\Leftrightarrow x>1\)
Kết hợp đk vậy x > 1 ; x khác 4
16 tháng 8 2021
b4 :
\(a,x-1=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(b,x-5=\left(\sqrt{x}-\sqrt{5}\right)\left(\sqrt{x}+\sqrt{5}\right)\)
\(c,x+2\sqrt{xy}+y=\left(\sqrt{x}+\sqrt{y}\right)^2\)
\(d,x-4\sqrt{x}\sqrt{y}+4y=\left(\sqrt{x}-2\sqrt{y}\right)^2\)
b5:
\(a,ĐK:x\ge1\)
\(\sqrt{9\left(x-1\right)}+\sqrt{4\left(x-1\right)}-\frac{4}{5}\sqrt{25\left(x-1\right)}=1\)
\(\Leftrightarrow3\sqrt{x-1}+2\sqrt{x-1}-4\sqrt{x-1}=1\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x=2\left(tm\right)\)
\(b,ĐK:x\ge5\)
\(\frac{1}{3}\sqrt{9\left(x-5\right)}+\frac{1}{2}\sqrt{4\left(x-5\right)}-\frac{7}{5}\sqrt{25\left(x-5\right)}=2\)
\(\Leftrightarrow\sqrt{x-5}+\sqrt{x-5}-7\sqrt{x-5}=2\)
\(\Leftrightarrow-5\sqrt{x-5}=2\)
\(\Leftrightarrow\sqrt{x-5}=-\frac{2}{5}\left(voli\right)\)
\(c,ĐK:x>0\)
\(\sqrt{x}+\frac{9}{\sqrt{x}}=6\)
\(\Leftrightarrow x+9=6\sqrt{x}\)
\(\Leftrightarrow x-6\sqrt{x}+9=0\)
\(\Leftrightarrow\left(\sqrt{x}-3\right)^2=0\)
\(\Leftrightarrow x=9\left(tm\right)\)
19 tháng 6 2016
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18 tháng 8 2015
Đề đúng phải như sau: Cho \(a,x,y>0\) thỏa mãn \(\sqrt{x^2+\sqrt[3]{x^4y^2}}+\sqrt{y^2+\sqrt[3]{x^2y^4}}=a\). Tính \(P=\sqrt[3]{x^2}+\sqrt[3]{y^2}\) theo \(a.\)
Với đề này có thể thấy \(\sqrt{P}\cdot P=a\Leftrightarrow P=\sqrt[3]{a^2}.\)
2 tháng 9 2017
a. Ta có:\(\frac{x}{y}\sqrt{\frac{y^2}{x^4}=}\) \(\frac{x}{y}.\frac{\left|y\right|}{x^2}=\frac{x.y}{x^2y}\)\(=\frac{1}{x}\)(Vì \(x\ne0;y>0\))
2 tháng 9 2017
b \(3x^2\sqrt{\frac{8}{x^2}}=3x^2\frac{2\sqrt{2}}{\left|x\right|}=\frac{6x^2\sqrt{2}}{-x}=-6x\sqrt{2}\)( Vì \(x< 0\))
2 tháng 7 2019
\(a,\frac{\sqrt{108x^3}}{\sqrt{12x}}=\frac{\sqrt{36.3.x^3}}{\sqrt{3.4.x}}=\frac{6\sqrt{3}.\sqrt{x}^3}{2\sqrt{3}.\sqrt{x}}=3\sqrt{x}^2=3x\)
\(b,\frac{\sqrt{13x^4y^6}}{\sqrt{208x^6y^6}}=\frac{\sqrt{13}.\sqrt{x^4}.\sqrt{y^6}}{\sqrt{16.13}.\sqrt{x^6}.\sqrt{y^6}}=\frac{\sqrt{13}.x^2y^3}{4\sqrt{13}x^3y^3}=\frac{1}{4x}\)
\(c,\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}+\sqrt{y}\right)^2\)
\(=\frac{\sqrt{x}^3+\sqrt{y}^3}{\sqrt{x}+\sqrt{y}}-\left(x+2\sqrt{xy}+y\right)\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x-2\sqrt{xy}-y\)
\(=x-\sqrt{xy}+y-x-2\sqrt{xy}-y=-3\sqrt{xy}\)
2 tháng 7 2019
\(d,\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\frac{\sqrt{\left(\sqrt{x}-1\right)^2}}{\sqrt{\left(\sqrt{x}+1\right)^2}}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
Đk chỗ này là \(\sqrt{x}-1\ge0\Rightarrow\sqrt{x}\ge\sqrt{1}\Rightarrow x\ge1\)nhé
\(e,\frac{x-1}{\sqrt{y}-1}.\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}-1}.\frac{y-2\sqrt{y}+1}{\left(x-1\right)^2}\)
\(=\frac{\left(x-1\right)\left(\sqrt{y}-1\right)^2}{\left(\sqrt{y}-1\right)\left(x-1\right)^2}=\frac{\sqrt{y}-1}{x-1}\)
\(a,\frac{\sqrt{7x^2y^4}}{\sqrt{28x^4y^4}}\)
\(\frac{\sqrt{7}xy^2}{2\sqrt{7}x^2y^2}=\frac{1}{2x}\)
\(b,\sqrt{2x+2\sqrt{2x-1}}+\sqrt{2x+2\sqrt{2x-1}}\)
\(\sqrt{2x-1+2\sqrt{2x-1}+1}+\sqrt{2x-1+2\sqrt{2x-1}+1}\)
\(\sqrt{\left(\sqrt{2x-1}+1\right)^2}+\sqrt{\left(\sqrt{2x-1}+1\right)^2}\)
\(\left|\sqrt{2x-1}+1\right|+\left|\sqrt{2x-1}+1\right|\)
\(\sqrt{2x-1}+1+\sqrt{2x-1}+1\)
\(2\sqrt{2x-1}+2\)
\(c,\frac{1}{3}\sqrt{9x-27}+\sqrt{2x-6}-\sqrt{4x-12}=2-\sqrt{2}\)
\(\sqrt{x-3}+\sqrt{2}\sqrt{x-3}-2\sqrt{x-3}=2-\sqrt{2}\)
\(\sqrt{x-3}\left(1+\sqrt{2}-2\right)=2-\sqrt{2}\)
\(\sqrt{x-3}\left(\sqrt{2}-1\right)=\sqrt{2}\left(\sqrt{2}-1\right)\)
\(\sqrt{x-3}=\sqrt{2}\)
\(x-3=2< =>x=5\)
a) \(\frac{\sqrt{7\left(-x^2\right)y^4}}{\sqrt{28x^4y^4}}=\frac{\sqrt{7}xy^2}{2\sqrt{7}x^2y^2}=\frac{1}{2x}\)(vì x > 0)
b) \(\sqrt{2x+2\sqrt{2x-1}}+\sqrt{2x+2\sqrt{2x-1}}\)
\(=\sqrt{2x-1+2\sqrt{2x-1}+1}+\sqrt{2x-1+2\sqrt{2x-1}}\)
\(=\sqrt{\left(\sqrt{2x-1}+1\right)^2}+\sqrt{\left(\sqrt{2x+1}+1\right)^2}=\sqrt{2x-1}+1+\sqrt{2x+1}+1\)
\(=2\sqrt{2x-1}+2\)
c) ĐK: x \(\ge\)3
Ta có:: \(\frac{1}{3}\sqrt{9x-27}+\sqrt{2x-6}-\sqrt{4x-12}=2-\sqrt{2}\)
<=> \(\sqrt{x-3}+\sqrt{2}.\sqrt{x-3}-2\sqrt{x-3}=2-\sqrt{2}\)
<=> \(\sqrt{x-3}.\left(\sqrt{2}-1\right)=\sqrt{2}\left(\sqrt{2}-1\right)\)
<=> \(\sqrt{x-3}=\sqrt{2}\) <=> x - 3 = 2 <=> x = 5 (tm)