Rút gọn biêu thức A = \(\frac{\sqrt{63y^3}}{\sqrt{7y}}\left(y>0\right)\)
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30 tháng 5 2017
a. \(\sqrt{\dfrac{63y^3}{7y}}\)=\(\sqrt{9y^2}\)=3y
b.\(\sqrt{\dfrac{48x^3}{3x^5}}\)=\(\sqrt{16\cdot\dfrac{1}{X^2}}\)= \(\sqrt{16}\cdot\sqrt{\dfrac{1}{X^2}}\)=\(4\cdot\dfrac{1}{X}=\dfrac{4}{X}\)
c.\(\sqrt{\dfrac{45mn^2}{20m}}=\sqrt{\dfrac{9n^2}{4}}=\dfrac{\sqrt{9n^2}}{\sqrt{4}}=\dfrac{3n}{2}\)
d. \(\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\dfrac{1}{2\sqrt{2}a}\)
19 tháng 9 2017
a) \(\dfrac{\sqrt{63y^3}}{\sqrt{7y}}=\sqrt{\dfrac{63y^3}{7y}}=\sqrt{9y^2}=3y\)
b) \(\dfrac{\sqrt{48x^3}}{\sqrt{3x^5}}=\sqrt{\dfrac{48x^3}{3x^5}}=\sqrt{\dfrac{16}{x^2}}=\dfrac{4}{x}\)
c) \(\dfrac{\sqrt{45mn^2}}{\sqrt{20m}}=\sqrt{\dfrac{45mn^2}{20m}}=\sqrt{\dfrac{9n^2}{4}}=\dfrac{3n}{2}\)
d) \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\dfrac{1}{2\left|a\right|\sqrt{2}}=\dfrac{-1}{2a\sqrt{2}}\)
1 tháng 12 2021
\(1,=0,9\left|x\right|\\ 2,Sửa:\dfrac{\sqrt{63y^3}}{\sqrt{7y}}=\sqrt{\dfrac{63y^3}{7y}}=\sqrt{9y^2}=3\left|y\right|=-3y\)
7 tháng 9 2016
\(2y+\sqrt{\frac{63y^3}{7y}}=2y+\sqrt{9y^2}=2y+3y=5y\)
\(\frac{3\sqrt{3\left(a-2\right)^2}}{27}=\frac{\sqrt{3\left(a-2\right)^2}}{9}=\frac{\sqrt{3}\left(2-a\right)}{\left(\sqrt{3}\right)^4}=\frac{2-a}{3\sqrt{3}}\)
\(x-4+\sqrt{16-8x+x^2}=x-4+x-4=2x-8\)
16 tháng 8 2016
a)\(\frac{\sqrt{63y^3}}{\sqrt{7}y}=\frac{\sqrt{7\cdot3^2\cdot y^2\cdot y}}{\sqrt{7}y}=\frac{\sqrt{7}\cdot\sqrt{3^2}\cdot\sqrt{y^2}\cdot\sqrt{y}}{\sqrt{7}y}=\frac{\sqrt{7}\cdot3\cdot y\cdot\sqrt{y}}{\sqrt{7}y}=3\sqrt{y}\)
b)\(\frac{\sqrt{48x^3}}{\sqrt{3x^5}}=\frac{\sqrt{4^2\cdot3\cdot x^2\cdot x}}{\sqrt{3\cdot x^2\cdot x^3}}=\frac{\sqrt{4^2}\cdot\sqrt{3}\cdot\sqrt{x^3}}{\sqrt{3}\cdot\sqrt{x^2}\cdot\sqrt{x^3}}=\frac{4}{x}\)
c)\(\frac{\sqrt{45mn^2}}{\sqrt{20m}}=\frac{\sqrt{5\cdot3^2\cdot m\cdot n^2}}{\sqrt{5\cdot2^2\cdot m}}=\frac{\sqrt{5}\cdot\sqrt{3^2}\cdot\sqrt{m}\cdot\sqrt{n^2}}{\sqrt{5}\cdot\sqrt{2^2}\cdot\sqrt{m}}=\frac{3\left|n\right|}{2}\)
d)\(\frac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\frac{\sqrt{4^2\cdot a^2\cdot a^2\cdot b^2\cdot b^2\cdot b^2}}{\sqrt{4^2\cdot8\cdot a^2\cdot a^2\cdot a^2\cdot b^2\cdot b^2\cdot b^2}}=\frac{\sqrt{4^2}\cdot\sqrt{a^2}\cdot\sqrt{a^2}\cdot\sqrt{b^2}\cdot\sqrt{b^2}\cdot\sqrt{b^2}}{\sqrt{4^2}\cdot\sqrt{8}\cdot\sqrt{a^2}\cdot\sqrt{a^2}\cdot\sqrt{a^2}\cdot\sqrt{b^2}\cdot\sqrt{b^2}\cdot\sqrt{b^2}}=\frac{4\cdot a^2\cdot b^3}{4\cdot\sqrt{8}\cdot\left|a\right|^3\cdot b^3}=\frac{a^2}{\sqrt{8}\left|a\right|^3}\)
2 tháng 9 2021
Ta có: \(B=\left(\dfrac{2x+1}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{x}{x+\sqrt{x}+1}\right)\)
\(=\dfrac{2x\sqrt{x}-2x+\sqrt{x}-1-x\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}:\dfrac{x+\sqrt{x}+1-x}{x+\sqrt{x}+1}\)
\(=\dfrac{x\sqrt{x}-2x+\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(x+1\right)\cdot\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2\cdot\left(x-\sqrt{x}+1\right)}\)
15 tháng 8 2023
\(=\dfrac{xy\left(x^{\dfrac{1}{2}}+y^{\dfrac{1}{2}}\right)}{x^{\dfrac{1}{2}}+y^{\dfrac{1}{2}}}=xy\)
23 tháng 8 2023
\(A=\dfrac{x^{\dfrac{3}{2}}y+xy^{\dfrac{3}{2}}}{\sqrt{x}+\sqrt{y}}=\left(x+y\right).\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\).
21 tháng 7 2019
\(\(A=\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right):\frac{\sqrt{x}}{\sqrt{x}+1}\left(x\ge0;x\ne1\right)\)\)
\(\(=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\frac{\sqrt{x}}{\sqrt{x}+1}\)\)
\(\(=\frac{\left(\sqrt{x}-1\right).\left(\sqrt{x}+2\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}:\frac{\sqrt{x}}{\sqrt{x}+1}\)\)
\(\(=\frac{x+2\sqrt{x}-\sqrt{x}-2-\left(x+\sqrt{x}-2\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}:\frac{\sqrt{x}}{\sqrt{x}+1}\)\)
\(=\frac{x+2\sqrt{x}-\sqrt{x}-2-x-\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}:\frac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{2}{x-1}\)
Vậy \(A=\frac{2}{x-1}vs\left(x\ge0;x\ne1\right)\)
_Ko chắc , đag bận nên còn phần b , tí mk giải nối_
_Minh ngụy_
B
21 tháng 7 2019
\(ĐK:x\ge0;x\ne1\)
\(a,A=\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right):\frac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\left(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right):\frac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\left(\frac{x-\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\frac{x+\sqrt{x}-2\sqrt{x}-2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right):\frac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\frac{x-\sqrt{x}+2\sqrt{x}-2-x-\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)\sqrt{x}}\)
\(=\frac{2}{x-1}\)
Vậy với \(x\ge0;x\ne1\)thì \(A=\frac{2}{x-1}\)
\(b,\)Ta có:\(A=\frac{2}{x-1}\)
Để A nhận giá trị nguyên \(\Leftrightarrow2⋮x-1\)
Vì \(x\in Z\Rightarrow x-1\inƯ_{\left(2\right)}=\left\{\pm1;\pm2\right\}\)
Ta có bảng sau:
| \(x-1\) | \(1\) | \(-1\) | \(2\) | \(-2\) |
| \(x\) | \(2\left(TM\right)\) | \(0\left(TM\right)\) | \(3\left(TM\right)\) | \(-1\left(L\right)\) |
Vậy để A nhận giá trị nguyên \(x\in\left\{2;0;3\right\}\)
\(A=\sqrt{\frac{63y^3}{7y}}=\sqrt{9y^2}=\sqrt{\left(3y\right)^2}=\left|3y\right|=3y\)( y > 0)