Help me !!!
1. Rút gọn
a) \(2\left(x+y\right)\sqrt{\frac{1}{x^2+2xy+y^2}}\left(x+y>0\right)\)
b) \(\frac{3x}{7y}\sqrt{\frac{49y^2}{9x^2}}\left(x>0,y< 0\right)\)
\(\frac{3x}{7y}\sqrt{\frac{49y^2}{9x^2}\left(x>0,y< 0\right)}\)
\(\frac{3x}{7y}\sqrt{\frac{49y^2}{9x^2}}\) \(=\frac{3x}{7y}|\frac{7y}{3x}|\left(1\right)\)
mà \(x>0,y< 0\)
=>\(\left(1\right)\) = \(\frac{3x.\left(-7y\right)}{7y.3x}=-1\)
chúc bn học tốt
\(\frac{3x}{7y}\sqrt{\frac{49y^2}{9x^2}}\)
\(=\frac{3x}{7y}\sqrt{\frac{\left(7y\right)^2}{\left(3x\right)^2}}\)\(=\frac{3x}{7y}\cdot\frac{\left|7y\right|}{\left|3x\right|}\)
mak ta có \(x>0;y< 0\)
\(\Rightarrow\frac{3x}{7y}\cdot\frac{-7y}{3x}\)\(\Rightarrow\frac{3x\cdot-7y}{7x\cdot3x}=\left(-1\right)\)
\(\Rightarrow\frac{3x}{7y}\sqrt{\frac{49y^2}{9x^2}}=\left(-1\right)\)
rút gọn Bt
a)\(\frac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
b)\(\frac{x-y}{\sqrt{y}-1}.\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\left(x\ne1,y\ne1,y>0\right)\)
a) \(\frac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\left(\sqrt{x}-\sqrt{y}\right)}-\left(\sqrt{x}-\sqrt{y}\right)^2=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}-x+2\sqrt{xy}-y\)
\(=3\sqrt{xy}\)
b) \(\frac{x-y}{\sqrt{y}-1}.\sqrt{\frac{\left(\sqrt{y}-1\right)^4}{\left(x-1\right)^4}}=\frac{x-y}{\sqrt{y}-1}.\frac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^2}=\frac{\left(x-y\right)\left(\sqrt{y}-1\right)}{\left(x-1\right)^2}\)
a) \(=\frac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\sqrt{x}-\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=x+\sqrt{xy}+y-x+2\sqrt{xy}-y=3\sqrt{xy}\)
Rút gọn các biểu thức sau:
a)\(\frac{\sqrt{108x^3}}{\sqrt{12x}}\left(x>0\right)\)
b)\(\frac{\sqrt{13x^4y^6}}{\sqrt{208x^6y^6}}\left(x< 0;y\ne0\right)\)
c)\(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}+\sqrt{y}\right)^2\)
d) \(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\left(x\ge\right)\)
e)\(\frac{x-1}{\sqrt{y}-1}.\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\left(y>0;x\ne1;y\ne1\right)\)
\(a,\frac{\sqrt{108x^3}}{\sqrt{12x}}=\frac{\sqrt{36.3.x^3}}{\sqrt{3.4.x}}=\frac{6\sqrt{3}.\sqrt{x}^3}{2\sqrt{3}.\sqrt{x}}=3\sqrt{x}^2=3x\)
\(b,\frac{\sqrt{13x^4y^6}}{\sqrt{208x^6y^6}}=\frac{\sqrt{13}.\sqrt{x^4}.\sqrt{y^6}}{\sqrt{16.13}.\sqrt{x^6}.\sqrt{y^6}}=\frac{\sqrt{13}.x^2y^3}{4\sqrt{13}x^3y^3}=\frac{1}{4x}\)
\(c,\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}+\sqrt{y}\right)^2\)
\(=\frac{\sqrt{x}^3+\sqrt{y}^3}{\sqrt{x}+\sqrt{y}}-\left(x+2\sqrt{xy}+y\right)\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x-2\sqrt{xy}-y\)
\(=x-\sqrt{xy}+y-x-2\sqrt{xy}-y=-3\sqrt{xy}\)
\(d,\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\frac{\sqrt{\left(\sqrt{x}-1\right)^2}}{\sqrt{\left(\sqrt{x}+1\right)^2}}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
Đk chỗ này là \(\sqrt{x}-1\ge0\Rightarrow\sqrt{x}\ge\sqrt{1}\Rightarrow x\ge1\)nhé
\(e,\frac{x-1}{\sqrt{y}-1}.\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}-1}.\frac{y-2\sqrt{y}+1}{\left(x-1\right)^2}\)
\(=\frac{\left(x-1\right)\left(\sqrt{y}-1\right)^2}{\left(\sqrt{y}-1\right)\left(x-1\right)^2}=\frac{\sqrt{y}-1}{x-1}\)
Linh ơi, câu a,b,c bạn làm đều đúng hết kết quả cách làm đều đúng nhưng mà ở chỗ câu c): \(\sqrt{x}^3+\sqrt{y}^3\)
không phải vậy đâu, mặc dù mình biết bạn hiểu, hay do sơ suất, nhưng mà chỗ đó là \(\sqrt{x^3}+\sqrt{y^3}\)nha! Dù sao cũng cảm ơn bạn nha!
giải hệ phương trình
a) \(\left\{{}\begin{matrix}\sqrt{2x^2+2y^2}+\sqrt{\frac{4}{3}\left(x^2+xy+y^2\right)}=2\left(x+y\right)\\\sqrt{3x+1}+\sqrt{5x+4}=3xy-y+3\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\\\sqrt{x+2y+1}+2\sqrt[3]{12x+7y+8}=2xy+x+5\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}x^2+xy+x+3=0\\\left(x+1\right)^2+3\left(y+1\right)+2\left(xy-\sqrt{x^2y+2y}\right)=0\end{matrix}\right.\)
b)\(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\)
\(\Rightarrow\left(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}\right)^2=\left(3\left(x+y\right)\right)^2\)
\(\Leftrightarrow\sqrt{\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)}=x^2+7xy+y^2\)
\(\Rightarrow\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)=\left(x^2+7xy+y^2\right)^2\)
\(\Leftrightarrow9\left(x-y\right)^2\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)
\(\rightarrow\left(x;y\right)\in\left\{\left(0;0\right),\left(1;1\right)\right\}\)
caau a) binh phuong len ra no x=y tuong tu
c)
ĐK $y \geqslant 0$
Hệ đã cho tương đương với
$\left\{\begin{matrix} 2x^2+2xy+2x+6=0\\ (x+1)^2+3(y+1)+2xy=2\sqrt{y(x^2+2)} \end{matrix}\right.$
Trừ từng vế $2$ phương trình ta được
$x^2+2+2\sqrt{y(x^2+2)}-3y=0$
$\Leftrightarrow (\sqrt{x^2+2}-\sqrt{y})(\sqrt{x^2+2}+3\sqrt{y})=0$
$\Leftrightarrow x^2+2=y$
Rút gọn
a.\(\left(2\sqrt{x}+\sqrt{2x}\right)\left(\sqrt{x}-\sqrt{2x}\right)\)
b. \(\left(\sqrt{3x}+\sqrt{2x}\right)\left(3\sqrt{x}-\sqrt{6x}\right)\)
c.\(\left(\frac{4}{3}\sqrt{3}+\sqrt{2}\sqrt{3\frac{1}{3}}\right)\left(\sqrt{1,2}+\sqrt{2}-4\sqrt{\frac{1}{3}}\right)-2\)
d.\(\left(2\sqrt{x}+\sqrt{y}\right)\left(3\sqrt{x}-2\sqrt{y}\right)\)(x,y lớn hơn hoặc bằng 0)
e.\(\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{x}\sqrt{y}+\sqrt{y}\right)\) (x,y lớn hơn hoặc bằng 0)
a) \(Q=\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+2x\sqrt{x}+y\sqrt{y}}{x\sqrt{x}+y\sqrt{y}}\left(x>0,y>0\right)\)
Rút Gọn
b) \(M=\frac{x^2-\sqrt{2}}{x^4+\left(\sqrt{3}-\sqrt{2}\right)x^2-\sqrt{6}}\)
Rút Gọn
Rút gọn
a) \(\frac{x}{y}\sqrt{\frac{y^2}{x^4}}\left(x\ne0;y>0\right)\) b) \(3x^2\sqrt{\frac{8}{x^2}}\left(x< 0\right)\) c) \(2x^3y^3\sqrt{\frac{4}{x^8y^6}}\left(x\ne0;y< 0\right)\)
d)\(\frac{\sqrt{4x^4y^6}}{\sqrt{196x^6y^6}}\left(x< 0;y\ne0\right)\)
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 \(x\ne0;y>0\))
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 \(x< 0\))
Rút gọn
\(B=\frac{2}{x^2-y^2}\sqrt{\frac{9\left(x^2+2xy+y^2\right)}{4}}\) với x > -y
\(C=\sqrt{\frac{2a}{3}}.\sqrt{\frac{3a}{8}}\) với a >hoặc= 0
\(\frac{1}{a-b}\sqrt{a^4\left(a-b\right)^2}\) với a > 0
\(B=\frac{2}{x^2-y^2}\cdot\sqrt{\frac{9\left(x^2+2xy+y^2\right)}{4}}=\frac{2}{\left(x-y\right)\left(x+y\right)}\cdot\sqrt{\frac{9\left(x+y\right)^2}{4}}\)
\(=\frac{2}{\left(x-y\right)\left(x+y\right)}\cdot\frac{\sqrt{9\left(x+y\right)^2}}{\sqrt{4}}=\frac{2}{\left(x-y\right)\left(x+y\right)}\cdot\frac{3\left(x+y\right)}{2}\)(vì x > -y <=> x + y > 0)
\(=\frac{3}{x-y}\)
\(C=\sqrt{\frac{2a}{3}}.\sqrt{\frac{3a}{8}}=\sqrt{\frac{2a}{3}\cdot\frac{3a}{8}}=\sqrt{\frac{6a^2}{24}}=\sqrt{\frac{a^2}{4}}=\frac{a}{2}\)(vì a > = 0)
\(D=\frac{1}{a-b}\cdot\sqrt{a^4\left(a-b\right)^2}=\frac{1}{a-b}\cdot a^2\left(a-b\right)=a^2\)(a > b > 0)
câu cuối điều kiện là a>b
\(\frac{1}{a-b}\sqrt{a^4\left(a-b\right)^2}=\frac{a^2\left|a-b\right|}{a-b}=\frac{a^2\left(a-b\right)}{a-b}=a^2\) (vì a>b)
Rút gọn và tính giá trị biểu thức: a, \(\frac{x+\sqrt{xy}}{y+\sqrt{xy}}\)
b, \(\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\)
c, \(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
d,\(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\left(x\ge0\right)\)
e,\(\frac{x-1}{\sqrt{y}-1}\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\left(x\ne1,y\ne1,y>0\right)\)
\(\(b)\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(a,b\ge0;a,b\ne1\right)\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{a}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab+1}\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{ab}-1\right)}\left(a,b\ge0.a,b\ne1\right)\)\)
_Minh ngụy_
\(\(c)\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)\)( tự ghi điều kiện )
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)^2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(x\sqrt{x}+x\sqrt{y}-2x\sqrt{y}-2y\sqrt{x}+y\sqrt{x}+y\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)\)( phá ngoặc và tính )
\(\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\sqrt{xy}\)\)
_Minh ngụy_
\(\(d)\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\left(x\ge0\right)\)\)
\(\(=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}\)\)
\(\(=\frac{|\sqrt{x}-1|}{|\sqrt{x}+1|}\)\)
\(\(=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)\)( vì \(\(x\ge0\)\))
_Minh ngụy_