Mn giúp e vs ạ.. e đg cần gấp, cảm ơn mn nhiều lắm ạ
Q=\(\dfrac{x-3\sqrt{ }x-4}{x-\sqrt{ }x-12}\)
\(P=\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{ }x-\sqrt{ }y}\)
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AH
Akai Haruma
Giáo viên
8 tháng 7 2018
Lời giải:
Ta có: \(xy+yz+xz=1\)
\(\Rightarrow \left\{\begin{matrix} x^2+1=x^2+xy+yz+xz=(x+y)(x+z)\\ y^2+1=y^2+xy+yz+xz=(y+z)(y+x)\\ z^2+1=z^2+xy+yz+xz=(z+x)(z+y)\end{matrix}\right.\)
Do đó:
\(\sqrt{\frac{(y^2+1)(z^2+1)}{x^2+1}}=\sqrt{\frac{(y+z)(y+x)(z+x)(z+y)}{(x+y)(x+z)}}=\sqrt{(y+z)^2}=y+z\)
\(\Rightarrow x\sqrt{\frac{(y^2+1)(z^2+1)}{x^2+1}}=x(y+z)\)
Hoàn toàn tt:
\(y\sqrt{\frac{(z^2+1)(x^2+1)}{y^2+1}}=y(x+z)\); \(z\sqrt{\frac{(x^2+1)(y^2+1)}{z^2+1}}=z(x+y)\)
Do đó:
\(A=x(y+z)+y(x+z)+z(x+y)=2(xy+yz+xz)=2\)
XO
9 tháng 7 2023
Có : \(x-2y-\sqrt{xy}+\sqrt{x}-2\sqrt{y}=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)+\sqrt{x}-2\sqrt{y}=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}+1\right)=0\)
\(\Leftrightarrow\sqrt{x}=2\sqrt{y}\) (Do \(\sqrt{x}+\sqrt{y}+1>0,\forall x;y>0\))
\(\Leftrightarrow x=4y\)
Khi đó \(P=\dfrac{7y}{\left(2\sqrt{y}+3\sqrt{y}\right).\left(\sqrt{x}+2\sqrt{y}\right)}\)
\(=\dfrac{7y}{5\sqrt{y}.4\sqrt{y}}=\dfrac{7}{20}\)
HS
1 tháng 8 2018
\(\dfrac{1}{\sqrt{x-1}-\sqrt{x}}+\dfrac{1}{\sqrt{x-1}+\sqrt{x}}+\dfrac{\sqrt{x^3}-x}{\sqrt{x-1}}\)
\(\Leftrightarrow-\left(\sqrt{x-1}+\sqrt{x}\right)-\left(\sqrt{x-1}-\sqrt{x}\right)+\dfrac{x\sqrt{x}-x}{\sqrt{x-1}}\)
\(\Leftrightarrow-\sqrt{x-1}-\sqrt{x}-\sqrt{x-1}+\sqrt{x}+\dfrac{x\sqrt{x}-x}{\sqrt{x-1}}\)
\(\Leftrightarrow-2\sqrt{x-1}+\dfrac{x\sqrt{x}-x}{\sqrt{x-1}}\)
\(\Leftrightarrow\dfrac{-2\left(x-1\right)+x\sqrt{x}-x}{\sqrt{x-1}}\)
\(\Leftrightarrow\dfrac{-2x+2+x\sqrt{x}-x}{\sqrt{x-1}}\)
\(\Leftrightarrow\dfrac{-3x+2+x\sqrt{x}}{\sqrt{x-1}}\)
19 tháng 7 2022
a: \(=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}=\sqrt{ab}-\sqrt{ab}=0\)
b: \(=\dfrac{\left(\sqrt{x}-2\sqrt{y}\right)^2}{\sqrt{x}-2\sqrt{y}}+\dfrac{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{x}-2\sqrt{y}+\sqrt{y}=\sqrt{x}-\sqrt{y}\)
c: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)
\(=\sqrt{x}+2-\sqrt{x}-2=0\)
27 tháng 12 2018
a) \(\left\{{}\begin{matrix}\dfrac{3x}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\) (ĐKXĐ: \(x\ne-1;y\ne-4\))
Đặt \(\dfrac{x}{x+1}=a;\dfrac{1}{y+4}=b\left(a\ne0;b\ne0\right)\)
Hệ phương trình đã cho trở thành
\(\left\{{}\begin{matrix}3a-2b=4\left(1\right)\\2a-5b=9\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow2a=9+5b\Leftrightarrow a=\dfrac{9+5b}{2}\)
Thay \(a=\dfrac{9+5b}{2}\) vào \(\left(1\right)\), ta có:
\(\dfrac{3\left(9+5b\right)}{2}-2b=4\)
\(\Leftrightarrow27+15b-4b=8\)
\(\Leftrightarrow11b=-19\Leftrightarrow b=\dfrac{-19}{11}\)
Thay \(b=\dfrac{-19}{11}\) vào \(\left(2\right)\), ta có:
\(2a-5\cdot\dfrac{-19}{11}=9\)
\(\Leftrightarrow a=\dfrac{2}{11}\)
Với \(a=\dfrac{2}{11}\Rightarrow\dfrac{x}{x+1}=\dfrac{2}{11}\)
\(\Leftrightarrow11x=2x+2\Leftrightarrow x=\dfrac{2}{9}\)
Với \(b=\dfrac{-19}{11}\Rightarrow\dfrac{1}{y+4}=\dfrac{-19}{11}\)
\(\Leftrightarrow-19y-76=11\)
\(\Leftrightarrow y=\dfrac{-90}{19}\)
27 tháng 12 2018
b,Ta có:
\(PT\Leftrightarrow7+3.\sqrt[3]{2+x}.\sqrt[3]{5-x}\left(\sqrt[3]{2+x}+\sqrt[3]{5-x}\right)=1\)
Thay \(\sqrt[3]{2+x}+\sqrt[3]{5-x}=1\) vào PT
\(\Rightarrow\) \(3.\sqrt[3]{2+x}.\sqrt[3]{5-x}=-6\)
\(\Leftrightarrow\sqrt[3]{2+x}.\sqrt[3]{5-x}=-2\)
\(\Leftrightarrow\left(2+x\right)\left(5-x\right)=-8\)
\(\Leftrightarrow x^2-3x-18=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=6\end{matrix}\right.\)
Thử lại thấy x= - 3, x=6 thỏa mãn
Vậy x= -3, x = 6
AH
Akai Haruma
Giáo viên
11 tháng 8 2021
Lời giải:
\(A=\frac{x^2}{\sqrt{x^4+8xy^3}}+\frac{2y^2}{\sqrt{y^4+y(x+y)^3}}\)
Xét:
\(x^4+8xy^3-(x^2+2y^2)^2=8xy^3-4y^4-4x^2y^2\)
\(=-4y^2(x^2-2xy+y^2)=-4y^2(x-y)^2\leq 0\)
\(\Rightarrow x^4+8xy^3\leq (x^2+2y^2)^2\)
\(\Rightarrow \frac{x^2}{\sqrt{x^4+8xy^3}}\geq \frac{x^2}{x^2+2y^2}(*)\)
Mặt khác:
\(y^4+y(x+y)^3-(x^2+2y^2)^2=x^3y+3xy^3-2y^4-x^4-x^2y^2\)
\(=x^3(y-x)+3y^3(x-y)+y^4-x^2y^2\)
\(=x^3(y-x)+3y^3(x-y)+y^2(y-x)(y+x)\)
\(=(y-x)(x^3-2y^3+xy^2)\)
\(=(y-x)[(x-y)(x^2+xy+y^2)+y^2(x-y)]\)
\(=-(x-y)^2(x^2+xy+2y^2)\leq 0\)
\(\Rightarrow y^4+y(x+y)^3\leq (x^2+2y^2)^2\Rightarrow \frac{2y^2}{\sqrt{y^4+y(x+y)^3}}\geq \frac{2y^2}{x^2+2y^2}(**)\)
Từ $(*); (**)\Rightarrow A\geq 1$
29 tháng 5 2022
\(=\left(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}-\dfrac{\sqrt{y}\left(x-y\right)}{x-y}\right):\dfrac{x+2\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\)\(=\left(\sqrt{x}+\sqrt{y}-\sqrt{y}\right)\cdot\dfrac{\sqrt{x}+\sqrt{y}}{\left(\sqrt{x}+\sqrt{y}\right)^2}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)
AT
1 tháng 8 2018
rút gọn hở bạn?
đkxđ: x>0 ; x≠1
\(S=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right)+\left(x-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)+\dfrac{x-1}{\sqrt{x}}\left(\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{x-1}\right)\)
\(=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}+\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{\left(\sqrt{x}+1-\sqrt{x}+1\right)\left(\sqrt{x}+1+\sqrt{x}-1\right)}{x-1}\)
\(\dfrac{2\sqrt{x}}{\sqrt{x}}+\dfrac{2\cdot2\sqrt{x}}{\sqrt{x}}=\dfrac{6\sqrt{x}}{\sqrt{x}}=6\)
20 tháng 11 2022
\(B=\dfrac{xy}{xy}+\dfrac{\left(x-y\right)x}{x\left(x-y\right)}-\dfrac{y\left(x-y\right)}{y\left(x-y\right)}=1\)
\(Q=\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\left(ĐK:x\ge0;x\ne16\right)\\ =\dfrac{x-4\sqrt{x}+\sqrt{x}-4}{x-4\sqrt{x}+3\sqrt{x}-12}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-4\right)+\left(\sqrt{x}-4\right)}{\sqrt{x}\left(\sqrt{x}-4\right)+3\left(\sqrt{x}-4\right)}\\ =\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-4\right)}\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
\(P=\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}=\sqrt{xy}\)