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a/ \(\hept{\begin{cases}\sqrt{xy}+\sqrt{1-y}=\sqrt{y}\left(1\right)\\2\sqrt{xy-y}-\sqrt{y}=-1\left(2\right)\end{cases}}\)
Điều kiện: \(\hept{\begin{cases}x\ge1\\0\le y\le1\end{cases}}\)
Xét phương trình (1) ta đễ thấy y = 0 không phải là nghiệm:
\(\sqrt{xy}+\sqrt{1-y}=\sqrt{y}\)
\(\Leftrightarrow\sqrt{y}\left(1-\sqrt{x}\right)=\sqrt{1-y}\)
\(\Leftrightarrow1-\sqrt{x}=\frac{\sqrt{1-y}}{\sqrt{y}}\)
\(\Rightarrow1-\sqrt{x}\ge0\)
\(\Leftrightarrow x\le1\)
Kết hợp với điều kiện ta được x = 1 thê vô PT (2) ta được y = 1
b/ \(\hept{\begin{cases}\sqrt{\frac{2x}{y}}+\sqrt{\frac{2y}{x}}=3\left(1\right)\\x-y+xy=3\left(2\right)\end{cases}}\)
Xét pt (1) ta có
\(\sqrt{\frac{2x}{y}}+\sqrt{\frac{2y}{x}}=3\)
Đặt \(\sqrt{\frac{x}{y}}=a\left(a>0\right)\)thì pt (1) thành
\(\sqrt{2}a+\frac{\sqrt{2}}{a}=3\)
\(\Leftrightarrow a^2+1=\frac{3}{\sqrt{2}}\)
Tới đây đơn giản rồi làm tiếp nhé
\(a,\dfrac{x+2\sqrt{x}-3}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{x+3\sqrt{x}-\sqrt{x}-3}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+3\right)-\left(\sqrt{x}+3\right)}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(\Rightarrow\sqrt{x}+3\)
\(b,\dfrac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{4y+7\sqrt{y}-4\sqrt{y}-7}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{\sqrt{y}.\left(4\sqrt{y}\right)-\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{\left(4\sqrt{y}+7\right).\left(\sqrt{y}-1\right)}{4\sqrt{y}+7}\)
\(\Rightarrow\sqrt{y}-1\)
\(c,\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(\Leftrightarrow\dfrac{\sqrt{xy}.\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)
\(\Rightarrow\sqrt{xy}\)
\(d,\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}-4\sqrt{x}-4}{x+3\sqrt{x}-4\sqrt{x}-12}\)
\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)}{\sqrt{x}.\left(x+3\right)-4\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right).\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-4\right)}\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
\(\Rightarrow\dfrac{x-2\sqrt{x}-3}{x-9}\)
\(e,\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+\sqrt{4}}\)
\(\Leftrightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+2}\)
\(\Rightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{3}\)
Bài 1:
Áp dụng BĐT AM-GM:
\(9=x+y+xy+1=(x+1)(y+1)\leq \left(\frac{x+y+2}{2}\right)^2\)
\(\Rightarrow 4\leq x+y\)
Tiếp tục áp dụng BĐT AM-GM:
\(x^3+4x\geq 4x^2; y^3+4y\geq 4y^2\)
\(\frac{x}{4}+\frac{1}{x}\geq 1; \frac{y}{4}+\frac{1}{y}\geq 1\)
\(\Rightarrow x^3+y^3+x^2+y^2+5(x+y)+\frac{1}{x}+\frac{1}{y}\geq 5(x^2+y^2)+\frac{3}{4}(x+y)+2\)
Mà:
\(5(x^2+y^2)\geq 5.\frac{(x+y)^2}{2}\geq 5.\frac{4^2}{2}=40\)
\(\frac{3}{4}(x+y)\geq \frac{3}{4}.4=3\)
\(\Rightarrow A= x^3+y^3+x^2+y^2+5(x+y)+\frac{1}{x}+\frac{1}{y}\geq 40+3+2=45\)
Vậy \(A_{\min}=45\Leftrightarrow x=y=2\)
Bài 2:
\(B=\frac{a^2}{a-1}+\frac{2b^2}{b-1}+\frac{3c^2}{c-1}\)
\(B-24=\frac{a^2}{a-1}-4+\frac{2b^2}{b-1}-8+\frac{3c^2}{c-1}-12\)
\(=\frac{a^2-4a+4}{a-1}+\frac{2(b^2-4b+4)}{b-1}+\frac{3(c^2-4c+4)}{c-1}\)
\(=\frac{(a-2)^2}{a-1}+\frac{2(b-2)^2}{b-1}+\frac{3(c-2)^2}{c-1}\geq 0, \forall a,b,c>1\)
\(\Rightarrow B\geq 24\)
Vậy \(B_{\min}=24\Leftrightarrow a=b=c=2\)
a, \(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\dfrac{\sqrt{3}.\sqrt{5}-\sqrt{3}.\sqrt{2}}{\sqrt{5}.\sqrt{7}-\sqrt{7}.\sqrt{2}}\)
\(=\dfrac{\sqrt{3}.\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}.\left(\sqrt{5}-\sqrt{2}\right)}=\dfrac{\sqrt{3}}{\sqrt{7}}\)
b, \(\dfrac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}\)
\(=\dfrac{2.\sqrt{5}.\sqrt{3}-2.\sqrt{2}.\sqrt{5}-\sqrt{3}.\sqrt{3}+\sqrt{2}.\sqrt{3}}{2.\sqrt{5}-2.\sqrt{2}.\sqrt{5}-\sqrt{3}+\sqrt{2}.\sqrt{3}}\)
\(=\dfrac{2\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{3}.\left(\sqrt{3}-\sqrt{2}\right)}{2\sqrt{5}.\left(1-\sqrt{2}\right)-\sqrt{3}.\left(1-\sqrt{2}\right)}\)
\(=\dfrac{\left(2\sqrt{5}+\sqrt{3}\right).\left(\sqrt{3}-\sqrt{2}\right)}{\left(2\sqrt{5}-\sqrt{3}\right).\left(1-\sqrt{2}\right)}=\dfrac{\sqrt{3}-\sqrt{2}}{1-\sqrt{2}}\)
c, \(\dfrac{x+\sqrt{xy}}{y+\sqrt{xy}}=\dfrac{\sqrt{x}.\sqrt{x}+\sqrt{x}.\sqrt{y}}{\sqrt{y}.\sqrt{y}+\sqrt{x}.\sqrt{y}}\)
\(=\dfrac{\sqrt{x}.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{y}.\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{\sqrt{x}}{\sqrt{y}}\)
Chúc bạn học tốt!!!
d) \(\dfrac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\) = \(-\dfrac{\sqrt{a}\left(1+\sqrt{ab}\right)-\sqrt{b}\left(1+\sqrt{ab}\right)}{1-ab}\)
= \(-\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(1+\sqrt{ab}\right)}{\left(1+\sqrt{ab}\right)\left(1-\sqrt{ab}\right)}\) = \(-\dfrac{\sqrt{a}-\sqrt{b}}{1-\sqrt{ab}}\) = \(\dfrac{\sqrt{b}-\sqrt{a}}{1-\sqrt{ab}}\)
\(A,ĐKXĐ:x;y\ge0\)
\(A=\sqrt{xy}-2\sqrt{y}-5\sqrt{x}+10\)
\(=\sqrt{y}\left(\sqrt{x}-2\right)-5\left(\sqrt{x}-2\right)\)
\(=\left(\sqrt{x}-2\right)\left(\sqrt{y}-5\right)\)
\(ĐKXĐ:x;y\ge0\)
\(B=a\sqrt{x}+b\sqrt{y}-\sqrt{xy}-ab\)
\(=\left(a\sqrt{x}-\sqrt{xy}\right)+\left(b\sqrt{y}-ab\right)\)
\(=\sqrt{x}\left(a-\sqrt{y}\right)+b\left(\sqrt{y}-a\right)\)
\(=\sqrt{x}\left(a-\sqrt{y}\right)-b\left(a-\sqrt{y}\right)\)
\(=\sqrt{x}\left(a-\sqrt{y}\right)-b\left(a-\sqrt{y}\right)\)
\(=\left(a-\sqrt{y}\right)\left(\sqrt{x}-b\right)\)
f)\(\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}:\frac{1}{\sqrt{x}-\sqrt{y}}\)
\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}.\left(\sqrt{x}-\sqrt{y}\right)\)
\(=x-y\)
Lời giải:
Bổ sung ĐK $x,y\geq 0$ để các biểu thức có nghĩa.
a)
\(A=x+y-8\sqrt{x}-2\sqrt{y}-2019=(x-8\sqrt{x}+16)+(y-2\sqrt{y}+1)-2036\)
\(=(\sqrt{x}-4)^2+(\sqrt{y}-1)^2-2036\)
Ta thấy \((\sqrt{x}-4)^2\geq 0; (\sqrt{y}-1)^2\geq 0\) với mọi \(x,y\geq 0\)
Do đó: \(A=(\sqrt{x}-4)^2+(\sqrt{y}-1)^2-2036\geq -2036\)
Vậy GTNN của $A$ là $-2036$ khi \(\left\{\begin{matrix} \sqrt{x}-4=0\\ \sqrt{y}-1=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=16\\ y=1\end{matrix}\right.\)
b)
\(B=x+y+12\sqrt{x}-4\sqrt{y}+19=(x+12\sqrt{x})+(y-4\sqrt{y}+4)+15\)
\(=x+12\sqrt{x}+(\sqrt{y}-2)^2+15\)
Ta thấy: \(x+12\sqrt{x}\geq 0; (\sqrt{y}-2)^2\geq 0, \forall x,y\geq 0\)
\(\Rightarrow B\ge 0+0+15=15\)
Vậy GTNN của $B$ là $15$ khi \(\left\{\begin{matrix} x+12\sqrt{x}=0\\ \sqrt{y}-2=0\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=0\\ y=4\end{matrix}\right.\)
c)
\(C=2x+y-10\sqrt{x}-6\sqrt{y}+2\sqrt{xy}+8\)
\(=(x+y+2\sqrt{xy})+x-10\sqrt{x}-6\sqrt{y}+8\)
\(=(\sqrt{x}+\sqrt{y})^2-6(\sqrt{x}+\sqrt{y})+(x-4\sqrt{x})+8\)
\(=(\sqrt{x}+\sqrt{y})^2-6(\sqrt{x}+\sqrt{y})+9+(x-4\sqrt{x}+4)-5\)
\(=(\sqrt{x}+\sqrt{y}-3)^2+(\sqrt{x}-2)^2-5\)
\(\geq 0+0-5=-5\) với mọi $x,y\ge 0$
Vậy GTNN của $C$ là $-5$ đạt tại \(\left\{\begin{matrix} \sqrt{x}+\sqrt{y}-3=0\\ \sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} y=1\\ x=4\end{matrix}\right.\)
d)
\(D=2y+x-2\sqrt{x}-2\sqrt{y}+2\sqrt{xy}+2\)
\(=(y+x+2\sqrt{xy})+y-2\sqrt{x}-2\sqrt{y}+2\)
\(=(\sqrt{x}+\sqrt{y})^2-2(\sqrt{x}+\sqrt{y})+1+y+1\)
\(=(\sqrt{x}+\sqrt{y}-1)^2+y+1\)
\(\geq 0+0+1=1\) với mọi $x,y\geq 0$
Vậy GTNN của $D$ là $1$ khi \(\left\{\begin{matrix} \sqrt{x}+\sqrt{y}-1=0\\ y=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} y=0\\ x=1\end{matrix}\right.\)
Lời giải:
Bổ sung ĐK $x,y\geq 0$ để các biểu thức có nghĩa.
a)
\(A=x+y-8\sqrt{x}-2\sqrt{y}-2019=(x-8\sqrt{x}+16)+(y-2\sqrt{y}+1)-2036\)
\(=(\sqrt{x}-4)^2+(\sqrt{y}-1)^2-2036\)
Ta thấy \((\sqrt{x}-4)^2\geq 0; (\sqrt{y}-1)^2\geq 0\) với mọi \(x,y\geq 0\)
Do đó: \(A=(\sqrt{x}-4)^2+(\sqrt{y}-1)^2-2036\geq -2036\)
Vậy GTNN của $A$ là $-2036$ khi \(\left\{\begin{matrix} \sqrt{x}-4=0\\ \sqrt{y}-1=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=16\\ y=1\end{matrix}\right.\)
b)
\(B=x+y+12\sqrt{x}-4\sqrt{y}+19=(x+12\sqrt{x})+(y-4\sqrt{y}+4)+15\)
\(=x+12\sqrt{x}+(\sqrt{y}-2)^2+15\)
Ta thấy: \(x+12\sqrt{x}\geq 0; (\sqrt{y}-2)^2\geq 0, \forall x,y\geq 0\)
\(\Rightarrow B\ge 0+0+15=15\)
Vậy GTNN của $B$ là $15$ khi \(\left\{\begin{matrix} x+12\sqrt{x}=0\\ \sqrt{y}-2=0\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=0\\ y=4\end{matrix}\right.\)
c)
\(C=2x+y-10\sqrt{x}-6\sqrt{y}+2\sqrt{xy}+8\)
\(=(x+y+2\sqrt{xy})+x-10\sqrt{x}-6\sqrt{y}+8\)
\(=(\sqrt{x}+\sqrt{y})^2-6(\sqrt{x}+\sqrt{y})+(x-4\sqrt{x})+8\)
\(=(\sqrt{x}+\sqrt{y})^2-6(\sqrt{x}+\sqrt{y})+9+(x-4\sqrt{x}+4)-5\)
\(=(\sqrt{x}+\sqrt{y}-3)^2+(\sqrt{x}-2)^2-5\)
\(\geq 0+0-5=-5\) với mọi $x,y\ge 0$
Vậy GTNN của $C$ là $-5$ đạt tại \(\left\{\begin{matrix} \sqrt{x}+\sqrt{y}-3=0\\ \sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} y=1\\ x=4\end{matrix}\right.\)
d)
\(D=2y+x-2\sqrt{x}-2\sqrt{y}+2\sqrt{xy}+2\)
\(=(y+x+2\sqrt{xy})+y-2\sqrt{x}-2\sqrt{y}+2\)
\(=(\sqrt{x}+\sqrt{y})^2-2(\sqrt{x}+\sqrt{y})+1+y+1\)
\(=(\sqrt{x}+\sqrt{y}-1)^2+y+1\)
\(\geq 0+0+1=1\) với mọi $x,y\geq 0$
Vậy GTNN của $D$ là $1$ khi \(\left\{\begin{matrix} \sqrt{x}+\sqrt{y}-1=0\\ y=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} y=0\\ x=1\end{matrix}\right.\)