1. a)Tìm x , biết \(\sqrt{4x^2-4x+1}=3\)
b) Chứng minh \(\left(\sqrt{x}+\sqrt{y}\right)\left(\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}\right)=x-y\) với x > 0 ; y> 0
1. Tính:
\(\sqrt{\dfrac{x-1+\sqrt{2x-3}}{x+2-\sqrt{2x+3}}}\)
2. Chứng minh:
a) \(\dfrac{\left(3\sqrt{xy}-6y.2x\sqrt{y}+4y\sqrt{x}\right)\left(3\sqrt{y}+2\sqrt{xy}\right)}{y\left(\sqrt{x}-2\sqrt{y}\right)\left(y-4x\right)}=1\)
b) \(\left(\sqrt{x}-\sqrt{y}-\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right)\left(\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}+\dfrac{y}{\sqrt{x}-\sqrt{y}}-\dfrac{2\sqrt{xy}}{xy}\right)=\sqrt{x}+\sqrt{y}\)
1.
\(\sqrt{\dfrac{x-1+\sqrt{2x-3}}{x+2-\sqrt{2x+3}}}\Leftrightarrow\)\(\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\sqrt{\dfrac{\left(\sqrt{2x-3}+1\right)^2}{\left(\sqrt{2x+3}-1\right)^2}}\end{matrix}\right.\)\(\Leftrightarrow\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\sqrt{2x-3}+1}{\sqrt{2x+3}-1}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\left(\sqrt{2x-3}+1\right)\left(\sqrt{2x+3}+1\right)}{2\left(x+1\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\sqrt{4x^2-9}+\sqrt{2x-3}+\sqrt{2x+3}+1}{2\left(x+1\right)}\end{matrix}\right.\)
hết tối giải rồi
Giải hệ pt
1/\(\left\{{}\begin{matrix}4x\sqrt{y+1}+8x=\left(4x^2-4x-3\right)\sqrt{x+1}\\\dfrac{x}{x+1}+x^2=\left(y+2\right)\sqrt{\left(x+1\right)\left(y+1\right)}\end{matrix}\right.\)
2/\(\left\{{}\begin{matrix}x\sqrt{y^2+6}+y\sqrt{x^2+3}=7xy\\x\sqrt{x^2+3}+y\sqrt{y^2+6}=x^2+y^2+2\end{matrix}\right.\)\(\left\{{}\begin{matrix}x\sqrt{y^2+6}+y\sqrt{x^2+3}=7xy\\x\sqrt{x^2+3}+y\sqrt{y^2+6}=x^2+y^2+2\end{matrix}\right.\)
3/\(\left\{{}\begin{matrix}\left(2x+y-1\right)\left(\sqrt{x+3}+\sqrt{xy}+\sqrt{x}\right)=8\sqrt{x}\\\left(\sqrt{x+3}+\sqrt{xy}\right)^2+xy=2x\left(6-x\right)\end{matrix}\right.\)\(\left\{{}\begin{matrix}\left(2x+y-1\right)\left(\sqrt{x+3}+\sqrt{xy}+\sqrt{x}\right)=8\sqrt{x}\\\left(\sqrt{x+3}+\sqrt{xy}\right)^2+xy=2x\left(6-x\right)\end{matrix}\right.\)
4/\(\left\{{}\begin{matrix}\sqrt{xy+x+2}+\sqrt{x^2+x}-4\sqrt{x}=0\\xy+x^2+2=x\left(\sqrt{xy+2}+3\right)\end{matrix}\right.\)\(\left\{{}\begin{matrix}\sqrt{xy+x+2}+\sqrt{x^2+x}-4\sqrt{x}=0\\xy+x^2+2=x\left(\sqrt{xy+2}+3\right)\end{matrix}\right.\)
m.n giúp e mấy bài này vs ạ!!
35Cho biểu thức
P=\(\left[\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)\frac{2}{\sqrt{x}+\sqrt{y}}+\frac{1}{x}+\frac{1}{y}\right]:\frac{\sqrt{x^3}+y\sqrt{x}+x\sqrt{y}+\sqrt{y^3}}{\sqrt{xy^3}+\sqrt{x^3y}}\)
a) Rút gọn P
b)Cho xy=16 . Tìm Min P
34 Cho biểu thức
P=\(\frac{x}{\sqrt{xy}-2y}-\frac{2\sqrt{x}}{x+\sqrt{x}-2\sqrt{xy}-2\sqrt{y}}-\frac{1-x}{1-\sqrt{x}}\)
a) Rút gọn P
b)Tính P biết 2x^2+y^2-4x-2xy+4=0
1. Chứng minh \(\sqrt[3]{3+\sqrt[3]{3}}+\sqrt[3]{3-\sqrt[3]{3}}< 2\sqrt[3]{3}\)
2. a) Tính \(A=\frac{2b.\sqrt{x^2-1}}{x-\sqrt{x^2-1}}\) với \(x=\frac{1}{2}\left(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\right)\left(a,b>0\right) \)
b) Tính \(B=\frac{xy-\sqrt{x^2-1}.\sqrt{y^2-1}}{xy+\sqrt{x^2-1}.\sqrt{y^2-1}}\) với \(x=\frac{1}{2}\left(a+\frac{1}{a}\right);y=\frac{1}{2}\left(b+\frac{1}{b}\right)\left(a,b\ge1\right)\)
3. Cho x,y thỏa mãn \(xy\ge0\). Tính \(B=\left(\left|\sqrt{xy}+\frac{x}{2}+\frac{y}{2}\right|-\left|x\right|\right)+\left(\left|\sqrt{xy}-\frac{x}{2}-\frac{y}{2}\right|-\left|y\right|\right)\)
4. Cho \(\frac{2x+2\sqrt{x}+13}{\left(\sqrt{x}-2\right)\left(x+1\right)^2}=\frac{A}{\sqrt{x}-2}+\frac{B\sqrt{x}+C}{x+1}+\frac{D\sqrt{x}+E}{\left(x+1\right)^2}\). Tìm các số A,B,C,D,E để đẳng thức trên là đúng với mọi x
Rút gọn:
a/ \(\frac{\left(\sqrt{x^2+9}-3\right)\left(\sqrt{x^2+9}+3\right)\left(x+\sqrt{xy}+y\right)\sqrt{x-2\sqrt{xy}+y}}{x\left(x\sqrt{x}-y\sqrt{y}\right)}\) (với x>0, y\(\ge\)0, x\(\ne\)y
b/ \(\left[\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right).\frac{2}{\sqrt{x}+\sqrt{y}}+\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right]:\frac{\sqrt{x^3}+y\sqrt{x}+x\sqrt{y}+\sqrt{y^3}}{\sqrt{x^3y}+\sqrt{xy^3}}\)(với x>0 và x\(\ne\)1
c/ \(\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{xy}+\sqrt{x}}{1-\sqrt{xy}}+1\right):\left(1-\frac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-\frac{\sqrt{x}+1}{\sqrt{xy}+1}\right)\)(với x>0 và x\(\ne\)1
ai giúp t với
1:\(\left\{\begin{matrix}x\sqrt{12-y}+\sqrt{y\left(12-x^2\right)}=12\\x^3-8x-1=2\sqrt{y-2}\end{matrix}\right.\)
2:\(\left\{\begin{matrix}\left(1-y\right)\sqrt{x-y}+x=2+\left(x-y-1\right)\sqrt{y}\\2y^2-3x+6y+1=2\sqrt{x-2y}-\sqrt{4x-5y-3}\end{matrix}\right.\)
3:\(\left\{\begin{matrix}y\left(x^2+2x+2\right)=x\left(y^2+6\right)\\\left(y-1\right)\left(x^2+2x+7\right)=\left(x+1\right)\left(y^2+1\right)\end{matrix}\right.\)
4:\(\left\{\begin{matrix}x-2\sqrt{y+1}=3\\x^3-4x^2\sqrt{y+1}-9x-8y=-52-4xy\end{matrix}\right.\)
5:\(\left\{\begin{matrix}\frac{y-2x+\sqrt{y}-x}{\sqrt{xy}}+1=0\\\sqrt{1-xy}+x^2-y^2=0\end{matrix}\right.\)
Cho x,y thỏa mãn x>1, y<0 và \(\frac{\left(x+y\right)\left(x^3-y^3\right)\sqrt{4x-2\sqrt{4x-1}}}{\left(1-\sqrt{4x-1}\right)\left(x^2y^2+xy^3+y^4\right)}=-8\). Vậy \(\frac{x}{y}=\)
\(A=\frac{\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)\sqrt{4x-1-2\sqrt{4x-1}+1}}{-\left(\sqrt{4x-1}-1\right).y^2\left(x^2+xy+y^2\right)}=\frac{\left(x^2-y^2\right)\sqrt{\left(\sqrt{4x-1}-1\right)^2}}{-\left(\sqrt{4x-1}-1\right).y^2}\)
Do \(x>1\Rightarrow4x-1>1\Rightarrow\sqrt{4x-1}>1\Rightarrow\sqrt{4x-1}-1>0\)
\(\Rightarrow A=\frac{\left(x^2-y^2\right)\left(\sqrt{4x-1}-1\right)}{-\left(\sqrt{4x-1}-1\right).y^2}=\frac{x^2-y^2}{-y^2}=1-\left(\frac{x}{y}\right)^2\)
\(A=-8\Rightarrow1-\left(\frac{x}{y}\right)^2=-8\Rightarrow\left(\frac{x}{y}\right)^2=9\)
Do \(\left\{{}\begin{matrix}x>1\\y< 0\end{matrix}\right.\) \(\Rightarrow\frac{x}{y}< 0\Rightarrow\frac{x}{y}=-3\)
Tìm max
\(A=3\sqrt{2x-1}+x\sqrt{5-4x^2}\left(\frac{1}{2}\le x\le\frac{\sqrt{5}}{2}\right)\)
\(B=\frac{xyz\left(x+y+z+\sqrt{x^2+y^2+z^2}\right)}{\left(x^2+y^2+z^2\right)\left(xy+yz+zx\right)}\left(x,y,z>0\right)\)
A
Áp dụng BĐT cosi ta có
\(\sqrt{\left(2x-1\right).1}\le\frac{2x-1+1}{2}=x\)
\(x\sqrt{5-4x^2}\le\frac{x^2+5-4x^2}{2}=\frac{-3x^2+5}{2}\)
Khi đó
\(A\le3x+\frac{-3x^2+5}{2}=\frac{-3x^2+6x+5}{2}=\frac{-3\left(x-1\right)^2}{2}+4\le4\)
MaxA=4 khi \(\hept{\begin{cases}2x-1=1\\x^2=5-4x^2\\x=1\end{cases}\Rightarrow}x=1\)
B
Áp dụng BĐT cosi ta có :
\(x^2+y^2+z^2\ge\frac{1}{3}\left(x+y+z\right)^2\)
=> \(x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}\)
=> \(B\le\frac{xyz.\left(\sqrt{3\left(x^2+y^2+z^2\right)}+\sqrt{x^2+y^2+z^2}\right)}{\left(x^2+y^2+z^2\right)\left(xy+yz+xz\right)}=\frac{xyz.\left(\sqrt{3}+1\right)}{\left(xy+yz+xz\right)\sqrt{x^2+y^2+z^2}}\)
Lại có \(x^2+y^2+z^2\ge3\sqrt[3]{x^2y^2z^2}\); \(xy+yz+xz\ge3\sqrt[3]{x^2y^2z^2}\)
=> \(\sqrt{x^2+y^2+z^2}\left(xy+yz+xz\right)\ge3\sqrt[3]{x^2y^2z^2}.\sqrt{3\sqrt[3]{x^2y^2z^2}}=3\sqrt{3}.xyz\)
=> \(B\le\frac{\sqrt{3}+1}{3\sqrt{3}}=\frac{3+\sqrt{3}}{9}\)
\(MaxB=\frac{3+\sqrt{3}}{9}\)khi x=y=z
Chứng minh (với những giá trị của biến làm cho biểu thức có nghĩa)
a) \(\dfrac{\left(3\sqrt{xy}-6y-2x\sqrt{y}+4y\sqrt{x}\right)\left(3\sqrt{y}+2\sqrt{xy}\right)}{y\left(\sqrt{x}-2\sqrt{y}\right)\left(y-4x\right)}=1\)
b) \(\left(\sqrt{x}-\sqrt{y}-\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right)\left(\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}+\dfrac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}+\dfrac{2\sqrt{xy}}{x-y}\right)=\sqrt{x}+\sqrt{y}\)
So sánh:
\(A=\sqrt{\dfrac{37}{4}-\sqrt{49+12\sqrt{5}}}\) với \(B=\sqrt{5}-\dfrac{3}{2}\)
Giúp với mình sắp cần rồi