Tính giá trị biểu thức A= \(\frac{xy-\sqrt{x^2-1}.\sqrt{y^2-1}}{xy+\sqrt{x^2-1}.\sqrt{y^2-1}}\) với \(x=\frac{1}{2}\left(a+\frac{1}{a}\right)\) và \(y=\frac{1}{2}\left(b+\frac{1}{b}\right)\)với a, b \(\ge\)1
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_
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
Tính giá trị biểu thức P=\(\frac{xy-\sqrt{x^2-1}.\sqrt{y^2-1}}{xy+\sqrt{x^2-1}.\sqrt{y^2-1}}\). Với x=\(\frac{1}{2}\left(a+\frac{1}{a}\right)\) và y=\(\frac{1}{2}\left(b+\frac{1}{b}\right)\) với \(a,b\ge1\)
1)Tính giá trị của biểu thức A=\(\frac{xy-\sqrt{\left(x^2-1\right)}\cdot\sqrt{y^2-1}}{xy+\sqrt{x^2-1}\cdot\sqrt{y^2-1}}\)với x=\(\frac{1}{2}\left(a+\frac{1}{a}\right)\)và y=\(\frac{1}{2}\left(b+\frac{1}{b}\right)\)với a\(\ge\)1 ,
b\(\ge\)1
2)Cho ba số a,b,c thỏa mãn \(0\le a,b,c\le2\)và a+b+c=3. Chứng minh \(\sqrt{ab}+\sqrt{bc}\sqrt{ca}\ge\sqrt{2}\)
giúp mình với . Cảm ơn
Cho biểu thức:
\(A=\left[\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)\frac{2}{\sqrt{x}+\sqrt{y}}+\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right]:\frac{\sqrt{x^3}+y\sqrt{x}+x\sqrt{y}+\sqrt{y^3}}{\sqrt{x^3y}+\sqrt{xy^3}}\)
a, Rút gọn A
b, Biết xy=6. Tìm giá trị của x,y để A có GTNN
Tính giá trị của các biểu thức sau:
a) \(\frac{\left(1+\sqrt{x}\right)^2-4\sqrt{x}}{1-\sqrt{x}}\) với x = 2
b) \(\frac{\left(\sqrt{x}-\sqrt{y}\right)+4\sqrt{xy}}{1+\sqrt{xy}}\) với x = 2, y = 3
c) \(\frac{x+y}{y}\). \(\sqrt{\frac{x^2y^2+2x^2y^3+xy^4}{x^2+3xy+y^2}}\) với x =2, y =1
cho biểu thức: \(P=\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{xy}+\sqrt{x}}{1-\sqrt{xy}}+1\right):\left(1-\frac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-\frac{\sqrt{x}+1}{\sqrt{xy}+1}\right)\) \(P=\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{xy}+\sqrt{x}}{1-\sqrt{xy}}+1\right):\left(1-\frac{\sqrt{xy}+1}{\sqrt{xy}-1}-\frac{\sqrt{x}+1}{\sqrt{xy}+1}\right).\backslash\ \)với \(x,y\ge0;x,y\ne1\)
a) Rút gọn P
b) Tính P khi \(x=\sqrt[3]{4-2\sqrt{6}}+\sqrt[3]{4+2\sqrt{6}}\)và \(y=x^2+6\)
a/ \(P=\frac{1}{\sqrt{xy}}\)
b/ \(x^3=8-6x\)
\(\Rightarrow P=\frac{1}{\sqrt{x\left(x^2+6\right)}}=\frac{1}{\sqrt{x^3+6x}}=\frac{1}{\sqrt{8-6x+6x}}=\frac{1}{2\sqrt{2}}\)
Cho x,y,z>0 và xy+yz+zx=1
a, tính giá trị biểu thức:
\(P=x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}+y\sqrt{\frac{\left(1+x^2\right)\left(1+z^2\right)}{1+y^2}}+z\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}\)
b, CMR:
\(\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}=\frac{2xy}{\sqrt{\left(1+x^2\right)\left(1+y^2\right)\left(1+z^2\right)}}\)
Ta có \(1+x^2=x^2+xy+yz+xz=\left(x+y\right)\left(x+z\right)\)
Tương tự \(1+y^2=\left(x+y\right)\left(y+z\right)\)
\(1+z^2=\left(x+z\right)\left(y+z\right)\)
Thay vào A ta được
\(P=x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)
=2(xy+xz+yz)=2
\(b,VT=VP\)
\(\Leftrightarrow\frac{x}{xy+yz+zx+x^2}+\frac{y}{xy+yz+zx+y^2}+\frac{z}{xy+yz+zx+z^2}\)
\(=\frac{2xyz}{\sqrt{\left(xy+yz+zx+x^2\right)\left(xy+yz+zx+y^2\right)\left(xy+yz+zx+z^2\right)}}\)
\(\Leftrightarrow\frac{x}{\left(x+y\right)\left(x+z\right)}+\frac{y}{\left(x+y\right)\left(y+z\right)}+\frac{z}{\left(x+z\right)\left(y+z\right)}\)
\(=\frac{2xyz}{\sqrt{\left(x+y\right)\left(x+z\right)\left(y+z\right)\left(y+x\right)\left(z+x\right)\left(y+z\right)}}\)
\(\Leftrightarrow\frac{x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{2xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(\Leftrightarrow xy+xz+xy+yz+xz+yz=2xyz\)
\(\Leftrightarrow2=2xyz\)
\(\Leftrightarrow xyz=1\)
Đù =)))
Xác định gt các bt sau:
\(a.A=\frac{xy-\sqrt{x^2-1}.\sqrt{y^2-1}}{xy+\sqrt{x^2-1}.\sqrt{y^2-1}}\) với \(x=\frac{1}{2}\left(a+\frac{1}{a}\right),y=\frac{1}{2}\left(b+\frac{1}{b}\right)\) (a>1; b>1)
\(b.B=\frac{\sqrt{a+bx}+\sqrt{a-bx}}{\sqrt{a+bx}-\sqrt{a-bx}}\) với \(x=\frac{2am}{b\left(1+m^2\right)},\left|m\right|< 1\)
\(x^2-1=\frac{1}{4}\left(a^2+\frac{1}{a^2}+2\right)-1=\frac{1}{4}\left(a-\frac{1}{a}\right)^2\)
\(\Rightarrow\sqrt{x^2-1}=\frac{1}{2}\left(a-\frac{1}{a}\right)\)
Tương tự \(\sqrt{y^2-1}=\frac{1}{2}\left(b-\frac{1}{b}\right)\)
\(A=\frac{\frac{1}{4}\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)-\frac{1}{4}\left(a-\frac{1}{a}\right)\left(b-\frac{1}{b}\right)}{\frac{1}{4}\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)+\frac{1}{4}\left(a-\frac{1}{a}\right)\left(b-\frac{1}{b}\right)}=\frac{ab+\frac{a}{b}+\frac{b}{a}+\frac{1}{ab}-ab-\frac{1}{ab}+\frac{a}{b}+\frac{b}{a}}{ab+\frac{a}{b}+\frac{b}{a}+\frac{1}{ab}+ab+\frac{1}{ab}-\frac{a}{b}-\frac{b}{a}}\)
\(=\frac{\frac{a}{b}+\frac{b}{a}}{ab+\frac{1}{ab}}=\frac{a^2+b^2}{a^2b^2+1}\)
b/ \(B=\frac{\left(\sqrt{a+bx}+\sqrt{a-bx}\right)^2}{a+bx-\left(a-bx\right)}=\frac{a+\sqrt{a^2-b^2x^2}}{bx}\)
\(a^2-b^2x^2=a^2-\frac{4a^2m^2}{\left(1+m^2\right)^2}=\frac{a^2\left(m^4+2m^2+1\right)-4a^2m^2}{\left(1+m^2\right)^2}=\frac{a^2\left(1-m^2\right)^2}{\left(1+m^2\right)^2}\)
\(\Rightarrow B=\left(a+\frac{a\left(1-m^2\right)}{1+m^2}\right).\left(\frac{1+m^2}{2am}\right)=\frac{a+am^2+a-am^2}{2am}=\frac{1}{m}\)
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