\(y=\frac{1}{x^2+\sqrt{x}}\)
Những câu hỏi liên quan
rút gọn:a)left(frac{1}{2+2sqrt{x}}+frac{1}{2-2sqrt{x}}-frac{x^2+1}{1-x^2}right)timesleft(1+frac{1}{x}right)b)left(frac{2sqrt{xy}}{x-y}+frac{sqrt{x}-sqrt{y}}{2sqrt{x}+sqrt{y}}right)timesfrac{2sqrt{x}}{sqrt{x}+sqrt{y}}+frac{sqrt{y}}{sqrt{y}-sqrt{x}}c)left(frac{x-1}{sqrt{x}-1}+frac{xsqrt{x}-1}{1-x}right)divfrac{left(sqrt{x}-1right)^2+sqrt{x}}{sqrt{x}+1}
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rút gọn:
a)\(\left(\frac{1}{2+2\sqrt{x}}+\frac{1}{2-2\sqrt{x}}-\frac{x^2+1}{1-x^2}\right)\times\left(1+\frac{1}{x}\right)\)
b)\(\left(\frac{2\sqrt{xy}}{x-y}+\frac{\sqrt{x}-\sqrt{y}}{2\sqrt{x}+\sqrt{y}}\right)\times\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}+\frac{\sqrt{y}}{\sqrt{y}-\sqrt{x}}\)
c)\(\left(\frac{x-1}{\sqrt{x}-1}+\frac{x\sqrt{x}-1}{1-x}\right)\div\frac{\left(\sqrt{x}-1\right)^2+\sqrt{x}}{\sqrt{x}+1}\)
a, dk \(x\ge0.x\ne1\)
\(\left(\frac{1+\sqrt{x}+1-\sqrt{x}}{2\left(1-x\right)}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)=\(\left(\frac{1}{1-x}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)
=\(\left(\frac{1+x-x^2-1}{1-x^2}\right)\left(\frac{x+1}{x}\right)=\frac{x\left(1-x\right)\left(x+1\right)}{x\left(1-x\right)\left(1+x\right)}=1\)
phan b,c ban tu lam not nhe dai lam mk ko lam dau mk co vc ban rui
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x+y4Rightarrowfrac{x+y}{2}2Rightarrowsqrt{frac{x+y}{2}}sqrt{2}P.sqrt{frac{x+y}{2}}sqrt{2}sqrt{x^2+frac{1}{x^2}}+sqrt{2}sqrt{x^2+frac{1}{x^2}}Leftrightarrowsqrt{2}Psqrt{1+1}sqrt{x^2+frac{1}{x^2}}+sqrt{1+1}sqrt{x^2+frac{1}{x^2}}Leftrightarrowsqrt{2}Pge x+frac{1}{x}+y+frac{1}{y}x+frac{1}{x}left(frac{1}{x}+4xright)-3xge4-3xy+frac{1}{y}left(frac{1}{y}+4yright)-3yge4-3yRightarrowsqrt{2}Pge8-3left(x+yright)8-3.4-4đến đay sau răng
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\(x+y=4\Rightarrow\frac{x+y}{2}=2\Rightarrow\sqrt{\frac{x+y}{2}}=\sqrt{2}\)
\(P.\sqrt{\frac{x+y}{2}}=\sqrt{2}\sqrt{x^2+\frac{1}{x^2}}+\sqrt{2}\sqrt{x^2+\frac{1}{x^2}}\)
\(\Leftrightarrow\sqrt{2}P=\sqrt{1+1}\sqrt{x^2+\frac{1}{x^2}}+\sqrt{1+1}\sqrt{x^2+\frac{1}{x^2}}\)
\(\Leftrightarrow\sqrt{2}P\ge x+\frac{1}{x}+y+\frac{1}{y}\)
\(x+\frac{1}{x}=\left(\frac{1}{x}+4x\right)-3x\ge4-3x\)
\(y+\frac{1}{y}=\left(\frac{1}{y}+4y\right)-3y\ge4-3y\)
\(\Rightarrow\sqrt{2}P\ge8-3\left(x+y\right)=8-3.4=-4\)
đến đay sau răng
Cho x,y,z là các số dương. Chứng minh rằng:
\(\frac{1}{\sqrt{x}+3\sqrt{y}}+\frac{1}{\sqrt{y}+3\sqrt{z}}+\frac{1}{\sqrt{z}+3\sqrt{x}}\ge\frac{1}{\sqrt{x}+2\sqrt{y}+\sqrt{z}}+\frac{1}{\sqrt{y}+2\sqrt{z}+\sqrt{x}}+\frac{1}{\sqrt{z}+2\sqrt{x}+\sqrt{y}}\)
Rút gọn:
\(A=\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\left(\frac{1}{x}+\frac{1}{y}\right).\frac{1}{x+y+2\sqrt{xy}}+\frac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}.\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)\right]\)
\(x=\sqrt{2-\sqrt{3}};y=\sqrt{2+\sqrt{3}}\)
\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\left(\frac{1}{x}+\frac{1}{y}\right).\frac{1}{x+y+2\sqrt{xy}}+\frac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}.\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)\right)\)rút gọn biết x=2-\(\sqrt{3}\)và y =\(2+\sqrt{3}\)
Ta có :
Đặt A=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\left(\frac{x+y}{xy}\right).\frac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}+\frac{2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}.\left(\sqrt{x}+\sqrt{y}\right)^3}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\frac{x+y}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}+\frac{2\sqrt{xy}}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\frac{1}{xy}\)
=\(\frac{xy.\left(\sqrt{x}-\sqrt{y}\right)}{xy\sqrt{xy}}\)
=\(\frac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)
=\(\frac{\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}}{\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}\)
=\(\frac{\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}}{\sqrt{4-3}}\)
=\(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
=> \(A^2=\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)^2\)
=\(2-\sqrt{3}-2\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+2+\sqrt{3}\)
=\(4-2\sqrt{4-3}\)
=\(4-2\)
=\(2\)
=>\(A=\sqrt{2}\)
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bài 1: rút gọn:
C=\(\left(\frac{x-y}{\sqrt{x}-\sqrt{y}}+\frac{x\sqrt{x}-y\sqrt{y}}{y-x}\right):\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
bài 2 :rút gọn
E=\(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
35Cho biểu thứcPleft[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}+ysqrt{x}+xsqrt{y}+sqrt{y^3}}{sqrt{xy^3}+sqrt{x^3y}}a) Rút gọn Pb)Cho xy16 . Tìm Min P 34 Cho biểu thức Pfrac{x}{sqrt{xy}-2y}-frac{2sqrt{x}}{x+sqrt{x}-2sqrt{xy}-2sqrt{y}}-frac{1-x}{1-sqrt{x}}a) Rút gọn Pb)Tính P biết 2x^2+y^2-4x-2xy+40
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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
bài 5ĐK:x2,y1frac{36}{sqrt{x-2}}+frac{4}{sqrt{y-1}}28-4sqrt{x-2}-sqrt{y-1}..Leftrightarrowfrac{36}{sqrt{x-2}}+4sqrt{x-2}+frac{4}{sqrt{y-1}}+sqrt{y-1}28Áp dụng AM-GM ta có:frac{36}{sqrt{x-2}}+4sqrt{x-2}ge2sqrt{frac{144sqrt{x-2}}{sqrt{x-2}}}24frac{4}{sqrt{y-1}}+sqrt{y-1}ge2sqrt{frac{4sqrt{y-1}}{sqrt{y-1}}}4.Rightarrowfrac{36}{sqrt{x-2}}+4sqrt{x-2}+frac{4}{sqrt{y-1}}+sqrt{y-1}ge28.Dấu xảy ra khi frac{36}{sqrt{x-2}}4sqrt{x-2}Leftrightarrow x11left(nright).frac{4}{sqrt{y-1}}sqrt{y-1}Leftrightarrow y5...
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bài 5
ĐK:\(x>2,y>1\)
\(\frac{36}{\sqrt{x-2}}+\frac{4}{\sqrt{y-1}}=28-4\sqrt{x-2}-\sqrt{y-1}..\)\(\Leftrightarrow\frac{36}{\sqrt{x-2}}+4\sqrt{x-2}+\frac{4}{\sqrt{y-1}}+\sqrt{y-1}=28\)
Áp dụng AM-GM ta có:
\(\frac{36}{\sqrt{x-2}}+4\sqrt{x-2}\ge2\sqrt{\frac{144\sqrt{x-2}}{\sqrt{x-2}}}=24\)
\(\frac{4}{\sqrt{y-1}}+\sqrt{y-1}\ge2\sqrt{\frac{4\sqrt{y-1}}{\sqrt{y-1}}}=4.\)
\(\Rightarrow\frac{36}{\sqrt{x-2}}+4\sqrt{x-2}+\frac{4}{\sqrt{y-1}}+\sqrt{y-1}\ge28.\)
Dấu \(=\)xảy ra khi \(\frac{36}{\sqrt{x-2}}=4\sqrt{x-2}\Leftrightarrow x=11\left(n\right).\)
\(\frac{4}{\sqrt{y-1}}=\sqrt{y-1}\Leftrightarrow y=5\left(n\right).\)
Vậy \(x=11,y=5\)
<br class="Apple-interchange-newline"><div id="inner-editor"></div>x>2;y>1
Khi đó Pt ⇔36√x−2 +4√x−2+4√y−1 +√y−1=28
theo BĐT Cô si ta có 36√x−2 +4√x−2≥2.√36√x−2 .4√x−2=24
và 4√y−1 +√y−1≥2√4√y−1 .√y−1=4
Pt đã cho có VT>= 28 Dấu "=" xảy ra ⇔
36√x−2 =4√x−2⇔x=11
và 4√y−1 =√y−1⇔y=5
Đối chiếu với ĐK thì x=11; y=5 là nghiệm của PT
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cho x,y,z là các số thực dương thỏa mãnxy+yz+zx1. Chứng minh rằng text{x/căn(1+x^2)+y/căn(1+y^2)+z/căn(1+z^2)+1/x^2+1/y^2+1/z^221/2}frac{x}{sqrt{1+x^2}}+frac{y}{sqrt{1+y^2}}+frac{z}{sqrt{1+z^2}}+frac{1}{x^2}+frac{1}{y^2}+frac{1}{z^2}gefrac{21}{2}frac{x}{sqrt{1+x^2}}+frac{y}{sqrt{1+y^2}}+frac{z}{sqrt{1+z^2}}+frac{1}{x^2}+frac{1}{y^2}+frac{1}{z^2}gefrac{21}{2}
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cho x,y,z là các số thực dương thỏa mãn\(xy+yz+zx=1\). Chứng minh rằng \(\text{x/căn(1+x^2)+y/căn(1+y^2)+z/căn(1+z^2)+1/x^2+1/y^2+1/z^2>=21/2}\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\ge\frac{21}{2}\)
\(\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\ge\frac{21}{2}\)
Đặt \(P=\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
Do x,y,z là các số thực dương nên ta biến đổi \(P=\frac{1}{\sqrt{1+\frac{1}{x^2}}}+\frac{1}{\sqrt{1+\frac{1}{y^2}}}+\frac{1}{\sqrt{1+\frac{1}{z^2}}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
Đặt \(a=\frac{1}{x^2};b=\frac{1}{y^2};c=\frac{1}{z^2}\left(a,b,c>0\right)\)thì \(xy+yz+zx=\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}=1\)và \(P=\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}+\frac{1}{\sqrt{1+c}}+a+b+c\)
Biến đổi biểu thức P=\(\left(\frac{1}{2\sqrt{a+1}}+\frac{1}{2\sqrt{a+1}}+\frac{a+1}{16}\right)+\left(\frac{1}{2\sqrt{b+1}}+\frac{1}{2\sqrt{b+1}}+\frac{b+1}{16}\right)\)\(+\left(\frac{1}{2\sqrt{c+1}}+\frac{1}{2\sqrt{c+1}}+\frac{c+1}{16}\right)+\frac{15a}{16}+\frac{15b}{16}+\frac{15c}{b}-\frac{3}{16}\)
Áp dụng Bất Đẳng Thức Cauchy ta có
\(P\ge3\sqrt[3]{\frac{a+1}{64\left(a+1\right)}}+3\sqrt[3]{\frac{b+1}{64\left(b+1\right)}}+3\sqrt[3]{\frac{c+1}{64\left(c+1\right)}}+\frac{15a}{16}+\frac{15b}{16}+\frac{15c}{16}-\frac{3}{16}\)
\(=\frac{33}{16}+\frac{15}{16}\left(a+b+c\right)\ge\frac{33}{16}+\frac{15}{16}\cdot3\sqrt[3]{abc}\)
Mặt khác ta có \(1=\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\ge3\sqrt[3]{\frac{1}{abc}}\Leftrightarrow abc\ge27\)
\(\Rightarrow P\ge\frac{33}{16}+\frac{15}{16}\cdot3\sqrt[3]{27}=\frac{33}{16}+\frac{15}{16}\cdot9=\frac{21}{2}\)
Dấu "=" xảy ra khi a=b=c hay \(x=y=z=\frac{\sqrt{3}}{3}\)
1)begin{cases}x+sqrt{x^2+1}y+sqrt{y^2-1}left(1right)3sqrt{y-1}+sqrt{x}2sqrt{y+1}left(2right)end{cases} nhân liên hợp pt 1 đc (left(x^2-y^2+1right)left(frac{1}{x+sqrt{y^2-1}}+frac{1}{sqrt{x^2+1}+y}right) thì TH1 x^2-y^2+1 lm ntn 2begin{cases}sqrt{x^2+xy+2y^2}+sqrt{xy}3ysqrt{y-1}+sqrt{x-1}+x+y6end{cases}3begin{cases}frac{sqrt{x^2+5}}{x}+frac{sqrt{y^2+3}}{y}frac{7}{2}xsqrt{x^2+5}+ysqrt{y^2+3}3+x^2+y^2end{cases}
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1)\(\begin{cases}x+\sqrt{x^2+1}=y+\sqrt{y^2-1}\left(1\right)\\3\sqrt{y-1}+\sqrt{x}=2\sqrt{y+1}\left(2\right)\end{cases}\) nhân liên hợp pt 1 đc (\(\left(x^2-y^2+1\right)\left(\frac{1}{x+\sqrt{y^2-1}}+\frac{1}{\sqrt{x^2+1}+y}\right)\) thì TH1 \(x^2-y^2+1\) lm ntn
2\(\begin{cases}\sqrt{x^2+xy+2y^2}+\sqrt{xy}=3y\\\sqrt{y-1}+\sqrt{x-1}+x+y=6\end{cases}\)
3\(\begin{cases}\frac{\sqrt{x^2+5}}{x}+\frac{\sqrt{y^2+3}}{y}=\frac{7}{2}\\x\sqrt{x^2+5}+y\sqrt{y^2+3}=3+x^2+y^2\end{cases}\)