P=\(\dfrac{1}{2\left(1+\sqrt{a}\right)}+\dfrac{1}{2\left(1-\sqrt{a}\right)}-\dfrac{a^2+2}{1-a^2}\)
a rut gon P
b tìm a để P =\(\dfrac{7}{8}\)
Những câu hỏi liên quan
A = \(\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-3.\left(\dfrac{\sqrt{x}+3}{x-9}\right)\right):\left(\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}-1\right)\)
a) Rut gon A
b) Tim GTNN cua A
P = \(\dfrac{\sqrt{a}-1}{3\sqrt{a}+\left(\sqrt{a}-1\right)^2}-\dfrac{6-2\left(\sqrt{a}-1\right)^2}{a\sqrt{a}-1}+\dfrac{2}{\sqrt{a}-1}\)
Rút gon P
Tìm x để P=1
Tính P tại x=\(7-2\sqrt{6}\)
ĐKXĐ: \(\left\{{}\begin{matrix}a\ge0\\a\ne1\end{matrix}\right.\)
a) Ta có: \(P=\dfrac{\sqrt{a}-1}{3\sqrt{a}+\left(\sqrt{a}-1\right)^2}-\dfrac{6-2\left(\sqrt{a}-1\right)^2}{a\sqrt{a}-1}+\dfrac{2}{\sqrt{a}-1}\)
\(=\dfrac{\sqrt{a}-1}{a+\sqrt{a}+1}-\dfrac{-2a+4\sqrt{a}+4}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}+\dfrac{2}{\sqrt{a}-1}\)
\(=\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}-\dfrac{-2a+4\sqrt{a}+4}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}+\dfrac{2\left(a+\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)
\(=\dfrac{a-2\sqrt{a}+1+2a-4\sqrt{a}-4+2a+2\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)
\(=\dfrac{5a-4\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)
\(=\dfrac{5a-5\sqrt{a}+\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)
\(=\dfrac{5\sqrt{a}\left(\sqrt{a}-1\right)+\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)
\(=\dfrac{\left(\sqrt{a}-1\right)\left(5\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)
\(=\dfrac{5\sqrt{a}+1}{a+\sqrt{a}+1}\)
b) Để P=1 thì \(5\sqrt{a}+1=a+\sqrt{a}+1\)
\(\Leftrightarrow a+\sqrt{a}+1-5\sqrt{a}-1=0\)
\(\Leftrightarrow a-4\sqrt{a}=0\)
\(\Leftrightarrow\sqrt{a}\left(\sqrt{a}-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{a}=0\\\sqrt{a}-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=0\left(nhận\right)\\a=16\left(nhận\right)\end{matrix}\right.\)
Vậy: Để P=1 thì \(a\in\left\{0;16\right\}\)
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\(P=\left(\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{\sqrt{x}}\right).\left(\dfrac{1}{1-\sqrt{x}}-1\right)\)
a, rut gon
b, tim x de P=\(\sqrt{x}\)
a: Sửa đề; \(P=\left(\dfrac{3x+3\sqrt{x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\right)\cdot\left(\dfrac{1}{1-\sqrt{x}}-1\right)\)
\(=\dfrac{3x+3\sqrt{x}-3-x+1+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{1-1+\sqrt{x}}{1-\sqrt{x}}\)
\(=\dfrac{3x+3\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}}{1-\sqrt{x}}=\dfrac{3\sqrt{x}}{1-\sqrt{x}}\)
b: Để \(P=\sqrt{x}\) thì \(3\sqrt{x}=\sqrt{x}-x\)
\(\Leftrightarrow x+2\sqrt{x}=0\)
hay x=0
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cho P=\(\left(\dfrac{3}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{x-1}\right):\left(\dfrac{x+2}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\\ \)
a, rut gon
b, tim x de P=\(\sqrt{x}-1\)
a/ ĐKXĐ: \(x\ge0,x\ne1\)
\(P=\left(\dfrac{3}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{x-1}\right):\left(\dfrac{x+2}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\)
= \(\dfrac{3\left(\sqrt{x}+1\right)+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
= \(\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
= \(\dfrac{4\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
= \(\dfrac{4\sqrt{x}}{\sqrt{x}+1}\)
b/ Với \(x\ge0,x\ne1\)
Để \(P=\sqrt{x}-1\Leftrightarrow\dfrac{4\sqrt{x}}{\sqrt{x}+1}=\sqrt{x}-1\)
\(\Leftrightarrow4\sqrt{x}=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow x-4\sqrt{x}-1=0\)
\(\Leftrightarrow\left(\sqrt{x}-2+\sqrt{5}\right)\left(\sqrt{x}-2-\sqrt{5}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-2+\sqrt{5}=0\\\sqrt{x}-2-\sqrt{5}=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2-\sqrt{5}\left(ktm\right)\\\sqrt{x}=2+\sqrt{5}\left(tm\right)\end{matrix}\right.\)
\(\Leftrightarrow x=9+4\sqrt{5}\)
Vậy để \(P=\sqrt{x}-1\) thì \(x=9+4\sqrt{5}\)
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Cho A = \(\left(\dfrac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}+\dfrac{\sqrt{x}-\sqrt{y}}{1+\sqrt{xy}}\right):\left(1+\dfrac{x+y+2xy}{1-xy}\right)\)
a, Rut gon bieu thuc A
b, Tinh gia tri cua A khi x = \(\dfrac{1}{1+\sqrt{2}}\)
c, Tim Max A
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a: \(A=\dfrac{\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}+\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}}{1-xy}:\dfrac{1-xy+x+y+2xy}{1-xy}\)
\(=\dfrac{2\sqrt{x}+2y\sqrt{x}}{x+y+xy+1}\)
\(=\dfrac{2\sqrt{x}\left(y+1\right)}{\left(x+1\right)\left(y+1\right)}=\dfrac{2\sqrt{x}}{x+1}\)
b: \(x=\dfrac{1}{\sqrt{2}+1}=\sqrt{2}-1\)
\(A=\dfrac{2\sqrt{\sqrt{2}-1}}{\sqrt{2}-1+1}=\sqrt{2\left(\sqrt{2}-1\right)}\)
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Câu 1:
Q left(dfrac{1}{sqrt{a}-1}-dfrac{1}{sqrt{a}}right):left(dfrac{sqrt{a}+1}{sqrt{a}-1}-dfrac{sqrt{a}+2}{sqrt{a}-1}right)left(age0,ane4,ane1right)
a) Rút gon Q
b) Tìm giá trị của a để Q dương
Caau2:
B left(dfrac{2x+1}{sqrt{x^3}-1}-dfrac{sqrt{x}}{x+sqrt{x}+1}right)left(dfrac{1+sqrt{x^3}}{1+sqrt{x}}-sqrt{x}right)left(xge0,xne1right)
a) rút gon B
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Câu 1:
Q= \(\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\left(a\ge0,a\ne4,a\ne1\right)\)
a) Rút gon Q
b) Tìm giá trị của a để Q dương
Caau2:
B= \(\left(\dfrac{2x+1}{\sqrt{x^3}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\right)\left(x\ge0,x\ne1\right)\)
a) rút gon B
Câu 1:
a: \(Q=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
b: Để Q>0 thì \(\sqrt{a}-2>0\)
=>a>4
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Rut gon A= \(\dfrac{\sqrt{1-\sqrt{1-x^2}}.\left\{\sqrt{\left(x+1\right)^3}+\sqrt{\left(1-x\right)^3}\right\}}{2-\sqrt{1-x^2}}\)
Rut gon bieu thuc sau:
\(A=\left(\dfrac{a-\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}+1}{a+\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{a}\right)\)
điều kiện : \(x>0;x\ne1\)
ta có : \(A=\left(\dfrac{a-\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}+1}{a+\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{a}\right)\)
\(\Leftrightarrow A=\left(\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right):\left(\dfrac{\sqrt{a}+1}{a}\right)\)
\(\Leftrightarrow A=\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\left(\dfrac{a}{\sqrt{a}+1}\right)=\left(\dfrac{a-1}{\sqrt{a}}\right)\left(\dfrac{a}{\sqrt{a}+1}\right)\)\(\Leftrightarrow A=\left(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}}\right)\left(\dfrac{a}{\sqrt{a}+1}\right)=\sqrt{a}\left(\sqrt{a}-1\right)=a-\sqrt{a}\)
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từ giả thiết, ta có dfrac{1}{xy}+dfrac{1}{yz}+dfrac{1}{zx}1
đặt left(dfrac{1}{xy};dfrac{1}{yz};dfrac{1}{zx}right)left(a;b;cright)Rightarrow a+b+c1 left(dfrac{ac}{b};dfrac{ab}{c};dfrac{bc}{a}right)left(dfrac{1}{x^2};dfrac{1}{y^2};dfrac{1}{z^2}right)
ta có VTdfrac{1}{sqrt{1+dfrac{1}{x^2}}}+dfrac{1}{sqrt{1+dfrac{1}{y^2}}}+dfrac{1}{sqrt{1+dfrac{1}{z^1}}}sqrt{dfrac{1}{1+dfrac{ac}{b}}}+sqrt{dfrac{1}{1+dfrac{ab}{c}}}+sqrt{dfrac{1}{1+dfrac{bc}{a}}}
dfrac{1}{sqrt{dfrac{b+ac}{b}}}+dfrac{1}{sqrt{dfrac...
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từ giả thiết, ta có \(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)
đặt \(\left(\dfrac{1}{xy};\dfrac{1}{yz};\dfrac{1}{zx}\right)=\left(a;b;c\right)\Rightarrow a+b+c=1\) =>\(\left(\dfrac{ac}{b};\dfrac{ab}{c};\dfrac{bc}{a}\right)=\left(\dfrac{1}{x^2};\dfrac{1}{y^2};\dfrac{1}{z^2}\right)\)
ta có VT=\(\dfrac{1}{\sqrt{1+\dfrac{1}{x^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{y^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{z^1}}}=\sqrt{\dfrac{1}{1+\dfrac{ac}{b}}}+\sqrt{\dfrac{1}{1+\dfrac{ab}{c}}}+\sqrt{\dfrac{1}{1+\dfrac{bc}{a}}}\)
=\(\dfrac{1}{\sqrt{\dfrac{b+ac}{b}}}+\dfrac{1}{\sqrt{\dfrac{a+bc}{a}}}+\dfrac{1}{\sqrt{\dfrac{c+ab}{c}}}=\sqrt{\dfrac{a}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\dfrac{b}{\left(b+c\right)\left(b+a\right)}}+\sqrt{\dfrac{c}{\left(c+a\right)\left(c+b\right)}}\)
\(\le\sqrt{3}\sqrt{\dfrac{ac+ab+bc+ba+ca+cb}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=\sqrt{3}.\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)
ta cần chứng minh \(\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\le\dfrac{3}{2}\Leftrightarrow\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{9}{4}\Leftrightarrow8\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
<=>\(8\left(a+b+c\right)\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\) (luôn đúng )
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