Cho \(P=\left(\dfrac{3\sqrt{a}}{a+\sqrt{ab}+b}-\dfrac{3a}{a\sqrt{a}-b\sqrt{b}}+\dfrac{1}{\sqrt{a}-\sqrt{b}}\right):\left(\dfrac{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}{2a+2\sqrt{ab}+2b}\right)\)
Tìm \(a\in Z\) để \(P\in Z\)
Những câu hỏi liên quan
3.P=\(\left(\dfrac{3\sqrt{a}}{a+\sqrt{ab}+b}-\dfrac{3a}{a\sqrt{a}-b\sqrt{b}}+\dfrac{1}{\sqrt{a}-\sqrt{b}}\right)\):\(\left(\dfrac{\left(a-1\right).\left(\sqrt{a}-\sqrt{b}\right)}{2a+2\sqrt{ab}+2b}\right)\)
a)Rút gọn P
b)Tìm những giá trị nguyên của a để P có giá trị nguyên
M = \(\left(\dfrac{3\sqrt{a}}{a+\sqrt{ab}+b}-\dfrac{3a}{a\sqrt{a}-b\sqrt{b}}+\dfrac{1}{\sqrt{a}-\sqrt{b}}\right):\dfrac{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}{2a+2\sqrt{ab}+2b}\)
a) Rút gọn M
b) Tìm những GT nguyên của A để M có GT nguyên
!!Help 
a: ĐKXĐ: a>=0; b>=0; ab<>0; a<>1\(M=\dfrac{3\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)-3a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{2\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(a-1\right)}\)
\(=\dfrac{3a-3\sqrt{ab}-3a+a+\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{1}{\left(\sqrt{a}-\sqrt{b}\right)\left(a-1\right)}\)
\(=\dfrac{a-2\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\cdot\dfrac{1}{a-1}=\dfrac{1}{a-1}\)
b: M nguyên khi a-1 thuộc {1;-1}
=>a thuộc {2;0}
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Cho biểu thức:
\(P=\left(\dfrac{3\sqrt{a}}{a+\sqrt{ab}+b}-\dfrac{3a}{a\sqrt{a}-b\sqrt{b}}+\dfrac{1}{\sqrt{a}-\sqrt{b}}\right):\left(\dfrac{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}{2a+2\sqrt{ab}+2b}\right)\)
a) Rút gọn P
b) Tìm những giá trị nguyên của a để P có giá trị nguyên
a: \(P=\dfrac{3a-3\sqrt{ab}-3a+a+\sqrt{ab}+b}{a\sqrt{a}-b\sqrt{b}}:\dfrac{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}{2\left(a+\sqrt{ab}+b\right)}\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{2\left(a+\sqrt{ab}+b\right)}{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
=2/a-1
b: Để P nguyên thì \(a-1\in\left\{1;-1;2;-2\right\}\)
=>\(a\in\left\{2;0;3\right\}\)
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Chứng minh :
a) dfrac{3x}{2y}+dfrac{3}{2}sqrt{dfrac{3}{5}}-sqrt{dfrac{3}{4}}dfrac{3sqrt{x}}{2}.left(dfrac{sqrt{x}}{y}+sqrt{dfrac{3}{5x}}-sqrt{dfrac{1}{3}}right)
b)ab.sqrt{1+dfrac{1}{a^2b^2}}-sqrt{a^2b^2+1}0 , với a ; b 0
c) left(dfrac{3}{a}sqrt{dfrac{a^3}{b}}-dfrac{1}{2}sqrt{dfrac{4}{ab}}-2sqrt{dfrac{b}{a}}right):sqrt{dfrac{1}{ab}}3a-2b-1 với a, b 0
d)left(sqrt{dfrac{16a}{b}}+3sqrt{4ab}-asqrt{dfrac{36b}{a}}+2sqrt{ab}right):left(sqrt{ab}+dfrac{a}{b}sqrt{dfrac{b}{a}}+sqrt{dfrac{a}{b}}rig...
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Chứng minh :
a) \(\dfrac{3x}{2y}+\dfrac{3}{2}\sqrt{\dfrac{3}{5}}-\sqrt{\dfrac{3}{4}}=\dfrac{3\sqrt{x}}{2}.\left(\dfrac{\sqrt{x}}{y}+\sqrt{\dfrac{3}{5x}}-\sqrt{\dfrac{1}{3}}\right)\)
b)\(ab.\sqrt{1+\dfrac{1}{a^2b^2}}-\sqrt{a^2b^2+1}=0\) , với a ; b > 0
c) \(\left(\dfrac{3}{a}\sqrt{\dfrac{a^3}{b}}-\dfrac{1}{2}\sqrt{\dfrac{4}{ab}}-2\sqrt{\dfrac{b}{a}}\right):\sqrt{\dfrac{1}{ab}}=3a-2b-1\) với a, b >0
d)\(\left(\sqrt{\dfrac{16a}{b}}+3\sqrt{4ab}-a\sqrt{\dfrac{36b}{a}}+2\sqrt{ab}\right):\left(\sqrt{ab}+\dfrac{a}{b}\sqrt{\dfrac{b}{a}}+\sqrt{\dfrac{a}{b}}\right)=2\) Với a, b >0
Mọi người giúp tớ với ạ !!!!!! Mình thật sự cần gấp vào ngày mai !!!!
b)CM: \(ab\sqrt{1+\dfrac{1}{a^2b^2}}-\sqrt{a^2b^2+1}=0\)
\(VT=ab\sqrt{\dfrac{a^2b^2+1}{\left(ab\right)^2}}-\sqrt{a^2b^2+1}\)
\(VT=ab\dfrac{\sqrt{a^2b^2+1}}{ab}-\sqrt{a^2b^2+1}\)
\(VT=\sqrt{a^2b^2+1}-\sqrt{a^2b^2+1}\)
\(VT=0=VP\)
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Cho a, b, x là những số dương. Đơn giản các biểu thức sau :
a) Aleft[dfrac{2a+left(abright)^{dfrac{1}{2}}}{3a}right]^{-1}left[dfrac{a^{dfrac{3}{2}}-b^{dfrac{3}{2}}}{a-left(abright)^{dfrac{1}{2}}}-dfrac{a-b}{sqrt{a}+sqrt{b}}right]
b) Bleft(dfrac{sqrt{a}-sqrt{x}}{sqrt{a+x}}-dfrac{sqrt{a+x}}{sqrt{a}+sqrt{x}}right)^{-2}-left(dfrac{sqrt{a}-sqrt{x}}{sqrt{a+x}}-dfrac{sqrt{a+x}}{sqrt{a}-sqrt{x}}right)^{-2}
c) Csqrt{16^{dfrac{1}{log_74}}+81^{dfrac{1}{log_69}}+15}
d) D49^{1-log_72}+5^{-log_54}
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Cho a, b, x là những số dương. Đơn giản các biểu thức sau :
a) \(A=\left[\dfrac{2a+\left(ab\right)^{\dfrac{1}{2}}}{3a}\right]^{-1}\left[\dfrac{a^{\dfrac{3}{2}}-b^{\dfrac{3}{2}}}{a-\left(ab\right)^{\dfrac{1}{2}}}-\dfrac{a-b}{\sqrt{a}+\sqrt{b}}\right]\)
b) \(B=\left(\dfrac{\sqrt{a}-\sqrt{x}}{\sqrt{a+x}}-\dfrac{\sqrt{a+x}}{\sqrt{a}+\sqrt{x}}\right)^{-2}-\left(\dfrac{\sqrt{a}-\sqrt{x}}{\sqrt{a+x}}-\dfrac{\sqrt{a+x}}{\sqrt{a}-\sqrt{x}}\right)^{-2}\)
c) \(C=\sqrt{16^{\dfrac{1}{\log_74}}+81^{\dfrac{1}{\log_69}}+15}\)
d) \(D=49^{1-\log_72}+5^{-\log_54}\)
Cho \(a,b,c>0\) thỏa mãn \(ab+bc+ca=3\) . CMR : \(\sqrt[3]{\dfrac{a}{b\left(b+2c\right)}}+\sqrt[3]{\dfrac{b}{c\left(c+2a\right)}}+\sqrt[3]{\dfrac{c}{a\left(a+2b\right)}\ge\dfrac{3}{\sqrt[3]{3}}}\)
Cho biểu thức I = \(\left(\dfrac{1}{\sqrt{a}+\sqrt{b}}+\dfrac{3\sqrt{ab}}{a\sqrt{a}+b\sqrt{b}}\right)\).\(\left[\left(\dfrac{1}{\sqrt{a}-\sqrt{b}}+\dfrac{3\sqrt{ab}}{a\sqrt{a}-b\sqrt{b}}\right):\dfrac{a-b}{a+\sqrt{ab}+b}\right]\)
Rút gọn I
a) Tính giá trị của I với a = 16, b = 4
\(I=\dfrac{a-\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\cdot\left[\left(\dfrac{a+\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\right)\cdot\dfrac{a+\sqrt{ab}+b}{a-b}\right]\)
\(=\dfrac{a+2\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\cdot\left(\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{a+\sqrt{ab}+b}{a-b}\right)\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\cdot\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a-b\right)}\)
\(=\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)^2\cdot\left(a-\sqrt{ab}+b\right)}\)
Khi a=16 và b=4 thì \(I=\dfrac{16+4+4\cdot\sqrt{16\cdot4}}{\left(4-2\right)^2\cdot\left(16-\sqrt{16\cdot4}+4\right)}=\dfrac{20+4\cdot8}{4\cdot12}\)
\(=\dfrac{20+32}{48}=\dfrac{52}{48}=\dfrac{13}{12}\)
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Thực hiện phép tính.
a) \(\left(\sqrt{ab}+2\sqrt{\dfrac{b}{a}}-\sqrt{\dfrac{a}{b}+\sqrt{\dfrac{1}{ab}}}\right)\sqrt{ab}\)
b) \(\left(\dfrac{am}{b}\sqrt{\dfrac{n}{m}}-\dfrac{ab}{n}\sqrt{mn}+\dfrac{a^2}{b^2}\sqrt{\dfrac{m}{n}}\right).a^2b^2.\sqrt{\dfrac{n}{m}}\)
Giải chi tiết ra hộ mình với ạ, mình cảm ơn ạ.
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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