Rút gọn :
\(D=\left(\frac{a-b}{a^{\frac{3}{4}}+a^{\frac{1}{2}}.b^{\frac{1}{4}}}-\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}}\right):\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)^{-1}\sqrt{\frac{a}{b}}\)
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2 tháng 10 2019
\(K=\sqrt{\frac{1}{a^2+b^2}+\frac{1}{\left(a+b\right)^2}+\sqrt{\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{\left(a^2+b^2\right)^2}}}\)
\(=\sqrt{\frac{1}{a^2+b^2}+\frac{1}{\left(a+b\right)^2}+\sqrt{\left(\frac{1}{a^2}+\frac{1}{b^2}\right)^2-\frac{2}{a^2+b^2}\left(\frac{1}{a^2}+\frac{1}{b^2}\right)+\frac{1}{\left(a^2+b^2\right)^2}}}\)
\(=\sqrt{\frac{1}{a^2+b^2}+\frac{1}{\left(a+b\right)^2}+\sqrt{\left(\frac{1}{a^2}+\frac{1}{b^2}-\frac{1}{a^2+b^2}\right)^2}}\)
\(=\sqrt{\frac{1}{\left(a+b\right)^2}+\frac{1}{a^2}+\frac{1}{b^2}}\)
\(=\sqrt{\frac{1}{\left(a+b\right)^2}+\left(\frac{1}{a}+\frac{1}{b}\right)^2-\frac{2}{\left(a+b\right)}\left(\frac{1}{a}+\frac{1}{b}\right)}\)
\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right)^2}=\left|\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right|\)
Chúc bạn học tốt !!!
H9
HT.Phong (9A5)
CTVHS
18 tháng 8 2023
a) \(a^{\dfrac{1}{3}}\cdot a^{\dfrac{1}{2}}\cdot a^{\dfrac{7}{6}}=a^{\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{7}{6}}=a^2\)
b) \(a^{\dfrac{2}{3}}\cdot a^{\dfrac{1}{4}}:a^{\dfrac{1}{6}}=a^{\dfrac{2}{3}+\dfrac{1}{4}-\dfrac{1}{6}}=a^{\dfrac{3}{4}}\)
c) \(\left(\dfrac{3}{2}a^{-\dfrac{3}{2}}\cdot b^{-\dfrac{1}{2}}\right)\left(-\dfrac{1}{3}a^{\dfrac{1}{2}}b^{\dfrac{2}{3}}\right)=\left(\dfrac{3}{2}\cdot-\dfrac{1}{3}\right)\left(a^{-\dfrac{3}{2}}\cdot a^{\dfrac{1}{2}}\right)\left(b^{-\dfrac{1}{2}}\cdot b^{\dfrac{2}{3}}\right)\)
\(=-\dfrac{1}{2}a^{-1}b^{-\dfrac{1}{3}}\)
23 tháng 3 2017
Ta có:
\(A=\frac{1}{\left(x+y\right)^3}\left(\frac{1}{x^4}-\frac{1}{y^4}\right)=\frac{1}{\left(x+y\right)^3}.\frac{\left(y^2+x^2\right)\left(x+y\right)\left(y-x\right)}{x^4y^4}=\frac{\left(x^2+y^2\right)\left(y-x\right)}{\left(x+y\right)^2x^4y^4}\)
\(B=\frac{1}{\left(x+y\right)^4}.\left(\frac{1}{x^3}-\frac{1}{y^3}\right)=\frac{\left(y-x\right)\left(y^2+xy+x^2\right)}{\left(x+y\right)^4x^3y^3}\)
\(C=\frac{1}{\left(x+y\right)^5}\left(\frac{1}{x^2}-\frac{1}{y^2}\right)=\frac{y-x}{\left(x+y\right)^4x^2y^2}\)
\(\Rightarrow A+B+C=\frac{\left(x^2+y^2\right)\left(y-x\right)}{\left(x+y\right)^2x^4y^4}+\frac{\left(y-x\right)\left(x^2+xy+y^2\right)}{\left(x+y\right)^4x^3y^3}+\frac{\left(y-x\right)}{\left(x+y\right)^4x^2y^2}\)
\(=\frac{y^3-x^3}{x^4y^4\left(x+y\right)^2}\)
b/ Thế vô rồi tính nhé
23 tháng 3 2017
Đoạn gần cuối thay y-x= 1 luôn
\(A+B+C=\frac{x^2+y^2}{\left(x+y\right)^2x^4y^4}+\left(\frac{\left(x+y\right)^2}{\left(x+y\right)^4\left(xy\right)^3}\right)\\ \)
\(A+B+C=\frac{x^2+y^2}{\left(x+y\right)^2\left(xy\right)^4}+\frac{1}{\left(x+y\right)^2\left(xy\right)^3}\)
\(A+B+C=\frac{x^2+y^2+xy}{\left[\left(x+y\right)xy\right]^2\left(xy\right)^2}\) giờ mới thay không biết đã tối giản chưa
11 tháng 9 2017
xét \(\sqrt{\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{\left(a^2+b^2\right)^2}}=\sqrt{\frac{b^4\left(a^2+b^2\right)^2+a^4\left(a^2+b^2\right)^2+a^4b^4}{a^4b^4\left(a^2+b^2\right)^2}}=\sqrt{\frac{a^8+b^8+2a^2b^6+a^4b^4+a^4b^4+2a^6b^2+a^4b^4}{\left[a^2b^2\left(a^2+b^2\right)\right]^2}}\)=\(\sqrt{\frac{\left(a^4+b^4\right)^2+2a^2b^2\left(a^4+b^4\right)+a^4b^4}{\left[a^2b^2\left(a^2+b^2\right)\right]^2}}=\sqrt{\frac{\left(a^4+b^4+a^2b^2\right)^2}{\left[a^2b^2\left(a^2+b^2\right)\right]^2}}\)
ML
1 tháng 7 2015
\(A=\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{\left(a^2+b^2\right)^2}=\frac{\left(a^4+b^4\right)\left(a^2+b^2\right)^2+a^4b^4}{a^4b^4\left(a^2+b^2\right)^2}\)
\(=\frac{\left(a^4+b^4\right)\left(a^4+b^4+2a^2b^2\right)+a^4b^4}{\left[a^2b^2\left(a^2+b^2\right)\right]^2}=\frac{\left(a^4+b^4\right)^2+2a^2b^2\left(a^4+b^4\right)+\left(a^2b^2\right)^2}{\left[a^2b^2\left(a^2+b^2\right)\right]^2}\)
\(=\frac{\left(a^4+b^4+a^2b^2\right)^2}{\left[a^2b^2\left(a^2+b^2\right)\right]^2}\)
\(\Rightarrow B=\frac{1}{a^2+b^2}+\frac{1}{\left(a+b\right)^2}+\sqrt{A}\)\(=\frac{1}{a^2+b^2}+\frac{1}{\left(a+b\right)^2}+\frac{\left(a^2+b^2\right)^2-a^2b^2}{a^2b^2\left(a^2+b^2\right)}\)
\(=\frac{1}{a^2+b^2}+\frac{1}{\left(a+b\right)^2}+\frac{a^2+b^2}{a^2.b^2}-\frac{1}{a^2+b^2}\)
\(=\)\(\frac{\left(a^2+b^2\right)\left(a+b\right)^2+a^2b^2}{a^2b^2\left(a+b\right)^2}=\frac{\left(a^2+b^2\right)\left(a^2+b^2+2ab\right)+a^2b^2}{\left[ab\left(a+b\right)\right]^2}\)
\(=\frac{\left(a^2+b^2\right)^2+2\left(a^2+b^2\right).ab+\left(ab\right)^2}{\left[ab\left(a+b\right)\right]^2}\)
\(=\frac{\left(a^2+b^2+ab\right)^2}{\left[ab\left(a+b\right)\right]^2}=\left[\frac{a^2+b^2+ab}{ab\left(a+b\right)}\right]^2\)
\(\Rightarrow\sqrt{B}=\left|\frac{a^2+b^2+ab}{ab\left(a+b\right)}\right|=\frac{a^2+b^2+ab}{\left|ab\left(a+b\right)\right|}\)
\(D=\left(\frac{a-b}{a^{\frac{3}{4}}+a^{\frac{1}{2}}.b^{\frac{1}{4}}}-\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}}\right):\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)^{-1}\sqrt{\frac{a}{b}}\)
\(=\left[\frac{a-b}{a^{\frac{1}{2}}\left(a^{\frac{1}{4}}+b^{\frac{1}{4}}\right)}-\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}}\right]:\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)^{-1}\sqrt{\frac{b}{a}}\)
\(=\frac{a-b-a+a^{\frac{1}{2}}.b^{\frac{1}{2}}}{a^{\frac{1}{2}}\left(a^{\frac{1}{4}}+b^{\frac{1}{4}}\right)}.\frac{1}{\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)}=\frac{b^{\frac{1}{2}}}{a^{\frac{1}{2}}}\frac{\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)}{\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)}\sqrt{\frac{a}{b}}.\sqrt{\frac{a}{b}}=1\)