Rút gọn A=\(\frac{a^2+2ab+b^2}{a}.\left(\frac{a}{a^2+2ab+b^2}-\frac{a}{a^2-b^2}\right)+\frac{5a-3b}{a-b}\)
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Rút gọn : A = \(\frac{a}{b-a}+\left(\frac{2ab}{a^2-b^2}+\frac{a-b}{2a+2b}\right):\frac{a+b}{2a}\)
Rút gọn biểu thức
\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{a+b}\left(\frac{1}{a}+\frac{1}{b}\right):\frac{a^2+b^2+2ab}{ab}\)
Ai nhanh mk K cho
Cho a+b+c=0 và a,b,c khác 0.Rút gọn biểu thức
M=\(\frac{2ab}{a^2+\left(b+c\right)\left(b-c\right)}+\frac{2bc}{b^2+\left(c+a\right)\left(c-a\right)}+\frac{2ca}{c^2+\left(a+b\right)\left(a-b\right)}\)
ta có : a+b+c=0=>a+b=-c ; b+c=-a ; a+c=-b
ta có: M= \(\frac{2ab}{a^2+\left(b+c\right)\left(b-c\right)}+\frac{2bc}{b^2+\left(c+a\right)\left(c-a\right)}+\frac{2ca}{c^2+\left(a+b\right)\left(a-b\right)}\)
M=\(\frac{2ab}{a^2-a\left(b-c\right)}+\frac{2bc}{b^2-b\left(c-a\right)}+\frac{2ca}{c^2-c\left(a-b\right)}\)
M=\(\frac{2ab}{a\left(a-b+c\right)}+\frac{2bc}{b\left(b-c+a\right)}+\frac{2ca}{c\left(c-a+b\right)}\)
M=\(\frac{2ab}{-ab+\left(a+c\right)}+\frac{2bc}{-bc+\left(a+b\right)}+\frac{2ac}{-ac+\left(b+c\right)}\)
M=\(\frac{2ab}{-2ab}+\frac{2bc}{-2bc}+\frac{2ca}{-2ca}\)
M=-1-1-1=-3
Vậy với a+b+c=0 thì M=-3
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cho a+b+c=0 và a, b, c đều khác 0. Rút gọn biểu thức:
\(\frac{2ab}{a^2+\left(b+c\right)\left(b-c\right)}+\frac{2bc}{b^2+\left(c+a\right)\left(c-a\right)}+\frac{2ca}{c^2+\left(a+b\right)\left(a-b\right)}\)
Rút gọn:a,Asqrt{9left(a+bright)}-2sqrt{16left(a+bright)}-3sqrt{a+b}+frac{1}{5}sqrt{25left(a+bright)}b,Bfrac{2absqrt{a}}{sqrt{a}+sqrt{b}}+frac{2absqrt{b}}{sqrt{a}+sqrt{b}}c,Cfrac{2ab}{a+sqrt{ab}}+frac{2ab}{b+sqrt{ab}}d,Dfrac{frac{2absqrt{a}}{sqrt{a}+sqrt{b}}+frac{2absqrt{b}}{sqrt{a}+sqrt{b}}}{frac{2ab}{a+sqrt{ab}}+frac{2ab}{b+sqrt{ab}}}Giúp mình với.Thanks
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Rút gọn:
\(a,A=\sqrt{9\left(a+b\right)}-2\sqrt{16\left(a+b\right)}-3\sqrt{a+b}+\frac{1}{5}\sqrt{25\left(a+b\right)}\)
\(b,B=\frac{2ab\sqrt{a}}{\sqrt{a}+\sqrt{b}}+\frac{2ab\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(c,C=\frac{2ab}{a+\sqrt{ab}}+\frac{2ab}{b+\sqrt{ab}}\)
\(d,D=\frac{\frac{2ab\sqrt{a}}{\sqrt{a}+\sqrt{b}}+\frac{2ab\sqrt{b}}{\sqrt{a}+\sqrt{b}}}{\frac{2ab}{a+\sqrt{ab}}+\frac{2ab}{b+\sqrt{ab}}}\)
Giúp mình với.Thanks
cho anh tao đê
a,A= -7\(\sqrt{a+b}\)
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Cho biểu thức sau:
\(P=\left(\frac{1}{ab-2}+\frac{1}{ab+2}+\frac{2ab}{a^2b^2+4}+\frac{4a^3b^3}{a^4b^4+16}\right).\frac{a^4b^4+16}{a^4b^4}\)
a, Rút gọn biểu thức P
b, Tính giá trị của P khi \(\frac{a^2+4}{b^2+9}=\frac{a^2}{9}\)
ĐKXĐ : \(\hept{\begin{cases}ab-2\ne0\\ab+2\ne0\\a^4b^4\ne0\end{cases}}\Rightarrow ab\ne\pm2;a\ne0;b\ne0\)
\(P=\left(\frac{1}{ab-2}+\frac{1}{ab+2}+\frac{2ab}{a^2b^2+4}+\frac{4a^3b^3}{a^4b^4+16}\right).\frac{a^4b^4+16}{a^4b^4}\)
\(=\left(\frac{2ab}{a^2b^2-4}+\frac{2ab}{a^2b^2+4}+\frac{4a^3b^3}{a^4b^4+16}\right).\frac{a^4b^4+16}{a^4b^4}\)
\(=\left(\frac{4a^3b^3}{a^4b^4-16}+\frac{4a^3b^3}{a^4b^4+16}\right).\frac{a^4b^4+16}{a^4b^4}\)
\(=\frac{8a^5b^5}{a^8b^8-16^2}.\frac{a^4b^4+16}{a^4b^4}=\frac{8a^5b^5\left(a^4b^4+16\right)}{\left(a^4b^4-16\right)\left(a^4b^4+16\right).a^4b^4}\)
\(=\frac{8ab}{a^4b^4-16}\)
b) Khi \(\frac{a^2+4}{b^2+9}=\frac{a^2}{9}\)
=> (a2 + 4).9 = a2(b2 + 9)
=> 9a2 + 36 = a2b2 + 9a2
=> a2b2 = 36
=> (ab)2 = 36
=> \(\orbr{\begin{cases}ab=6\left(tm\right)\\ab=-6\left(tm\right)\end{cases}}\)
Khi ab = 6 => P = \(\frac{8ab}{\left(ab\right)^4-16}=\frac{8.6}{6^4-16}=\frac{48}{1280}=\frac{3}{80}\)
Khi ab = -6 => P = \(\frac{8ab}{\left(ab\right)^4-16}=\frac{8.\left(-6\right)}{\left(-6\right)^4-16}=-\frac{3}{80}\)
Rút gọn các phân thức :
a ) \(\frac{\left(a+b\right)^2-c^2}{a+b+c}\) b.\(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}\)
a) \(\frac{\left(a+b\right)^2-c^2}{a+b+c}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{a+b+c}=a+b-c\)
b ) \(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\frac{a^2+2ab+b^2-c^2}{a^2+ac+c^2-b^2}\)
\(=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+c+b\right)\left(a+c-b\right)}=\frac{a+b-c}{a-b+c}\)
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Rút gọn các phân thức sau :
a ) \(\frac{\left(a+b\right)^2-c^2}{a+b+c}\) b ) \(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}\)
a ) \(\frac{\left(a+b\right)^2-c^2}{a+b+c}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{a+b+c}=a+b-c\)
b ) \(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\frac{a^2+2ab+b^2-c^2}{a^2+2ac+c^2-b^2}\)
\(=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+c+b\right)\left(a+c-b\right)}=\frac{a+b-c}{a-b+c}\)
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a) \(\frac{\left(a+b\right)^2-c^2}{a+b+c}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{a+b+c}=a+b-c\)
b) \(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\frac{\left(a^2+2ab+b^2\right)-c^2}{\left(a^2+2ac+c^2\right)-b^2}=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+c+b\right)\left(a+c-b\right)}=\frac{a+b-c}{a+c-b}\)
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a) Rút gọn Q = \(\left(\frac{a}{a+b}+\frac{a^2}{b^2-a^2}\right):\left(\frac{a^2}{a+b}-\frac{a^2}{a^2+b^2+2ab}\right)\)
b) Tính Q tại a = 1+ \(\sqrt{2}\); b = 1 - \(\sqrt{2}\)