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\(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(=\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\)
\(=\frac{2}{7}:\frac{1}{\frac{7}{2}}=\frac{2}{7}:\frac{2}{7}=1\)
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a.
\(\left(2\frac{5}{6}+1\frac{4}{9}\right):\left(10\frac{1}{12}-9\frac{1}{2}\right)\)
\(=\left(2+1\right)+\left(\frac{5}{6}+\frac{4}{9}\right):\left(10-9\right)+\left(\frac{1}{12}-\frac{1}{2}\right)\)
\(=3+\left(\frac{15}{18}+\frac{8}{18}\right):1+\left(\frac{1}{12}-\frac{6}{12}\right)\)
\(=\left(3+\frac{23}{18}\right):\left(1+\frac{-5}{12}\right)\)
\(=\frac{77}{18}:\frac{7}{12}\)
\(=\frac{77}{18}.\frac{12}{7}=\frac{22}{3}\)
Chúc bạn học tốt!!!
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
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A = -10/3 + 19/6.7/5 - 19/3.1/10 + 19/10.4/3
A = 1/30.(-10.10 + 19.7 - 19.1 + 19.4)
A = 1/30.(-100 + 19.11 - 19)
A = 1/30.(-100 + 19.10)
A = 1/30.(-100 + 190)
A = 1/30.90
A = 3
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2) 1\26+1\27+1\28+........+1\50=1+1\2+1\3+......+1\50 -( 1+1\2+1\3+.....+1\25)=1+1\2+1\3+....+1\50-2.(1\2+1\4+1\6+....+1\50)=1-1\2+1\3-1\4+.....+1\49-1\50=vế phải(đpcm)
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\(A=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)..............\left(1-\frac{1}{780}\right)\)
\(A=\frac{2}{3}.\frac{5}{6}.......\frac{779}{780}\)
\(A=\frac{4}{6}.\frac{10}{12}........\frac{1558}{1560}\)
\(A=\frac{1.4.2.5...........38.41}{2.3.3.4..........39.40}\)
\(A=\frac{\left(1.2.......38\right).\left(4.5..........41\right)}{\left(2.3.......39\right).\left(3.4............40\right)}\)
\(A=\frac{41}{39}\)
\(A=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right).......\left(1-\frac{1}{780}\right)\)
\(=\left(\frac{3}{3}-\frac{1}{3}\right)\left(\frac{6}{6}-\frac{1}{6}\right)..........\left(\frac{780}{780}-\frac{1}{780}\right)\)
\(=\frac{2}{3}.\frac{5}{6}..............\frac{779}{780}\)
sao nữa??
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\(\frac{\frac{2}{3}+\frac{2}{5}-\frac{2}{9}}{\frac{4}{3}+\frac{4}{5}-\frac{4}{9}}\) _ \(\frac{3-\frac{3}{11}-\frac{3}{17}}{5-\frac{5}{11}-\frac{5}{17}}\)
=\(\frac{2\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{9}\right)}{4\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{9}\right)}\)_ \(\frac{3\left(1-\frac{1}{11}-\frac{1}{17}\right)}{5\left(1-\frac{1}{11}-\frac{1}{17}\right)}\)= \(\frac{2}{4}-\frac{3}{5}\)= \(\frac{-1}{10}\)