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$=\frac{2008+\frac{2007}{2}+\frac{2006}{3}+\frac{2005}{4}+...+\frac{2}{2007}+\frac{1}{2008}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}$
$1+\left(1+\frac{2007}{2}\right)+\left(1+\frac{2006}{3}\right)+...+\left(1+\frac{1}{2008}\right)$
$\frac{2009}{2009}+\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2008}$
$2009.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}\right)$
A=$\frac{2009.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}}$
A=2009
\(\frac{2}{3}.\frac{4}{5}+\frac{1}{3}.\frac{4}{5}=\frac{4}{5}\left(\frac{2}{3}+\frac{1}{3}\right)=\frac{4}{5}.\frac{3}{3}=\frac{4}{5}.1=\frac{4}{5}\)
\(\frac{1}{2}:\frac{3}{4}+\frac{1}{6}:\frac{3}{4}=\frac{3}{4}:\left(\frac{1}{2}+\frac{1}{6}\right)=\frac{3}{4}:\frac{2}{3}=\frac{9}{8}\)
\(\frac{2}{3}.\frac{4}{5}-\frac{1}{3}.\frac{4}{5}=\frac{4}{5}\left(\frac{2}{3}-\frac{1}{3}\right)=\frac{4}{5}.\frac{1}{3}=\frac{4}{15}\)
\(\frac{1}{2}:\frac{3}{4}-\frac{1}{6}:\frac{3}{4}=\frac{3}{4}:\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{3}{4}:\frac{1}{3}=\frac{9}{4}\)
\(\frac{2}{3}.\frac{4}{5}+\frac{1}{3}.\frac{4}{5}=\left(\frac{2}{3}+\frac{1}{3}\right).\frac{4}{5}=1.\frac{4}{5}=\frac{4}{5}\)
\(\frac{1}{2}:\frac{3}{4}+\frac{1}{6}:\frac{3}{4}=\frac{1}{2}.\frac{4}{3}+\frac{1}{6}.\frac{4}{3}=\left(\frac{1}{2}+\frac{1}{6}\right).\frac{4}{3}=\frac{2}{3}.\frac{4}{3}=\frac{8}{9}\)
c,d tương tự
\(a,8\frac{3}{4}+4\frac{1}{5}-3\frac{3}{4}\)
\(=\frac{35}{4}+\frac{21}{5}-\frac{15}{4}\)
\(=\frac{175+84-75}{20}\)
\(=\frac{184}{20}=\frac{46}{5}\)
\(b,3\frac{1}{2}\div\frac{1}{2}+3\frac{1}{2}\div\frac{1}{4}\)
\(=\frac{7}{2}\div\frac{1}{2}+\frac{7}{2}\div\frac{1}{4}\)
\(=\frac{7}{2}\div\left(\frac{1}{2}+\frac{1}{4}\right)\)
\(=\frac{7}{2}\div\frac{3}{4}\)
\(=\frac{7}{2}\times\frac{4}{3}\)
\(=\frac{14}{3}\)
\(1\frac{1}{3}+1\frac{1}{5}.y-\frac{4}{5}=2\frac{4}{5}\)
\(\Rightarrow\frac{4}{3}+\frac{6}{5}.y=2\frac{4}{5}+\frac{4}{5}\)
\(\Rightarrow\frac{4}{3}+\frac{6}{5}.y=2\frac{8}{5}=\frac{18}{5}\)
\(\Rightarrow\frac{6}{5}y=\frac{18}{5}-\frac{4}{3}\)
\(\Rightarrow\frac{6}{5}y=\frac{34}{15}\)
\(\Rightarrow y=\frac{34}{15}:\frac{6}{5}\)
\(\Rightarrow y=\frac{34}{15}.\frac{5}{6}=\frac{17}{9}\)
\(y\div2\frac{1}{3}=1\frac{3}{4}\div2\frac{1}{3}\)
\(1\frac{3}{4}=\frac{7}{4};2\frac{1}{3}=\frac{7}{3}\)
\(y\div2\frac{1}{3}=1\frac{3}{4}\div2\frac{1}{3}\)
\(\Rightarrow y\div\frac{7}{3}=\frac{3}{4}\)
\(\Rightarrow y=\frac{3}{4}\times\frac{7}{3}\)
\(\Rightarrow y=\frac{7}{4}\)
~ Ủng hộ nhé ~
\(y:2\frac{1}{3}=1\frac{3}{4}:2\frac{1}{3}\)
\(y:\frac{7}{3}=\frac{7}{4}:\frac{7}{3}\)
\(y:\frac{7}{3}=\frac{3}{4}\)
\(y=\frac{3}{4}\times\frac{7}{3}\)
\(y=\frac{7}{4}\)
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)
em có máy tính FX-570VNPLUS ko?
có thì ấn SHIFT rồi SLOVE cuối cùng nhấn = sẽ có ngay kết quả!
\(\frac{3}{4}\div y+\frac{1}{2}\div\frac{1}{4}=4\)
\(\frac{3}{4}\div y+\frac{1}{4}=4\)
\(\frac{3}{4}\div y=4-\frac{1}{4}\)
\(\frac{3}{4}\div y=\frac{15}{4}\)
\(y=\frac{3}{4}\div\frac{15}{4}=\frac{1}{5}\)