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\(\frac{11}{17}\) : \(\frac{33}{34}\) = \(\frac{11}{17}\) x \(\frac{34}{33}\) = \(\frac{374}{561}\) = \(\frac{2}{3}\)
a) \(\frac{9}{22}.\frac{33}{18}=\frac{9.33}{22.18}=\frac{297}{396}=\frac{3}{4}\)
b) \(\frac{12}{35}:\frac{36}{25}=\frac{12}{35}.\frac{25}{36}=\frac{12.25}{35.36}=\frac{300}{1260}=\frac{5}{21}\)
c) \(\frac{19}{17}:\frac{76}{51}=\frac{19}{17}.\frac{51}{76}=\frac{19.51}{17.76}=\frac{969}{1292}=\frac{3}{4}\)
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)
\(\frac{21}{11}\times\frac{22}{17}:\frac{42}{34}\)
\(=\frac{44}{17}:\frac{42}{34}\)
\(=\frac{44}{21}\)
Hok tốt
\(\frac{21}{11}.\frac{22}{17}\div\frac{42}{34}\)
\(=\frac{21}{11}.\frac{22}{17}.\frac{34}{42}\)
\(=\frac{21.22.34}{11.17.42}\)
\(=\frac{21.11.2.17.2}{11.17.2.21}\)
\(=2\)
\(\frac{33}{2}+\frac{33}{6}+\frac{33}{18}+\frac{33}{54}+\frac{33}{162}+\frac{33}{486}\)
\(=\frac{33.3+33.3+33.3+33.3+33.3}{486}\)
\(=\frac{99.5}{486}\)
\(=\frac{495}{486}\)
Gọi \(A=\frac{33}{2}+\frac{33}{6}+...+\frac{33}{486}\)
\(A=33.\left[\left(\frac{1}{1.2}+\frac{1}{2.3}\right)+\left(\frac{1}{3.6}+\frac{1}{6.9}\right)\left(\frac{1}{9.18}+\frac{1}{18.27}\right)\right]\)
\(A=33.\left[\frac{2}{3}+\frac{2}{9}+\frac{2}{27}\right]\)
\(A=66.\left[\frac{9}{27}+\frac{3}{27}+\frac{1}{27}\right]\)
\(A=66.\frac{13}{27}\)
\(A=\frac{286}{9}\)
sai hay đúng cx ko biết nha
\(\frac{9}{22}\cdot55=\frac{9\cdot55}{22}=\frac{495}{22}=\frac{45}{2}\)
\(\frac{11}{17:\frac{33}{34}}=\frac{11}{17\cdot\frac{34}{33}}=\frac{11}{\frac{578}{33}}=\frac{363}{578}\)