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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)
\(3\frac{3}{4}\times4=\frac{3\times4+3}{4}\times4=\frac{15}{4}\times4=\frac{15\times4}{4}=15\)
\(4\frac{3}{2}\div1\frac{1}{2}=\frac{4\times2+3}{2}\div\frac{1\times2+1}{2}=\frac{11}{2}\div\frac{3}{2}=\frac{11}{2}\times\frac{2}{3}=\frac{11}{3}\)
\(4\frac{3}{4}+2\frac{3}{2}=\frac{4\times4+3}{4}+\frac{2\times2+3}{2}=\frac{19}{4}+\frac{7}{2}=\frac{19}{4}+\frac{14}{4}=\frac{33}{4}\)
\(3\frac{8}{6}-4\frac{1}{3}=\frac{3\times8+6}{6}-\frac{4\times3+1}{3}=\frac{30}{6}-\frac{13}{3}=\frac{30}{6}-\frac{26}{6}=\frac{4}{6}=\frac{2}{3}\)
\(3\frac{2}{4}\)x 4 = \(\frac{14}{4}\)x 4 = \(\frac{56}{4}\)
\(4\frac{3}{2}\): \(1\frac{1}{2}\)= \(\frac{11}{2}\): \(\frac{3}{2}\)= \(\frac{11}{2}\)x \(\frac{2}{3}\)= \(\frac{22}{6}\)= \(\frac{11}{3}\)
\(4\frac{3}{4}\)+ \(2\frac{3}{2}\)= \(\frac{11}{4}\)+ \(\frac{7}{2}\)= \(\frac{25}{4}\)= 6,25
\(3\frac{8}{6}\)- \(4\frac{1}{3}\)= \(\frac{13}{3}\)- \(\frac{13}{3}\)= 0
\(=\frac{2}{2.\left(1+2\right)}+\frac{2}{2\left(1+2+3\right)}+...+\)\(\frac{2}{2\left(1+2+...+50\right)}\)
\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{2250}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{50.51}\)
\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{50.51}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{51}\right)\)
\(=2.\frac{49}{102}\)
\(=\frac{49}{51}\)
Đặt A = \(\frac{\frac{1}{2}}{1+2}+\frac{\frac{1}{2}}{1+2+3}+...+\frac{\frac{1}{2}}{1+2+3+....+100}\)
= \(\frac{1}{2}\left(\frac{1}{2.3:2}+\frac{1}{3.4:2}+\frac{1}{4.5:2}+...+\frac{1}{100.101:2}\right)\)
= \(\frac{1}{2}\left(\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{100.101}\right)\)
= \(\frac{1}{2}.2\left(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{100.101}\right)\)
= 1\(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{100}-\frac{1}{101}\right)\)
= \(\frac{1}{2}-\frac{1}{101}=\frac{101}{202}-\frac{2}{202}=\frac{99}{202}\)
cộng hết tất cả 1/1+2+3+.....+10 thì ta chỉ cần cộng 1+2+3+4+5+6+7+8+9+10 là xong rồi tự tính
\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+.............+\frac{1}{1+2+3+......+10}\)
= \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+..............+\frac{1}{45}\)
Đến đây bạn làm tiếp nhé
Đây mà toán lớp 5 à.
Áp dụng công thức
\(\frac{1}{1+2+...+n}=\frac{1}{\frac{n\left(n+1\right)}{2}}=\frac{2}{n\left(n+1\right)}\) ta được
\(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+....+50}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{50.51}\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{51}\right)=\frac{49}{51}\)
Ta có : \(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+.......+\frac{1}{1+2+3+......+50}\)
\(=\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+......+\frac{1}{\frac{50.51}{2}}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+......+\frac{2}{50.51}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{50.51}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{50}-\frac{1}{51}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{51}\right)\)
\(=2.\frac{1}{2}-2.\frac{1}{51}\)
\(=1-\frac{2}{51}=\frac{49}{51}\)