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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)
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 : \(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+.......+\frac{1}{1+2+3+......+50}\)
\(=\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+......+\frac{1}{\frac{50.51}{2}}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+......+\frac{2}{50.51}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{50.51}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{50}-\frac{1}{51}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{51}\right)\)
\(=2.\frac{1}{2}-2.\frac{1}{51}\)
\(=1-\frac{2}{51}=\frac{49}{51}\)
\(a.\)\(1\frac{2}{3}:\frac{2}{3}-\frac{3}{4}\cdot\frac{2}{3}+5\frac{3}{7}\)
\(=\frac{5}{3}:\frac{2}{3}-\frac{3}{4}\cdot\frac{2}{3}+\frac{38}{7}\)
\(=\frac{5}{3}\cdot\frac{3}{2}-\frac{3}{4}\cdot\frac{2}{3}+\frac{38}{7}\)
\(=\frac{5}{2}-\frac{1}{2}+\frac{38}{7}\)
\(=\frac{4}{2}+\frac{38}{7}\)
\(=2+\frac{38}{7}\)
\(=\frac{14}{7}+\frac{38}{7}\)
\(=\frac{52}{7}\)
\(b.1\frac{1}{3}-1\frac{1}{4}:1\frac{1}{2}+2\frac{3}{4}\cdot3\frac{2}{3}\)
\(=\frac{4}{3}-\frac{5}{4}:\frac{3}{2}+\frac{11}{4}\cdot\frac{11}{3}\)
\(=\frac{4}{3}-\frac{5}{4}\cdot\frac{2}{3}+\frac{11}{4}\cdot\frac{11}{3}\)
\(=\frac{4}{3}-\frac{5}{6}+\frac{121}{12}\)
\(=\frac{16}{12}-\frac{10}{12}+\frac{121}{12}\)
\(=\frac{6}{12}+\frac{121}{12}\)
\(=\frac{127}{12}\)
\(c.7\cdot\frac{2}{3}-\frac{2}{5}:\frac{1}{2}-\frac{2}{3}\)
\(=7\cdot\frac{2}{3}-\frac{2}{5}\cdot\frac{2}{1}-\frac{2}{3}\)
\(=7\cdot\frac{2}{3}-\frac{4}{5}-\frac{2}{3}\)
\(=\frac{14}{3}-\frac{4}{5}-\frac{2}{3}\)
\(=\frac{70}{15}-\frac{12}{15}-\frac{10}{15}\)
\(=\frac{58}{15}-\frac{10}{15}\)
\(=\frac{48}{15}=\frac{16}{5}\)
\(\frac{5}{3}:\frac{2}{3}-\frac{3}{4}\cdot\frac{2}{3}+\frac{38}{7}\)
\(\frac{5}{2}-\frac{1}{2}+\frac{38}{7}\)
\(2+\frac{38}{7}\)
\(\frac{52}{7}\)
\(=\frac{2}{2.\left(1+2\right)}+\frac{2}{2\left(1+2+3\right)}+...+\)\(\frac{2}{2\left(1+2+...+50\right)}\)
\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{2250}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{50.51}\)
\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{50.51}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{51}\right)\)
\(=2.\frac{49}{102}\)
\(=\frac{49}{51}\)
ở dãy 1 thì số đứng sau bằng tổng hai số đứng trước
ta có 5 số tiếp theo la 40,74, 136,...
\(\dfrac{3}{4}\times\dfrac{8}{5}:1\dfrac{1}{6}\)
=\(\dfrac{6}{5}:\) \(\dfrac{7}{6}\)
=\(\dfrac{6}{5}\times\dfrac{6}{7}=\dfrac{36}{35}\)
2\(\dfrac{1}{3}\) x 1\(\dfrac{1}{4}\) -\(\dfrac{7}{5}\)
\(\dfrac{7}{3}\times\dfrac{5}{4}-\) \(\dfrac{7}{5}\)
\(\dfrac{35}{12}-\dfrac{7}{5}\)
\(\dfrac{175}{60}-\dfrac{84}{60}=\dfrac{91}{60}\)
4\(\dfrac{2}{3}+1\dfrac{1}{4} +2\dfrac{1}{3}+2\dfrac{3}{7}\)
(4 +2) + \(\left(\dfrac{2}{3}+\dfrac{1}{3}\right)\) +1\(\dfrac{1}{4}\) + \(2\dfrac{3}{7}\)
6 + 1 + \(\dfrac{5}{4}\) + \(\dfrac{17}{7}\)
7 + \(\dfrac{103}{28}\)
\(\dfrac{299}{28}\)
\(8\frac{7}{10}+2\frac{3}{4}=\frac{87}{10}+\frac{11}{4}=\frac{174}{20}+\frac{55}{20}=\frac{229}{20}\)
Bạn chỉ cần đưa về phân số xong tính bình thường. Muốn đổi từ hỗn số sang phân số, ta chỉ cần lấy phần nguyên nhân cho mẫu rồi cộng với tử là xong. Chứ bạn cứ hỏi mấy bài dễ như thế này thì k giỏi đc đâu!!!
1/2 + 1/3 = 2/6 + 3/6 = 5/6
1/2 + 1/3 + 1/4 = 5/6 + 1/4 = 20/24 + 6/24 = 13/12
1/2 + 1/3 + 1/4 + 1/5 = 13/12 + 1/5 =65/60 + 12/60 = 77/60
\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+50}\)
= \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{1275}\)
= \(2\times\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{2550}\right)\)
= \(2\times(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{50.51})\)
= \(2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{50}-\frac{1}{51}\right)\)
= \(2\times\left(\frac{1}{2}-\frac{1}{51}\right)\)
= \(2\times\frac{49}{102}\)
= \(\frac{49}{51}\)
A=1/1+2 + 1/1+2+3 + 1/1+2+3+4 +... + 1/1+2+3+...+50
A = 1/3 + 1/6 + 1/10 + 1/15 + ...+1/1275
Nhân cả hai vế với 1/2, ta có:
A/2 = 1/6 + 1/12 + 1/20 + 1/30 + ... + 1/2550
A/2 = 1/2x3 + 1/3x4 + 1/4x5 + 1/5x6 + ... + 1/50x51
A/2 = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +..... + 1/50 - 1/51
A/2 = 1-1/51
A/2 = 49/102
A = 49/51
\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+50}\)
\(=\frac{2}{2(1+2)}+\frac{2}{2(1+2+3)}+\frac{2}{2(1+2+3+4)}+...+\frac{2}{2(1+2+3+...+50)}\)
\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{2\left[\frac{(1+50)\cdot50}{2}\right]}\)
\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{2550}\)
\(=2\left[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{2550}\right]\)
\(=2\left[\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{50\cdot51}\right]\)
\(=2\left[\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right]\)
\(=2\left[\frac{1}{2}-\frac{1}{51}\right]=\frac{49}{51}\)
\(\frac{1}{1+2}\)+\(\frac{1}{1+2+3}\)+\(\frac{1}{1+2+3+4}\)+....+\(\frac{1}{1+2+3+4+....+50}\)
B=\(\frac{1}{\left[1+2\right]\cdot2:2}\)+\(\frac{1}{\left[1+3\right]\cdot3:2}\)+\(\frac{1}{\left[1+4\right]\cdot4:2}\)+....+\(\frac{1}{\left[1+50\right]\cdot50:2}\)
B=\(\frac{1}{3\cdot2:2}\)+\(\frac{1}{4\cdot3:2}\)+\(\frac{1}{5\cdot4:2}\)+....+\(\frac{1}{51\cdot50:2}\)
B*\(\frac{1}{2}\)=\(\frac{1}{3\cdot2}\)+\(\frac{1}{4\cdot3}\)+\(\frac{1}{5\cdot4}\)+....+\(\frac{1}{51\cdot50}\)
B*\(\frac{1}{2}\)=\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{5}\)+...+\(\frac{1}{50}\)-\(\frac{1}{51}\)
B*\(\frac{1}{2}\)=\(\frac{1}{2}\)-\(\frac{1}{51}\)
B*\(\frac{1}{2}\)=\(\frac{49}{102}\)
B=\(\frac{49}{102}\):\(\frac{1}{2}\)
B=\(\frac{49}{51}\)
Dấu . là Dấu nhân
Đặt \(A=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+50}\)
\(\Rightarrow A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{1275}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{2550}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{50\times51}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{50}-\frac{1}{51}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+...+\left(\frac{1}{50}-\frac{1}{50}\right)-\frac{1}{51}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{51}=\frac{49}{102}\)
\(\Rightarrow A=\frac{49}{102}:\frac{1}{2}=\frac{49}{51}\)
~ Hok tốt ~