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\(y\div2\frac{1}{3}=1\frac{3}{4}\div2\frac{1}{3}\)
\(1\frac{3}{4}=\frac{7}{4};2\frac{1}{3}=\frac{7}{3}\)
\(y\div2\frac{1}{3}=1\frac{3}{4}\div2\frac{1}{3}\)
\(\Rightarrow y\div\frac{7}{3}=\frac{3}{4}\)
\(\Rightarrow y=\frac{3}{4}\times\frac{7}{3}\)
\(\Rightarrow y=\frac{7}{4}\)
~ 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)
\(y:\frac{5}{2}=\frac{7}{4}:\frac{7}{3}\)
\(y:\frac{5}{2}=\frac{3}{4}\)
\(y=\frac{3}{4}.\frac{5}{2}\)
\(y=\frac{15}{8}\)
Vậy \(y=\frac{15}{8}\)
Chúc bạn zui ~^^
\(y:\frac{5}{2}=\frac{7}{4}:\frac{7}{3}\)
\(y:\frac{5}{2}=\frac{3}{4}\)
\(y=\frac{3}{4}\cdot\frac{5}{2}\)
\(y=1.875\)
Vậy y = 1.875
\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+50}\)
= \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{1275}\)
= \(2\times\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{2550}\right)\)
= \(2\times(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{50.51})\)
= \(2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{50}-\frac{1}{51}\right)\)
= \(2\times\left(\frac{1}{2}-\frac{1}{51}\right)\)
= \(2\times\frac{49}{102}\)
= \(\frac{49}{51}\)
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
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 : \(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+.......+\frac{1}{1+2+3+......+50}\)
\(=\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+......+\frac{1}{\frac{50.51}{2}}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+......+\frac{2}{50.51}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{50.51}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{50}-\frac{1}{51}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{51}\right)\)
\(=2.\frac{1}{2}-2.\frac{1}{51}\)
\(=1-\frac{2}{51}=\frac{49}{51}\)
y:\(\frac{7}{3}=\frac{7}{4}:\frac{7}{3}\)
y.\(\frac{3}{7}=\frac{7}{4}.\frac{3}{7}\)
y.3/7=3/4
y=3/4:3/7
y=7/3
Vậy y=7/3
\(y=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+....+\frac{1}{1+2+3+...+49+50}=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{1275}\)\(=2\cdot\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{2550}\right)=2\cdot\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{50.51}\right)\)\(=2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right)=2\cdot\left(\frac{1}{2}-\frac{1}{51}\right)=2\cdot\frac{49}{102}=\frac{49}{51}\)