Sửa đề:

\(A=\dfrac{4}{2.5}+\dfrac{4}{5.8}+\dfrac{4}{8.11}+...+\dfrac{4}{65.68}\)

\(A=4.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)

\(A=4.\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\)

\(A=4.\left(\dfrac{34}{68}-\dfrac{1}{68}\right)\)

\(A=4.\dfrac{33}{68}\)

\(A=\dfrac{33}{17}\)

A = \(\dfrac{4}{2.5}\) + \(\dfrac{4}{5.8}\)+ \(\dfrac{4}{8.11}\)+...+ \(\dfrac{4}{65.68}\)

A = \(\dfrac{4}{3}\).( \(\dfrac{3}{2.5}\) + \(\dfrac{3}{5.8}\)+ \(\dfrac{3}{8.11}\)+....+ \(\dfrac{3}{65.68}\))

A = \(\dfrac{4}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{11}\)+...+ \(\dfrac{1}{65}\)- \(\dfrac{1}{68}\)

A = \(\dfrac{4}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{68}\))

A = \(\dfrac{4}{3}\). \(\dfrac{33}{68}\)

A = \(\dfrac{11}{17}\)

bạn kiểm tra lại đề nhé.

\(A=\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{2001\cdot2005}\)

\(A=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-...+\dfrac{1}{2001}-\dfrac{1}{2005}\)

\(A=1-\dfrac{1}{2005}=\dfrac{2004}{2005}\)

\(B=\dfrac{3}{10\cdot12}+\dfrac{3}{12\cdot14}+...+\dfrac{3}{998\cdot1000}\)

\(\dfrac{2}{3}B=\dfrac{2}{10\cdot12}+...+\dfrac{2}{998\cdot1000}\)

\(\dfrac{2}{3}B=\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-...+\dfrac{1}{998}-\dfrac{1}{1000}\)

\(\dfrac{2}{3}B=\dfrac{1}{10}-\dfrac{1}{1000}=\dfrac{99}{1000}\)

\(B=\dfrac{99}{1000}:\dfrac{2}{3}=\dfrac{297}{2000}\)

\(A=\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{2001.2005}\)

\(\Rightarrow A=4\left(\dfrac{1}{1.5}+\dfrac{1}{5.9}+...+\dfrac{1}{2001.2005}\right)\)

\(\Rightarrow A=4.\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{2001}-\dfrac{1}{2005}\right)\)

\(\Rightarrow A=1-\dfrac{1}{2005}\)

\(\Rightarrow A=\dfrac{2004}{2005}\)

\(B=\dfrac{3}{10.12}+\dfrac{3}{12.14}+...+\dfrac{3}{998.1000}\)

\(\Rightarrow B=3\left(\dfrac{1}{10.12}+\dfrac{1}{12.14}+...+\dfrac{1}{998.1000}\right)\)

\(\Rightarrow B=3.\dfrac{1}{2}\left(\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{12}+...+\dfrac{1}{998}-\dfrac{1}{1000}\right)\)

\(\Rightarrow B=\dfrac{3}{2}\left(\dfrac{1}{10}-\dfrac{1}{1000}\right)\)

\(\Rightarrow B=\dfrac{3}{2}.\dfrac{99}{1000}\)

\(\Rightarrow B=\dfrac{297}{2000}\)

\(\dfrac{3}{2}A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\)

\(\dfrac{3}{2}A=\dfrac{4-1}{1.4}+\dfrac{7-4}{4.7}+\dfrac{10-7}{7.10}+...+\dfrac{97-94}{94.97}\)

\(\dfrac{3}{2}A=\dfrac{4}{1.4}-\dfrac{1}{1.4}+\dfrac{7}{4.7}-\dfrac{4}{4.7}+\dfrac{10}{7.10}-\dfrac{7}{7.10}+...+\dfrac{97}{94.97}-\dfrac{94}{94.97}\)

\(\dfrac{3}{2}A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\)

\(\dfrac{3}{2}A=1-\dfrac{1}{97}=\dfrac{96}{97}\)

⇒ A = \(\dfrac{96}{97}:\dfrac{3}{2}=\dfrac{64}{97}\)

Câu B cách làm tương tự, thắc mắc gì bạn cứ hỏi nhé.

Lời giải:
Hiệu đáy lớn và đáy bé là:
$141\times 2: 23=\frac{282}{23}$ (m) 

Đáy bé hình thang:

$\frac{282}{23}: (5-3).3=\frac{423}{23}$ (m)

Đáy lớn hình thang:
$\frac{282}{23}: (5-3).5=\frac{705}{23}$ (m)

Diện tích hình thang lúc đầu:

$(\frac{423}{23}+\frac{705}{23}).23:2=564$ (m²)

\(\left(x-5\right)\left(2x+3\right)-2x\left(x-3\right)+x+7\)

\(=2x^2+3x-10x-15-2x^2+6x+x+7\) ( Nhân phân phối )

\(=\left(2x^2-2x^2\right)+\left(3x-10x+6x+x\right)+\left(-15+7\right)\)

\(=0+0x-8\)

\(=-8\)

Vậy biểu thức có giá trị không phụ thuộc vào giá trị của biến.

ko cần làm bài 2 đâu.chỉ cần bài 1 hoy nha

cau hoi cung de thoi

lafm đi le anh thu

cách 2:

a=\(6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)

a=(6-5-3)-(2/3+5/3-7/3)+(1/2+3/2-5/2)

a=-2-1/2

a=-5/2

Cach 1 jdujsoidfojaiuh

a) \(\frac{42\times15\times11}{33\times30\times35}=\frac{7\times6\times3\times5\times11}{11\times3\times6\times5\times7\times5}\)

                                \(=\frac{1}{5}\)

b) \(\frac{132\times34\times7}{24\times35\times22}=\frac{11\times12\times17\times2\times7}{12\times2\times7\times5\times11\times2}\)

                              \(=\frac{1}{10}\)

bạn tự gạch bỏ những chữ số giống nhau nha

ai giúp em với:((

(x làdấu nhân) là a)\(\frac{42x15x11}{33x30x35}\)=  \(\frac{42x15x11}{3x11x15x2x35}\)= \(\frac{42}{3x2x35}\)= \(\frac{42}{210}\)

  (bài b bạn tự làm)

1.

\(B=20182018.2017-20172017.2018\)

\(B=2018.10001.2017-2017.10001.2018\)

\(B=0\)

1 B=20182018.2017-20172017.2018

B=2018.10001.2017-2017.10001.2018

B=0

2 C=1²+2²+3²+...+100²

C=1(1+0)+2(1+1)+3(1+2)+...+100(1+99)

C=1+2+1.2+3+2.3+...+100+99.100

C=(1+2+3+...+100)+(1.2+2.3+...+99.100)

C=[(1+100).100:2]+[(99.100.101):3]

C=5050+333300

C=338350

1) \(B=20182018.2017-20172017.2018\)

\(B=2018.10001.2017-2017.10001.2018\)

\(B=0-0\)

\(B=0\)

3) \(D=\left(3.4.2^{16}\right)^2:11.2^{13}.4^{11}-16^9\)

\(D=3^2.2^4.2^{32}:11.2^{13}.2^{22}-2^{36}\)

\(D=3^2.2^{71}:11-2^{36}\)

\(D=2^{36}.\left(9.11.2^{35}-1\right)\)