A= 1+3+3²+3³+...+3⁹⁹

A= (1+3+3²+3³)+...+(3⁹⁶+3⁹⁷+3⁹⁸+3⁹⁹)

A= (1+3+3²+3³)+...+3⁹⁶(1+3+3²+3³)

A= 40 + ... + 3⁹⁹.40

Vì 40 chia hết cho 40 nên A chia hết cho 40

Chúc bn học tốt

\(A=1+3+3^2+...+3^{99}=\left(1+3+3^2+3^3\right)+...+3^{96}\left(1+3+3^2+3^3\right)\)

\(=40+...+3^{99}.40=40\left(1+3^{99}\right)⋮40\)

Vậy ta có đpcm 

c: =>2/3x=1/10+1/2=1/10+5/10=6/10=3/5

hay \(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)

d: \(\Leftrightarrow\dfrac{4}{9}:x=\dfrac{2}{3}-\dfrac{3}{5}=\dfrac{1}{15}\)

hay \(x=\dfrac{4}{9}:\dfrac{1}{15}=\dfrac{4}{9}\cdot15=\dfrac{20}{3}\)

f: (x+1/2)(2/3-2x)=0

=>x+1/2=0 hoặc 2/3-2x=0

=>x=-1/2 hoặc x=1/3

Lê Minh Phương tham khảo bài mình nhé

\(a,\frac{9}{-7}< x>\frac{7}{2}\)

\(\Leftrightarrow\frac{-9}{7}< x>\frac{7}{2}\)

\(\Leftrightarrow\frac{-18}{14}< x>\frac{49}{14}\)

\(\Leftrightarrow-18< x>49\)

\(\Leftrightarrow x\in\left\{-17;-16;-15;...;50\right\}\)

Còn bài kia tương tự

\(a,\frac{9}{-7}< x< \frac{7}{2}\)

\(\Rightarrow\frac{9.2}{-7.2}< x< \frac{7.7}{2.7}\)

\(\Rightarrow\frac{-18}{14}< x< \frac{49}{14}\)

\(\text{vì}x\in Z\Rightarrow x=-\frac{14}{14};\frac{0}{14};\frac{14}{14};\frac{28}{14};\frac{42}{14}\)

\(\text{hay }x=\left\{-1;0;1;2;3\right\}\)

Vì 13 là lẻ \(\Rightarrow\) 13, 13², 13³, 13⁴, 13⁵, 13⁶ là lẻ.

Mà lẻ + lẻ + lẻ + lẻ + lẻ + lẻ = chẵn nên 13 + 13² + 13³ + 13⁴ + 13⁵ + 13⁶ là chẵn. \(\Rightarrow\) 13 + 13² + 13³ + 13⁴ + 13⁵ + 13⁶ \(⋮\) 2

\(\Rightarrow\) ĐPCM

\(\left(x+1\right)+\left(x+2\right)+...+\left(x+51\right)=0\)

\(x+1+x+2+...+x+51=0\)

\(51x+\left(1+2+...+51\right)=0\)

số số hạng trong dãy số 1...51

\(\left(51-1\right):1+1=51\)

tổng dãy số trên là

\(\left(51+1\right).51:2=1326\)

TA THAY VÀO

\(51x+1326=0\)

\(51x=-1326\)

\(\Rightarrow x=-26\)

Câu a sai đề à bạn???

a: 32+x⋮2

=>x⋮2

mà x∈{6;13;15;28;33}

nên x∈{6;28}

b: 12-x⋮3

mà 12⋮3

nên x⋮3

mà x∈{18;25;36;47;54}

nên x∈{18;36;54}

c: 18-x⋮9

mà 18⋮9

nên x⋮9

mà x∈{8;27;35;49;56}

nên x=27

m

1) Tìm x

\(\frac{11}{2}.x+\frac{1}{3}.x=1\)

\(\Rightarrow x\left(\frac{11}{2}+\frac{1}{3}\right)=1\)

\(\Rightarrow x\left(\frac{33}{6}+\frac{2}{6}\right)=1\)

\(\Rightarrow x.\frac{35}{6}=1\)

\(\Rightarrow x=\frac{6}{35}\)

2) So sánh

\(\frac{59}{40}< \frac{50}{31}\)( cái này bạn quy đồng là ra, mik chỉ ghi kq, bạn tự tính )

3)\(\frac{1}{3}+\frac{4}{7}-\frac{5}{14}-\frac{1}{2}-\frac{2}{3}\)

\(=\left(\frac{1}{3}-\frac{2}{3}\right)+\left(\frac{4}{7}-\frac{5}{14}\right)-\frac{1}{2}\)

\(=-\frac{1}{3}+\frac{3}{14}-\frac{1}{2}\)

\(=-\frac{13}{21}\)

1)\(\frac{11}{2}.x+\frac{1}{3}.x=1\)

\(x.\left(\frac{11}{2}+\frac{1}{3}=1\right)\)

\(x.\frac{35}{6}=1\)

\(x=1:\frac{35}{6}\)

\(x=\frac{6}{35}\)

2) Ta có:

\(\frac{59}{40}=\frac{1829}{1240}\)

\(\frac{50}{31}=\frac{2000}{1240}\)

Vì \(2000>1829\Rightarrow\frac{2000}{1240}>\frac{1829}{1240}\Rightarrow\frac{50}{31}>\frac{59}{40}\)

3)\(\frac{1}{3}+\frac{4}{7}-\frac{5}{14}-\frac{1}{2}-\frac{2}{3}\)

\(=\left(\frac{1}{3}-\frac{2}{3}\right)+\left(\frac{4}{7}-\frac{5}{14}-\frac{1}{2}\right)\)

\(=-\frac{1}{3}+\left(\frac{8}{14}-\frac{5}{14}-\frac{7}{14}\right)\)

\(=\frac{-1}{3}+\frac{-4}{14}\)

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

\(=\frac{-7}{21}+\frac{-6}{21}\)

\(=\frac{-13}{21}\)

MỚI LÀM LÚC TỐI,HÊN QUÁ:

\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)

\(3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)

\(2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(6A=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)

\(4A=3-\left(\frac{101}{3^{99}}-\frac{100}{3^{100}}\right)\)

\(4A=3-\frac{203}{3^{100}}\)

\(A=\frac{3}{4}-\frac{203}{3^{100}\cdot4}< \frac{3}{4}\)

a) Ta có 

\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\)

\(2A=1+\frac{1}{2}+...+\frac{1}{2^6}\)

\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^6}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\right)\)

\(A=1-\frac{1}{2^7}\)

Do \(1-\frac{1}{2^7}< 1\Rightarrow A< 1\left(đpcm\right)\)