học ngu tự mà lm đi hỏi con mẹ mày

Ta có : 1+3+5+7+9+...+x=100

   <=> (x+1).[(x-1):2+1]:2=100

                (x+1).[x-1/2+1]=200

                     (x+1).x+1/2=200

                      (x+1).(x+1)=400

                              (x+1)²=400

   => \(\orbr{\begin{cases}x+1=20\\x+1=-20\end{cases}}\)

Mà 1+3+5+7+9+...+x=100

Nên x+1=20

  =>      x=19

\(a)\) \(\frac{x-1}{3}=\frac{x+2}{4}\)

\(\Leftrightarrow\)\(\frac{4\left(x-1\right)}{12}=\frac{3\left(x+2\right)}{12}\)

\(\Leftrightarrow\)\(4\left(x-1\right)=3\left(x+2\right)\)

\(\Leftrightarrow\)\(4x-4=3x+6\)

\(\Leftrightarrow\)\(4x-3x=6+4\)

\(\Leftrightarrow\)\(x=10\)

Vậy \(x=10\)

Câu b) thiếu đề nhé bạn 

Chúc bạn học tốt ~ 

Thanks bạn Phùng Minh Quân nha!

Mk chỉnh lại cau b nè:

b, Tìm x € Z để \(B=\frac{2x+9}{x-1}\)có giá trị nguyên.

a) \(\frac{22}{7}\div\left(11-\chi\right)=\frac{7}{5}-\frac{2}{3}\)

\(\frac{22}{7}\div\left(11-\chi\right)=\frac{11}{15}\)

\(\left(11-\chi\right)=\frac{22}{7}\div\frac{11}{15}\)

\(\left(11-\chi\right)=\frac{30}{7}\)

\(\chi=11-\frac{30}{7}\)

\(\chi=\frac{47}{7}\)

b) (x+1)+(x+2)+(x+3)+...+(x+100)=5550

Từ 1 đến 100 có 100 số hạng => Có 100 x

(x + x + x + .... + x) + (1 + 2 + 3 + .. + 100) = 5550

Áp dụng tính chất cộng dãy số cách đều, ta có

(100.x) + 5050 = 5550

100.x = 5550 - 5050

100.x = 500

x = 500 : 100

x = 5 

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

\(\frac{1}{3}:\left(2x-1\right)=-5-\frac{1}{4}\)

\(\frac{1}{3}:\left(2x-1\right)=-\frac{20}{4}-\frac{1}{4}\)

\(\frac{1}{3}:\left(2x-1\right)=-\frac{21}{4}\)

\(\left(2x-1\right)=\frac{1}{3}:-\frac{21}{4}\)

\(\left(2x-1\right)=\frac{1}{3}.-\frac{4}{21}\)

\(\left(2x-1\right)=-\frac{4}{63}\)

2x= -4/63 + 1

2x = 59/63

x = 59/63 : 2

x = 59/126

1/3:(2.x-1)=-5-1/4

1/3:(2.x-1)=-21/4

2.x-1=1/3:-21/4

2.x-1=-4/63

2.x=-4/63+1

2.x=\(3\frac{59}{63}\)

x=\(3\frac{59}{63}\):2

x=\(1\frac{61}{63}\)

Sửa đề \(\frac{11}{13}\)chứ không phải \(\frac{11}{3}\)

\(\frac{2,75-2,2+\frac{11}{7}+\frac{11}{13}}{0,75-0,6+\frac{3}{7}+\frac{3}{13}}-x-\frac{1}{9}=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)

+) Đặt \(A=\frac{2,75-2,2+\frac{11}{7}+\frac{11}{13}}{0,75-0,6+\frac{3}{7}+\frac{3}{13}}\)

\(A=\frac{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}\)

\(A=\frac{11\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)

\(A=\frac{11}{3}\)(1)

+) Đặt \(B=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)

\(B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}\)

\(B=\frac{2}{2}\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}\right)\)

\(B=\frac{2}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}\right)\)

\(B=\frac{2}{2}\left(1-\frac{1}{9}\right)=1\cdot\frac{8}{9}=\frac{8}{9}\)(2)

Từ (1) và (2) => \(A-x-\frac{1}{9}=B\)

=> \(\frac{11}{3}-x-\frac{1}{9}=\frac{8}{9}\)

=> \(\frac{11}{3}-x=1\)

=> \(x=\frac{11}{3}-1=\frac{8}{3}\)

Vậy x = 8/3

\(\frac{x-5}{3}=\frac{6}{5}\)

\(x-5=\frac{6}{5}.3\)

\(x-5=\frac{18}{5}\)

\(x=\frac{18}{5}+5\)

\(\Rightarrow x=\frac{43}{5}\)

Vậy ...

\(\frac{1}{3}=\frac{2-x}{4}\)

\(\frac{4}{3}=2-x\)

\(x=2-\frac{4}{3}\)

\(\Rightarrow x=\frac{2}{3}\)

Vậy ...

x - 5 = 6 * 3 / 5 => x - 5 = 3.6 => x = 8.6

2 - x = 4 / 3 => x = 2 / 3

x-[17/2-6/35]=-1/3

x-583/70=-1/3

x=-1/3+583/70

x=1679/210

vậy x=1769/210

[2/3-(x-7/4)]=9/2+5/4

[2/3-(x-7/4)]=23/4

(x-7/4)=23/4+2/3

(x-7/4)=77/12

x=77/12+7/4

x=49/6

vậy x=49/6

ko ghi đề

\(=25,97+\left(6,54+103,46\right)\)

\(=25,97+110\)

\(=135,97\)

a/135,97

b/1500

c/10

\(9^{x-1}=9^2\Rightarrow x-1=2\Rightarrow x=3\)

\(8^{x-1}=8^2\Rightarrow x-1=2\Rightarrow x=3\)

\(7^{x-1}=7^2\Rightarrow x-1=2\Rightarrow x=3\)

ủng hộ nha bà con cô bác

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