Những câu hỏi liên quan
22- Nhật Minh 6/5
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Nguyễn Lê Phước Thịnh
21 tháng 12 2021 lúc 21:08

g: =>3x-10=200

hay x=70

Thien Phuc
31 tháng 12 2021 lúc 15:01

c) (7x + x – 28) :100 – 38 = 9

<=> 8x - 28 = 4700

<=> x = 591

Thien Phuc
31 tháng 12 2021 lúc 15:02

d) 936 – (4x + 5x ) = 900

<=> 9x = 36

<=> x = 4

Nguyễn Đỗ Ngọc Minh
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Nguyễn Đỗ Ngọc Minh
5 tháng 5 2021 lúc 22:23

Sửa đề: 3x1⋅5+3x5⋅9+3x9⋅13+...+3x81⋅85=4153x1⋅5+3x5⋅9+3x9⋅13+...+3x81⋅85=415

⇔x⋅34(1−15+15−19+19−113+...+181−185)=415⇔x⋅34(1−15+15−19+19−113+...+181−185)=415

⇔x⋅6385=415⇔x⋅6385=415

hay x=68189

 

Nguyễn Lê Phước Thịnh
5 tháng 5 2021 lúc 22:30

Sửa đề: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)

a) Ta có: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)

\(\Leftrightarrow\dfrac{3x}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{81}-\dfrac{1}{85}\right)=\dfrac{4}{15}\)

\(\Leftrightarrow\dfrac{3x}{4}\cdot\dfrac{84}{85}=\dfrac{4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{63}{85}=\dfrac{4}{15}\)

hay \(x=\dfrac{68}{189}\)

Vậy: \(x=\dfrac{68}{189}\)

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Nguyễn Lê Phước Thịnh
5 tháng 5 2021 lúc 21:27

Sửa đề: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)

a) Ta có: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)

\(\Leftrightarrow\dfrac{3x}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{81\cdot85}\right)=\dfrac{4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{81}-\dfrac{1}{85}\right)=\dfrac{4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{85}\right)=\dfrac{4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{63}{85}=\dfrac{4}{15}\)

hay \(x=\dfrac{68}{189}\)

Vậy: \(x=\dfrac{68}{189}\)

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Nguyễn Lê Phước Thịnh
5 tháng 5 2021 lúc 21:20

Sửa đề: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)

a) Ta có: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)

\(\Leftrightarrow\dfrac{3x}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{81\cdot85}\right)=\dfrac{4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{81}-\dfrac{1}{85}\right)=\dfrac{4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{85}\right)=\dfrac{4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{63}{85}=\dfrac{4}{15}\)

hay \(x=\dfrac{68}{189}\)

Vậy: \(x=\dfrac{68}{189}\)

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Nguyễn Lê Phước Thịnh
5 tháng 5 2021 lúc 22:33

Sửa đề: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)

a) Ta có: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)

\(\Leftrightarrow\dfrac{3x}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{81}-\dfrac{1}{85}\right)=\dfrac{4}{15}\)

\(\Leftrightarrow\dfrac{3x}{4}\cdot\dfrac{84}{85}=\dfrac{4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{63}{85}=\dfrac{4}{15}\)

hay \(x=\dfrac{68}{189}\)

Vậy: \(x=\dfrac{68}{189}\)

Xem chi tiết
Nguyễn Lê Phước Thịnh
5 tháng 5 2021 lúc 21:20

Sửa đề: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)

a) Ta có: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)

\(\Leftrightarrow\dfrac{3x}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{81\cdot85}\right)=\dfrac{4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{81}-\dfrac{1}{85}\right)=\dfrac{4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{85}\right)=\dfrac{4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{63}{85}=\dfrac{4}{15}\)

hay \(x=\dfrac{68}{189}\)

Vậy: \(x=\dfrac{68}{189}\)

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Nguyễn Lê Phước Thịnh
5 tháng 5 2021 lúc 22:30

Sửa đề: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)

a) Ta có: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)

\(\Leftrightarrow\dfrac{3x}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{81}-\dfrac{1}{85}\right)=\dfrac{4}{15}\)

\(\Leftrightarrow\dfrac{3x}{4}\cdot\dfrac{84}{85}=\dfrac{4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{63}{85}=\dfrac{4}{15}\)

hay \(x=\dfrac{68}{189}\)

Vậy: \(x=\dfrac{68}{189}\)

Xem chi tiết
Nguyễn Lê Phước Thịnh
5 tháng 5 2021 lúc 21:27

Sửa đề: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)

a) Ta có: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)

\(\Leftrightarrow\dfrac{3x}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{81\cdot85}\right)=\dfrac{4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{81}-\dfrac{1}{85}\right)=\dfrac{4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{85}\right)=\dfrac{4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{63}{85}=\dfrac{4}{15}\)

hay \(x=\dfrac{68}{189}\)

Vậy: \(x=\dfrac{68}{189}\)

Pham Trong Bach
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Cao Minh Tâm
1 tháng 9 2018 lúc 15:13

2.x – 138 = 23.32

2x – 138 = 8.9

2x – 138 = 72

2x = 72 + 138

2x = 210

x = 210 : 2

x = 105

Vậy x = 105

Nguyen thi bích ngọc
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Nguyễn Lê Phước Thịnh
15 tháng 8 2021 lúc 20:06

a: Ta có: \(814-\left(x-305\right)=712\)

\(\Leftrightarrow x-305=102\)

hay x=407

b: Ta có: \(2x-138=2^3\cdot2^2\)

\(\Leftrightarrow2x=32+138=170\)

hay x=85

c: Ta có: \(20-\left[7\left(x-3\right)+4\right]=2\)

\(\Leftrightarrow7\left(x-3\right)+4=18\)

\(\Leftrightarrow7\left(x-3\right)=14\)

\(\Leftrightarrow x-3=2\)

hay x=5