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Tớ biết làm đúng 100%:
\((x\cdot1+x\cdot\frac{7}{9})\left(x\cdot1+x\cdot\frac{7}{20}\right)...\left(x\cdot1+x\cdot\frac{7}{9200}\right)=\frac{186}{25}\)
\(x\cdot\left(1+\frac{7}{9}\right)\cdot x\left(1+\frac{7}{20}\right)\cdot...\cdot x\left(1+\frac{7}{9200}\right)=\frac{186}{25}\)
\(\left(x\cdot x\cdot...\cdot x\right)(\frac{16}{9}+\frac{27}{20}+...+\frac{9207}{9200})=\frac{186}{25}\)
\(\left(x\cdot x\cdot...\cdot x\right)\left(\frac{2\cdot8}{1\cdot9}+\frac{3\cdot9}{2\cdot10}+...+\frac{93\cdot99}{92\cdot100}\right)=\frac{186}{25}\)
\(x^{92}\cdot\frac{2\cdot8\cdot3\cdot9\cdot...\cdot93\cdot99}{1\cdot9\cdot2\cdot10\cdot...\cdot92\cdot100}=\frac{186}{25}\)
\(x^{92}\cdot\frac{\left(2\cdot3\cdot...\cdot93\right)\cdot\left(8\cdot9\cdot...\cdot99\right)}{\left(1\cdot2\cdot...\cdot92\right)\cdot\left(9\cdot10\cdot...\cdot100\right)}=\frac{186}{25}\)
\(x^{92}\cdot\frac{93\cdot8}{100}=\frac{186}{25}\)
\(x^{92}\cdot\frac{186}{25}=\frac{186}{25}\)
\(x^{92}=\frac{186}{25}:\frac{186}{25}\)
\(x^{92}=1\Rightarrow x=1\)
cô tớ giải rồi . x=1 (đúng 100%)
7x.(2+x)-7x.(x+3)=14
7x.[(2+x)-(x+3)]=14
7x.[x+2-x-3]=14
7x.(-1) =14
7x =14:(-1)
7x =-14
x =-14:7
x =-2
Chúc bn học tốt
\(7x\cdot\left(2+x\right)-7x\cdot\left(x+3\right)=14\)
\(7x\cdot\left(2+x-x-3\right)=14\)
\(7x\cdot\left(-1\right)=14\)
\(7x=-14\)
\(x=-2\)
\(7x.\left(2+x\right)-7x.\left(x+3\right)=14\)
\(7x.\left(2+x-x-3\right)=14\)
\(7x.\left(-1\right)=14\)
\(7x=14:\left(-1\right)\)
\(7x=-14\)
\(x=\left(-14\right):7\)
\(x=-2\)
7x.( 2 + x ) - 7x( x + 3 ) = 14
<=> 7x( 2 + x - x - 3 ) = 14
<=> 7x.(-1) = 14
<=> -7x = 14
<=> x = -2
7x(2+x )- 7x (x+3) = 14
<=> 14x + 7x2 - 7x2 - 21x = 14
<=> -7x = 14
<=> x = -2
2|x-6|=|x-6|+7x+2-7x
2|x-6|=|x-6|+2
|x-6|=|x-6|+1
minh moi nghi den do thoi
2|x-6| + 7x -2 =|x-6| + 7x
=>2|x-6| -|x-6| - 2=7x-7x
=>|x-6|-2=0
=>|x-6|=2
=>x-6=2 hoặc x-6=-2
Phần sau tự giải
do các phân số ở hàng số thứ 2 đã tối giản nên x=0=>7x=0 =>tổng các phân số sau đều tối giản
\(7\frac{x}{2.5}+7\frac{x}{5.8}+.....+7.\frac{x}{17.20}=\frac{21}{10}\)
\(7\left(\frac{x}{2.5}+\frac{x}{5.8}+...+\frac{x}{17.20}\right)=\frac{21}{10}\)
\(\frac{x}{2.5}+\frac{x}{5.8}+...+\frac{x}{17.20}=\frac{21}{70}\)
\(\frac{x.3}{2.5.3}+\frac{x.3}{5.8.3}+...+\frac{x.3}{17.20.3}=\frac{21}{70}\)
\(x.\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{17.20}\right)=\frac{21}{70}\)
\(x.\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{21}{70}\)
\(x.\frac{1}{3}.\frac{9}{20}=\frac{21}{70}\)
=> \(x=2\)
\(x=\frac{7x}{2}\)\(-\frac{7x}{5}+\)\(\frac{7x}{5}\)\(-\frac{7x}{8}\)\(+\frac{7x}{8}\)\(-\frac{7x}{11}\)\(+\frac{7x}{11}\)\(-\frac{7x}{14}\)\(+\frac{7x}{14}\)\(-\frac{7x}{17}+\)\(\frac{7x}{17}\)\(-\frac{7x}{20}\)\(=\frac{21}{10}\)
\(x=\frac{7x}{2}\)\(-\frac{7x}{20}\)\(=\frac{21}{10}\)
\(x=\frac{7x.10}{20}\)\(+\frac{7x}{20}\)\(=\frac{21}{10}\)
\(x=\frac{7x.10+7x}{20}\)\(=\frac{21}{10}\)
\(x=\frac{7x.\left(10+2\right)}{20.2}\)\(=\frac{7x.12}{40}\)\(=\frac{21}{10}\)
\(=>\frac{7x.12:4}{40:4}=\)\(\frac{21}{10}\)
\(=>x=1\)
7x.(2+x)-7x.(x+3)=14
7x.[(2+x)-(3+x)]=14
7x.[2+x-3-x]=14
7x.(-1) =14
7x =14:(-1)
7x =-14
x =-14:7
x =-2
Chúc bn học tốt
\(7x\left(2+x\right)-7x\left(x+3\right)=14\)
\(\Leftrightarrow14x+7x^2-7x^2-21x=14\)
\(\Leftrightarrow-7x=14\)\(\Leftrightarrow x=-2\)
Vậy \(x=-2\)
7x.(2+x)-7x.(x+3)=14
7x.(2+x-x-3)=14
7x.(-1)=14
7x=14:(-1)
7x=-14
x=-14:7
x=2
Vậy x=2
tk nha!