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Đề sai r, từ \(x_1;x_2;x_3;...;x_{51}\) có 51 số, chia thành 2 nhóm sẽ thừa 1 số, mà đề bài said: \(x_1+x_2=x_3+x_4=x_5+x_6=...=x_{50}+x_{51}\) vẫn đủ số ->vô cmn lí
Sửa đề
Ta có : x1 + x2 + x3 + x4 + ... + x51 = 0
Xét tổng : \((x_1+x_2)+(x_3+x_4)+...+(x_{49}+x_{50})=1\cdot25=25\)\((\)có 25 cặp\()\)
Tổng x1 + x2 + x3 + x4 + ... + x51 = 0
Hay 25 + x51 = 0
x51 = 0 - 25
x51 = -25
Khi đó x50 + x51 = 1 thay bằng x50 + -25 = 1
=> x50 = 1 - \((\)- 25\()\)
=> x50 = 26
Vậy x50 = 26
Chúc bạn học tốt :>
Tìm x
a) 4/5 ( x - 1/3) - \(3\dfrac{1}{2}\)= 50%
b) \(8\dfrac{3}{5}\cdot x-2\dfrac{1}{5}\cdot x=16\%\)
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a) \(\dfrac{4}{5}\left(x-\dfrac{1}{3}\right)-3\dfrac{1}{2}=50\%\)
\(\dfrac{4}{5}x-\dfrac{4}{15}-\dfrac{7}{2}=\dfrac{1}{2}\)
\(\dfrac{4}{5}x=\dfrac{1}{2}+\dfrac{7}{2}+\dfrac{4}{15}\)
\(\dfrac{4}{5}x=\dfrac{64}{15}\)
\(x=\dfrac{64}{15}:\dfrac{4}{5}\\ x=\dfrac{16}{3}\)
b) \(8\dfrac{3}{5}.x-2\dfrac{1}{5}.x=16\%\)
\(\dfrac{43}{5}x-\dfrac{11}{5}x=\dfrac{4}{25}\)
\(\dfrac{32}{5}x=\dfrac{4}{25}\)
\(x=\dfrac{4}{25}:\dfrac{32}{5}\)
\(x=\dfrac{1}{40}\)
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\(x-\dfrac{1}{50}+x-\dfrac{5}{96}+x+\dfrac{9}{30}=7\\ \Leftrightarrow3x\simeq6,772\\ \Leftrightarrow x\simeq2,257\)
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a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3
=>x=-1/3+3/4=-4/12+9/12=5/12
b: =>x(1/2-5/6)=7/2
=>-1/3x=7/2
hay x=-21/2
c: (4-x)(3x+5)=0
=>4-x=0 hoặc 3x+5=0
=>x=4 hoặc x=-5/3
d: x/16=50/32
=>x/16=25/16
hay x=25
e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4
=>2x=-7/4+3=5/4
hay x=5/8
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`(2/5-x)^3*50%=1/2`
`=>(2/5-x)^3*1/2=1/2`
`=>(2/5-x)^3=1/2:1/2`
`=>(2/5-x)^3=1`
`=>2/5-x=1`
`=>x=2/5-1`
`=>x=2/5-5/5`
`=>x=-3/5`
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Chú ý: \(a^2-1=\left(a-1\right)\left(a+1\right)\)
Áp dụng:
\(A=\frac{2.4}{3^2}.\frac{3.5}{4^2}.\frac{4.6}{5^2}...\frac{49.51}{50^2}=\frac{2.3.4^2.5^2...49^2.50.51}{3^2.4^2.5^2...50^2}=\frac{2.51}{3.50}=\frac{51}{75}\)
(-50): ( x+ 1) = -5
(x+1 ) = (-50) : (-5)
x+1 = 10
x = 10-1
x= 9
giúp mik vs