tìm x biết: (5x+1)^2-(5x+3)(5x-3)=30
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NT
1
20 tháng 7 2015
1 , <=> 25x^2 + 10x + 1 - ( 25x^2 - 9) = 30
<=> 25x^2 + 10x + 1 - 25x^2 + 9 = 30
<=> 10x + 10 = 30
<=> 10 ( x + 1) = 30
<=> x + 1 = 3
<=> x = 2
2, ( x + 3)(x^2 - 3x + 9 ) - x(x+2)(x-2) = 15
<=> x^3 - 27 - x(x^2 - 4) = 15
<=> x^3 - 27 - x^3 + 4x = 15
<=> 4x -27 = 15
<=> 4x = 15 + 27
<=> 4x =42
<=> x = 42/4 = 21/2
******************
22 tháng 10 2020
a) ( 5x + 1 )2 - ( 5x + 3 )( 5x - 3 ) = 30
⇔ 25x2 + 10x + 1 - ( 25x2 - 9 ) = 30
⇔ 25x2 + 10x + 1 - 25x2 + 9 = 30
⇔ 10x + 10 = 30
⇔ 10x = 20
⇔ x = 2
b) ( x + 3 )2 + ( x - 2 )( x + 2 ) - 2( x - 1 )2 = 7
⇔ x2 + 6x + 9 + x2 - 4 - 2( x2 - 2x + 1 ) = 7
⇔ 2x2 + 6x + 5 - 2x2 + 4x - 2 = 7
⇔ 10x + 3 = 7
⇔ 10x = 4
⇔ x = 4/10 = 2/5
T
1
29 tháng 6 2016
\(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)
\(=25x^2+1+10x-\left(25x^2-9\right)=30\)
\(25x^2+1+10x-25x^2+9=30\)
\(10x+10=30\)
\(10x=30-10=20\)
\(x=\frac{20}{10}=2\)
NH
10 tháng 4 2020
a, \(\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)
\(\Leftrightarrow x^2-4x+4-\left(x^2+6x+9\right)-4x-4=5\)
\(\Leftrightarrow x^2-4x+4-x^2-6x-9-4x-4=5\)
\(\Leftrightarrow-14x-9=5\)
\(\Leftrightarrow-14x=14\)
\(\Leftrightarrow x=-1\)
Vậy....
b, \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)
\(\Leftrightarrow\left(2x\right)^2-3^2-\left(x^2-2x+1\right)-3x^2+15x=-44\)
\(\Leftrightarrow4x^2-9-x^2+2x-1-3x^2+15x=-44\)
\(\Leftrightarrow-10+17x=-44\)
\(\Leftrightarrow17x=-34\)
\(\Leftrightarrow x=-2\)
Vậy....
c, \(\left(5x+1\right)^2-\left(5x+3\right)\left(5x-3\right)=30\)
\(\Leftrightarrow\left(5x\right)^2+10x+1-\left[\left(5x\right)^2-3^2\right]=30\)
\(\Leftrightarrow\left(5x\right)^2+10x+1-\left(5x\right)^2+9=30\)
\(\Leftrightarrow10x+10=30\)
\(\Leftrightarrow10x=20\)
\(\Leftrightarrow x=2\)
Vậy....
d, \(\left(x+3\right)^2+\left(x-2\right)\left(x+2\right)-2\left(x-2\right)^2=7\)
\(\Leftrightarrow x^2+6x+9+x^2-4-2\left(x^2-4x+4\right)=7\)
\(\Leftrightarrow2x^2+6x+5-2x^2+8x-8=7\)
\(\Leftrightarrow14x-3=7\)
\(\Leftrightarrow14x=10\)
\(\Leftrightarrow x=\frac{10}{14}=\frac{5}{7}\)
Vậy...
YN
2 tháng 10 2021
a) \(x^2-2x+1=25\)
\(\Rightarrow x^2-2x+1-25=0\)
\(\Rightarrow x^2-2x.1+1^2-5^2=0\)
\(\Rightarrow\left(x-1\right)^2-5^2=0\)
\(\Rightarrow\left(x-1-5\right)\left(x-1+5\right)=0\)
\(\Rightarrow\left(x-6\right)\left(x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-6=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=\left(-4\right)\end{cases}}\)
b) \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)30\)
\(\Rightarrow\left(5x\right)^2+2.5x.1+1^2-\left(25x^2-9\right)-30=0\)
\(\Rightarrow25x^2+10x+1-25x^2+9-30=0\)
\(\Rightarrow\left(25x^2-25x^2\right)+10x+\left(1+9-30\right)=0\)
\(\Rightarrow10x-20=0\)
\(\Rightarrow10x=20\)
\(\Rightarrow x=2\)
DT
2
26 tháng 9 2016
a) (5x+1)2-(5x+3).(5x-3)=30
\(\Leftrightarrow25x^2+10x+1-25x^2+9-30=0\)
\(\Leftrightarrow10x-20=0\)
\(\Leftrightarrow10x=20\)
\(\Leftrightarrow x=2\)
b) (x-3).(x2+3x+9)+x.(x+2).(2-x)=1
\(\Leftrightarrow x^3-3^3+x\left(4-x^2\right)-1=0\)
\(\Leftrightarrow x^3-27+4x-x^3-1=0\)
\(\Leftrightarrow4x-28=0\)
\(\Leftrightarrow4x=28\)
\(\Leftrightarrow x=7\)
TL
0
\(\left(5x+1\right)^2-\left(5x+3\right)\left(5x-3\right)=30\)
\(25x^2+10x+1-\left(25x^2-9\right)=30\)
\(25x^2+10x+1-25x^2+9=30\)
\(10x+10=30\)
\(10x=20\)
\(x=2\)
T_T tôi chưa hc lp 8