(5x+1)-(5x-3)(5x+3)=30
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\(\left(5x+1\right)^2-\left(5x+3\right)\left(5x-3\right)=30\)
\(25x^2+10x+1-\left(25x^2-9\right)=30\)
\(25x^2+10x+1-25x^2+9=30\)
\(10x+10=30\)
\(10x=20\)
\(x=2\)
Ta có: (5x+1)^2 - (25x^2 - 9)= 30
(5x+1 - 5x)(5x+1+5x) + 9 = 30
10x +10 = 30
=> x=2
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
\(\Leftrightarrow x^3-3^3+x\left(4-x^2\right)-1=0\)
\(\Leftrightarrow x^3-27+4x-x^3-1=0\)
\(\Leftrightarrow4x-28=0\)
\(\Leftrightarrow4x=28\)
\(\Leftrightarrow x=7\)
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
******************
(x+2)^2= 9
=> (x+2)^2= 3^2=(-3)^2
TH1: x+2=3
=> x=3-2=1
TH2: x+2=-3
=> x=(-3)-2=-5
Bài làm :
\(a,\left(x+2\right)^2-9=0\)
\(\Leftrightarrow\left(x+2\right)^2=9\)
\(\Leftrightarrow\left(x+2\right)^2=3^2\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=3\\x+2=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)
Vậy x = 1 hoặc x = -5 .
\(b,\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)
\(\Leftrightarrow25x^2+10x+1-\left(25x^2-3^2\right)=30\)
\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)
\(\Leftrightarrow\left(25x^2-25x^2\right)+10x=30-9-1\)
\(\Leftrightarrow10x=20\)
\(\Leftrightarrow x=2\)
Vậy x = 2 .
\(c,\left(x-1\right)\left(x^2+x+1\right)+x\left(x+2\right)\left(2-x\right)=5\)
\(\Leftrightarrow x^3+x^2+x-x^2-x-1+\left(x^2+2x\right)\left(2-x\right)=5\)
\(\Leftrightarrow x^3-1+2x^2-x^3+4x-2x^2=5\)
\(\Leftrightarrow\left(x^3-x^3\right)+\left(2x^2-2x^2\right)+4x=5+1\)
\(\Leftrightarrow4x=6\)
\(\Leftrightarrow x=\frac{3}{2}\)
Vậy x = 3/2 .
Học tốt nhé
\(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)
\(=25x^2+1+10x-\left(25x^2-9\right)=30\)
\(25x^2+1+10x-25x^2+9=30\)
\(10x+10=30\)
\(10x=30-10=20\)
\(x=\frac{20}{10}=2\)
a: Ta có: \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)
\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)
\(\Leftrightarrow10x=20\)
hay x=2
b: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)
\(\Leftrightarrow x^3-1-x^3+4x=5\)
\(\Leftrightarrow4x=6\)
hay \(x=\dfrac{3}{2}\)
Hình như bạn ghi sai đề. Đề đúng: \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)
\(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)
\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)
\(\Leftrightarrow10x+10=30\)
\(\Leftrightarrow10x=20\)
\(\Leftrightarrow x=2\)