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\(\left(5x+1\right)^2=\left(2x-3\right)^2\)
\(\Rightarrow\left(5x+1\right)^2-\left(2x-3\right)^2=0\)
\(\Rightarrow\left(5x+1+2x-3\right)\left(5x+1-2x+3\right)=0\)
\(\Rightarrow\left(7x-2\right)\left(3x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7x-2=0\\3x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{2}{7}\\x=\frac{-4}{3}\end{cases}}}\)
Vậy.......
\(a,x^3-4x^2+8x-8\)
\(=\left(x^3-8\right)-\left(4x^2-8x\right)\)
\(=\left(x^3-2^3\right)-4x\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+4x+4\right)-4x\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+4x+4-x+2\right)\)
\(=\left(x-2\right)\left(x^2+3x+6\right)\)
\(b,\left(2xy+1\right)^2-\left(2x+y\right)^2\)
\(=\left(2xy+1+2x+y\right)\left(2xy+1-2x-y\right)\)
\(=\left[\left(2xy+2x\right)+\left(y+1\right)\right]\cdot\left[\left(2xy-2x\right)-\left(y-1\right)\right]\)
\(=\left[2x\left(y+1\right)+1\cdot\left(y+1\right)\right]\cdot\left[2x\left(y-1\right)-1\cdot\left(y-1\right)\right]\)
\(=\left[\left(y+1\right)\left(2x+1\right)\right]\cdot\left[\left(y-1\right)\left(2x-1\right)\right]\)
\(=\left(y+1\right)\left(y-1\right)\left(2x-1\right)\left(2x+1\right)\)
\(=\left(y^2-1\right)\left(4x^2-1\right)\)
\(c,1+6x-6x^2-x^3\)
\(=\left(1-x^3\right)-\left(6x^2-6x\right)\)
\(=\left(1^3-x^3\right)-6x\left(x-1\right)\)
\(=\left(1-x\right)\left(1+2x+x^2\right)+6x\left(1-x\right)\)
\(=\left(1-x\right)\left(1+2x+x^2+6x\right)\)
\(=\left(1-x\right)\left(1+8x+x^2\right)\)
\(a,x^2-9-2\left(x+3\right)^2=0\)
\(\Rightarrow\left(x^2-3^2\right)-2\left(x+3\right)^2=0\)
\(\Rightarrow\left(x-3\right)\left(x+3\right)-2\left(x+3\right)\left(x+3\right)=0\)
\(\Rightarrow\left(x+3\right)\left[x-3-2\cdot\left(x+3\right)\right]=0\)
\(\Rightarrow\left(x+3\right)\left[x-3-2x-6\right]=0\)
\(\Rightarrow\left(x+3\right)\left(-x-9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\-x-9=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\-x=9\Rightarrow x=-9\end{cases}}\)
\(b,\left(5x+1\right)^2=\left(2x-3\right)^2\)
\(\Rightarrow\left(5x+1\right)^2\div\left(2x-3\right)^2=0\)
\(\Rightarrow\left[\left(5x+1\right)\div\left(2x-3\right)\right]^2=0\)
\(\Rightarrow\left(5x+1\right)\div\left(2x-3\right)=0\)
\(\Rightarrow5x+1=0\)
\(\Rightarrow x=-\frac{1}{5}\)
Ik mk nha, hôm nay ngày mai, ngày kia mk ik 3 lần lại cho bạn (thành 9 lần)
Nhớ kb với mìn lun nha!! Mk rất vui đc làm quen vs bạn, cảm ơn mn nhìu lắm
\(1a,8x^2y^2-12x^3+6x^2\)
\(=2\left(4x^2y^2-13x^3+3x^2\right)\)
\(b,5x\left(x-y\right)-\left(x-y\right)\)( sai đề hả )
\(=\left(x-y\right)\left(5x-1\right)\)
\(c,4x\left(x-2\right)-\left(2-x\right)^2\)
\(=4x\left(x-2\right)-\left(x-2\right)^2\)
\(=\left(x-2\right)\left(4x-x+2\right)=\left(x-2\right)\left(3x+2\right)\)
\(2,\)\(x\left(x-3\right)-\left(3-x\right)=0\)
\(\Rightarrow x\left(x-3\right)+\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+1=0\end{cases}\Rightarrow\hept{\begin{cases}x=3\\x=-1\end{cases}}}\)
phần b làm theo đề thôi nhé ko biết đầu bài đúng ko
\(5x\left(x-y\right)-\left(y-y\right)\)
\(=5x\left(x-y\right)\)
HA ha ngắn gọn vãi
1a/ x3+x2+x+1=0
x2(x+1).(x+1)=0
=> x2(x+1)=0 x =1
hoặc =>[
x+1=0 x=-1
b/(x+2)2=x+2
x2+2.x.2+22 =x+2
x+x+4x+4=x+2
6x+4=x+2
....
c/(x+1)(6x2+2x)+(x-1)(6x2+2x)=0
x2-12 + (6x2+2x)2=0
=> x2-1 = 0 x=1
hoặc => [
(6x2+2x)2=0 x= 0
a) 4x2-8x=0
(2x)2-2.2.2x+4-4=0
(2x-2)2 =4
2x-2=2
2x =4
x=2
Nhớ k cho mk nha
b) \(\left(5x+1\right)^2=\left(2x-3\right)^2\)
\(\Leftrightarrow\left(5x+1\right)^2-\left(2x-3\right)^2=0\)
\(\Leftrightarrow\left(5x+1-2x+3\right).\left(5x+1+2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+4=0\\7x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=-4\\7x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{4}{3}\\x=\frac{2}{7}\end{matrix}\right.\)
a) x2 - 9 - 2(x+3)2 = 0
⇔ \(x^2-9-2\left(x^2+6x+9\right)=0\)
⇔ \(x^2-9-2x^2-12x-18=0\)
\(\Leftrightarrow-x^2-12x^{ }-27=0\)
⇔ \(x^2+12x+27=0\)
⇔ \(x^2+9x+3x+27=0\)
⇔ x( x + 9 ) + 3 ( x + 9 ) =0
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-9\end{matrix}\right.\)