Bài 1: tìm x
1, (3x-1)(2x+7)-(x+1)(6x-5)=10
Cái chỗ 10 đấy mk ko biết nóa là 10 hay 16
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\(e,x^2-y^2+2x+1=\left(x^2+2x+1\right)-y^2\)
\(=\left(x+1\right)^2-y^2=\left(x+1-y\right)\left(x+1+y\right)\)
\(f,x^3+2x^2+2x+1=\left(x^3+1\right)+\left(2x^2+2x\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1+2x\right)\)
\(=\left(x+1\right)\left(x^2+x+1\right)\)
\(x^2-y^2+2x+1\)
\(=\left(x^2+2x+1\right)-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x-y+1\right)\left(x+y+1\right)\)
hk tốt
^^
\(5\left(2x-1\right)^2+4\left(x-1\right)\left(x+3\right)-2\left(5-3x\right)^2\)
\(=5\left(4x^2-4x+1\right)+4\left(x^2+3x-x-3\right)-2\left(25-30x+9x^2\right)\)
\(=20x^2-20x+5+4x^2+12x-4x-12-50+60x-18x^2\)
\(=6x^2+48x-57\)
\(5\left(2x-1\right)^2+4\left(x-1\right)\left(x+3\right)-2\left(5-3x\right)^2\)
\(=5\left(4x^2-4x+1\right)+4\left(x-1\right)\left(x+3\right)-2\left(5-3x\right)^2\)
\(=5\left(4x^2-4x+1\right)+4\left(x-1\right)\left(x+3\right)-2\left(25-3x+9x^2\right)\)
\(=20x^2-20x+5+4\left(x-1\right)\left(x+3\right)-2\left(25-30x+9x^2\right)\)
\(=20x^2-20x+5+4x^2+8x-12-50+60x-18x^2\)
\(=6x^2+48x-57\)
\(\left(ab+1\right)^2-\left(a+b\right)^2=\left(ab+1+a+b\right)\left(ab+1-a-b\right).\)
\(=\left(a+1\right)\left(b+1\right)\left(a-1\right)\left(b-1\right)\)
\\(x^2-2x-4y^2-4y=\left(x^2-2x+1\right)-\left(4y^2+4y+1\right)=\left(x-1\right)^2-\left(2y+1\right)^2\)
\(=\left(x+1+2y+1\right)\left(x+1-2y-1\right)=\left(x+2y+2\right)\left(x-2y\right)\)
Dùng hằng đẳng thức là xong
a, \(\left(x+y\right)^3-x^3-y^3=x^3+3x^2y+3xy^2+y^3-x^3-y^3\)
\(=3x^2y+3xy^2=3xy\left(x+y\right)\)
b, \(x^2+6xy+9y^2=\left(x+3y\right)^2\)
\(x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\)
Câu này dễ mà
Nhân hết ra là xong, ez ghê
\(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=10\)
\(\Leftrightarrow6x^2+21x-2x-7-6x^2+5x-6x+5=10\)
\(\Leftrightarrow18x-2=10\Leftrightarrow x=\frac{2}{3}\)