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\(x^2+5x-6=x^2-x+6x-6=x\left(x-1\right)+6\left(x-1\right)=\left(x+6\right)\left(x-1\right)\)
\(5x^2+5xy-x-y=5x\left(x+y\right)-\left(x+y\right)=\left(5x-1\right)\left(x+y\right)\)
\(7x-6x^2-2=-\left(6x^2-7x+2\right)=-\left[\left(6x^2-3x\right)-\left(4x+2\right)\right]=-\left[3x\left(2x-1\right)-2\left(2x-1\right)\right]=-\left[\left(3x-2\right)\left(2x-1\right)\right]\)
d) \(x^2+4x+3=x^2+x+3x+3=x\left(x+1\right)+3\left(x+1\right)=\left(x+3\right)\left(x+1\right)\)
nhiều quá, các bn ngại làm, chia nhỏ ra,mk làm cho 2 câu
a) x2 +5x -6 = x2 -x +x + 5x -6
= x2 -x +6x -6
= x( x-1) + 6(x-1) = (x-1)(x+6)
b) 5x2 +5xy -x-y = 5x(x+y) -(x+y)
= (x+y)(5x-1)
c) 7x -6x2 -2 = 6( x+2/3)(x+1/2)
d) x2 +4x +3 = x2 +x +3x +3
= x(x+1) + 3(x+1)
= (x+1)(x+3)
a)\(=\left(x-2\right)\left(5x-3x^2\right)\)
\(=x\left(x-2\right)\left(5-3x\right)\)
b)\(=\left(4x-x^2-4\right)\left(4x+x^2+4\right)\)
\(=-\left(x^2-4x+4\right)\left(x+2\right)^2\)
=\(-\left(x-2\right)^2\left(x+2\right)^2\)
c)\(\left(=5x^2-5xy\right)+\left(7y-7x\right)\)
\(=-5x\left(y-x\right)+7\left(y-x\right)\)
\(=\left(y-x\right)\left(7-5x\right)\)
d)\(=\left(3x^2-6x\right)-\left(2x-4\right)\)
\(=3x\left(x-2\right)-2\left(x-2\right)\)
\(=\left(x-2\right)\left(3x-2\right)\)
e)\(=[\left(x^2\right)^2+8^2+2.x^2.8]-2.8.x^2\)
\(=\left(x^2+8\right)^2-16x^2\)
\(=\left(x^2+8-4x\right)\left(x^2+8+4x\right)\)
__Chúc bạn hc tốt ......
f: Ta có: \(16x^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\)
a: \(\Leftrightarrow\dfrac{3x-2}{\left(x-2\right)\left(x-10\right)}-\dfrac{4x+3}{\left(x+8\right)\left(x-2\right)}=\dfrac{8x+11}{\left(x-10\right)\left(x+8\right)}\)
=>(3x-2)(x+8)-(4x+3)(x-10)=(8x+11)(x-2)
=>3x^2+24x-2x-16-4x^2+40x-3x+30=8x^2-16x+11x-22
=>-x^2+59x+14-8x^2+5x+22=0
=>-9x^2+54x+36=0
=>x^2-6x-4=0
=>\(x=3\pm\sqrt{13}\)
b: \(\Leftrightarrow\dfrac{2x-5}{\left(x+9\right)\left(x-4\right)}-\dfrac{x-6}{\left(x+7\right)\left(x-4\right)}=\dfrac{x+8}{\left(x+9\right)\left(x+7\right)}\)
=>(2x-5)(x+7)-(x-6)(x+9)=(x+8)(x-4)
=>2x^2+14x-5x-35-x^2-9x+6x+54=x^2+4x-32
=>x^2+6x+19=x^2+4x-32
=>2x=-51
=>x=-51/2
câu 1/ 5x(\(4x^2\)-2x+1) - 2x(\(10x^2\)-5x-2)
= 5x.\(4x^2\)-5x.2x+ 5x.1 - ( 2x.\(10x^2\)-2x.5x-2x.2)
= 9\(x^3\)-10\(x^2\)+5x - 20\(x^3\)+10\(x^2\)+4x
= (9\(x^3\)-\(20x^3\)) + (-10\(x^2\)+10\(x^2\)) + (5x+4x)
= \(-11x^3\) + 9x
à cj ơi, e 2k6, đọc phần lí thuyết r lm, nên có lỗi sai j mong cj thông cảm
a: 3x-5>15-x
=>4x>20
hay x>5
b: \(3\left(x-2\right)\left(x+2\right)< 3x^2+x\)
=>3x2+x>3x2-12
=>x>-12
a) x(x-5)+2(x-5) = (x-5)(x+2)
b) (x-7)(x-2)
c) (x+2)(x^2+2x+4)+5y(x+2) = (x+2)(x^2+2x+4+5y)
d) (x^2+8)^2 -16x^2 = (x^2+8-4x)(x^2+8+4x)