b: \(2x^2-7xy+3y^2+x-3y\)

\(=2x^2-6xy-xy+3y^2+x-3y\)

\(=2x\left(x-3y\right)-y\left(x-3y\right)+\left(x-3y\right)\)

\(=\left(x-3y\right)\left(2x-y+1\right)\)

Lời giải:
a.

Đặt $2a^2+5ab-3b^2-7b-2=(a+mb+n)(2a+pb+k)$ với $m,n,p,k$ nguyên 

$\Leftrightarrow 2a^2+5ab-3b^2-7b-2=2a^2+ab(2m+p)+mpb^2+a(k+2n)+b(km+np)+kn$ 

Đồng nhất hệ số:

\(\left\{\begin{matrix} 2m+p=5\\ mp=-3\\ k+2n=0\\ km+np=-7\\ kn=-2\end{matrix}\right.\)

Giải hpt này ta thu được $m=3; n=1; p=-1; k=-2$

Vậy $2a^2+5ab-3b^2-7b-2=(a+3b+1)(2a-b-2)$

b. Đa thức không phân tích được thành nhân tử

lm theo pp đồng nhất hệ số ạ

b: Ta có: \(2x^2-7xy+3y^2+x-3y\)

\(=2x^2-6xy-xy+3y^2+x-3y\)

\(=2x\left(x-3y\right)-y\left(x-3y\right)+\left(x-3y\right)\)

\(=\left(x-3y\right)\left(2x-y+1\right)\)

somebody help me 

\(1,2x^2-3x-2\) 

\(=2x^2-4x+x-2\)

\(=2x\left(x-2\right)+\left(x-2\right)\) 

\(=\left(2x+1\right)\left(x-2\right)\) 

\(2,4x^2-7x-2\)

\(=4x^2-8x+x-2\) 

\(=4x\left(x-2\right)+x-2\)

\(\left(4x+1\right)\left(x-2\right)\)

a: =4x^2+8x-3x-6

=4x(x+2)-3(x+2)

=(x+2)(4x-3)

b: =3(3x^2-2x-1)

=3(3x^2-3x+x-1)

=3(x-1)(3x+1)

c: =2x^2-4x+x-2

=2x(x-2)+(x-2)

=(x-2)(2x+1)

d: =3x^2+3x-2x-2

=3x(x+1)-2(x+1)

=(x+1)(3x-2)

e: =3x^2+9x+x+3

=3x(x+3)+(x+3)

=(x+3)(3x+1)

a) \(4x^2+5x-6\)

\(=4x^2+8x-3x-6\)

\(=\left(4x^2+8x\right)-\left(3x+6\right)\)

\(=4x\left(x+2\right)-3\left(x+2\right)\)

\(=\left(x+2\right)\left(4x-3\right)\)

b) \(9x^2-6x-3\)

\(=3\left(3x^2-2x-1\right)\)

\(=3\left(3x^2-3x+x-1\right)\)

\(=3\left[3x\left(x-1\right)+\left(x-1\right)\right]\)

\(=3\left(x-1\right)\left(3x+1\right)\)

c) \(2x^2-3x-2\)

\(=2x^2-4x+x-2\)

\(=\left(2x^2-4x\right)+\left(x-2\right)\)

\(=2x\left(x-2\right)+\left(x-2\right)\)

\(=\left(2x+1\right)\left(x-2\right)\)

d) \(3x^2+x-2\)

\(=3x^2+3x-2x-2\)

\(=\left(3x^2+3x\right)-\left(2x+2\right)\)

\(=3x\left(x+1\right)-2\left(x+1\right)\)

\(=\left(x+1\right)\left(3x-2\right)\)

e) \(3x^2+10x+3\)

\(=3x^2+9x+x+3\)

\(=3x\left(x+3\right)+\left(x+3\right)\)

\(=\left(x+3\right)\left(3x+1\right)\)

\(A=\left(2x-1\right)^2+9\ge9\\ A_{min}=9\Leftrightarrow x=\dfrac{1}{2}\\ B=2\left(x^2-2\cdot\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{1}{8}=2\left(x-\dfrac{3}{4}\right)^2+\dfrac{1}{8}\ge\dfrac{1}{8}\\ B_{min}=\dfrac{1}{8}\Leftrightarrow x=\dfrac{3}{4}\\ C=\left(4x^2+4xy+y^2\right)+2\left(2x+y\right)+1+\left(y^2+4y+4\right)-4\\ C=\left[\left(2x+y\right)^2+2\left(2x+y\right)+1\right]+\left(y+2\right)^2-4\\ C=\left(2x+y+1\right)^2+\left(y+2\right)^2-4\ge-4\\ C_{min}=-4\Leftrightarrow\left\{{}\begin{matrix}2x=-1-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-2\end{matrix}\right.\)

\(D=\left(3x-1-2x\right)^2=\left(x-1\right)^2\ge0\\ D_{min}=0\Leftrightarrow x=1\\ G=\left(9x^2+6xy+y^2\right)+\left(y^2+4y+4\right)+1\\ G=\left(3x+y\right)^2+\left(y+2\right)^2+1\ge1\\ G_{min}=1\Leftrightarrow\left\{{}\begin{matrix}3x=-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-2\end{matrix}\right.\)

\(H=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(2y^2+4y+2\right)+2\\ H=\left(x-y\right)^2+\left(x+1\right)^2+2\left(y+1\right)^2+2\ge2\\ H_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=-1\\y=-1\end{matrix}\right.\Leftrightarrow x=y=-1\)

Ta luôn có \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\\ \Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\\ \Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz\ge3xy+3yz+3xz\\ \Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\\ \Leftrightarrow\dfrac{3^2}{3}\ge xy+yz+xz\\ \Leftrightarrow K\le3\\ K_{max}=3\Leftrightarrow x=y=z=1\)

a: Ta có: \(x^2-4y^2-2x-4y\)

\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x-2y-2\right)\)

c: Ta có: \(x^3+2x^2y-x-2y\)

\(=x^2\left(x+2y\right)-\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x-1\right)\left(x+1\right)\)

d: Ta có: \(3x^2-3y^2-2\cdot\left(x-y\right)^2\)

\(=3\left(x-y\right)\left(x+y\right)-2\cdot\left(x-y\right)^2\)

\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)

\(=\left(x-y\right)\left(x+5y\right)\)

e: Ta có: \(x^3-4x^2-9x+36\)

\(=x^2\left(x-4\right)-9\left(x-4\right)\)

\(=\left(x-4\right)\left(x-3\right)\left(x+3\right)\)

f: Ta có: \(x^2-y^2-2x-2y\)

\(=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y-2\right)\)

Trong SGK có chỉ đó b

Ko hiểu thì kb vs mik

mik chỉ thêm cho

Chỉ mình với nha