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

\(12x-9-4x^2=-\left(2x-3\right)^2\\ Sửa:x^3-6x^2y+12xy^2-8y^3=\left(x-2y\right)^3\)

nếu bạn đăng nhiều lần thì cứ xóa đi anh

cái này thì giáo viên toán đã đề cập từ lâu rồi ạ

\(a,PT\Leftrightarrow\left(1-2m\right)x=m+4\)

Bậc nhất \(\Leftrightarrow1-2m\ne0\Leftrightarrow m\ne\dfrac{1}{2}\)

\(b,x=2\Leftrightarrow2-4m-m-4=0\Leftrightarrow m=-\dfrac{2}{5}\\ c,m=5\Leftrightarrow-9x-9=0\Leftrightarrow x=-1\)

Ta có \(\dfrac{1}{3^3}< \dfrac{1}{2.3.4}=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)\)

\(\dfrac{1}{5^3}< \dfrac{1}{4.5.6}=\dfrac{1}{2}\left(\dfrac{1}{4.5}-\dfrac{1}{5.6}\right)\\ ...\\ \dfrac{1}{2021^3}< \dfrac{1}{2020.2021.2022}=\dfrac{1}{2}\left(\dfrac{1}{2020.2021}-\dfrac{1}{2021.2022}\right)\)

Cộng VTV ta được

\(VT< \dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{2021.2022}\right)=\dfrac{1}{12}-\dfrac{1}{2\left(2021.2022\right)}< \dfrac{1}{12}\)

\(a,=\left(6x+1-6x+1\right)^2=2^2=4\\ b,=3x^2-6x-5x+5x^2-8x^2+24=24-11x\\ c,=14x^2+7x-6x-3-5x^2-20x+2x+8-9x^2+17x=5\\ d,=6x^2+48x-5x-40-6x^2-9x+2x+3-36x+27=-10\\ e,=-3x\left(x^2+4x+4\right)+\left(x+3\right)\left(x^2-1\right)-4x^2+12x-9\\ =-3x^3-12x^2-4x+x^3-x+3x^2-3x-4x^2+12x-9\\ =-2x^3-13x^2+4x-9\\ f,=2x\left(x^2-8x+16\right)-\left(x+5\right)\left(x^2-4\right)+2\left(x^2-10x+25\right)-x^2+2x-1\\ =2x^3-16x^2+32x-x^3+4x-5x^2+20+2x^2-20x+50-x^2+2x-1\\ =x^3-20x^2+18x+69\)

Gọi số hs khối 6,7,8 lần lượt là a,b,c∈N*

Áp dụng tc dtsbn:

\(2a=3b=4c\Rightarrow\dfrac{a}{6}=\dfrac{b}{4}=\dfrac{c}{3}=\dfrac{a+b+c}{6+4+3}=\dfrac{130}{13}=10\\ \Rightarrow\left\{{}\begin{matrix}a=60\\b=40\\c=30\end{matrix}\right.\)

Vậy ...

\(g,=x^3-8-x^3+3x^2-3x+1+7=3x^2-3x\\ h,=x\left(x^2-4\right)+x^3+27=x^3-4x+x^3+27=2x^3-4x+27\\ i,=27x^3+54x^2+36x+8-54x^2-36x+x^3-3x^2+3x-1-28x^3+3x^2-3x=7\)

sai bét