3/2x^4y^3

tính j v?

tính cái gì ??????

a, \(3x-5\ge0\Leftrightarrow x\ge\frac{5}{3}\)

b, \(\left(2x-6\right)\left(4x-20\right)\ge0\Leftrightarrow8\left(x-3\right)\left(x-5\right)\ge0\)

TH1 : \(\hept{\begin{cases}x-3\ge0\\x-5\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge3\\x\ge5\end{cases}}\Leftrightarrow x\ge5}\)

TH2 : \(\hept{\begin{cases}x-3\le0\\x-5\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le3\\x\le5\end{cases}}\Leftrightarrow x\le3\)

c, \(4-7x\ge0\Leftrightarrow-7x\ge-4\Leftrightarrow x\le\frac{4}{7}\)

d, \(\frac{4-2x}{3x+9}\ge0\Leftrightarrow\frac{2x-4}{3x+9}\le0\)

Vì \(2x-4< 3x+9\)

\(\hept{\begin{cases}2x-4\le0\\3x+9\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le2\\x\ge-3\end{cases}\Leftrightarrow-3\le x\le2}\)

\(x\left(1-3x\right)\left(4-3x\right)-\left(x-4\right)\left(3x+5\right)\)

\(=x\left(4-3x-12x+9x^2\right)-\left(3x^2+5x-12x-20\right)\)

\(=4x-15x^2+9x^3-3x^2+7x+20=9x^3-18x^2+11x+20\)

2.

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

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

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

k) \(\left(x+1\right)^3-\left(3z\right)^3=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)

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

Bạn làm nốt cho mình bài 3 với ạ !