a)

\(-x^2+2x-3=-\left(x^2-2x+1\right)-2\\ =-\left(x-1\right)^2-2\le-2< 0\)

vậy\(2x-x^2-3< 0\)

32,64,96

\(a.\:\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\\ 4x^2+12x+9-4x^2+4=49\\ 12x=49-9\\ x=\dfrac{40}{12}=\dfrac{10}{3}\)

\(a.\:3x-15^7=1\\ 3x=1+15^7\\ x=\dfrac{1+15^7}{3}\approx56953125\)

\(b.\:4^{2x}=64\\ 4^{2x}=4^3\\ \Rightarrow2x=3\\ x=\dfrac{3}{2}\)

\(c.\:x^5-x^3=0\\ x^3\left(x^2-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^3=0\\x^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)

đưa đề đi

\(c.\:\left(3x+4\right)^2-\left(3x+1\right)\left(3x-1\right)\\ =9x^2+24x+16-9x^2+1\\ 40x=-1\\ x=-\dfrac{1}{40}\)

\(d.\:\left(3x-1\right)^2-\left(3x-2\right)^2=0\\ \left(3x-1+3x-2\right)\left(3x-1-3x+2\right)=0\\ \left(6x-3\right)=0\\ x=\dfrac{1}{2}\)

\(g.\:\left(2x+1\right)^2-\left(x-1\right)^2=0\\ \left(2x+1+x-1\right)\left(2x+1-x+1\right)=0\\ 3x\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

câu 9: k=-3

vì có 1 đẳng thức này nè: \(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

câu 10 thì tui ko bt

Bạn xét chữ số tận cùng   

a, \(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{524}+1+\frac{x+329}{5}+\frac{20}{5}-4=0\)

\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)

=> x+329=0  => x= -329

b. tương tụ

c,  x=0, x=4