đùa thoii, bn này cụ mik=))

#323

`->` bạn này có ghệ r, là mình nha='))

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

`= x^4 -x^3y+x^2y^2+xy^3 +x^3y-x^2y^2 +xy^3 +y^4`

`= (-x^3y +x^3y) +(x^2y^2-x^2y^2) +(xy^3 +xy^3) +x^4+y^4`

`= 2xy^3+x^4+y^4`

Đề có sai ko c...

`6A`

\(a,\left(x-2,5\right)^2=9\\ \Rightarrow\left(x-2,5\right)^2=3^2\\ \Rightarrow\left[{}\begin{matrix}x-2,5=3\\x-2,5=-3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5,5\\x=-0,5\end{matrix}\right.\)

\(b,\left(1-x\right)^3=-8\\ \Rightarrow\left(1-x\right)^3=-2^3\\ \Rightarrow1-x=-2\\ \Rightarrow x=1+2\\ \Rightarrow x=3\\ c,2^{4-x}=16\\ \Rightarrow2^{4-x}=2^4\\ \Rightarrow4-x=4\\ \Rightarrow x=4-4\\ \Rightarrow x=0\)

Bài `6B`

\(a,\left(5+x\right)^3=-27\\ \Rightarrow\left(5+x\right)^3=-3^3\\ \Rightarrow5+x=-3\\ \Rightarrow x=-3-5\\ \Rightarrow x=-8\)

\(b,\left(1-x\right)^2=\dfrac{1}{9}\\ \Rightarrow\left(1-x\right)^2=\left(\dfrac{1}{3}\right)^2\\ \Rightarrow\left[{}\begin{matrix}1-x=\dfrac{1}{3}\\1-x=-\dfrac{1}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)

\(c,5^{3+x}=125\\ \Rightarrow5^{3+x}=5^2\\ \Rightarrow3+x=2\\ \Rightarrow x=2-3\\ \Rightarrow x=-1\)

c tách ra nhé^^

\(3\dfrac{1}{2}:y=\dfrac{8}{5}\\ \dfrac{7}{2}:y=\dfrac{8}{5}\\ y=\dfrac{7}{2}:\dfrac{8}{5}\\ y=\dfrac{7}{2}\times\dfrac{5}{8}\\ y=\dfrac{35}{16}\)

Vậy `y=35/16`

\(\left|\dfrac{1}{2}x\right|=3-2x\\ \Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=3-2x\\\dfrac{1}{2}x=-3+2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x+2x=3\\\dfrac{1}{2}x-2x=-3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\dfrac{5}{2}x=3\\-\dfrac{3}{2}x=-3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=2\end{matrix}\right.\)

`8^n : 4=128`

`=> 8^n= 128*4`

`=> 8^n=512`

`=> 8^n=8^3`

`=>n=3`

Vậy `n=3`