Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
a. ĐKXĐ: x \(\ne\pm3\)
b. M = \(\frac{3}{x-3}+\frac{6x}{x^2-9}+\frac{x}{x+3}\)
= \(\frac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{6x}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
= \(\frac{3x+9+6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}\) = \(\frac{9+6x+x^2}{\left(x-3\right)\left(x+3\right)}\)= \(\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}=\frac{x+3}{x-3}\)
c. M = 0 hay \(\frac{x+3}{x-3}=0\) => x + 3 = 0 <=> x = -3 (Loại)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) C = c + d + 2 ( c − d ) 3 = ( 3 c − d ) 3 .
b) D = m − n ( n + p ) 3 = ( m − 2 n − p ) 3 .
![](https://rs.olm.vn/images/avt/0.png?1311)
a)xm+4+xm+3-x-1
=(xm+4-x)+(xm+3-1)
=x(xm+3-1)+(xm+3-1)
=(x+1)(xm+3-1)
Với x=-2 ta có:... bn tự thay
b)x6-x4+2x3+2x2=x6-2x5+2x4+2x5-4x4+4x3+x4-2x3+2x2
=x4(x2-2x+2)+2x3(x2-2x+2)+x2(x2-2x+2)
=(x4+2x3+x2)(x2-2x+2)
=[x2(x2+2x+1)](x2-2x+2)
=x2(x+1)2(x2-2x+2)
Với x=-2 bn tự thay nhé h mk bận
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(m+n\right)^2+\left(m-n\right)^2+2\left(m+n\right)\left(m-n\right)\)
'\(=\left[\left(m+n\right)+\left(m-n\right)\right]^2=4m^2\)
b) \(\left(y-3\right)\left(y+3\right)\left(y^2+9\right)-\left(y^2+2\right)\left(y^2-2\right)\)
\(=\left(y^2-9\right)\left(y^2+9\right)-\left(y^4-4\right)\)
\(=y^4-81-y^4+4=-77\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x^m+7y^n\right)^2=4x^{2m}+28x^my^n+49y^{2n}\)
\(\left[\left(x-3\right)-z\right]^2=\left(x-3\right)^2-2\left(x-3\right)z+z^2=x^2-6x+9-2xz+6z+z^2\)
\(\left(4a^2-3b^2\right)\left(3b^2+4a^2\right)=\left(4a^2\right)^2-\left(3b^2\right)^2=16a^4-9b^4\)
Tham khảo nhé~
lớp 6 đó nhưng mình cho lớp 8
thách thức nhau là đây chứ đâu -.-