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\(48-4y^2-4y=-\left(4y^2+4y-48\right)\)
\(=-\left[\left(2y\right)^2+2.2y+1-49\right]\)
\(=-\left[\left(2y+1\right)^2-7^2\right]\)
\(=-\left(2y-6\right)\left(2y+8\right)\)
\(48-4y^2-4y\)
\(=-\left(4y^2+4y-48\right)\)
\(=-\left(4y^2+4y+1-49\right)\)
\(=-\left[\left(2y+1\right)^2-7^2\right]\)
\(=-\left(2y+1-7\right)\left(2y+1+7\right)\)
\(=-\left(2y-6\right)\left(2y+8\right)\)
\(=-4\left(y-3\right)\left(y+4\right)\)
- (5 + 4y) (5 - 4y) = - [52 - (4y)2] = - (25 - 16y2) = - 25 + 16y2 = 16y2 - 25
\(z\left(3x-5z\right)-x\left(5z-3x\right)\)z(3x-5z)-x(5z-3x)
=z(3x-5z)+x(3x-5z)
=(z+x)(3x-5z)
\(xy+3x-4y-12=0\)
\(\Leftrightarrow x\left(y+3\right)-4\left(y+3\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(y+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\y=-3\end{matrix}\right.\)
Bài 1:
a. \(=[(3x+(4y-5z)][3x-(4y-5z)]=(3x)^2-(4y-5z)^2\)
\(=9x^2-(16y^2-40yz+25z^2)=9x^2-16y^2+40yz-25z^2\)
b.
\(=(3a-1)^2+2(3a-1)(3a+1)+(3a+1)^2=[(3a-1)+(3a+1)]^2=(6a)^2=36a^2\)
Bài 2:
\((x+y+z)^3=[(x+y)+z]^3=(x+y)^3+3(x+y)^2z+3(x+y)z^2+z^3\)
\(=[x^3+y^3+3xy(x+y)]+3(x+y)z(x+y+z)+z^3\)
\(=x^3+y^3+z^3+3xy(x+y)+3(x+y)z(x+y+z)\)
\(=x^3+y^3+z^3+3(x+y)(xy+zx+zy+z^2)\)
\(=x^3+y^3+z^3+3(x+y)(z+x)(z+y)\) (đpcm)
\(=\left[3x+\left(4y-5z\right)\right]\left[3x-\left(4y-5z\right)\right]\)
Giờ bạn đã thấy hằng đẳng thức rồi chứ ạ?