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\(x\left(y+z\right)-y\left(x-z\right)=xy+xz-yx+yz\)
\(=xy-xy+\left(zx+zy\right)\)
\(=\left(x+y\right)z\)
b, \(\left(m-n\right)\left(m+n\right)=m^2+mn-nm-n^2\)
\(=m^2-n^2\)
a, \(a\left(b+c\right)-b\left(a-c\right)\)
\(=ab+ac-\left(ab-bc\right)\)
\(=ab+ac-ab+bc\)
\(=ac+bc\)
\(=\left(a+b\right)c\)
b,\(\left(a+b\right)\left(a-b\right)\)
\(=\left(aa+ab\right)-\left(ab+bb\right)\)
\(=aa+ab-ab-bb\)
\(=aa-bb\)
\(=a^2-b^2\)
VT=(a+b)(a-b)=a(a-b)+b(a-b)=a2-ab+ab-b2=a2-b2
ta có: VT=VP=>đpcm
VT = a + b 2 = a + b . a + b = a ( a + b ) + b ( a + b ) = a 2 + ab + ba + b 2 = a 2 + 2 ab + b 2 = VP ( dpcm )
VT = a − b . a + b = a ( a + b ) − b ( a + b ) = a 2 + ab − ba − b 2 = a 2 − b 2 = VP ( dpcm )
VT = ( a + b ) ( a − b ) = a 2 − a . b + b . a − b 2 = a 2 − b 2 + ab − ab = a 2 − b 2 + 0 = a 2 − b 2 = VP . Vậy ( a + b ) ( a − b ) = a 2 − b 2
\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)
\(=\left(x^2+x\right)\left(x+2\right)\left(x+3\right)+1\)
\(=\left(x^3+x^2+2x^2+2x\right)\left(x+3\right)+1\)
\(=\left(x^3+3x^2+2x\right)\left(x+3\right)+1\)
\(=x^4+3x^3+2x^2+3x^3+9x^2+6x+1\)
\(=x^4+\left(3x^3+3x^3\right)+\left(2x^2+9x^2\right)+6x+1\)
\(=x^4+6x^3+11x^2+6x+1\)
\(=\left(x^2+3x+1\right)^2\) (Bằng vế phải)
\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)
\(=\left[x\left(x+3\right)\right]\left[\left(x+1\right)\left(x+2\right)\right]+1\)
\(=\left(x^2+3x\right)\left(x^2+2x+x+2\right)+1\)
\(=\left(x^2+3x+1-1\right)\left(x^2+3x+1+1\right)+1\)
\(=\left(x^2+3x+1\right)^2-1^2+1\)
\(=\left(x^2+3x+x\right)^2\)