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a) \(\left(3x+2\right).\left(x-3\right)-3x.\left(x+\frac{1}{3}\right)\)
\(=3x^2-9x+2x-6-\left(3x^2+x\right)\)
\(=3x^2-9x+2x-6-3x^2-x\)
\(=\left(3x^2-3x^2\right)+\left(-9x+2x-x\right)-6\)
\(=-8x-6.\)
Chúc bạn học tốt!
\(B=\left(3x-2\right)^2-\left(x+2\right).\left(x-2\right)\)
\(=\left(3x-2\right)^2-\left(x^2-2^2\right)\)
\(=9x^2-12x+4-x^2+4\)
\(=8x-12x+8\)
\(C=\left(x+4\right)^2-7x.\left(x-2\right)\)
\(=x^2+8x+16-\left(7x^2-14x\right)\)
\(=x^2+8x+16-7x^2+14x\)
\(=-6x^2+22x+16\)
\(D=-4x.\left(2x-7\right)+\left(x+5\right)^2\)
\(=-8x^2+28x+x^2+10x+25\)
\(=-7x^2+38x+25\)
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c: \(\Leftrightarrow2x^3-6x^2+4x+x^2-3x+2+a-2⋮x^2-3x+2\)
=>a-2=0
=>a=2
d: \(\dfrac{5x^3+4x^2-6x-a}{5x-1}=\dfrac{5x^3-x^2+5x^2-x-5x+1-a-1}{5x-1}\)
\(=x^2+x-1+\dfrac{-a-1}{5x-1}\)
Để dư bằng -3 thì -a-1=-3
=>a+1=3
=>a=2
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a) \(\left(x-3y\right)\left(x+3y\right)=x^2-9y^2\)
b) \(\left(3-y\right)^2=9-6y+y^2\)
c) \(\left(3x+2y^2\right)^3=27x^3+54x^2y^2+36xy^4+8y^6\)
d) \(\left(x+y\right)^3+\left(x-y\right)^3\)
\(=\left(x+y+x-y\right)\left[\left(x+y\right)^2-\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)
Mình ko hiểu đề cho lắm tính cái gì ?? nên mình chỉ giải hằng đẳng thức
a) \(\left(x-3y\right)\left(x+3y\right)=x^2-\left(3y\right)^2=x^2-9y^2\)
b) \(\left(3-y\right)^2=3^2-2.3y+y^2=9-6y+y^2\)
c) \(\left(3x+2y^2\right)^3=\left(3x+2y^2\right)\left[\left(3x\right)^2-3x.2y^2+\left(2y^2\right)^2\right]\)
\(=\left(3x+2y^2\right)\left[9x^2-6xy^2+4y^4\right]\)
\(=8y^6+36xy^4+54x^2y^2+27x^3\)
d) \(\left(x+y\right)^3+\left(x-y\right)^3\)
\(=\left(x+y+x-y\right)\left[\left(x+y\right)^2-\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)
\(=2x\left(x^2+2xy+y^2-x^2+y^2+x^2-2xy+y^2\right)\)
\(=2x\left(x^2+3y^2\right)=2x^3+6xy^2\)