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a,(5x-2y)(x2-xy+1)=5x3-5x2+5x-2yx2+2xy2-2y
=5x3-7x2y+2xy2+5x-2y
b,(x-2)(x+2)(\(\dfrac{1}{2}\) x-5)=x2-4.\(\left(\dfrac{1}{2}x-5\right)\)
=\(\dfrac{1}{2}x^3-5x^2-2x+20\)
c,\(\left(x^2-2x+3\right)\left(\dfrac{1}{2}x-5\right)\)
=\(\dfrac{1}{2}x^3-5x^2-1x^2+10x+\dfrac{3}{2}x-15\)
=\(\dfrac{1}{2}x^3-6x^2+\dfrac{23}{2}x-15\)
d,\(\left(x^2-5\right)\left(x+3\right)+\left(x+4\right)\left(x-x^2\right)\)
=\(x^3+3x^2-5x-15+x^2-x^3+4x-4x^2\)
=\(-5x+4x-15\)
=\(-x-15\)
Chúc bạn học tốt(mỏi tay quá)
\(a,\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\)
\(=x^2+2x+1-\left(x^2-2x+1\right)-3\left(x^2-1\right)\)
\(=x^2+2x+1-x^2+2x-1-3x^2+2=-3x^2+4x+2\)\(b,5\left(x+2\right)\left(x-2\right)-\left(2x-3\right)^2-x^2+17\)
\(=5\left(x^2-4\right)-\left(4x^2-12x+9\right)-x^2+17\)
\(=5x^2-20-4x^2+12x-9-x^2+17=12x-12\)
\(S=\dfrac{\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc}{2a^2+2b^2+2c^2-2ab-2bc-2ac}\)
\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)}{2a^2+2b^2+2c^2-2ab-2bc-2ac}\)
\(=\dfrac{3\cdot\left(2a^2+2b^2+2c^2-2ab-2bc-2ac\right)\cdot\dfrac{1}{2}}{2a^2+2b^2+2c^2-2ab-2bc-2ac}=\dfrac{3}{2}\)
\(A=\left|2x-5\right|.\left(4-\left|2x-5\right|\right)\\ A=4.\left|2x-5\right|-\left|2x-5\right|^2\\ A=-\left(\left|2x-5\right|^2-4.\left|2x-5\right|+4\right)+4\\ A=-\left(\left|2x-5\right|-2\right)^2+4\)
Vì \(\left(\left|2x+5\right|-2\right)^2\ge0\) với mọi x nên \(-\left(\left|2x+5\right|-2\right)^2\le0\)
=> \(-\left(\left|2x+5\right|-2\right)^2+4\le4\) hay \(A\le4\)
Dấu ''='' xảy ra khi và chỉ khi \(\left(\left|2x+5\right|-2\right)^2=0\)
=> \(\left|2x-5\right|=2\)
=> \(\left[\begin{array}{nghiempt}2x-5=2\\2x-5=-2\end{array}\right.\) => \(\left[\begin{array}{nghiempt}2x=7\\2x=3\end{array}\right.\) => \(\left[\begin{array}{nghiempt}x=3,5\\x=1,5\end{array}\right.\)
Chúc bạn làm bài tốt