Phạm

Gì vậy mẹ

cẹp cẹp

\(a.\left(2x-1\right)^2-2\left(2x-1\right)\left(x+3\right)+\left(x+3\right)^2\)

\(=4x^2-4x+1-4x^2-10x+6+x^2+6x+9\)

\(=x^2-8x+16.\)

\(b.\left(4x+3\right)^2-2x\left(x+6\right)-5\left(x-2\right)\left(x+2\right)-10\)

\(=16x^2+24x+9-2x^2-12x-5x^2+20-10\)

\(=9x^2+12x+19.\)

\(c.\left(3x+1\right)\left(9x^2-3x+1\right)-\left(x-2\right)\left(x^2+2x+4\right)\)

\(=27x^3+1-x^3+8=26x^3+9.\)

\(d.\left(2x-1\right)^3-4x^2\left(2x-3\right)\)

\(=8x^3-12x^2+6x-1-8x^2+12x^2=6x-1.\)

\(a.8x^3+\dfrac{1}{64}=\left(2x\right)^3+\left(\sqrt[3]{\dfrac{1}{64}}\right)^3=\left(2x+\dfrac{1}{4}\right)\left(2^2x^2-\dfrac{1}{4}\cdot2x+\left(\dfrac{1}{4}\right)^2\right)\)

\(=\left(2x+\dfrac{1}{4}\right)\left(4x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right).\)

\(b.-x^2-16+8x=-x^2+8x-16=-\left(x^2-8x+16\right)\)

\(=-\left(x-4\right)\left(x-4\right)=-\left(x-4\right)^2.\)

\(c.\left(2x+y\right)^3-\left(2x-y\right)^3\)

\(=2y\left[\left(2x+y\right)^2+\left(2x+y\right)\left(2x-y\right)+\left(2x-y\right)^2\right]\)

\(=2y\left(12x^2+y^2\right).\)

\(d.x^3-6x^2+12x-8=x^3-3\cdot x^2\cdot2+3\cdot x\cdot2^2-2^3\)

\(=\left(x-2\right)^3.\)

\(e.\dfrac{1}{4}x^2-\dfrac{1}{9}y^2=\left(\dfrac{1}{2}x\right)^2-\left(\dfrac{1}{3}y\right)^2=\left(\dfrac{1}{2}x+\dfrac{1}{3}y\right)\left(\dfrac{1}{2}x-\dfrac{1}{3}y\right).\)

\(a.\)

\(14\) \(chiahếtcho7\)

\(98\) \(chiahếtcho7\)

\(\Rightarrow14+98\) \(chiahếtcho7\)

\(b.\)

\(28\) \(chiahếtcho7\)

\(42\) \(chiahếtcho7\)

\(210\) \(chiahếtcho7\)

\(\Rightarrow28+42+210\) \(chaihếtcho7.\)

\(c.\)

\(35\) \(chiahếtcho7\)

\(25\) \(kochiahếtcho7\)

\(280\) \(chiahếtcho7\)

\(\Rightarrow28-25+280\) \(kochiahếtcho7\)

Câu d. thì anh ko biết cách làm nên xin lỗi nhé