\(a=\left(x-2\right)\left(x+2\right)\left(x+3\right)-\left(x+1\right)^3=\left(x^2-4\right)\left(x+3\right)-\left(x^3+3x^2+3x+1\right)=x^3+3x^2-4x-12-x^3-3x^2-3x-1=-7x-13\)\(b=\left(x+2\right)^3-\left(x-2\right)^3-12x^3=\left(x+2-x+2\right)\left[\left(x-2\right)^2+\left(x-2\right)\left(x+2\right)+\left(x+2\right)^2\right]-12x^3\)\(=4.\left(x^2-2x+1+x^2-4+x^2+2x+1\right)-12x^3=4\left(3x^2-2-3x^3\right)\)

\(\left(4x+1\right)\left(-4x+1\right)-16x\left(-5\right)=17\)

\(\Leftrightarrow\left(1+4x\right)\left(1-4x\right)+80x-17=0\)

\(\Leftrightarrow1-16x^2+80x-17=0\)

\(\Leftrightarrow-16x^2-16+80x=-16\left(x^2-5x+1\right)=0\Leftrightarrow-16\left[\left(x^2-5x+\dfrac{25}{4}\right)-\dfrac{21}{4}\right]=0\Leftrightarrow-16\left(x-\dfrac{5}{2}\right)^2+84=0\Rightarrow\left(x-\dfrac{5}{2}\right)^2=\dfrac{21}{4}\) \(\Rightarrow\left[{}\begin{matrix}x-\dfrac{5}{2}=\sqrt{\dfrac{21}{4}}\\x-\dfrac{5}{2}=-\sqrt{\dfrac{21}{4}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{21}{4}}+\dfrac{5}{2}\\x=-\sqrt{\dfrac{21}{4}}+\dfrac{5}{2}\end{matrix}\right.\)

olm -_-3 câu được -150đ ! Nản r

\(x^2+2x+1-4x^2=\left(x+1\right)^2-\left(2x\right)^2=\left(x+1+2x\right)\left(x+1-2x\right)=\left(3x+1\right)\left(1-x\right)\)

\(\left(2016x-2017y\right)-\left(2016x-2018z\right)+\left(2017y-2018z\right)=2018\Leftrightarrow2016x-2017y-2016x+2018z+2017y-2018z=2018\Leftrightarrow0=2018\)(vô lí )

Vậy : không tìm được các số nguyên x , y , z ...

Vì tam giác ABC đồng dạng với tam giác DEF\(\Rightarrow\dfrac{AB}{DE}=\dfrac{BC}{EF}=\dfrac{AC}{DF}\Leftrightarrow\dfrac{4}{3}=\dfrac{20}{EF}=\dfrac{AC}{DF}\Rightarrow EF=\dfrac{3.20}{4}=15cm\)Ta có :

\(\dfrac{AC}{DF}=\dfrac{4}{3}\) và \(AC-DF=6\Rightarrow AC=\dfrac{6}{4-3}.4=24cm;DF=24-6=18cm\)

a)

x(x^2-9)=0

x(x^2-3^2)=0

x(x-3)(x+3)

b) x^2-6x+x-6=0

x(x-6)+(x-6)=0

(x-6)(x+1)=0

Ta có :

\(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3+c^3-3abc-3a^2b-3ab^2\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\) Thay vào biểu thức ta được:

\(A=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)}{a^2+b^2+c^2-ab-bc-ac}=a+b+c\)

\(1,\) \(Al_2\left(SO_4\right)_3\) + \(6AgNO_3\) --> \(2Al\left(NO_3\right)_3\) + \(3Ag_2SO_4\)

Số phân tử \(Al_2\left(SO_4\right)_3\) : Số phân tử \(AgNO_3\) : Số phân tử \(Al\left(NO_3\right)_3\): số phân tử \(Ag_2SO_4\) = 1:6:2:3

\(a,x^3-6x^2y+12xy^2-8x^3=\left(x-2y\right)^3\)

\(b,x^2+2x+1-4x^2=\left(x+1\right)^2-\left(2x\right)^2=\left(x+1+2x\right)\left(x+1-2x\right)=\left(3x+1\right)\left(1-x\right)\)