\(y=\left|x^2+x+2016\right|+\left|x^2+x-6\right|\\ =\left|\left(x^2+x\right)+2016\right|+\left|6-\left(x^2+x\right)\right|\)
Áp dụng bđt: \(\left|A\right|+\left|B\right|\ge\left|A+B\right|\) ta được:
\(y\ge\left|x^2+x+2016+6-x^2-x\right|=2022\)
Vậy min y là 2022 khi \(-3\le x\le2\)
Tìm min \(A=\left|x-2016\right|+\left|x-2017\right|\)
a) Tìm min : \(\left|x-5\right|+\left|x+6\right|\)
b) Tìm max : \(\left|3x-1\right|-\left(3x-1\right)^2\)
Rút gọn : \(\frac{1}{\left(x+y\right)^3}.\left(\frac{1}{x^3}+\frac{1}{y^3}\right)+\frac{3}{\left(x+y\right)^5}\left(\frac{1}{x^2}+\frac{1}{y^2}\right)+\frac{6}{\left(x+y\right)^5}\left(\frac{1}{x}+\frac{1}{y}\right)\)
Tìm x :
a ) \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6.\)
b ) \(4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)
a) Chứng minh nếu \(a,b\in Z\) và \(a+b⋮3\) thì \(a^3+b^3⋮3^2\)
b) Tìm min \(\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
a) Tìm x,y biết: x4+x2-y2+y+10=0
b) Tính giá trị biểu thức: \(\frac{\left(1+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(29^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(30^4+\frac{1}{4}\right)}\)
Rút gọn
a) \(x.\left(x+4\right).\left(x-4\right)-\left(x^2+1\right).\left(x-1\right)\)
b) \(\left(y-3\right).\left(y+3\right).\left(y^2+9\right)-\left(y^2+2\right).\left(y^2-2\right)\)
Tính
a) \(\left(x+3\right).\left(x^2-3x+9\right)-x.\left(x-2\right)\left(x+2\right)\)
b) \(\left(x+y\right)^3-x.\left(x+3y\right)^2+y\left(y-3x\right)^2\)
Phân tích đa thức thành nhân tử :
\(a,\left(x+y\right)^5-x^5-y^5\)
\(b,\left(x^2+y^2\right)^3+\left(z^2-x^2\right)^3+\left(y^2+z^2\right)^3\)
\(c,x^9-x^7-x^6-x^5+x^4+x^3+x^2+1\).