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\(1,\\ a,A=4x^2\left(-3x^2+1\right)+6x^2\left(2x^2-1\right)+x^2\\ A=-12x^4+4x^2+12x^2-6x^2+x^2=-x^2=-\left(-1\right)^2=-1\\ b,B=x^2\left(-2y^3-2y^2+1\right)-2y^2\left(x^2y+x^2\right)\\ B=-2x^2y^3-2x^2y^2+x^2-2x^2y^3-2x^2y^2\\ B=-4x^2y^3-4x^2y^2+x^2\\ B=-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^3-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^2+\left(0,5\right)^2\\ B=\dfrac{1}{8}-\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{8}\)
\(2,\\ a,\Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ b,\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3=8=-2^3\\ \Leftrightarrow x=2\\ c,\Leftrightarrow4x^2\left(4x-2\right)-x^3+8x^2=15\\ \Leftrightarrow16x^3-8x^2-x^3+8x^2=15\\ \Leftrightarrow15x^3=15\\ \Leftrightarrow x^3=1\Leftrightarrow x=1\)
a: Ta có: \(2x\left(x-3\right)+x-3=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
b: Ta có: \(x^2\left(x-6\right)-x^2+36=0\)
\(\Leftrightarrow\left(x-6\right)\left(x^2-x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=3\\x=-2\end{matrix}\right.\)
`x^2+x+1=x^2+x+1/4+3/4=(x+1/2)^2 +3/4`
Vì `(x+1/2)^2 >= 0` với mọi `x`
`=>(x+1/2)^2 +3/4 >= 3/4` với mọi `x`
`=>` Biểu thức Min `=3/4<=>x=-1/2`
_____________
`(x-3)(x+5)+4=x^2+2x-11=x^2+2x+1-12=(x+1)^2-12`
Vì `(x+1)^2 >= 0` với mọi `x`
`=>(x+1)^2-12 >= -12` với mọi `x`
`=>` Biểu thức Min `=-1/2<=>x=-1`
Bài 8:
\(F=x^2-2x+1+x^2-6x+9=2x^2-8x+10\\ F=2\left(x^2-4x+4\right)+2=2\left(x-2\right)^2+2\ge2\\ F_{min}=2\Leftrightarrow x=2\)
Bài 9:
\(A=-x^2+2x-1+5=-\left(x-1\right)^2+5\le5\\ A_{max}=5\Leftrightarrow x=1\\ B=-x^2+10x-25+2=-\left(x-5\right)^2+2\le2\\ B_{max}=2\Leftrightarrow x=5\\ C=-x^2+6x-9+9=-\left(x-3\right)^2+9\le9\\ C_{max}=9\Leftrightarrow x=3\)
\(a,=x^2-4-x^2+2x+3=2x-1\\ b,=x^3+3x^2-5x-15+x^2-x^3+4x-4x^2=-x-15\\ c,=2x^2+3x-10x-15-2x^2+6x+x+7=-8\\ d,=\left(2x+1+3x-1\right)^2=25x^2\)
\(a,=x^2-4-x^2-2x-1=-2x-5\\ b,=8x^3-1-8x^3-1=-2\\ 3,\\ a,\Rightarrow x^3+8-x^3+2x=15\\ \Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\\ b,\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\\ \Rightarrow7x=14\Rightarrow x=2\)
Bài 2:
a) \(=x^2-4-x^2-2x-1=-2x-5\)
b) \(=8x^3-1-8x^3-1=-2\)
Bài 3:
a) \(\Rightarrow x^3+8-x^3+2x=15\)
\(\Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\)
b) \(\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\)
\(\Rightarrow7x=14\Rightarrow x=2\)
Bài 4.
\(A=2x^3+(x+1)^3-3x(x-2)(x+2)-3(x^2+5x+9)\\=2x^3+(x^3+3x^2+3x+1)-3x(x^2-4)-3x^2-15x-27\\=2x^3+x^3+3x^2+3x+1-3x^3+12x-3x^2-15x-27\\=(2x^3+x^3-3x^3)+(3x^2-3x^2)+(3x+12x-15x)+(1-27)\\=-26\\---\)
\(B=x(x-4x)+x(2-x)(x+2)+4(2x^2-5x+4)\\=x\cdot(-3x)+x(2-x)(2+x)+8x^2-20x+16\\=-3x^2+x(4-x^2)+8x^2-20x+16\\=-3x^2+4x-x^3+8x^2-20x+16\)
Bạn kiểm tra lại đề giúp mình!
\(C=(x-2y)(x^2+2xy+4y^2)-(x^3-8y^3+10)\) (sửa đề)
\(=x^3-(2y)^3-x^3+8y^2-10\\=x^3-8y^3-x^3+8y^3-10\\=(x^3-x^3)+(-8y^3+8y^3)-10\\=-10\)
Bài 5.
\(d)xy^2-3x^3y^2-2x(xy-3xy^2)\\=xy^2-3x^3y^2-2x^2y+6x^2y^2\\---\\f)(x-y)(2x+y)-2x^2+y^2+3xy\\=x(2x+y)-y(2x+y)-2x^2+y^2+3xy\\=2x^2+xy-2xy-y^2-2x^2+y^2+3xy\\=(2x^2-2x^2)+(xy-2xy+3xy)+(-y^2+y^2)\\=2xy\)
\(Toru\)
a: \(A=x^3+3x^2-5x-15+x^2-x^3+4x-4x^2\)
\(=-x-15\)
\(=-\left(-1\right)-15=1-15=-14\)