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\(A=x\left(9x^2-16\right)-9\left(x^3+8\right)+16x\\ A=9x^3-16x-9x^3-72+16x\\ A=-72\)
\(A=x\left(3x-4\right)\left(3x+4\right)-9\left(x+2\right)\left(x^2-2x+4\right)+16x\)
\(=x\left(9x^2-16\right)-9\left(x^3+8\right)+16x\)
\(=9x^3-16x-9x^3-72+16x=-72\)
a. \(\left(x+2\right)^{^2}-\left(x-4\right)^{^2}+x^{^2}-3x+1=x^{^2}+4x+4-x^{^2}+8x-16+x^{^2}-3x+1=x^{^2}+9x-11\)
b. \(\left(2x+2\right)^{^2}-4x\left(x+2\right)=4x^{^2}+8x+4-4x^{^2}-8x=4\)
Bài 1:
a: Ta có: \(A=\left(k-4\right)\left(k^2+4k+16\right)-\left(k^3+128\right)\)
\(=k^3-64-k^3-128\)
=-192
b: Ta có: \(B=\left(2m+3n\right)\left(4m^2-6mn+9n^2\right)-\left(3m-2n\right)\left(9m^2+6mn+4n^2\right)\)
\(=8m^3+27n^3-27m^3+8n^3\)
\(=-19m^3+35n^3\)
Bài 4:
a: Ta có: \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=16\)
\(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=16\)
\(\Leftrightarrow9x=9\)
hay x=1
b: ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-2\right)=15\)
\(\Leftrightarrow x^3+8-x^3+2x=15\)
\(\Leftrightarrow2x=7\)
hay \(x=\dfrac{7}{2}\)
a: Đặt A=|x-2|+|2x-1|
TH1: x<1/2
=>2x-1<0 và x-2<0
A=|x-2|+|2x-1|
=2-x+1-2x
=-3x+3
TH2: 1/2<=x<2
=>2x-1>=0 và x-2<0
=>A=2-x+2x-1=x+1
TH3: x>=2
=>2x-1>0 và x-2>=0
=>A=2x-1+x-2=3x-3
b: Đặt B=|4-3x|-|2x+1|
=|3x-4|-|2x+1|
TH1: x<-1/2
=>\(2x+1< 0;3x-4< 0\)
=>\(B=4-3x-\left(-2x-1\right)\)
\(=4-3x+2x+1\)
\(=5-x\)
TH2: \(-\dfrac{1}{2}< =x< \dfrac{4}{3}\)
=>\(2x+1>=0;3x-4< 0\)
=>\(B=4-3x-\left(2x+1\right)\)
\(=4-3x-2x-1=-5x+3\)
TH3: \(x>=\dfrac{4}{3}\)
=>\(3x-4>=0;2x+1>0\)
=>\(B=3x-4-\left(2x+1\right)\)
\(=3x-4-2x-1\)
=x-5
\(a,\left(x-5\right)\left(2x+1\right)-2x\left(x-3\right)\\ =x.2x-5.2x+x-5-2x.x-2x.\left(-3\right)\\ =2x^2-10x+x-5-2x^2+6x\\ =2x^2-2x^2-10x+x+6x-5\\ =-3x-5\)
\(b,\left(2+3x\right)\left(2-3x\right)+\left(3x+4\right)^2\\ =\left[2^2-\left(3x\right)^2\right]+\left[\left(3x\right)^2+2.3x.4+4^2\right]\\=4-9x^2+\left(9x^2+24x+16\right)\\ =24x+20\)
`a)(2x-1)^2+(x+3)^2-5(x-7)(x+7)`
`=4x^2-4x+1+x^2+6x+9-5(x^2-49)`
`=5x^2-5x^2-4x+6x+1+9+245`
`=2x+255`
`b)(x-2)(x^2+2x+4)-(25+x^3)`
`=x^3-8-x^3-25=-33`
Lời giải:
a.
$(2x-1)^2+(x+3)^2-5(x-7)(x+7)$
$=4x^2-4x+1+(x^2+6x+9)-5(x^2-49)$
$=5x^2+2x+10-(5x^2-245)=2x+255$
b.
$(x-2)(x^2+2x+4)-(25+x^3)=(x^3-2^3)-(25+x^3)$
$=-8-25=-33$
Bài 1:
\(a,6x^2-15x^3y\\ b,=-\dfrac{2}{3}x^2y^3+\dfrac{2}{3}x^4y-\dfrac{8}{3}xy\)
Bài 2:
\(a,=20x^3-10x^2+5x-20x^3+10x^2+4x=9x\\ b,=3x^2-6x-5x+5x^2-8x^2+24=24-11x\\ c,=x^5+x^3-2x^3-2x=x^5-x^3-2x\)
B = (x – 2)( x 2 + 2x + 4) – x(x – 1)(x + 1) + 3x
B = (x – 2)( x 2 + x.2 + 2 2 ) – x( x 2 – 1) + 3x
B = x 3 – 2 3 – x . x 2 + x . 1 + 3 x B = x 3 – 8 – x 3 + x + 3 x
B = 4x – 8
Đáp án cần chọn là: D
1: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54\right)\)
\(=x^3+27-x^3-54\)
=-27
2: Ta có: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3\)
\(=2y^3\)
\(1,=x^3+270-x^3-54=-27\\ 2,=8x^3+y^3-8x^3+y^3=2y^3\\ 3,=x^3-3x^2+3x-1-x^3-8+3x^2-48=3x-57\\ 4,=x^3-x-x^3-1=-x-1\\ 5,=8x^3-5\left(8x^3+1\right)=-32x^3-5\\ 6,=27+x^3-27=x^3\\ 7,làm.ở.câu.3\\ 8,=x^3-6x^2+12x-8+6x^2-12x+6-x^3-1+3x\\ =3x-3\)
Lời giải:
$A=x[(3x)^2-4^2]-9(x^3+2^3)+16x$
$=x(9x^2-16)-9(x^3+8)+16x$
$=9x^3-16x-9x^3-72+16x$
$=-72$
\(A=x\left(3x-4\right)\left(3x+4\right)-9\left(x+2\right)\left(x^2-2x+4\right)+16x\)
\(=9x^3-16x-9x^3-72+16x\)
=-72