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1) \(x-2y=3\Rightarrow\hept{\begin{cases}x=3+2y\\y=\frac{x-3}{2}\end{cases}}\)
\(\Rightarrow A=2x\left(x+2y-3\right)-y\left(6x-3y-10\right)+x-7+\left(x-3y\right)^2\)
\(=2x^2+4xy-6x-6xy+3y^2+10y+x-7+x^2-6xy+9y^2\)
\(=3x^2+12y^2-8xy-5x+10y-7\)
\(=3.\left(3+2y\right)^2+12y^2-8\left(3+2y\right).y-5\left(3+2y\right)+10y-7\)
\(=3\left(9+12y+4y^2\right)+12y^2-8\left(3y+2y^2\right)-15-10y+10y-7\)
\(=27+36y+12y^2+12y^2-24y-16y^2-15-10y+10y-7\)
\(=8y^2+12y+5\)
\(M=\left(x^2-2x+1\right)\left(1+2x\right)-\left(x^2+2x+1\right)\left(1-3x\right)-\left(3-6x\right)\left(x^2+3x+2\right)\)
\(=x^2+2x^3-2x-4x^2+1+2x-x^2+3x^8-2x+6x^2-1+3x-3x^2-9x-6+6x^8\)\(+18x^2+12x=11x^3+17x^2+4x-6\)
\(M=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\left(a+b\right)\)
\(=\left(a+b\right)\left[\left(a+b\right)^2-3ab\right]+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\left(a+b\right)\)
\(=1-ab+3ab\left(1-2ab\right)+6a^2b^2\)
\(=1-3ab+3ab-6a^2b^2+6a^2b^2=1\)
Vậy M=1
M = a3 + b3 + 3ab( a2 + b2 ) + 6a2b2( a + b )
= ( a + b )3 - 3ab( a + b ) + 3ab[ ( a + b )2 - 2ab ] + 6a2b2( a + b )
= 13 - 3ab.1 + 3ab( 12 - 2ab ) + 6a2b2.1
= 1 - 3ab + 3ab - 6a2b2 + 6a2b2
= 1
Bài 1:
a) \(x^3-5x^2+8x-4\)
\(=x^3-4x^2+4x-x^2+4x-4\) \(=x\left(x^2-4x+4\right)-\left(x^2-4x+4\right)\)\(=\left(x-1\right)\left(x-2\right)^2\)
b) Ta có: \(\frac{A}{M}=\frac{10x^2-7x-5}{2x-3}=5x+4+\frac{7}{2x-3}\)
Với \(x\in Z\)thì \(A⋮M\)khi \(\frac{7}{2x-3}\in Z\)\(\Rightarrow7⋮\left(2x-3\right)\)\(\Rightarrow2x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(\Rightarrow=\left\{1;5;\pm2\right\}\)thì khi đó \(A⋮M\)
Các bài làm này có đúng ko ạ, ai đó duyệt giúp em, em cảm ơn.
Bài 1:
a)x3-5x2+8x-4=x3-4x2+4x-x2+4x-4
=x(x2-4x-4)-(x2-4x+4)
=(x-1) (x-2)2
b)Xét:
\(\frac{a}{b}-\frac{10x^2-7x-5}{2x-3}\)
=\(5x+4+\frac{7}{2x-3}\)
Với x thuộc Z thì A /\ B khi \(\frac{7}{2x-3}\) thuộc Z => 7 /\ (2x-3)
Mà Ư(7)={-1;1;-7;7} => x=5;-2;2;1 thì A /\ B
c)Biến đổi \(\frac{x}{y^3-1}-\frac{x}{x^3-1}=\frac{x^4-x-y^4+y}{\left(y^3-1\right)\left(x^3-1\right)}\)
=\(\frac{\left(x^4-y^4\right)\left(x-y\right)}{xy\left(y^2+y+1\right)\left(x^2+x+1\right)}\)(do x+y=1=>y-1=-x và x-1=-y)
=\(\frac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)-\left(x-y\right)}{xy\left[x^2y^2+y^2x+y^2+xy^2+xy+y+x^2+x+1\right]}\)
=\(\frac{\left(x-y\right)\left(x^2+y^2-1\right)}{xy\left[x^2y^2+xy\left(x+y\right)+x^2+y^2+xy+2\right]}\)
=\(\frac{\left(x-y\right)\left(x^2-x+y^2-y\right)}{xy\left[x^2y^2+\left(x+y\right)^2+2\right]}=\frac{\left(x-y\right)\left[x\left(x-1\right)+y\left(y-1\right)\right]}{xy\left(x^2y^2+3\right)}\)
=\(\frac{\left(x-y\right)\left[x\left(-y\right)+y\left(-x\right)\right]}{xy\left(x^2y^2+3\right)}=\frac{\left(x-y\right)\left(-2xy\right)}{xy\left(x^2y^2+3\right)}\)
=\(\frac{-2\left(x-y\right)}{x^2y^2+3}\)Suy ra điều phải chứng minh
Bài 2 )
a)(x2+x)2+4(x2+x)=12 đặt y=x2+x
y2+4y-12=0 <=>y2+6y-2y-12=0
<=>(y+6)(y-2)=0 <=> y=-6;y=2
>x2+x=-6 vô nghiệm vì x2+x+6 > 0 với mọi x
>x2+x=2 <=> x2+x-2=0 <=> x2+2x-x-2=0
<=>x(x+2)-(x+2)=0 <=>(x+2)(x-1) <=> x=-2;x-1
Vậy nghiệm của phương trình x=-2;x=1
b)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}+\frac{x+4}{2005}+\frac{x+5}{2004}\)\(+\frac{x+6}{2003}\)
=\(\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)+\left(\frac{x+4}{2005}+1\right)\)\(+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)
<=>\(\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}\)\(+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)
<=>\(\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}\)\(-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)
Nhờ OLM xét giùm em vs ạ !
\(\text{Vì }a^2-b^2=c^2-d^2=1\Leftrightarrow\hept{\begin{cases}a^2=b^2+1\left(1\right)\\d^2=c^2-1\left(2\right)\end{cases}}\)
\(\text{Ta có: }a^2d^2-a^2d^2=0\)
\(\Rightarrow a^2.\left(c^2-1\right)-d^2.\left(b^2+1\right)=0\)
\(\Rightarrow a^2c^2-b^2d^2-a^2-d^2=0\)
\(\Rightarrow a^2c^2-b^2d^2=a^2+d^2\)
Vậy \(a^2c^2-b^2d^2=a^2+d^2\)
a/x +b/y +c/z =0 ->ayz+bxz+cxz=0
x/a + y/b + z/c=1 ->(x/a +y/b +z/c)^2=1
x^2/a^2 + y^2/b^2 + z^2/c^2 +2(xy/ab +yz/bc +xz/ac)=1
x^2/a^2 + y^2/b^2 + z^2/c^2 =1- 2* ayz+bxz+cxz/abc=1-2*0=1-0=1 =>ĐPCM
k hộ mik nha
#)Giải :
\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\rightarrow ayz+bxz+cxy=0\)
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\rightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\)
\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1-2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1-2\frac{ayz+bxz+cxy}{abc}=1-2.0=1\left(đpcm\right)\)
#~Will~be~Pens~#