Rút gọn
( x + y - z ) 2 + ( y - z ) 2 + 2( x - y + z) (z - y )
Tìm x
x3 + 3x 2 + 3x + 28 = 0
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a: \(Y=\dfrac{3\left(x^2-x-1\right)-x^2+1}{\left(x+2\right)\left(x-1\right)}+\dfrac{x-2}{x}\cdot\dfrac{1-1+x}{1-x}\)
\(=\dfrac{2x^2-3x-2}{\left(x+2\right)\left(x-1\right)}+\dfrac{x-2}{x}\cdot\dfrac{-x}{x-1}\)
\(=\dfrac{2x^2-3x-2}{\left(x+2\right)\left(x-1\right)}-\dfrac{x-2}{x-1}\)
\(=\dfrac{2x^2-3x-2-x^2+4}{\left(x+2\right)\left(x-1\right)}=\dfrac{x^2-3x+2}{\left(x+2\right)\left(x-1\right)}=\dfrac{x-2}{x+2}\)
b: Y=2
=>2x+4=x-2
=>x=-6(nhận)
c; Y nguyên
=>x+2-4 chia hết cho x+2
=>x+2 thuộc {1;-1;2;-2;4;-4}
Kết hợp ĐKXĐ, ta được: x thuộc {-1;-3;-4;-6}
a/ \(\frac{3x^2-11x+8}{2x^2-9x+7}=\frac{\left(x-1\right)\left(3x-8\right)}{\left(x-1\right)\left(2x-7\right)}=\frac{3x-8}{2x-7}\)
câu b,c tương tự nha ^^
Bài 1:
a. \(=[(3x+(4y-5z)][3x-(4y-5z)]=(3x)^2-(4y-5z)^2\)
\(=9x^2-(16y^2-40yz+25z^2)=9x^2-16y^2+40yz-25z^2\)
b.
\(=(3a-1)^2+2(3a-1)(3a+1)+(3a+1)^2=[(3a-1)+(3a+1)]^2=(6a)^2=36a^2\)
Bài 2:
\((x+y+z)^3=[(x+y)+z]^3=(x+y)^3+3(x+y)^2z+3(x+y)z^2+z^3\)
\(=[x^3+y^3+3xy(x+y)]+3(x+y)z(x+y+z)+z^3\)
\(=x^3+y^3+z^3+3xy(x+y)+3(x+y)z(x+y+z)\)
\(=x^3+y^3+z^3+3(x+y)(xy+zx+zy+z^2)\)
\(=x^3+y^3+z^3+3(x+y)(z+x)(z+y)\) (đpcm)
=a, (x-3)(x+3)-(x-7)(x+7)= x2 - 9 - x2 + 7
= -2
b, (4x-5)2+(3x-2)2-2(4x+5)(3x-2)= (4x-5)2 - 2(4x+5)(3x-2) + (3x-2)2
= ( 4x - 5 - 3x + 2 )2
= ( x - 3 )2
c, 2(3x-y)(3x+y)+(3x-y)2+(3x+y)2= 2(3x-y)(3x+y)+(3x-y)2+(3x+y)2
= (3x-y)2+ 2(3x-y)(3x+y)+ (3x+y)2
= ( 3x - y + 3x + y )2
= ( 6x )2
= 36x2
d, (x-y+z)2+(z-y)2+2(x-y+z+2(x-y+z)(y-z-y+z)(y-z)
1, rút gọn
a, (x-3)(x+3)-(x-7)(x+7)
= x^2 - 9 - (x^2 - 49)
= x^2 - 9 - x^2 + 49
= 40
b, (4x-5)2+(3x-2)2-2(4x+5)(3x-2)
= 16x^2 - 40x + 25 + 9x^2 - 12x + 4 - 2(12x^2 - 8x + 15x - 10)
= 25x^2 - 52x + 29 - 24x^2 + 16x - 30x + 20
= x^2 - 66x + 49
c, 2(3x-y)(3x+y)+(3x-y)2+(3x+y)2
= 2(9x^2 - y^2) + 9x^2 - 6xy + y^2 + 9x^2 + 6xy + y^2
= 18x^2 - 2y^2 + 18x^2 + 2y^2
= 36x^2
d, (x-y+z)2+(z-y)2+2(x-y+z+2(x-y+z)(y-z-y+z)(y-z)
= dài vl
Ta có: x+y+z=0
\(\Leftrightarrow\left(x+y+z\right)^2=0\)
\(\Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz=0\)(1)
Ta có: \(K=\dfrac{x^2+y^2+z^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{x^2+y^2+z^2}{x^2-2xy+y^2+y^2-2yz+z^2+z^2-2xz+x^2}\)
\(=\dfrac{x^2+y^2+z^2}{3x^2+3y^2+3z^2-x^2-y^2-z^2-2xy-2yz-2xz}\)
\(=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)-\left(x^2+y^2+z^2+2xy+2yz-2xz\right)}\)
\(=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)}=\dfrac{1}{3}\)
Vậy: \(K=\dfrac{1}{3}\)
\(K=\dfrac{x^2+y^2+z^2}{2\left(x^2+y^2+z^2\right)-2\left(xy+yz+zx\right)}\)
\(K=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)-\left(x+y+z\right)^2}=\dfrac{1}{3}\)