- Xét tử:

\(x^2+ax+ab+bx\)

\(=x\left(x+a\right)+b\left(x+a\right)\)

\(=\left(x+a\right)\left(x+b\right)\)

- Xét mẫu:

\(3bx-a^2-ax+3ab\)

\(=3bx+3ab-a^2-ax\)

\(=3b\left(x+a\right)-a\left(a+x\right)\)

\(=\left(x+a\right)\left(3b-a\right)\)

Vậy \(\dfrac{x^2+ax+ab+bx}{3bx-a^2-ax+3ab}=\dfrac{\left(x+a\right)\left(x+b\right)}{\left(x+a\right)\left(3b-a\right)}=\dfrac{x+b}{3b-a}\) với \(x\ne-a\)

Ta có: \(VT=\dfrac{x^2+ax+ab+bx}{3bx-a^2-ax+3ab}\)

\(=\dfrac{x\left(a+x\right)+b\left(a+x\right)}{3b\left(x+a\right)-a\left(a+x\right)}\)

\(=\dfrac{\left(b+x\right)\left(a+x\right)}{\left(3b-a\right)\left(a+x\right)}=\dfrac{b+x}{3b-a}=VP\)

\(\Rightarrowđpcm\)

Bài 1:

1) \(a\left(b-c\right)+b\left(c-a\right)+c\left(a-b\right)\)

\(=ab-ac+bc-ba+ca-cb\)

\(=0\)

2) \(a\left(bz-cy\right)+b\left(cx-az\right)+c\left(ay-bx\right)\)

\(=abz-acy+bcx-baz+cay-cbx\)

\(=0\)

Bài 2:

Ta có:

\(\dfrac{x^2+ax+ab+bx}{3bx-a^2-ax+3ab}\)

\(=\dfrac{\left(x^2+bx\right)+\left(ax+ab\right)}{\left(3bx-ax\right)+\left(3ab-a^2\right)}\)

\(=\dfrac{x\left(x+b\right)+a\left(x+b\right)}{x\left(3b-a\right)+a\left(3b-a\right)}\)

\(=\dfrac{\left(x+a\right)\left(x+b\right)}{\left(x+a\right)\left(3b-a\right)}\)

\(=\dfrac{x+b}{3b-a}\)

A = 4acx + 4bcx + 4ax + 4bx ( đã sửa '-' )

= 4x( ac + bc + a + b )

= 4x[ c( a + b ) + ( a + b ) ]

= 4x( a + b )( c + 1 )

B = ax - bx + cx - 3a + 3b - 3c

= x( a - b + c ) - 3( a - b + c )

= ( a - b + c )( x - 3 )

C = 2ax - bx + 3cx - 2a + b - 3c

= x( 2a - b + 3c ) - ( 2a - b + 3c )

= ( 2a - b + 3c )( x - 1 )

D = ax - bx - 2cx - 2a + 2b + 4c

= x( a - b - 2c ) - 2( a - b - 2c )

= ( a - b - 2c )( x - 2 )

E = 3ax² + 3bx² + ax + bx + 5a + 5b

= 3x²( a + b ) + x( a + b ) + 5( a + b )

= ( a + b )( 3x² + x + 5 )

F = ax² - bx² - 2ax + 2bx - 3a + 3b

= x²( a - b ) - 2x( a - b ) - 3( a - b )

= ( a - b )( x² - 2x - 3 )

= ( a - b )( x² + x - 3x - 3 )

= ( a - b )[ x( x + 1 ) - 3( x + 1 ) ]

= ( a - b )( x + 1 )( x - 3 )

\(2ax-bx+3cx-2a+b-3c\\ =x\left(2a-b+3c\right)-\left(2a-b+3c\right)\\ =\left(x-1\right)\left(2a-b+3c\right)\)

\(ax-bx-2cx-2a+2b+4c\\ =x\left(a-b-2c\right)-2\left(a-b-2c\right)\\ =\left(x-2\right)\left(a-b-2c\right)\)

\(3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\)

\(ax^2-bx^2-2ax+2bx-3a+3b\\ =x^2\left(a-b\right)-2x\left(a-b\right)-3\left(a+b\right)\\ =\left(x^2-2x-3\right)\left(a+b\right)\\ =\left(x+1\right)\left(x-3\right)\left(a+b\right)\)

Đây là bất đẳng thức Bunhia Cốpxki bạn, lên mạng tra cách giải là đc!

Sai đề rồi bn

khó quá

đúng là khó thiệt

quá khó

10. a) 

\(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\Leftrightarrow\frac{x^4}{a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\)

\(\Leftrightarrow\left(a+b\right)\left(x^4+y^4\right)=ab\left(x^2+y^2\right)^2\Leftrightarrow\left(bx^2-ay^2\right)^2=0\Leftrightarrow bx^2=ay^2\)

b) Từ \(ay^2=bx^2\Rightarrow\frac{y^2}{b}=\frac{x^2}{a}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)

\(\Rightarrow\frac{x^{2008}}{a^{1004}}=\frac{1}{\left(a+b\right)^{1004}}\); \(\frac{y^{2008}}{b^{1004}}=\frac{1}{\left(a+b\right)^{1004}}\)

\(\Rightarrow\frac{x^{2008}}{a^{1004}}+\frac{y^{2008}}{b^{1004}}=\frac{2}{\left(a+b\right)^{1004}}\)

25. Ta có \(\left(ax+by+cz\right)^2=0\Leftrightarrow a^2x^2+b^2y^2+c^2z^2=-2\left(abxy+bcyz+acxz\right)\)

Xét mẫu số của P : \(bc\left(y-z\right)^2+ac\left(x-z\right)^2+ab\left(x-y\right)^2=bc\left(y^2-2yz+z^2\right)+ac\left(x^2-2xz+z^2\right)+ab\left(x^2-2xy+y^2\right)\)

\(=y^2bc-2bcyz+bcz^2+acx^2-2xzac+acz^2+abx^2-2abxy+aby^2\)

\(=y^2bc+bcz^2+acx^2+acz^2+abx^2+aby^2-2\left(abxy+xzac+bcyz\right)\)

\(=y^2bc+bcz^2+acx^2+acz^2+abx^2+aby^2+a^2x^2+b^2y^2+c^2z^2\)

\(=c\left(ax^2+by^2+cz^2\right)+b\left(ax^2+by^2+cz^2\right)+a\left(ax^2+by^2+cz^2\right)=\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)\)

\(\Rightarrow P=\frac{ax^2+by^2+cz^2}{\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)}=\frac{1}{a+b+c}=\frac{1}{2007}\)

8. \(\frac{x^3}{a^3}+\frac{y^3}{b^3}=\left(\frac{x}{a}+\frac{y}{b}\right)^3-3.\frac{xy}{ab}\left(\frac{x}{a}+\frac{y}{b}\right)=1^3-3.\left(-2\right).1=7\)

8. \(\frac{x^3}{a^3}+\frac{y^3}{b^3}=\left(\frac{x}{a}+\frac{y}{b}\right)^3-3.\frac{xy}{ab}\left(\frac{x}{a}+\frac{y}{b}\right)=1^3-3.\left(-2\right).1=7\)

vào câu hỏi tương tự là có