bạn kiếm kiểu gì cx ko có ai giải đâu, đề này sai r, nãy mình sửa mới đúng

VÀ a, b c khác 0 nha

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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~#

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 !

\(\frac{a^3}{b+c}+\frac{b^3}{c+a}+\frac{c^3}{a+b}\)

\(=\frac{a^4}{ab+ac}+\frac{b^4}{cb+ba}+\frac{c^4}{ac+bc}\)

\(\ge\frac{\left(a^2+b^2+c\right)^2}{2\left(ab+bc+ca\right)}=\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)}{2\left(ab+bc+ca\right)}\)

Mà \(a^2+b^2+c^2\ge ab+bc+ca\Rightarrowđpcm\)

\(\frac{a^3}{b+c}+\frac{a^3}{b+c}+\frac{\left(b+c\right)^2}{8}\ge3\sqrt[3]{\frac{a^3}{b+c}.\frac{a^3}{b+c}.\frac{\left(b+c\right)^2}{8}}=\frac{3a^2}{2}\)

Rồi tương tự các kiểu:v

Suy ra \(2VT\ge\frac{3}{2}\left(a^2+b^2+c^2\right)-\frac{\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2}{8}\)

\(\ge\frac{3}{2}\left(a^2+b^2+c^2\right)-\frac{a^2+b^2+c^2}{2}=\left(a^2+b^2+c^2\right)\) (chú ý \(\left(a+b\right)^2\le2\left(a^2+b^2\right)\))

Không phải dùng tới Cauchy-Schwarz:D