Ta có 

\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\)

\(\Rightarrow\frac{ayz+bxz+cxy}{xyz}=0\)

\(\Rightarrow ayz+bxz+cxy=0\)

Ta có

\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\)

\(\Rightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\)

\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+\frac{2xy}{ab}+\frac{2yz}{bc}+\frac{2xz}{ac}=1\)

\(\Rightarrow\frac{2xy}{ab}+\frac{2yz}{bc}+\frac{2xz}{ac}=1-\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)\)

\(\Rightarrow\frac{2xy.abc^2+2yz.a^2bc+2xz.ab^2c}{a^2b^2c^2}=1-\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)\)

\(\Rightarrow\frac{2abc.\left(cxy+ayz+bxz\right)}{a^2b^2c^2}=1-\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)\)

 Ta có \(cxy+ayz+bxz=0\)

\(\Rightarrow\frac{2abc.\left(cxy+ayz+bxz\right)}{a^2b^2c^2}=1-\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)\)

\(\Rightarrow\frac{2abc.0}{a^2b^2c^2}=1-\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)\)

\(\Rightarrow1-\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)=0\)

\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\left(đpcm\right)\)

bài này bạn bình phương vế thứ 2 lên rồi phân k vế 1 là ra đấy

\(D=\frac{x^2-3x+3}{x^2-2x+1}=\frac{x^2-3\left(x-1\right)}{\left(x-1\right)^2}\)

Đặt: x-1=y=>x=y+1. Ta có:

\(D=\frac{\left(y+1\right)^2-3y}{y^2}=\frac{y^2-y+1}{y^2}=1-\frac{1}{y}+\frac{1}{y^2}\)

Đặt: \(\frac{1}{y}=t\Rightarrow D=1-t+t^2\ge\frac{3}{4}\\ D=\frac{3}{4}\Leftrightarrow\left(t-\frac{1}{2}\right)^2=0\Rightarrow t=\frac{1}{2}\)

\(t=\frac{1}{2}\Leftrightarrow\frac{1}{y}=\frac{1}{2}\Rightarrow y=2\Leftrightarrow x-1=2\Rightarrow x=3\)

Vậy minD=\(\frac{3}{4}\Leftrightarrow x=3\)

D=\(\frac{x.x-3x+3}{x.x-2x+1}\)

D=\(\frac{x.\left(x-3\right)+3}{x.\left(x-2\right)+1}\)

D=\(\frac{x-3+3}{x-2+2}\)(Chia cả tử và mẫu cho x lần)

D=\(\frac{x}{x}\)

D=1

A=(8xy-6x^2)/(12y^2-9xy)

A=2x(4y-3x)/3y(4y-3x)

A=2x/3y

B=(2x^3-18x)/(x^4-81)

B=2x(x^2-9)/(x^2-9)(x^2+9)

B=2x/(x^2+9)

C=(x^2-x-30)/(x^2-25)

C=(x^2+6x-5x-30)/(x^2-25)

C=(x(x+6)-5(x+6))/(x-5)(x+5)

C=(x+6)(x-5)/(x-5)(x+5)

C=(x+6)/(x+5)

2x^2-8x

=2x(x-4)

a.)(x+y+z)^2-(x-y-z)^2

=(x+y+z-x+y+z)(x+y+z+x-y-z)

=(2y+2z)2x

=2(y+z)2x

=4x(y+z)

b.) (2a+b)^2-(a+b)-3a^2

=4a^2+4ab+b^2-a-b-3a^2

=a^2+4ab+b^2-a-b

hình như đề sai thì phải hay sao ấy bạn 

uk cn b chép thiếu bạn ạ

\(2\left(3x-5\right)-4\left(2+3\left(x-1\right)\right)=3\left(x-5\right)\)

           \(6x-10-4\left(2+3x-3\right)=3x-15\)

              \(6x-10-8-12x+12=3x-15\)

                                  \(6x-12x-3x=-15+10+8-12\)

                                                      \(-9x=-9\)

                                                            \(x=1\)

\(a,3a+3b-a^2-ab\)

\(=\left(3a-a^2\right)+\left(3b-ab\right)\)

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

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

\(b,8y^2-8yz-13y+13z\)

\(=\left(8y^2-8yz\right)-\left(13y-13z\right)\)

\(=8y\left(y-z\right)-13\left(y-z\right)\)

\(=\left(y-z\right)\left(8y-13\right)\)

\(c,3b^2+3c^2-ab^2-ac^2+2a-6\)

\(=\left(3b^2-ab^2\right)+\left(3c^2-ac^2\right)+\left(2a-6\right)\)

\(=b^2\left(3-a\right)+c^2\left(3-a\right)-2\left(3-a\right)\)

\(=\left(3-a\right)\left(b^2+c^2-2\right)\)

900 nhé bạn

1000-100=900