Ta có:

\(\frac{xy}{x+y}=\frac{yz}{y+z}=\frac{zx}{z+x}\rightarrow\frac{x+y}{xy}=\frac{y+z}{yz}=\frac{z+x}{zx}\)

\(\Rightarrow\frac{1}{x}+\frac{1}{y}=\frac{1}{y}+\frac{1}{z}=\frac{1}{z}+\frac{1}{x}\Rightarrow\frac{1}{x}=\frac{1}{y}=\frac{1}{z}\Rightarrow x=y=z\)

Thay tất cả giá trị x,y,z vào M ta được:

\(M=\frac{2020x^3+2020y^3+2020z^3}{x^3+y^3+z^3}+\frac{2021x^5+2021y^5}{x^5+y^5}\)

\(\Rightarrow M=\frac{2020\left(x^3+y^3+z^3\right)}{x^3+y^3+z^3}+\frac{2021\left(x^5+y^5\right)}{x^5+y^5}\)

\(\Rightarrow M=2020+2021=4041\)

câu 6 dễ mà câu 7

A=3-3/4+3/4-3/7+3/7-3/10+...+3/94-3/97+3/97-3/100

A=3-3/100

A=300/100-3/100

A=297/100

Câu 6:

\(\frac{x+5}{2005}+\frac{x+6}{2004}+\frac{x+7}{2003}=-3\)

=>\(\left(\frac{x+5}{2005}+1\right)+\left(\frac{x+6}{2004}+1\right)+\left(\frac{x+7}{2003}+1\right)=-3+3=0\)

=>\(\frac{x+2010}{2005}+\frac{x+2010}{2006}+\frac{x+2010}{2007}=0\)

=>x+2010=0

=>x=-2010

Câu 7:

\(A=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\cdots+\frac{3}{97\cdot100}\)

\(=1-\frac14+\frac14-\frac17+\cdots+\frac{1}{97}-\frac{1}{100}\)

\(=1-\frac{1}{100}=\frac{99}{100}\)

Ta có 2 TH:

+ Th1: \(x-2=x\)

=>\(x-x=2\)

=>\(0=2\)( Vô lý, loại)

+ Th2: \(x-2=-x\)

=>\(x+x=2\)

=>\(2x=2\)

=>\(x=1\)

Vậy x=1

\(|x-2|=x\)

\(\Rightarrow TH1:x-2=x\)

            \(x-x=2\)

            \(0=2\)

          \(\Rightarrow x\in\varnothing\)

\(TH2:x-2=-x\)

        \(x+x=2\)

        \(2x=2\)

         \(\Rightarrow x=1\)

Vậy \(x\in\left\{\varnothing;1\right\}\)

a) Xét \(x\le\frac{3}{2}\) ta có : \(\left(3-2x\right)-x=\left(2-x\right)\)

\(\Leftrightarrow3-3x=2-x\Leftrightarrow-2x=-1\Rightarrow x=-\frac{1}{2}\left(TM\right)\)

Xét \(\frac{3}{2}\le x\le2\) ta có : \(\left(2x-3\right)-x=2-x\)

\(\Leftrightarrow x-3=2-x\Leftrightarrow2x=5\Rightarrow x=\frac{5}{2}\left(l\right)\)

Xét \(x\ge2\) ta có : \(\left(2x-3\right)-x=x-2\)

\(\Leftrightarrow x-3=x-2\Rightarrow-3=-2\left(l\right)\)

Vậy \(x=-\frac{5}{2}\)

b) \(VT=\left|x+3\right|+\left|x+1\right|\ge0\forall x\) nên \(VP=3x\ge0\Rightarrow x\ge0\)

\(\Rightarrow\left|x+3\right|+\left|x+1\right|=x+3+x+1=2x+4\)

Ta có \(2x+4=3x\Rightarrow x=4\)

Vậy \(x=4\)

Mơn nha, Đinh Đức Hùng

Trả lời :

Có : \(\frac{x}{3}=\frac{y}{4}\), x + y = 14

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{x}{3}=\frac{y}{4}=\frac{x+y}{3+4}=\frac{14}{7}=2\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{3}=2\\\frac{y}{4}=2\end{cases}\Rightarrow\hept{\begin{cases}x=6\\y=8\end{cases}}}\)

\(\text{Áp dụng tính chất dãy tỉ số bằng nhau ta có:}\)

\(\frac{x}{3}=\frac{y}{4}=\frac{x+y}{3+4}=2\Rightarrow x=2.3=6;y=2.4=8\)