Bài 1: áp dụng tính chất dãy tỉ số bằng nhau ta được:

(a+b-c)/c=(b+c-a)/a=(c+a-b)/b=(a+b-c+b+c-a+c+a-b)/(a+b+c)=(a+b+c)/(a+b+c0=1

Do đó: (a+b+c)/c=1 suy ra a+b+c=c suy ra a+b=c-c=0 nên a=b (1)

(b+c-a)/a=1 suy ra b+c-a=a suy ra a+c-a=a (b=a) suy ra c=a (2) Từ (1) và(2) ta có: a=b=c

Suy ra:P= (1+b/a).(1+c/b).(1+a/c)=(1+a/a).(1+a/a).(1+a/a)=(1+1).(1+1).(1+1)=2.2.2=8

Bài 2: bạn cũng áp dụng tính chất dãy tỉ bằng nhau rồi xét giống bài 1 là ra

Theo đề, ta có:   \(\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{z}{t}=\dfrac{t}{x}\) \(=\dfrac{x+y+z+t}{y+z+t+x}=1\) .

\(\Rightarrow x=y;y=z;z=t;t=x\)

\(\Rightarrow x=y=z=t\)

\(M=\dfrac{2x-y}{z+t}+\dfrac{2y-z}{t+x}+\dfrac{2z-t}{x+y}+\dfrac{2t-x}{y-z}\)

\(M=\dfrac{2x-x}{x+x}+\dfrac{2x-x}{x+x}+\dfrac{2x-x}{x+x}+\dfrac{2x-x}{x+x}\)

\(M=\dfrac{1}{2}.4\)

\(M=2\)

\(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}=x\left(\frac{x}{y+z}+1-1\right)+y\left(\frac{y}{x+z}+1-1\right)+z\left(\frac{z}{x+y}+1-1\right)\)

\(=x\left(\frac{x+y+z}{y+z}-1\right)+y\left(\frac{x+y+z}{x+z}-1\right)+z\left(\frac{x+y+z}{x+y}-1\right)\)

\(=\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\right)-\left(x+y+z\right)=0\)

\(M=2019\)

a) A = x(y - z) + 2(z - y) = x(y - z) - 2(y - z) = (x - 2)(y - z) = (2 - 2)(1,007 - (-0,006)] = 0

b) B = 2x(y - z) + (z - y)(x + t) = 2x(y - z)  - (y - z)(x + t) = (2x - x - t)(y - z) = (x - t)(y - z) = [18,3 - (-31,7)](24,6 - 10,6) = 50.14 = 700

c) C = (x - y)(y + z) + y(y - x) = (x - y)(y + z) - y(x - y) = (x - y)(y + z - y) = (x - y).z = (0,86 - 0,26).1,5 = 0,6.1,5 = 0,9

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

⇒(x+y+z)(\(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\))=x+y+z

⇔\(\frac{x^2+xy+xz}{y+z}+\frac{xy+y^2+yz}{x+z}+\frac{xz+yz+z^2}{x+y}=x+y+z\)

⇔\(\frac{x^2}{y+z}+\frac{x\left(y+z\right)}{y+z}+\frac{y^2}{x+z}+\frac{y\left(x+z\right)}{x+z}+\frac{z^2}{x+y}+\frac{z\left(x+y\right)}{x+y}=x+y+z\)

⇔\(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}+x+y+z=x+y+z\)

Hay M+x+y+z=x+y+z

=>M=0

Lời giải:

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

\(\Rightarrow \left\{\begin{matrix} \frac{x^2}{y+z}+\frac{xy}{z+x}+\frac{xz}{x+y}=x\\ \frac{xy}{y+z}+\frac{y^2}{z+x}+\frac{zy}{x+y}=y\\ \frac{xz}{y+z}+\frac{yz}{z+x}+\frac{z^2}{x+y}=z\end{matrix}\right.\)

Cộng theo vế cả 3 đẳng thức trên:

\(\Rightarrow \frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}+\frac{xy+yz}{x+z}+\frac{xz+yz}{x+y}+\frac{xy+xz}{y+z}=x+y+z\)

\(\Leftrightarrow \frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}+y+z+x=x+y+z\)

\(\Leftrightarrow \frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}=0\)

Vậy $M=0$

nhân cả 2 vế với 2 rồi bunhia

câu c là \(\dfrac{1}{2}\)(x+y+z) nhé, mih chép nhầm

Đặt \(\dfrac{1}{a}=\dfrac{1}{x+y},\dfrac{1}{b}=\dfrac{1}{y+z},\dfrac{1}{c}=\dfrac{1}{z+x}\)

Đề trở thành: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\), tính \(P=\dfrac{bc}{a^2}+\dfrac{ac}{b^2}+\dfrac{ab}{c^2}\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\) Tương đương \(ab+bc=-ac\)

\(P=\dfrac{b^3c^3+a^3c^3+a^3b^3}{a^2b^2c^2}=\dfrac{\left(ab+bc\right)\left(a^2b^2-ab^2c+b^2c^2\right)+a^3c^3}{a^2b^2c^2}=\dfrac{-ac\left(a^2b^2-ab^2c+b^2c^2\right)+a^3c^3}{a^2b^2c^2}\)

\(=\dfrac{a^2c^2-a^2b^2+ab^2c-b^2c^2}{ab^2c}=\dfrac{ac}{b^2}-\dfrac{a}{c}+1-\dfrac{c}{a}\)\(=ac\left(\dfrac{1}{a^2}+\dfrac{2}{ac}+\dfrac{1}{c^2}\right)-\dfrac{a}{c}+1-\dfrac{c}{a}\) (do \(\dfrac{1}{b}=-\dfrac{1}{a}-\dfrac{1}{c}\) tương đương \(\dfrac{1}{b^2}=\dfrac{1}{a^2}+\dfrac{2}{ac}+\dfrac{1}{c^2}\)) 

\(=3\)

Vậy P=3