ta thấy x=0;y=0 là 1 nghiệm

Ta có: \(\frac{x+y}{2014}\ne\frac{x-y}{2016}\)

\(\Leftrightarrow2016x+2016y=2014x-2014y\)

\(\Leftrightarrow2x=-4030y\)

\(\Leftrightarrow x=-2015y\)

Thay \(x=-2015y\)vào \(\frac{x+y}{2014}=\frac{xy}{2015}\)ta được:

\(\Leftrightarrow\frac{-2015+y}{2014}=\frac{-2015y}{2015}\)

\(\Leftrightarrow\frac{-2014y}{2014}=\frac{-2015y^2}{2015}\)

\(\Leftrightarrow-y=-y^2\)

\(\Leftrightarrow y-y^2=0\)

\(\Leftrightarrow y\left(1-y\right)=0\)

\(\Rightarrow\orbr{\begin{cases}y=0\\1-y=0\end{cases}}\Rightarrow\orbr{\begin{cases}y=0\\y=1\end{cases}}\)

Trường hợp \(y=0\):

\(y=0\Rightarrow x.y=-2015.0=0\)

Trường hợp \(y=1\):

\(y=1\Rightarrow x.y=-2015.1=-2015\)

bạn ơi đề thiếu

\(5^x\left(5^2+1\right)=650\)

\(5^x.26=650\)

\(5^x=25\)

\(x=2\)

5^x+(5^x+2)=650

(5^x+5^x)+2=650( tính chất kết hợp)

2.5^x=648

5^x=648:2=324

Điều này ko xảy ra vì 5^x luôn lẻ, còn 324 chẵn

\(\frac{x+2}{x-2}-\frac{x-2}{x+2}+\frac{16}{4-x^2}\)

\(=\frac{\left(x+2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{-16}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{\left(x+2\right)^2-\left(x-2\right)^2-16}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x^2+4x+4-x^2+4x-4-16}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{8x-16}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{8\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{8}{x+2}\)

bạn ơi, cho mình hỏi

4-x² thì sẽ = 2²-x² = (2-x)(2+x)

vậy sao bạn ghi là (x-2)(x+2) vậy???

Biến đổi như thế này nha bạn :

\(\frac{16}{4-x^2}=\frac{-1\cdot16}{-1\cdot\left(4-x^2\right)}\)

\(=\frac{-16}{-4+x^2}=\frac{-16}{x^2-4}\)

\(=\frac{-16}{\left(x-2\right)\left(x+2\right)}\)

Ta có: \(3x=2y\Rightarrow y=\frac{3}{2}x\)\(;\)\(3x=\frac{3}{2}z\Rightarrow z=\frac{3}{\frac{3}{2}}x\Rightarrow z=2x\)

\(\Rightarrow x+y+z=x+\frac{3}{2}x+2x=4,5x=18\Rightarrow x=4\)

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

(Dấu . là dấu nhân nha bạn)

Áp dụng tính chất của dãy tỉ số = nhau ta có:

\(\frac{b+c-a}{a}=\frac{c+a-b}{b}=\frac{a+b-c}{c}=\frac{\left(b+c-a\right)+\left(c+a-b\right)+\left(a+b-c\right)}{a+b+c}\)

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

\(\frac{1}{4}+\frac{1}{3}:\left(2x-1\right)=-5\)

\(\frac{1}{3}:\left(2x-1\right)=-5-\frac{1}{4}\)

\(\frac{1}{3}:\left(2x-1\right)=-\frac{20}{4}-\frac{1}{4}\)

\(\frac{1}{3}:\left(2x-1\right)=-\frac{21}{4}\)

\(\left(2x-1\right)=\frac{1}{3}:-\frac{21}{4}\)

\(\left(2x-1\right)=\frac{1}{3}.-\frac{4}{21}\)

\(\left(2x-1\right)=-\frac{4}{63}\)

2x= -4/63 + 1

2x = 59/63

x = 59/63 : 2

x = 59/126

1/3:(2.x-1)=-5-1/4

1/3:(2.x-1)=-21/4

2.x-1=1/3:-21/4

2.x-1=-4/63

2.x=-4/63+1

2.x=\(3\frac{59}{63}\)

x=\(3\frac{59}{63}\):2

x=\(1\frac{61}{63}\)

\(\frac{1}{4}+\frac{1}{3}\div\left(2x-1\right)=-5\)

\(\frac{1}{3}\div\left(2x-1\right)=-5-\frac{1}{4}\)

\(\frac{1}{3}\div\left(2x-1\right)=\frac{-20}{4}+\frac{-1}{4}\)

\(\frac{1}{3}\div\left(2x-1\right)=\frac{-21}{4}\)

\(2x-1=\frac{1}{3}\div\frac{-21}{4}\)

\(2x-1=\frac{1}{3}.\frac{-4}{21}\)

\(2x-1=\frac{-4}{63}\)

\(2x=\frac{-4}{63}+1\)

\(2x=\frac{-4}{63}+\frac{63}{63}\)

\(2x=\frac{59}{63}\)

\(x=\frac{59}{63}\div2\)

\(x=\frac{59}{63}.\frac{1}{2}\)

\(x=\frac{59}{126}\)