\(\left(4x-15\right)^{2016}=\left(4x-15\right)^{2015}\\ \Leftrightarrow\left[{}\begin{matrix}4x-15=0\\4x-15=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}4x=15\\4x=16\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{4}\\x=4\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{15}{4};4\right\}\)

a) => (4x-15).(4x-15)²⁰¹⁵=(4x-15)²⁰¹⁵

=> 4x-15=1

=> x=4

b) => 4.2^x+6-480= 0

=> 4.2^x-474=0

=> 4.2^x=474

=> 2^x= 118,5

ko có gt x thoả mãn đề bài

chả biết câu b trình bày đúng hay sai, hay là đầu bài chép nhầm nữa. Nếu sai ai đó chữa lại hộ cái nhé

                                                                              _HẾT_

b, 2^x+2^x+1+2^x+2+2^x+3-480=0

2^x.1+2^x.2+2^x.2^2+2^x.2^3=480

2^x.(1+2+2^2+2^3)=480

2^x.15=480

2^x=32

2^x=2^5

x=5

1) Áp dụng tích chất dãy tỉ số bằng nhau ta có:

\(\frac{x+y}{2015}=\frac{xy}{2016}=\frac{x-y}{2017}=\frac{x+y-x+y}{2015-2017}=\frac{2y}{-2}\)

\(=-y\)

\(\Rightarrow xy=-2016y;x+y=-2015y;\)

\(x-y=-2017y\)

\(\Rightarrow-2016y-xy=0\)

\(\Rightarrow y\left(-2016-x\right)=0\)

\(\Rightarrow\orbr{\orbr{\begin{cases}y=0\\-2016-x=0\end{cases}\Rightarrow}}\orbr{\begin{cases}y=0\\x=-2016\end{cases}}\)

\(+) \)\(y=0\Rightarrow0+x=-2015.0=0\Rightarrow x=0\)

\(+) \)\(x=-2016\Rightarrow-2016-y=-2017y\Rightarrow-2016\)

Vậy +) x=y=0

       +) x=-2016;y=1

2) Có: \(\frac{2x+2}{3}=\frac{x+1}{1,5};\frac{4z+2}{5}=\frac{z+0,5}{1,25};\frac{3y-1}{4}=\frac{y-\frac{1}{3}}{\frac{4}{3}}\)

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

\(\frac{x+1}{1,5}=\frac{y-\frac{1}{3}}{\frac{4}{3}}=\frac{z+0,5}{1,25}=\frac{x+y+z+\left(1-\frac{1}{3}+0,5\right)}{1,5+\frac{4}{3}+1,25}=\frac{7+\frac{7}{6}}{\frac{49}{12}}=2\)

Suy ra: \(x+1=2.1,5=3\Rightarrow x=2\)

             \(y-\frac{1}{3}=2.\frac{4}{3}=\frac{8}{3}\Rightarrow y=3\)

            \(z+0,5=2.1,25=2,5\Rightarrow z=2\)

Vậy x=2;y=3;z=2.

Câu 1 :

Áp dụng t/c dãy TSBN ta có : \(\frac{x+y}{2015}=\frac{xy}{2016}=\frac{x-y}{2017}=\frac{x+y+x-y}{2015+2017}=\frac{x}{2016}\)

\(\Rightarrow\frac{xy}{2016}=\frac{x}{2016}\)=> xy=x => xy-x=0 => x(y-1)=0 => x=0 hoặc y=1

+) Nếu x=0 => \(\frac{0+y}{2015}=\frac{0.y}{2016}\Rightarrow\frac{y}{2015}=0\Rightarrow y=0\)

+) Nếu y=1 => \(\frac{x+1}{2015}=\frac{x.1}{2016}\)=> 2016(x+1)=2015x => 2016x+2016 = 2015x => x=-2016

             Vậy ...

Câu 2 :

Áp dụng t/c dãy TSBN ta có : \(\frac{2x+2}{3}=\frac{3y-1}{4}=\frac{4z+2}{5}=\frac{6.\left(2x+2\right)+4.\left(3y-1\right)+3.\left(4z+2\right)}{3.6+4.4+5.3}\)

                                             \(=\frac{12\left(x+y+z\right)+14}{49}=\frac{12.7+14}{49}=2\)

Từ  \(\frac{2x+2}{3}=2\Rightarrow2x+2\Rightarrow6\Rightarrow2x=4\Rightarrow x=2\)

Tương tự tìm đc y=3 và z=2

            Vậy ...

1) 3x - 6 = 5x + 2

=> 3x - 5x = 2 + 6

=> -2x = 8 

=> x = -4

2) 15  - x = 4x - 5

=> 15 + 5 = 4x + x 

=> 20 = 5x

=> x = 4

3) x - 15 = 6 + 4x

=> x - 4x = 6 + 15

=> -3x = 21

=> x = -7

4) -12 + x = 5x - 20 

=> x - 5x = -20 + 12

=> -4x = -8

=> x = 2

5) 7x - 4 = 20 + 3x

=> 7x - 3x = 20 + 4

=> 4x = 24

=> x = 6

1) 3x-  6 = 5x + 2

5x - 3x = -6 - 2

2x = -8 => x = -4

Tương tự như trên 

2x-8/-3+x=-5x+6/4

a) 5 - x + 12 = 4 + x + 1

17 - x = 5 + x

x - (-x) = 17 - 5

2x = 12

x = 6 

b) 4x - 5 + (-15) = 3x - 10

4x - 20 = 3x - 10

3x - 4x = -20 + 10

-x = -10

x = 10

a)

5 - x + 12 = 4 + x + 1

17 - x = 5 + x x - (-x)

= 17 - 5 2x

= 12 x

= 6 b)

4x - 5 + (-15)

= 3x - 10 4x - 20

= 3x - 10 3x - 4x

= -20 + 10 -x

= -10 x = 10 

\(\dfrac{\left(4x-1\right)}{15}=\dfrac{\left(x+2\right)}{5}\)

\(\dfrac{\left(4x-1\right)}{15}-\dfrac{\left(x+2\right)}{5}=0\)

\(\dfrac{\left(4x-1\right)}{15}-\dfrac{3\left(x+2\right)}{3\times5}=0\)

\(\dfrac{4x-1}{15}-\dfrac{3x+6}{15}=0\)

\(4x-1-3x-6=0\)

\(x-7=0\)

\(x=7\)

\(\dfrac{4x-1}{15}=\dfrac{x+2}{5}\Rightarrow4x-1=\dfrac{15}{5}.\left(x+2\right)\)

\(\Rightarrow4x-1=3.\left(x+2\right)\)

\(\Rightarrow4x-1=3x+6\)

\(\Rightarrow x=7\)

\(\dfrac{4x-1}{15}=\dfrac{x+2}{5}\)

\(\Leftrightarrow\dfrac{4x-1}{15}=\dfrac{3\left(x+2\right)}{15}\)

\(\Leftrightarrow4x-1=3x+6\)

\(\Leftrightarrow4x-3x=6+1\)

\(\Leftrightarrow x=7\)

Vậy \(x=7\)

Ta có: (x+2015)^2016>=0(với mọi x)

|y-2017|>=0(với mọi y)

Do đó, (x+2015)^2016+|y-2017|>=0(với mọi x,y)

mà (x+2015)^2016+|y-2017|=0

nên (x+2015)^2016=0                 và |y-2017|=0

      x+2015=0                              y-2017=0

      x=0-2015                              y=0+2017

      x=-2015                               y=2017

Vậy x=-2015 và y=2017 thì x,y thỏa mãn đề

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

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

=> \(2x=-4030y\)

=>\(x=-2015y\)

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

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

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

\(-y=-y^2\)

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

   \(y\).(\(1-y\))\(=0\)

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

TH1 :\(y=0=>x.y=-2015.0=0\)

TH2 :\(y=1=>x.y=-2015.1=-2015\)

x=0 y=0

x=-2015 y=1

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 \(y=1\):

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