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

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

\(\left(4.x-15\right)^{2016}=\left(4.x-15\right)^{2015}\)

=>\(\left(4.x-15\right)^{2016}-\left(4.x-15\right)^{2015}=0\)

=>\(\left(4.x-15\right)^{2015}.\left[\left(4.x-15\right)-1\right]=0\)

=>\(\hept{\begin{cases}\left(4.x-15\right)^{2015}=0\\\left(4.x-15\right)-1=0\end{cases}}\)=>\(\hept{\begin{cases}4x-15=0\\4.x-15-1=0\end{cases}}\)=>\(\hept{\begin{cases}4x=15\\4x=16\end{cases}}\)

=>\(\hept{\begin{cases}x=\frac{15}{4}\\x=4\end{cases}}\)

Vậy.................

(4x - 15)²⁰¹⁶ = (4x - 15)²⁰¹⁵

(4x - 5)²⁰¹⁶ - (4x - 15)²⁰¹⁵ = 0

(4x - 5)²⁰¹⁵.(4x - 15) - (4x - 15)²⁰¹⁵ = 0

(4x - 5)²⁰¹⁵.[(4x - 15) - 1] = 0

=> \(\orbr{\begin{cases}\left(4x-5\right)^{2015}=0\\\left(4x-15\right)-1=0\end{cases}}\)=>\(\orbr{\begin{cases}4x-15=0\\4x-15=1\end{cases}}\)

=> \(\orbr{\begin{cases}4x=15\\4x=16\end{cases}}\)=>\(\orbr{\begin{cases}x=\frac{15}{4}\\x=4\end{cases}}\)

5-x-16=40+x

-x-x=40+16-5

(-x).2=51

-x=51/2=>x=51/2

1.x=40

2.x=10

3.x=

15-x = 4x -5 

15 + 5 = 4x + x 

5x = 20

x= 4

x - 15 = 6 + 4x 

-15 - 6 = 4x -x 

3x = -21 

x= -7

Tìm max B biết B=15−4x−x2B=15-4x-x^2B=15−4x−x2

DELL THỂ hiểu đc đề ghi đề như shi* vậy :(

x2 là J tui chưa thấy đề nào như zầy cả :)))

Ta có : \(-4x^2+4x-5=-\left(4x^2-4x+5\right)=-\left(2x-1\right)^2-4\le-4\)

\(\Rightarrow B\ge\dfrac{15}{-4}\)

Dấu ''='' xảy ra khi x = 1/2 

Vậy GTNN B là -15/4 khi x = 1/2 

a) -12+x=5x-20

-12+x  -5x = -20

-12+ ( -4x) = -20

-4x= -20-(-12)

-4x=-8

x= -8: (-4)

x= 2

Vậy....

Ko chắc nha

b) 4x-10=15-x

4x-10 +x=15

5x -10=15

5x= 15+10

5x= 25

x=25:5

x=5 

Vậy...

c) 7x-4=20+3x

7x -4 - 3x= 20

4x -4= 20

4x= 20+4

4x= 24

x= 24:4

x=6

Vậy...

Đặt \(x^2-4x=t\)

\(\Rightarrow t^2-8t+15=0\)

\(\Leftrightarrow t^2-3t-5t+15=0\)

\(\Leftrightarrow t\left(t-3\right)-5\left(t-3\right)=0\)

\(\Leftrightarrow\left(t-3\right)\left(t-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}t=5\\t=3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2-4x=5\\x^2-4x=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-5=0\\x^2-4x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-5x-5=0\\x^2-4x+4-7=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\left(x+1\right)-5\left(x+1\right)=0\\\left(x-2\right)^2-7=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)\left(x-5\right)=0\\\left(x-2-\sqrt{7}\right)\left(x-2+\sqrt{7}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=5\\x=2+\sqrt{7}\\x=2-\sqrt{7}\end{matrix}\right.\)

6.4^\(x\).4¹ + 4\(^x\).4\(^3\) = 2⁸.(7 + 15)

4\(^x\).(6.4 + 4³) = 2⁸.22

4\(^x\).(24 + 64) = 256.22

4\(^x\). 88 =       5632

4\(^x\)      = 532 : 88

4\(^x\)     = 64

4\(^x\)     = 4³

\(x\) = 3 

\(\left(4x+1\right)\left(x-3\right)-\left(x-7\right)\left(4x-1\right)=15\\ \Leftrightarrow4x^2+x-12x-3-\left(4x^2-28x-x+7\right)-15=0\\ \Leftrightarrow4x^2-11x-3-4x^2+29x-7-15=0\\ \Leftrightarrow18x=25\\ \Leftrightarrow x=\dfrac{25}{18}\)

Vậy \(x=\dfrac{25}{18}\)

Lời giải:

$(4x+1)(x-3)-(x-7)(4x-1)=15$

$\Leftrightarrow (4x^2-11x-3)-(4x^2-29x+7)=15$

$\Leftrightarrow 18x-10=15$

$\Leftrightarrow 18x=25$

$\Leftrightarrow x=\frac{25}{18}$

\(0,25\cdot\dfrac{7}{15}-\dfrac{1}{4}\cdot\left(-\dfrac{8}{15}\right)+2,75\)

\(=\dfrac{1}{4}\cdot\left(\dfrac{7}{15}+\dfrac{8}{15}\right)+2,75\)

\(=\dfrac{1}{4}\cdot1+2,75\)

\(=0,25+2,75\\ =3\)