\(\left(x-5\right)^4=\left(x-5\right)^6\)

\(\Rightarrow\left(x-5\right)^6-\left(x-5\right)^4=0\)

\(\Rightarrow\left(x-5\right)^4.\left[\left(x-5\right)^2-1\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(x-5\right)^4=0=0^4\\\left(x-5\right)^2-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\\left(x-5\right)^2=1=1^2=\left(-1\right)^2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=5\\x-5=1;x-5=-1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=5\\x=6;x=4\end{cases}}\)

Vậy \(x\in\left\{4;5;6\right\}\)

=> (x-5)⁴.(1-(x-5)²) =0(đảo vế và đặt thừa số chung)

=>1-(x-5)²=0      =>(x-5)²=1                    => x=6

                                                                      hoặc x=4

hoặc (x-5)⁴=0          hoặc x-5=0            hoặc x=5

(x-5)⁴ = (x-5)⁶

(x-5)⁴ - ( x-5)⁶ = 0

(x-5)⁴.{ 1 - (x-5)² }= 0

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

\(\left[{}\begin{matrix}x=5\\x-5=1\\x-5=-1\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)

\(x\in\) { 4; 5; 6}

1)    \(\frac{2}{3}x+\frac{3}{4}\left(x-5\right)=-\frac{5}{6}\)

      \(\frac{2}{3}x+\frac{3}{4}x-\frac{15}{4}=-\frac{5}{6}\)

       \(\frac{17}{12}x=-\frac{5}{6}+-\frac{15}{4}=-\frac{55}{12}\)

   \(x=-\frac{55}{12}:\frac{17}{12}=-\frac{55}{17}\)

a  tìm số nguyên x biết (x-5).(y-7)=1 
   (x-5).(y-7)=1 = 1.1 = -1.(-1) 
   TH1,
   x-5 = 1, y-7 = 1
   => x = 6, y = 8
   TH2

(x - 5)⁴ = (x - 5)⁶

⇒ (x - 5)⁶ - (x - 5)⁴ = 0

⇒ (x - 5)⁴.[(x - 5)² - 1]

⇒ (x - 5)⁴ (x - 6)(x - 4) = 0

⇒ x - 5 = 0 hoặc x - 6 = 0 hoặc x - 4 =0

⇒ x ϵ {5;6;4}

nếu có sai thì báo mình nhé, đây là đề ccuar cô chứ mình ko biết đau

1/ Phân tích đa thức thành nhân tử

a) x² – 5x – 6

b) x² – x – 6

c) 2x² – 3x + 1

d) 4 x⁴ – 21x²y² + y⁴

Giúp mk nha

Mk cần gấp lắm a) x2 - 5x - 6

= x2 + x - 6x - 6

=x(x + 1) - 6(x + 1)

= (x + 1)(x - 6)

b) x2 - x - 6

= x2 + 2x - 3x - 6

= x(x + 2) - 3(x + 2)

= (x + 2)(x - 3)

c) 2x2 - 3x + 1

= 2x2 - 2x - x + 1

= 2x(x - 1) - (x - 1)

= (x - 1)(2x - 1)

Ta có x(x + 2) = x² + 5x + 6

<=> x(x + 2) = x² + 2x + 3x + 6 

<=> x(x + 2) = x(x + 2) + 3(x + 2) 

<=> x(x + 2) = (x + 2)(x + 3)

<=> x(x + 2) - (x + 2)(x + 3) = 0

<=> (x + 2)(x - x - 3) = 0 

<=> (x + 2).(-3) = 0

<=> x + 2 = 0

<=> x = -2

Vậy x = -2 là giá trị  cần tìm 

Giải:

a) \(\dfrac{-5}{8}=\dfrac{x}{16}\) 

\(\Rightarrow x=\dfrac{16.-5}{8}=-10\) 

\(\dfrac{3x}{9}=\dfrac{2}{6}\) 

\(\Rightarrow3x=\dfrac{2.9}{6}=3\) 

\(\Rightarrow x=1\)

b) \(\dfrac{x+3}{15}=\dfrac{1}{3}\)  

\(\Rightarrow x+3=\dfrac{1.15}{3}=5\) 

\(\Rightarrow x=2\)

\(\dfrac{6}{2x+1}=\dfrac{2}{7}\) 

\(\Rightarrow2x+1=\dfrac{6.7}{2}=21\) 

\(\Rightarrow x=10\)

c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\) 

\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\) 

\(\Rightarrow x-6=\dfrac{18.4}{-12}=-6\) 

\(\Rightarrow x=0\) 

\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\) 

\(\Rightarrow y=\dfrac{-12.24}{18}=-16\) 

 \(\dfrac{3-x}{-12}=\dfrac{16}{y+1}=\dfrac{192}{-72}\) 

\(\Rightarrow\dfrac{3-x}{-12}=\dfrac{192}{-72}\) 

\(\Rightarrow3-x=\dfrac{192.-12}{-72}=32\) 

\(\Rightarrow x=-29\) 

\(\Rightarrow\dfrac{16}{y+1}=\dfrac{192}{-72}\) 

\(\Rightarrow y+1=\dfrac{16.-72}{192}=-6\) 

d) \(\dfrac{-2}{3}< \dfrac{x}{5}< \dfrac{-1}{6}\) 

\(\Rightarrow\dfrac{-20}{30}< \dfrac{6x}{30}< \dfrac{-5}{30}\) 

\(\Rightarrow6x\in\left\{-18;-12;-6\right\}\) 

\(\Rightarrow x\in\left\{-3;-2;-1\right\}\) 

\(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\) 

\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\) 

\(\Rightarrow5x\in\left\{-5;0;5;10\right\}\) 

\(\Rightarrow x\in\left\{-1;0;1;2\right\}\) 

e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\) 

\(\Rightarrow\dfrac{x+46}{20}=x+\dfrac{2}{5}\) 

\(\Rightarrow\dfrac{x+46}{20}=\dfrac{5x+2}{5}\) 

\(\Rightarrow5.\left(x+46\right)=20.\left(5x+2\right)\) 

\(\Rightarrow5x+230=100x+40\) 

\(\Rightarrow5x-100x=40-230\) 

\(\Rightarrow-95x=-190\) 

\(\Rightarrow x=-190:-95\) 

\(\Rightarrow x=2\) 

\(y\dfrac{5}{y}=\dfrac{86}{y}\) 

\(\Rightarrow y+\dfrac{5}{y}=\dfrac{86}{y}\) 

\(\Rightarrow\dfrac{y^2+5}{y}=\dfrac{86}{y}\) 

\(\Rightarrow y^2+5=86\) 

\(\Rightarrow y^2=86-5\) 

\(\Rightarrow y^2=81\) 

\(\Rightarrow\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\) 

Chúc bạn học tốt!