Sửa đề: \(\left(4,5-x\right)^4+\left(5,5-x\right)^4=1\)

=>\(\left(x-4,5\right)^4+\left(x-5,5\right)^4=1\)

=>\(\left\lbrack\left(x-5\right)+0,5\right\rbrack^4+\left\lbrack\left(x-5\right)-0,5\right\rbrack^4=1\) (1)

Đặt a=x-5; b=0,5

\(\left(a+b\right)^4+\left(a-b\right)^4\)

\(=\left(a^2+2ab+b^2\right)^2+\left(a^2-2ab+b^2\right)^2\)

\(=\left\lbrack\left(a^2+b^2\right)^2+2\cdot2ab\cdot\left(a^2+b^2\right)+\left(2ab\right)^2\right\rbrack+\left\lbrack\left(a^2+b^2\right)^2-2\cdot2ab\cdot\left(a^2+b^2\right)+\left(2ab\right)^2\right\rbrack\)

\(=2\left(a^4+2a^2b^2+b^4\right)+8a^2b^2=2a^4+12a^2b^2+2b^4\)

\(=2\left(x-5\right)^4+12\cdot\left(x-5\right)^2\cdot\left(0,5\right)^2+2\cdot\left(0,5\right)^4\)

\(=2\left(x-5\right)^4+3\left(x-5\right)^2+0,125\)

(1) sẽ trở thành: \(2\left(x-5\right)^4+3\left(x-5\right)^2+0,125=1\)

=>\(2\left(x-5\right)^4+3\left(x-5\right)^2-0,875=0\)

=>\(16\left(x-5\right)^4+24\left(x-5\right)^2-7=0\)

=>\(16\left(x-5\right)^4+28\left(x-5\right)^2-4\left(x-5\right)^2-7=0\)

=>\(\left\lbrack4\left(x-5\right)^2+7\right\rbrack\left\lbrack4\left(x-5\right)^2-1\right\rbrack=0\)

=>\(4\left(x-5\right)^2-1=0\)

=>\(\left(2x-10\right)^2=1\)

=>\(\left[\begin{array}{l}2x-10=1\\ 2x-10=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=11\\ 2x=9\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{11}{2}\\ x=\frac92\end{array}\right.\)

a)

b)
+\(x> 5,5\)
\(=> x - 4,5 > 1\)
\(=>(x -4,5)^4 > 1\)
=> pt vô nghiệm.
+\(x < 4,5 \)
\(​=> x - 5,5 < -1\)
\(=>(x - 5,5)^4 > 1\)
=> pt vô nghiệm

+\(4,5 < x < 5,5\)
\(=>(x - 4,5)^4 + (x - 5,5)^4 = (x -4,5)^4 + (5,5 -x)^4 < (x - 4,5 +5,5 -x)^4 = 1\)

vậy chung lại \(x = 4,5\) hoặc \(5,5\) là nghiệm

Câu b bạn giải đc cách đặt ẩn phụ kg

a: \(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2.15+3.75=1.6=\dfrac{8}{5}\)

=>x+4/15=8/5 hoặc x+4/15=-8/5

=>x=4/3 hoặc x=-28/15

c: =>x-y=0 và y+9/25=0

=>x=y=-9/25

d: =>-1/3<x-3/5<1/3

=>4/15<x<14/15

e: =>|x+5,5|>5,5

=>x+5,5>5,5 hoặc x+5,5<-5,5

=>x>0 hoặc x<-11

Nhân 2 vế với 2 rồi chuyển vế và rút gọn

Bạn Tên Là Long

a/ \(\Leftrightarrow2x^3+9x^2-27=0\)

\(\Leftrightarrow2x^3+12x^2+18x-3x^2-18x-27=0\)

\(\Leftrightarrow2x\left(x^2+6x+9\right)-3\left(x^2+6x+9\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(x+3\right)^2=0\)

\(\Leftrightarrow...\)

b/ \(\Leftrightarrow x^3-3x^2+3x-1+x^3+x^3+3x^2+3x+1=x^3+6x^2+12x+8\)

\(\Leftrightarrow x^3-3x^2-3x-4=0\)

\(\Leftrightarrow\left(x-4\right)\left(x^2+x+1\right)=0\)

c/ \(x\left(x+1\right)\left(x-1\right)\left(x+2\right)-24=0\)

\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)-24=0\)

Đặt \(x^2+x=t\)

\(t\left(t-2\right)-24=0\Leftrightarrow t^2-2t-24=0\Rightarrow\left[{}\begin{matrix}t=6\\t=-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2+x=6\\x^2+x=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+x-6=0\\x^2+x+4=0\end{matrix}\right.\)

d/ \(\Leftrightarrow\left(x-7\right)\left(x-2\right)\left(x-4\right)\left(x-5\right)-72=0\)

\(\Leftrightarrow\left(x^2-9x+14\right)\left(x^2-9x+20\right)-72=0\)

Đặt \(x^2-9x+14=0\)

\(t\left(t+6\right)-72=0\Leftrightarrow t^2+6t-72=0\Rightarrow\left[{}\begin{matrix}t=6\\t=-12\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2-9x+14=6\\x^2-9x+14=-12\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-9x+8=0\\x^2-9x+26=0\end{matrix}\right.\)