a, Mình nghĩ là đề sai .

b, Ta có : \(\frac{x-45}{55}+\frac{x-47}{45}=\frac{x-55}{45}+\frac{x-53}{47}\)

=> \(\frac{x-45}{55}-1+\frac{x-47}{45}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)

=> \(\frac{x-45}{55}-\frac{55}{55}+\frac{x-47}{53}-\frac{53}{53}=\frac{x-55}{45}-\frac{45}{45}+\frac{x-53}{47}-\frac{47}{47}\)

=> \(\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)

=> \(\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)

=> \(\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)

=> \(x-100=0\)

=> \(x=100\)

Vậy phương trình trên có tập nghiệm là \(S=\left\{100\right\}\)

c, Ta có : \(\frac{2-x}{2010}-1=\frac{1-x}{2011}-\frac{x}{2012}\)

=> \(\frac{2-x}{2010}-1=\frac{1-x}{2011}+\frac{-x}{2012}\)

=> \(\frac{2-x}{2010}+1=\frac{1-x}{2011}+1+\frac{-x}{2012}+1\)

=> \(\frac{2-x}{2010}+\frac{2010}{2010}=\frac{1-x}{2011}+\frac{2011}{2011}+\frac{-x}{2012}+\frac{2012}{2012}\)

=> \(\frac{2012-x}{2010}=\frac{2012-x}{2011}+\frac{2012-x}{2012}\)

=> \(\frac{2012-x}{2010}-\frac{2012-x}{2011}-\frac{2012-x}{2012}=0\)

=> \(\left(2012-x\right)\left(\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)=0\)

=> \(2012-x=0\)

=> \(x=2012\)

Vậy phương trình trên có tập nghiệm là \(S=\left\{2012\right\}\)

mấy câu này dễ mà :V câu a+c lấy mỗi phân số trừ cho 1 ra tử chung rút ra thì tính b+d thì cộng một tử chung rồi lại tính tiếp thôi

a) x+1/2004 + 1 + x+2/2003 +1 - x+3/2002 +1 - x+4/2001 +1

=> x+2005/2004 + x+2005/2003 - x+2005/2002 - x+2005/2001=0

=> (x + 2005) ( 1/2004+1/2003 - 1/2002 - 1/2001) =0

ta thấy 1/2004+1/2003-1/2002-1/2001 # 0

=> x+2005=0 => x=-2005

x^3 - x^2 - 21x + 45 = 0

=>x^3 + 5x^2 - 6x^2 - 30x + 9x + 45 = 0

=>  x^2(x + 5) - 6x(x + 5) + 9(x + 5) = 0

=> (x^2 - 6x + 9)(x + 5) = 0

=> (x - 3)^2(x + 5) = 0

=> x - 3 = 0 hoặc x + 5 = 0

=> x = 3 hoặc x = -5

Ta có: x3−x2+x−1=0

⇔x2(x−1)+(x−1)=0

⇔(x−1)(x2+1)=0(1)

Ta có: x2≥0∀x

⇒x2+1≥1≠0∀x(2)

Từ (1) và (2) suy ra x−1=0

⇔x=1Ta có: x3−x2+x−1=0

⇔x2(x−1)+(x−1)=0

⇔(x−1)(x2+1)=0(1)

Ta có: x2≥0∀x

⇒x2+1≥1≠0∀x(2)

Từ (1) và (2) suy ra x−1=0

⇔x=1

Ta có x/33-x/45=2/3

                   x/45=x/33-2/3

                   x/45=x/33-22/33

                   x/45=x-22/33

=>33x=(x-22)*45

=>33x=45x-990

=>33x-45x=-990

=>-12x=-990

=>x=82,5

cảm ơn nhá

\(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)

\(\Rightarrow\frac{x-45}{55}-1+\frac{x-47}{53}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)

\(\Rightarrow\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)

\(\Rightarrow\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)

\(\Rightarrow\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)

\(\Rightarrow x-100=0\).Do \(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\ne0\)

\(\Rightarrow x=100\)

\(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)

\(\frac{x-45}{55}-1-\frac{x-47}{53}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)

\(\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)

\(\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)

(x-100)(\(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}=0\)

-> x-100 = 0 -> x = 100

mà \(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\) khác 0

Vậy x = 100

\(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)

\(\Leftrightarrow\frac{x-45}{55}-1+\frac{x-47}{53}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)

\(\Leftrightarrow\frac{x-45-55}{55}+\frac{x-47-53}{47}-\frac{x-55-45}{45}-\frac{x-53-47}{47}=0\)

\(\Leftrightarrow\frac{x-100}{55}+\frac{x-100}{47}-\frac{x-100}{45}-\frac{x-100}{47}=0\)

\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)

\(\Leftrightarrow x-100=0\)

\(\Leftrightarrow x=100\)

Vậy pt có tập nghiệm S = { 100 }

Ta có : \(x^3-x^2-21x+45=0\)

=> \(x^3-3x^2+2x^2-6x-15x+45=0\)

=> \(x^2\left(x-3\right)+2x\left(x-3\right)-15\left(x-3\right)=0\)

=> \(\left(x^2+2x-15\right)\left(x-3\right)=0\)

=> \(\left(x^2+3x-5x-15\right)\left(x-3\right)=0\)

=> \(\left(x\left(x+3\right)-5\left(x+3\right)\right)\left(x-3\right)=0\)

=> \(\left(x-5\right)\left(x+3\right)\left(x-3\right)=0\)

=> \(\left[{}\begin{matrix}x-5=0\\x-3=0\\x+3=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=5\\x=3\\x=-3\end{matrix}\right.\)

Vậy phương trình có nghiệm là x = 5, x = -3, x = 3 .

Mk xin lỗi nha, câu c sai đề

c) (x+6)⁴ + (x+8)⁴ = 272