=>(3x-4)⁵-(3x-4)⁷=0

=>(3x-4)⁵-(3x-4)⁵.(3x-4)²=0

=>(3x-4)⁵.[1-(3x-4)²]=0

=>(3x-4)⁵=0=>3x=4=>x=4/3

hoặc 1-(3x-4)²=0=>(3x-4)²=1=>3x-4=1 hoặc 3x-4=-1=>x=5/3 hoặc x=1

vậy x=1

\(P=\dfrac{x^4-3x^3+5}{x-3}=\dfrac{x^3\left(x-3\right)+5}{x-3}=x^3+\dfrac{5}{x-3}\)

\(P\in Z\Rightarrow\dfrac{5}{x-3}\in Z\)

\(\Rightarrow x-3=Ư\left(5\right)=\left\{-5;-1;1;5\right\}\)

\(\Rightarrow x=\left\{-2;1;4;8\right\}\)

<=>(3x-4)^5-(3x-4)^7=0

<=>(3x-4)^5-(3x-4)^5.(3x-4)^2=0

<=>(3x-4)^5.[1-(3x-4)^2]=0

<=>(3x-4)^5=0<=>3x=4<=>x=4/3(loại)

 Hoặc 1-(3x-4)^2=0<=>(3x-4)^2=1<=>3x-4=1 hoặc 3x-4=-1<=>x=1 thoả mãn

 vậy x=1

\(\left(3x-4\right)^7-\left(3x-4\right)^5=0\)

\(\Rightarrow\left(3x-4\right)^2=0\)

\(\Leftrightarrow3x-4=0\)

\(x=\frac{4}{3}\)

ap dung t/c;      1ⁿ  = 0 ta có;

3x-4 = 1

x = 5/3

hoc toan la phai nam chac kien thuc co ban mk gop y vay thoi

a: \(\Leftrightarrow3x^3-2x^2+15x^2-10x+3x-2+7⋮3x-2\)

\(\Leftrightarrow3x-2\in\left\{1;-1;7;-7\right\}\)

hay \(x\in\left\{3;1\right\}\)

b: \(\Leftrightarrow2x^5-7x^3+4x^4-14x^2+14x^2-49x+49x-44⋮2x^2-7\)

\(\Leftrightarrow2401x^2-1936⋮2x^2-7\)

\(\Leftrightarrow4802x^2-3872⋮2x^2-7\)

\(\Leftrightarrow2x^2-7\inƯ\left(12935\right)\)

\(\Leftrightarrow2x^2-7\in\left\{1;5;13;65;199;995;2587;12935;-1;-5\right\}\)

\(\Leftrightarrow2x^2\in\left\{8;72;2\right\}\)

hay \(x\in\left\{2;-2;6;-6;1;-1\right\}\)

a) Ta thực hiện phép chia \(3x^3+13x^2-7x+5\) cho \(3x-2\). Khi đó ta có:

\(A=\frac{3x^3+13x^2-7x+5}{3x-2}=3x^2+5x+1+\frac{7}{3x-2}\)

Nếu x nguyên thì \(3x^2+5x+1\in\text{Z}\) nên để A nguyên thì \(\frac{7}{3x-2}\in Z\)

\(\Rightarrow3x-2\in\left\{-7;-1;1;7\right\}\)

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

b) Ta có: \(B=\frac{2x^5+4x^4-7x^3-44}{2x^2-7}=\left(x^3+2x^2+7\right)+\frac{5}{2x^2-7}\)

Để B nguyên thì \(\frac{5}{2x^2-7}\in Z\Rightarrow2x^2-7\in\left\{-5;-1;1;5\right\}\)

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

Ta có: \(M=\dfrac{x^5+3x^3-x^2+3x-7}{x^2+2}\)

\(=\dfrac{x^5+2x^3+x^3+2x-x^2-2+x-5}{x^2+2}\)

\(=\dfrac{x^3\left(x^2+2\right)+x\left(x^2+2\right)-\left(x^2+2\right)+\left(x-5\right)}{x^2+2}\)

\(=\dfrac{\left(x^2+2\right)\left(x^3+x-1\right)+\left(x-5\right)}{\left(x^2+2\right)}\)

\(=x^3+x-1+\dfrac{x-5}{x^2+2}\)

Để M nguyên thì \(x-5⋮x^2+2\)

\(\Leftrightarrow\left(x-5\right)\left(x+5\right)⋮x^2+2\)

\(\Leftrightarrow x^2-25⋮x^2+2\)

\(\Leftrightarrow x^2+2-27⋮x^2+2\)

mà \(x^2+2⋮x^2+2\)

nên \(-27⋮x^2+2\)

\(\Leftrightarrow x^2+2\inƯ\left(-27\right)\)

\(\Leftrightarrow x^2+2\in\left\{1;-1;3;-3;9;-9;27;-27\right\}\)

\(\Leftrightarrow x^2+2\in\left\{3;9;27\right\}\)(Vì \(x^2+2\ge2\forall x\))

\(\Leftrightarrow x^2\in\left\{1;7;25\right\}\)

hay \(x\in\left\{1;-1;\sqrt{7};-\sqrt{7};5;-5\right\}\)

Vậy: Để M nguyên thì \(x\in\left\{1;-1;\sqrt{7};-\sqrt{7};5;-5\right\}\)