Giải:

Gọi đa thức cần tìm là A

Ta có:

\(\dfrac{x^2}{x^2+7}=\dfrac{A}{x^4-49}\)

\(\Leftrightarrow\dfrac{x^2}{x^2+7}=\dfrac{A}{\left(x^7-7\right)\left(x^2+7\right)}\)

\(\Leftrightarrow\dfrac{x^2\left(x^2-7\right)}{\left(x^2+7\right)\left(x^2-7\right)}=\dfrac{A}{\left(x^2-7\right)\left(x^2+7\right)}\)

\(\Leftrightarrow x^2\left(x^2-7\right)=A\)

\(\Leftrightarrow A=x^4-7x^2\)

Vậy ...

b) Ta có:

\(f\left(x\right)=x^3-3x^2+3x-4\)

\(=x^3+2x-3x^2-6+x+2\)

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

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

Để f(x) \(⋮\) x² + 2 thì x + 2 \(⋮\) x² + 2

Đến đây tự làm

câu a sai đề k bn

\(\dfrac{1}{5}\)

ý bạn là nhân đa thức với đa thức hay sao ạ?

Chịu :)

S=n(n+1)mũ 2  trên   4

a: 7/-49=4/-28

7/4=-49/-28

-49/7=-28/4

4/7=-28/-49

b: -3/4=-15/20

-3/-15=4/20

4/-3=20/-15

-15/-3=20/4

c: -2/-9=-6/-27

-2/-6=-9/-27

-9/-2=-27/-6

-6/-2=-27/-9

a)

\(x+\frac{3}{5}=\frac{1}{4}\)

\(\Rightarrow x=-\frac{7}{20}\)

Vậy ........

b)

\(\frac{2}{3}-x=1\frac{4}{7}-2\frac{3}{4}\)

\(\Rightarrow\frac{2}{3}-x=\frac{11}{7}-\frac{11}{4}\)

\(\Rightarrow\frac{2}{3}-x=-\frac{33}{28}\)

\(\Rightarrow x=\frac{75}{28}\)

a) Theo quy tắc chuyển vế ta có:

\(x+\frac{3}{5}=\frac{1}{4}\Rightarrow x=\frac{1}{4}-\frac{3}{5}\)

\(x=\frac{1}{4}+\frac{\left(-3\right)}{5}=\frac{5+4.\left(-3\right)}{20}\\ \Rightarrow x=\frac{-7}{20}\)

b) Theo quy tắc chuyển vế ta có:

\(\frac{2}{3}-x=1\frac{4}{7}-2\frac{3}{4}\Rightarrow\frac{2}{3}=1\frac{4}{7}-2\frac{3}{4}+x\\ \Rightarrow x=\frac{2}{3}-1\frac{4}{7}+2\frac{3}{4}=\frac{2}{3}-\frac{11}{7}+\frac{11}{4}=\frac{56-132+231}{84}\\ x=\frac{155}{84}=1\frac{71}{84}\)

a) x + \(\frac{3}{5}\) = \(\frac{1}{4}\)

x = \(\frac{1}{4}-\frac{3}{5}\)

x = \(\frac{-7}{20}\)

b) \(\frac{2}{3}\) - x = \(1\frac{1}{4}\) - \(2\frac{3}{4}\)

\(\frac{2}{3}-x=\frac{-3}{2}\)

x = \(\frac{2}{3}-\frac{-3}{2}\)

x =\(\frac{13}{6}\)

Câu 1: 

a) Ta có: x-3 là ước của 13

\(\Leftrightarrow x-3\inƯ\left(13\right)\)

\(\Leftrightarrow x-3\in\left\{1;-1;13;-13\right\}\)

hay \(x\in\left\{4;2;16;-10\right\}\)(thỏa mãn)

Vậy: \(x\in\left\{4;2;16;-10\right\}\)

b) Ta có: \(x^2-7\) là ước của \(x^2+2\)

\(\Leftrightarrow x^2+2⋮x^2-7\)

\(\Leftrightarrow x^2-7+9⋮x^2-7\)

mà \(x^2-7⋮x^2-7\)

nên \(9⋮x^2-7\)

\(\Leftrightarrow x^2-7\inƯ\left(9\right)\)

\(\Leftrightarrow x^2-7\in\left\{1;-1;3;-3;9;-9\right\}\)

mà \(x^2-7\ge-7\forall x\)

nên \(x^2-7\in\left\{1;-1;3;-3;9\right\}\)

\(\Leftrightarrow x^2\in\left\{8;6;10;4;16\right\}\)

\(\Leftrightarrow x\in\left\{2\sqrt{2};-2\sqrt{2};-\sqrt{6};\sqrt{6};\sqrt{10};-\sqrt{10};2;-2;4;-4\right\}\)

mà \(x\in Z\)

nên \(x\in\left\{2;-2;4;-4\right\}\)

Vậy: \(x\in\left\{2;-2;4;-4\right\}\)

Câu 2: 

a) Ta có: \(2\left(x-3\right)-3\left(x-5\right)=4\left(3-x\right)-18\)

\(\Leftrightarrow2x-6-3x+15=12-4x-18\)

\(\Leftrightarrow-x+9+4x+6=0\)

\(\Leftrightarrow3x+15=0\)

\(\Leftrightarrow3x=-15\)

hay x=-5

Vậy: x=-5