dai the bn!

dài quá mình giúp ko nổi sorry nhá!

cái này mà lớp 7 trời

Bài 5: 

a: \(f\left(x\right)=4x^4-x^3-4x^2+x-1\)

\(g\left(x\right)=x^4+4x^3+x-5\)

b: \(f\left(x\right)-g\left(x\right)=4x^4-x^3-4x^2+x-1-x^4-4x^3-x+5\)

\(=3x^4-5x^3-4x^2+4\)

\(f\left(x\right)+g\left(x\right)=4x^4-x^3-4x^2+x-1+x^4+4x^3+x-5\)

\(=5x^4+3x^3-4x^2+2x-6\)

hơi mờ mong các pác thông cẻm :(

giúp tui đi mà

\(\dfrac{5^4.20^4}{25^5.4^5}=\dfrac{5^4.5^44^4}{5^4.5.5^4.5.4^4.4}=\dfrac{5^8.4^4}{5^4.5^4.5.5.4^4.4.4.}=\dfrac{5^8.4^4}{5^8.5^2.4^4.4}\)\(=\dfrac{1}{5^2.4}=\dfrac{1}{100}=0,01\)

c)

\(\dfrac{5^4.20^4}{25^5.4^5}\)

\(=\dfrac{5^4.\left(5.4\right)^4}{\left(5^2\right)^5.\left(2^2\right)^5}\)

\(=\dfrac{5^4.5^4.4^4}{5^{10}.2^{10}}\)

\(=\dfrac{5^8.\left(2^2\right)^4}{5^{10}.2^{10}}\)

\(=\dfrac{5^8.2^8}{5^{10}.2^{10}}\)

\(=\dfrac{1}{5^2.2^2}\)

\(=\dfrac{1}{25.4}\)

\(=\dfrac{1}{100}\)

\(a,\left(\dfrac{3}{7}+\dfrac{1}{2}\right)^2=\left(\dfrac{6}{14}+\dfrac{7}{14}\right)^2=\left(\dfrac{13}{14}\right)^2=\dfrac{169}{196}\)

\(b,\left(\dfrac{3}{4}-\dfrac{4}{5}\right)^2=\left(\dfrac{15}{20}-\dfrac{16}{20}\right)^2=\left(\dfrac{-1}{20}\right)^2=\dfrac{1}{400}\)

\(c,\dfrac{5^4\cdot20^4}{25^5\cdot4^5}=\dfrac{\left(5\cdot20\right)^4}{\left(25\cdot4\right)^5}=\dfrac{\left(100\right)^4}{\left(100\right)^5}=\dfrac{100000000}{10000000000}=\dfrac{1}{100}\)

\(d,\dfrac{\left(\dfrac{2}{3}\right)^3\cdot\left(\dfrac{-3}{4}\right)^2\cdot\left(-1\right)^{2003}}{\left(\dfrac{2}{5}\right)^2\cdot\left(-\dfrac{5}{12}\right)^3}=\dfrac{\dfrac{8}{27}\cdot\dfrac{9}{16}\cdot\left(-1\right)}{\dfrac{4}{25}\cdot\left(-\dfrac{125}{1728}\right)}\)

\(=\dfrac{\dfrac{72}{432}\cdot\left(-1\right)}{\left(-\dfrac{500}{43200}\right)}=\dfrac{\dfrac{1}{6}\cdot\left(-1\right)}{\left(-\dfrac{5}{432}\right)}=\dfrac{-\dfrac{1}{6}}{-\dfrac{5}{432}}=\dfrac{72}{5}\)

Bài 8:

a: Ta có: \(35⋮x+3\)

\(\Leftrightarrow x+3\in\left\{5;7;35\right\}\)

hay \(x\in\left\{2;4;32\right\}\)

b: Ta có: \(10⋮2x+1\)

\(\Leftrightarrow2x+1\in\left\{1;5\right\}\)

hay \(x\in\left\{0;2\right\}\)

Bài 1: 

a: Xét ΔAMB và ΔAMC có

AM chung

AB=AC

BM=CM

Do đó: ΔABM=ΔACM

ai giúp mik với :((