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

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

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

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

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)

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

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

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

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

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

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

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

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

Bài 1: 

a: Xét ΔAMB và ΔAMC có

AM chung

AB=AC

BM=CM

Do đó: ΔABM=ΔACM

ai giúp mik với :((

                                                         Bài giải

\(\left(\frac{3}{5}\right)^5\cdot x=\left(\frac{3}{4}\right)^7\)

\(\frac{3^5}{5^5}\cdot x=\frac{3^7}{4^7}\)

\(x=\frac{3^7}{4^7}\text{ : }\frac{3^5}{5^5}=\frac{3^7}{4^7}\cdot\frac{5^5}{3^5}=\frac{3^2\cdot5^5}{4^7}=\frac{28125}{16384}\)

a) Ta có: \(y=f\left(x\right)=4x^2-5\)

\(\Rightarrow\left\{{}\begin{matrix}f\left(3\right)=4.3^2-5=31\\f\left(-\dfrac{1}{2}\right)=4.\left(-\dfrac{1}{2}\right)^2-5=-4\end{matrix}\right.\)

b) Ta có: \(f\left(x\right)=-1\)

\(\Rightarrow4x^2-5=-1\)

\(\Leftrightarrow4x^2=4\)

\(\Leftrightarrow x^2=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Vậy \(x\in\left\{1;-1\right\}\) thì \(f\left(x\right)=-1\)

c) \(\forall x\in R,f\left(x\right)=f\left(-x\right)\Leftrightarrow f\left(-x\right)=4.\left(-x\right)^2-5=4x^2-5=f\left(x\right)\)

Vậy \(\forall x\in R\) thì \(f\left(x\right)=f\left(-x\right)\)

\(a.f\left(3\right)=4.3^2-5=31.\\ f\left(\dfrac{-1}{2}\right)=4.\left(\dfrac{-1}{2}\right)^2-5=-4.\)

\(b.f\left(x\right)=-1.\Rightarrow4x^2-5=-1.\\ \Leftrightarrow4x^2=4.\Leftrightarrow x^2=1.\\ \Leftrightarrow x=\pm1.\)

\(c.f\left(x\right)=f\left(-x\right).\\ \Rightarrow4x^2-5=4\left(-x\right)^2-5.\\ \Leftrightarrow4x^2-5=4x^2-5.\)

\(\Leftrightarrow0x=0\) (luôn đúng).

Vậy với mọi x ∈ R thì f (x)= f (-x).

làm gì??????????????????