thay x= -1/2 ; y= 4; z =6 vào biểu thức A

có: \(A=\left(\frac{-1}{2}\right)^2.4+\left(\left(\frac{-1}{2}\right)^3-3.\frac{-1}{2}.4.6\right)+\left(-14^{15}\right)^0\)

\(A=\frac{1}{4}.4+\left(\frac{-1}{8}-\left(-36\right)\right)+1\)

\(A=1+35\frac{7}{8}+1\)

\(A=37\frac{7}{8}\)

KL: \(A=37\frac{7}{8}\)  tại x= -1/2 ; y=4; z=6

CHÚC BN HỌC TỐT!!!

a/ \(M=x^4-xy^3+x^3y-y^4-1\)

\(\Leftrightarrow M=x^3\left(x+y\right)-y^3\left(x+y\right)-1\)

Mà \(x+y=0\)

\(\Leftrightarrow M=x^3.0-y^3.0-1\)

\(\Leftrightarrow M=-1\)

Vậy ...

cau b lam nhu the nao vay

Đặt \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=k\left(k\ne0\right)\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=5k\\z=7k\end{matrix}\right.\)

\(\Rightarrow A=\dfrac{2k-5k+7k}{2k+10k-7k}=\dfrac{k\left(2-5+7\right)}{k\left(2+10-7\right)}=\dfrac{4}{5}\)

Vậy A = \(\dfrac{4}{5}\)

Đặt \(\dfrac{x}{2}\) =\(\dfrac{y}{5}\) =\(\dfrac{z}{7}\) =k

=> A=\(\dfrac{x-y+x}{x+2y-z}\)=\(\dfrac{2k+5k-7k}{2k+10k-7k}\)=\(\dfrac{k\left(2-5+7\right)}{k\left(2+10-7\right)}\)=\(\dfrac{4}{5}\)

Vậy A= \(\dfrac{4}{5}\)

a. Tại x=\(\frac{-1}{2}\), ta có:

 \(\left(\frac{-1}{2}\right)^2+4.\left(\frac{-1}{2}\right)+3=\frac{1}{4}+\left(-2\right)+3=\frac{5}{4}\)

b. Ta có:

 \(x^2+4x+3=0\)

\(\Rightarrow x^2+x+3x+3=0\)

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

\(\Rightarrow x\left(x+1\right)+3\left(x+1\right)=0\)

\(\Rightarrow\left(x+1\right)\left(x+3\right)=0\)

\(\Rightarrow\hept{\begin{cases}x+1=0\\x+3=0\end{cases}\Rightarrow\hept{\begin{cases}x=-1\\x=-3\end{cases}}}\)

Vậy \(x=-1;x=-3\)