a) \(f\left(x\right)-g\left(x\right)=\left[x\left(x^2-2x+7\right)-1\right]-\left[x\left(x^2-2x-1\right)-1\right]\)

\(f\left(x\right)-g\left(x\right)=x^3-2x^2+7x-1-x^3+2x^2+x+1\)

\(f\left(x\right)-g\left(x\right)=8x\)

 \(f\left(x\right)+g\left(x\right)=x\left(x^2-2x+7\right)-1+x\left(x^2-2x-1\right)-1\)

 \(f\left(x\right)+g\left(x\right)=x^3-2x^2+7x-1+x^3-2x^2-x-1\)

 \(f\left(x\right)+g\left(x\right)=2x^3-4x^2+6x-2\)

b) 8x=0

=> x=0

=> Nghiệm đa thức f(x)-g(x)

c) Thay \(x=-\frac{3}{2}\)vào BT f(x)+g(x) ta được :

   \(2.\left(-\frac{3}{2}\right)^3-4\left(-\frac{3}{2}\right)^2+6\left(-\frac{3}{2}\right)-2\)

\(=6,75+9-9-2\)

\(=4,75\)

#H

1.

\(\frac{5}{12}-\frac{1}{12}x=\frac{3}{4}\)

\(\frac{1}{12}x=\frac{5}{12}-\frac{3}{4}\)

\(\frac{1}{12}x=-\frac{1}{3}\)

\(x=-\frac{1}{3}\div\frac{1}{12}\)

\(x=-4\)

b)\(|x-2|=6\)

\(\hept{\begin{cases}x-2=6\\x-2=-6\end{cases}\Rightarrow\hept{\begin{cases}x=8\\-4\end{cases}}}\)

Vậy \(x\in\left\{8;-4\right\}\)

\(x^2+y^2=z^2\)

Ta có: \(x^2+y^2-z^2-\left(x+y-z\right)=x\left(x-1\right)+y\left(y-1\right)-z\left(z-1\right)⋮2\)

nên \(\left(x^2+y^2-z^2\right)\equiv\left(x+y-z\right)\left(mod2\right)\)

suy ra \(x+y-z⋮2\Leftrightarrow x-y+3z⋮2\).

Mà \(x+3z-y>x+2z>2\)

Do đó \(x+3z-y\)là hợp số. 

\(\left|x\left(x-3\right)\right|=\frac{2}{3}x\)ĐK : \(x\ge0\)

TH1 : \(x^2-3x=\frac{2}{3}x\Leftrightarrow x^2-\frac{11}{3}=0\Leftrightarrow x=\sqrt{\frac{11}{3}}=\frac{\sqrt{33}}{3}\)

TH2 : \(x^2-3x=-\frac{2}{3}x\Leftrightarrow x^2-\frac{7}{3}=0\Leftrightarrow x=\sqrt{\frac{7}{3}}=\frac{\sqrt{21}}{3}\)

Cho tam giác ABC có AD là đường phân giác. Chứng minh rằng góc ADC - góc ADB = góc B - góc C

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