a: \(VP=a^3+b^3+c^3-3bac\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=VT\)

b: \(VT=\left(3a+2b-1\right)\left(a+5\right)-2b\left(a-2\right)\)

\(=3a^2+15a+2ab+10b-a-5-2ab+4b\)

\(=3a^2+14a+14b-5\)

\(VP=\left(3a+5\right)\left(a+3\right)+2\left(7b-10\right)\)

\(=3a^2+9a+5a+15+14b-20\)

\(=3a^2+14a+14b-5\)

=>VT=VP

c: \(VT=a\left(b-x\right)+x\left(a+b\right)\)

\(=ab-ax+ax+bx\)

\(=ab+bx=b\left(a+x\right)=VP\)

d: \(VT=a\left(b-c\right)-b\left(a+c\right)+c\left(a-b\right)\)

\(=ab-ac-ab-bc+ca-cb\)

\(=-2bc\)

=VP

a: \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)-3x^2=54\)

\(\Leftrightarrow x^3+9x^2+27x+27-x\left(9x^2+6x+1\right)+8x^3+1-3x^2=54\)

\(\Leftrightarrow9x^3+6x^2+27x+28-9x^3-6x^2-x=54\)

hay x=1

b: Ta có: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2+3x^2=-33\)

\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6x^2+12x+6+3x^2=-33\)

hay x=-1

1: TXĐ là D=R

Khi x∈D thì -x∈D
\(f\left(-x\right)=2\cdot\left(-x\right)+3\left(-x\right)^3\cdot cos\left(-2x\right)=-2x-3x^3\cdot cos2x\)

=-f(x)

=>f(x) là hàm số lẻ

2: ĐKXĐ: sin x<>0

=>\(x<>k\pi\)

=>TXĐ là D=R\{kπ}

Khi x∈D thì -x∈D

\(f\left(-x\right)=\frac{1}{\sin\left(-x\right)}=\frac{-1}{\sin x}=-f\left(x\right)\)

=>f(x) là hàm số lẻ

3: ĐKXĐ: tan x<>0 và \(x<>\frac{\pi}{2}+k\pi\)

=>\(x<>k\pi;x<>\frac{\pi}{2}+k\pi\)

=>\(x<>\frac{k\pi}{2}\)

=>TXĐ là D=R\{\(\frac{k\pi}{2}\) }

Khi x∈D thì -x∈D

\(f\left(-x\right)=\frac{\left.\left(-x\right)^2\right.}{3\cdot\tan\left(-x\right)}=\frac{x^2}{-3\cdot\tan x}=-f\left(x\right)\)

=>f(x) là hàm số lẻ

4: ĐKXĐ: \(3x<>\frac{\pi}{2}+k\pi\)

=>\(x<>\frac{\pi}{6}+\frac{k\pi}{3}\)

=>TXĐ là D=R\{\(\frac{\pi}{6}+\frac{k\pi}{3}\) }

Khi x∈D thì -x∈D

\(f\left(-x\right)=\sin\left(-x\right)+\tan\left(-3x\right)+cos\left(-3x\right)\)

=-sin x-tan 3x+cos3x

=>f(-x)<>-f(x) và f(-x)<>f(x)

=>f(x) không có tính chẵn lẻ

1 being 

2 to do

3 moving

4 to do

5 take/going

6 leaving 

7 repairing

8 playing

9 to use

10 speaking

AB<AC

=>góc B>góc C

=>90 độ-góc B<90 độ-góc C

=>góc HAB<góc HAC

\(AB< AC\\ \Rightarrow\widehat{C}< \widehat{B}\)

Xét tam giác \(AHB\) và \(AHC\) có

\(\left\{{}\begin{matrix}\widehat{HAB}=90^o-\widehat{B}\\\widehat{HAC}=90^o-\widehat{C}\end{matrix}\right.\)

mà \(\widehat{B}>\widehat{C}\)

\(\Rightarrow\widehat{HAB}< \widehat{HAC}\)

B A H C

a) Các góc kề bù nhau là: 

1. \(\widehat{xOy}\) và \(\widehat{xOt}\)

2. \(\widehat{yOz}\) và \(\widehat{zOt}\)    

b) Ta có: \(\widehat{yOt}\) là góc bẹt \(\Rightarrow\widehat{yOt}=180^o\)

Mà \(\widehat{xOy}\) và \(\widehat{xOt}\) kề bù \(\Rightarrow\widehat{xOy}+\widehat{xOt}=\widehat{yOt}\)

\(\Rightarrow\widehat{xOt}=\widehat{yOt}-\widehat{xOy}=180^o-45^o=135^o\) 

Ta có: \(\widehat{xOz}=\widehat{xOy}+\widehat{yOz}=45^o+30^o=75^o\)

Mà \(\widehat{yOz}\) và \(\widehat{zOt}\) kề bù \(\Rightarrow\widehat{yOz}+\widehat{zOt}=\widehat{yOt}=180^o\)

\(\Rightarrow\widehat{zOt}=\widehat{yOt}-\widehat{yOz}=180^o-30^o=150^o\)

sd

Chị có thể lên mạng để tham khảo ấy c. 

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mik lỡ tay đốt nó ròi ;-;;;