? 

Số phần biểu diễn số hs tb là :

\(\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{1}{6}\left(ph\text{ần}\right)\)

Số phần biều diễn `14` hs giỏi là :

`1-(1/2+1/6)=1/3(phần)`

Lớp `5b` có số hs :

`14:1/3=42(hs)`

Số hs tb  :

`42xx1/2=21(hs)`

Số hs khá : `42xx1/6=7`

.-. sửa thành 19 góc trong câu a ạ 

:v chữ thưa qá 

\(5)\\ \left(x+35\right)+\left(-81\right)=11\\ x+35=11+81\\ x+35=92\\ x=92-35\\ x=57\\ 6)\\ 18-\left(x+35\right)=-15\\x+35= 18+15\\ x+35=33\\ x=33-35\\ x=-2\)

\(7)\\ x+\left(-88\right)=-120\\ x=\left(-120\right)+88\\ x=-32\\ 8)\\ x-\left(-61\right)=120\\ x+61=120\\ x=59\)

\(9)\\ \left(-111\right)-x=38+\left(-12\right)\\ \left(-111\right)-x=26\\ x=-111-26\\ x=-137\\ 10)\\ 120+\left(x-55\right)=11-66\\ 120+\left(x-55\right)=-55\\ x-55=-55-120\\ x-55=-175\\ x=-175+55\\ x=-120\)

`a)` có tất cả `16` góc :

`b)` 

`@` Góc đỉnh `M`:

`hat(NMP);hat(NMA);hat(NMB);hat(AMB);hat(AMP);hat(BMP)`

`@` Góc đỉnh `N:`

\(\widehat{MNP};\widehat{MNA};\widehat{MNB}\)

`@` Góc đỉnh `A:`

\(\widehat{NAM};\widehat{NAP};\widehat{MAB}\)

`@` Góc đỉnh `B:`

\(\widehat{NBM};\widehat{NBP};\widehat{ABM};\widehat{ABP}\)

`@` Góc đỉnh `P:`

\(\widehat{NPM};\widehat{APM};\widehat{BPM}\)

Th1 :`x<0`

`=>4x<0`

Mà \(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|\ge0\\ \Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+3\right|=4x\left(v\text{ô}.l\text{ý}\right)\)

Th2:`x>0`

\(\Leftrightarrow\left(x+1\right)+\left(x+2\right)+\left(x+3\right)=4x\\ \Leftrightarrow x+x+x-4x=-1-2-3\\ \Leftrightarrow-x=-6\\ \Rightarrow x=6\left(tm\right)\)

bài 1:

\(a,3x\left(2x^2-2x+5\right)\\ =6x^3-6x^2+15\\ b,x^2\left(5x^3-x-\dfrac{1}{2}\right)\\ =x^2\cdot5x^3+x^2\cdot x+x^2\cdot\dfrac{-1}{2}\\ =5x^5-x^3-\dfrac{1}{2}x^2\)

~~~~~~~~~~~~~~~~~~~~~

Bài 2:

\(a,\left(2x^3y-\dfrac{1}{2}x^2+\dfrac{1}{5}xy\right)\cdot6xy^3\\ =3x^2y\cdot6xy^3-\dfrac{1}{2}x^2\cdot6xy^3+\dfrac{1}{5}xy\cdot6xy^3\\ =18x^3y^4-3x^3y^3+\dfrac{6}{5}x^2y^4\)