`A=(2/9-7/2) xx 4/3 +(16/9 - 5/12) xx 4/3`

`A=4/3 xx [(2/9-7/2)+(16/9-5/12)`

`A=4/3 xx [2/9 - 7/2 + 16/9 - 5/12]`

`A=4/3 xx [(2/9+16/9)+(-7/2-5/12)]`

`A=4/3 xx [18/9 + (-42/12 - 5/12)]`

`A=4/3 xx [2 -47/12]`

`A=4/3 xx [24/12 - 47/12]`

`A=4/3 xx (-23/12)`

`A=-23/9`

Vậy `A=-23/9`

A=(2/9-7/2)×4/3+(16/9-5/12)×4/3

A=[2/9-7/2+16/9-5/12]×4/3

A=[(2/9+16/9) - (7/2+5/12)] x4/3

A=[ 2 - 47/12] x 4/3

A= (-23/12) x 4/3

A= -23/9

Gọi a,b,c,d la so cây của 4 lớp 7a;7b;7c và 7d lần lượt tỉ lệ với 3;4;5;6 

a/3=b/4=c/5=d/6 va b-a=5

Áp dụng tính chất dãy tỉ số bằng nhau:

a/3=b/4=c/5=d/6=b-a/4-3=5/1=5

Suy ra :a/3=5=>a=5.3=15

b/4=5=>b=5.4=20

c/5=5=>c=5.5=25

d/6=5=>d=5.6=30

sửa đề

\(\left(4x^2-1\right)\left(x^3+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x^2-1=0\\x^3+8=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=\dfrac{1}{4}\\x^3=-8\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{1}{2}\\x=-2\end{matrix}\right.\)

vì \(\widehat{mOt}\) và \(\widehat{tOn}\) là hai góc kề bù

Nên \(\widehat{mOt}+\widehat{tOn}=180^o\)

       100⁰+ \(\widehat{tOn}=180^o\)

                  \(\widehat{tOn}=180^o-100^o\)

                  \(\widehat{tOn}=80^o\)

mặt khác:

\(\widehat{mOt}=\dfrac{\widehat{mOt}}{2}=\dfrac{100^o}{2}=50^o\) (tia phân giác)

Do đó: \(\widehat{xOn}=\widehat{tOn}+\widehat{xOt}=80^o+50^o=130^o\)