\(\left(-6-2\right).\left(-6+2\right)\)

\(=\left(-8\right).\left(-4\right)\)

\(=32\)

Bằng -32 nhé

\(x\div\frac{1}{2}+x\div\frac{1}{4}+x\div\frac{1}{8}+x\div\frac{1}{10}=96\)

\(2x+4x+8x+10x=96\)

\(24x=96\)

\(x=4\)

Ta có: \(x:\frac{1}{2}+x:\frac{1}{4}+x:\frac{1}{8}+x:\frac{1}{10}=96\)

\(\Rightarrow2x+4x+8x+10x=96\)

\(\Rightarrow24x=96\)

\(x=96:24=4\)

Vậy ...........................................

CHÚC BẠN HỌC TỐT

a) Có : \(\widehat{xOy}+\widehat{yOz}=\widehat{xOz}\)

\(\Rightarrow40^o+\widehat{yOz}=80^o\)

\(\Rightarrow\widehat{yOz}=40^o\)

b) Có : \(\widehat{xOt}=\frac{3}{2}\widehat{xOy}=\frac{3}{2}.40^o=60^o\)

- Có : \(\widehat{yOt}+\widehat{xOy}=\widehat{xOt}\)

\(\Rightarrow\widehat{yOt}+40^o=60^o\)

\(\Rightarrow\widehat{yOt}=20^o\)

- Có : \(\widehat{yOz}=2\widehat{yOt}\left(40^o=2.20^o\right)\)

=> Ot là tia phân giác góc yOz

#H

a) Đặt ab=cd=k⇒\hept{a=bkc=dkab=cd=k⇒\hept{a=bkc=dk

Khi đó (2a + 3c)(2b - 3d) 

= (2bk + 3dk)(2b - 3d)

= k(2b + 3d)(2b - 3d) (1)

(2a - 3c)(2b + 3d)

= (2bk - 2dk)(2b + 3d)

= k(2b - 3d)(2b + 3d) (2)

Từ (1)(2) => (2a + 3c)(2b - 3d) = (2a - 3c)(2b + 3d)

Thiếu chỗ : Đặt\(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)

-19x(16-25)+19x(-25-4)

=-19x(-9)+19x(-29)

=-19x(-9+29)

=-19x20

=-380

#H

\(\frac{x}{20}=\frac{7}{12}+\frac{11}{30}\)

\(\frac{x}{20}=\frac{35}{60}+\frac{22}{60}\)

\(\frac{x}{20}=\frac{57}{60}\)

\(\frac{x}{20}=\frac{19}{20}\)

\(\Rightarrow x=19\)