Mn giair nhanh giúp mik vs.Trình bày ra nhé
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( x2 - x + 7 ) ⋮ ( x + 1 )
=> ( x2 + x - 2x - 2 + 9 ) ⋮ ( x + 1 )
=> [ x( x + 1 ) - 2( x + 1 ) + 9 ] ⋮ ( x + 1 )
=> [ ( x + 1 )( x - 2 ) + 9 ] ⋮ ( x + 1 )
=> 9 ⋮ ( x + 1 )
=> ( x + 1 ) ∈ Ư(9) = { ±1 ; ±9 }
=> x ∈ { 0 ; -2 ; 8 ; -10 }

Bài 4:
a) Ta có: \(\widehat{yOz}+\widehat{xOy}=180^0\)(2 góc kề bù)
\(\Rightarrow\widehat{yOz}=180^0-\widehat{xOy}=180^0-50^0=130^0\)
b) Ta có: \(\widehat{zOt}=\widehat{yOt}=\dfrac{1}{2}\widehat{yOz}=\dfrac{1}{2}.130^0=65^0\)(do Ot là tia phân giác \(\widehat{yOz}\))
c) Ta có: \(\widehat{xOt}=\widehat{yOt}+\widehat{xOy}=65^0+50^0=115^0\)
Bài 5:
a) Ta có: \(\widehat{xOz}+\widehat{xOy}=180^0\)(2 góc kề bù)
\(\Rightarrow\widehat{xOz}=180^0-\widehat{xOy}=180^0-110^0=70^0\)
b) Ta có: \(\widehat{zOt}=\dfrac{1}{2}\widehat{xOz}=\dfrac{1}{2}.70^0=35^0\)( Ot là tia phân giác \(\widehat{xOz}\))
c) Ta có: \(\widehat{xOt}=\widehat{zOt}=35^0\)( Ot là tia phân giác \(\widehat{xOz}\))
Bài 4:
a: Ta có: \(\widehat{xOy}+\widehat{yOz}=180^0\)
\(\Leftrightarrow\widehat{yOz}=180^0-50^0\)
\(\Leftrightarrow\widehat{yOz}=130^0\)
b: \(\widehat{zOt}=\dfrac{\widehat{yOz}}{2}=65^0\)

36 x ( 10 + 1 )
= 36 x 11
= 396
36 x 10 + 36 x 1
= 360 + 36
= 396
Cách 1 :
36 x ( 10 + 1 )
= 36 x 11
= 396
Cách 2 :
36 x ( 10 + 1 )
= 36 x 10 + 36 x 1
= 360 + 36
= 396

Ta quy về bài toán tìm x như lớp 4.
\(x+\dfrac{1}{2}+\dfrac{3}{8}=\dfrac{17}{16}\)
\(x+\dfrac{8}{16}+\dfrac{6}{16}=\dfrac{17}{16}\)
\(x+\dfrac{14}{16}=\dfrac{17}{16}\)
\(x=\dfrac{17}{16}-\dfrac{14}{16}\)
\(x=\dfrac{3}{16}\)

a) \(x-\dfrac{3}{4}=-\dfrac{5}{8}\Rightarrow x=-\dfrac{5}{8}+\dfrac{3}{4}\Rightarrow x=\dfrac{1}{8}\)
b) \(x+\dfrac{5}{8}=-\dfrac{1}{4}\Rightarrow x=-\dfrac{1}{4}-\dfrac{5}{8}\Rightarrow x=-\dfrac{7}{8}\)
c) \(\dfrac{5}{6}+\dfrac{3}{4}x=\dfrac{5}{24}\Rightarrow x=\left(\dfrac{5}{24}-\dfrac{5}{6}\right):\dfrac{3}{4}\Rightarrow x=-\dfrac{5}{6}\)
d) \(\dfrac{3}{8}-\dfrac{2}{3}:x=-\dfrac{5}{12}\Rightarrow\dfrac{2}{3}:x=\dfrac{3}{8}+\dfrac{5}{12}\Rightarrow\dfrac{2}{3}:x=\dfrac{19}{24}\Rightarrow x=\dfrac{2}{3}:\dfrac{19}{24}=\dfrac{16}{19}\)
a) \(x-\dfrac{3}{4}=-\dfrac{5}{8}\\ \Rightarrow x=\dfrac{1}{8}\)
b) \(x+\dfrac{5}{8}=-\dfrac{1}{4}\\ \Rightarrow x=-\dfrac{7}{8}\)
c) \(\dfrac{5}{6}+\dfrac{3}{4}x=\dfrac{5}{24}\\ \Rightarrow\dfrac{3}{4}x=-\dfrac{5}{8}\\ \Rightarrow x=-\dfrac{5}{6}\)
d) \(\dfrac{3}{8}-\dfrac{2}{3}:x=-\dfrac{5}{12}\\ \Rightarrow\dfrac{2}{3}:x=\dfrac{19}{24}\\ \Rightarrow x=\dfrac{16}{19}\)
e) \(\left(6,5-2x\right):\dfrac{5}{13}=\dfrac{13}{10}\\ \Rightarrow6,5-2x=\dfrac{1}{2}\\ \Rightarrow2x=6\\ \Rightarrow x=3\)
f) \(\left|\dfrac{1}{3}x+\dfrac{1}{2}\right|-\dfrac{3}{4}=-\dfrac{1}{6}\\ \Rightarrow\left|\dfrac{1}{3}x+\dfrac{1}{2}\right|=\dfrac{7}{12}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{1}{3}x+\dfrac{1}{2}=\dfrac{7}{12}\\\dfrac{1}{3}x+\dfrac{1}{2}=-\dfrac{7}{12}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{13}{4}\end{matrix}\right.\)
g) \(\dfrac{x-3}{3}=\dfrac{2x+3}{5}\\ \Rightarrow5x-15=6x+9\\ \Rightarrow-x=24\\ \Rightarrow x=-24\)
h) \(\dfrac{x-5}{6}=\dfrac{6}{x-5}\\ \Rightarrow\left(x-5\right)^2=6^2\\ \Rightarrow\left[{}\begin{matrix}x-5=-6\\x-5=6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=11\end{matrix}\right.\)

\(\Rightarrow12+20x-60=45x-15\Leftrightarrow25x=-33\Leftrightarrow x=-\dfrac{33}{25}\)

Giải thích các bước giải:
13=13.1=(-1).(-13)
TH1: x-13=13
x=26
y+2=1
y=-1
TH2: x-13=1
x=14
y+2=13
y=11
TH3: x-3=-1
x=2
y+2=-13
y=-15
TH4: x-13=-13
x=0
y+2=-1
y=-3
Chúc bạn học tốt!!!!
Giải:
Vì x,y \(\in\)Z nên ta có bảng sau :
Vậy các cặp số (x , y) = { (26 ; -1 ) ; ( 14 ; 11 ) ; ( 0 ; -3 ) ; ( 12 ; - 15 )
Học tốt!!!