BT10 : Tìm x biết:
2) x + 4 \(⋮\) x
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Giải:
Ta có:
\(x+8⋮x+1\)
\(\Rightarrow\left(x+1\right)+7⋮x+1\)
\(\Rightarrow7⋮x+1\)
\(\Rightarrow x+1\in\left\{1;-1;7;-7\right\}\)
+) \(x+1=1\Rightarrow x=0\)
+) \(x+1=-1\Rightarrow x=-2\)
+) \(x+1=7\Rightarrow x=6\)
+) \(x+1=-7\Rightarrow x=-8\)
Vậy \(x\in\left\{0;-2;6;-8\right\}\)
\(\frac{x+8}{x+1}=\frac{x+7+1}{x+1}=\frac{x+1}{x+1}+\frac{7}{x+1}\)
\(\Rightarrow x+1\in\text{Ư}\left(7\right)=\left\{1;7;-1;-7\right\}\)
\(\Rightarrow x\in\left\{0;6;-2;-8\right\}\)
\(\frac{x+1+3}{x+1}=\frac{x+1}{x+1}+\frac{3}{x+1}\)
\(\Rightarrow x+1\in\text{Ư}\left(3\right)=\left\{1;-1;3;-3\right\}\)
\(\Rightarrow x\in\left\{0;-2;2;-4\right\}\)
Ta có :
\(x+1+3⋮x+1\)
=> \(x+1+3-\left(x+1\right)⋮x+1\)
=> \(3⋮x+1\)
=> \(x+1\inƯ_3\)
=> \(x+1\in\left\{1;3;-1;-3\right\}\)
=> \(x\in\left\{0;2;-2;-4\right\}\)
\(\frac{x+5}{x+1}=\frac{x+4+1}{x+1}=\frac{x+1}{x+1}+\frac{4}{x+1}\)
\(\Rightarrow x+1\in\text{Ư}\left(4\right)=\left\{1;2;4;-1;-2;-4\right\}\)
\(\Rightarrow x\in\left\{0;1;3;-2;-3;-5\right\}\)
x+5\(⋮\)x+1
x+1+4\(⋮\)x+1
Vì x+1\(⋮\)x+1
Buộc 4\(⋮\)x+1=>x+1ϵƯ(4)={1;2;4}
Với x+1=1=>x=0
x+1=2=>x=1
x+1=4=>x=3
Vậy xϵ{0;1;3}
Ta có :
\(9⋮2x+1\)
=> \(2x+1\inƯ_9\)
=> \(2x+1\in\left\{1;3;9\right\}\)
=> \(2x\in\left\{0;2;8\right\}\)
=> \(x\in\left\{0;1;4\right\}\)
Vậy \(x\in\left\{0;1;4\right\}\)
a: =-1/5x^5y^2
b: =-9/7xy^3
c: =7/12xy^2z
d: =2x^4
e: =3/4x^5y
f: =11x^2y^5+x^6
a: =-4xyz^2
b: =-9x^2y
c: =16x^2y^2
d: =1/6x^2y^3
e: =13/6x^3y^2
f: =7/12x^4y
a) -xyz² - 3xz.yz
= -xyz² - 3xyz²
= -4xyz²
b) -8x²y - x.(xy)
= -8x²y - x²y
= -9x²y
c) 4xy².x - (-12x²y²)
= 4x²y² + 12x²y²
= 16x²y²
d) 1/2 x²y³ - 1/3 x²y.y²
= 1/2 x²y³ - 1/3 x²y³
= 1/6 x²y³
e) 3xy(x²y) - 5/6 x³y²
= 3x³y² - 5/6 x³y²
= 13/6 x³y²
f) 3/4 x⁴y - 1/6 xy.x³
= 3/4 x⁴y - 1/6 x⁴y
= 7/12 x⁴y
bài 1 : a,ta có 3/x-1 =4/y-2=5/z-3 => x-1/3=y-2/4=z-3/5
áp dụng .... => x-1+y-2+z-3 / 3+4+5 = x+y+z-1-2-3/3+4+5 = 12/12=1
do x-1/3 = 1 => x-1 = 3 => x= 4 ( tìm y,z tương tự
Bài 1:
a) Ta có: 3/x - 1 = 4/y - 2 = 5/z - 3 => x - 1/3 = y - 2/4 = z - 3/5 áp dụng ... =>x - 1 + y - 2 + z - 3/3 + 4 + 5 = x + y + z - 1 - 2 - 3/3 + 4 + 5 = 12/12 = 1 do x - 1/3 = 1 => x - 1 = 3 => x = 4 ( tìm y, z tương tự )
\(\frac{x+4}{x}=\frac{x}{x}+\frac{4}{x}\)
\(\Rightarrow x\in\text{Ư}\left(4\right)=\left\{1;2;4;-1;-2;-4\right\}\)
Ta có :
\(x+4⋮x\)
=> \(x+4-x⋮x\)
=> \(4⋮x\)
=> \(x\inƯ_4\)
=> x\(\in\left\{1;2;4;-1;-2;-4\right\}\)