a) a . b = 9000 : BCNN ( a,b ) = 900
nhanh nha giúp mk với đang cần gấp
cảm ơn nhìu
K
Khách
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L
20 tháng 1 2022
You can refer this one:
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5 tháng 12 2015
Vì a*b=BCNN(a;b)*UCLN(a;b)
Suy ra: UCLN(a;b)=320/160=2
a=2*m
b=2*n
UCLN(m;n)=1
Ta có a*b=(2*m)*(12*n)=320
m*n*4=320
m*n=80
Nếu m=80 => a=160
n=1 => b=2
Nếu m=40 =>a=80
n=2 => b=4
Nếu m=20 => a=40
n=4 => b=8
Nếu m=16 => a=32
n=5 => b=10
H9
HT.Phong (9A5)
CTVHS
28 tháng 6 2023
Bài 2:
1) \(x^2-4=x^2-2^2=\left(x-2\right)\left(x+2\right)\)
2) \(1-4x^2=1^2-\left(2x\right)^2=\left(1-2x\right)\left(1+2x\right)\)
3) \(4x^2-9=\left(2x\right)^2-3^2=\left(2x+3\right)\left(2x-3\right)\)
4) \(9-25x^2=3^2-\left(5x\right)^2=\left(3-5x\right)\left(3+5x\right)\)
5) \(4x^2-25=\left(2x\right)^2-5^2=\left(2x+5\right)\left(2x-5\right)\)
6) \(9x^2-36=\left(3x\right)^2-6^2=\left(3x-6\right)\left(3x+6\right)\)
7) \(\left(3x\right)^2-y^2=\left(3x-y\right)\left(3x+y\right)\)
8) \(x^2-\left(2y\right)^2=\left(x-2y\right)\left(x+2y\right)\)
9) \(\left(2x\right)^2-y^2=\left(2x-y\right)\left(2x+y\right)\)
10) \(\left(3x\right)^2-9y^4=\left(3x\right)^2-\left(3y^2\right)^2=\left(3x-3y^2\right)\left(3x+3y^2\right)\)
H9
HT.Phong (9A5)
CTVHS
28 tháng 6 2023
Bài 2:
21) \(\left(\dfrac{x}{3}-\dfrac{y}{4}\right)\left(\dfrac{x}{3}+\dfrac{y}{4}\right)=\left(\dfrac{x}{3}\right)^2-\left(\dfrac{y}{4}\right)^2=\dfrac{x^2}{9}-\dfrac{y^2}{16}\)
22) \(\left(\dfrac{x}{y}-\dfrac{2}{3}\right)\left(\dfrac{x}{y}+\dfrac{2}{3}\right)=\left(\dfrac{x}{y}\right)^2-\left(\dfrac{2}{3}\right)^2=\dfrac{x^2}{y^2}-\dfrac{4}{9}\)
23) \(\left(\dfrac{x}{2}+\dfrac{y}{3}\right)\left(\dfrac{x}{2}-\dfrac{y}{3}\right)=\left(\dfrac{x}{2}\right)^2-\left(\dfrac{y}{3}\right)^2=\dfrac{x^2}{4}-\dfrac{y^2}{9}\)
24) \(\left(2x-\dfrac{2}{3}\right)\left(\dfrac{2}{3}+2x\right)=\left(2x-\dfrac{2}{3}\right)\left(2x+\dfrac{2}{3}\right)=\left(2x\right)^2-\left(\dfrac{2}{3}\right)^2=4x^2-\dfrac{4}{9}\)
25) \(\left(2x+\dfrac{3}{5}\right)\left(\dfrac{3}{5}-2x\right)=\left(\dfrac{3}{5}+2x\right)\left(\dfrac{3}{5}-2x\right)=\left(\dfrac{3}{5}\right)^2-\left(2x\right)^2=\dfrac{9}{25}-4x^2\)
26) \(\left(\dfrac{1}{2}x-\dfrac{4}{3}\right)\left(\dfrac{4}{3}+\dfrac{1}{2}x\right)=\left(\dfrac{1}{2}x-\dfrac{4}{3}\right)\left(\dfrac{1}{2}x+\dfrac{4}{3}\right)=\left(\dfrac{1}{2}x\right)^2-\left(\dfrac{4}{3}\right)^2=\dfrac{1}{4}x^2-\dfrac{16}{9}\)
27) \(\left(\dfrac{2}{3}x^2-\dfrac{y}{2}\right)\left(\dfrac{2}{3}x^2+\dfrac{y}{2}\right)=\left(\dfrac{2}{3}x^2\right)^2-\left(\dfrac{y}{2}\right)^2=\dfrac{4}{9}x^4-\dfrac{y^2}{4}\)
28) \(\left(3x-y^2\right)\left(3x+y^2\right)=\left(3x\right)^2-\left(y^2\right)^2=9x^2-y^4\)
29) \(\left(x^2-2y\right)\left(x^2+2y\right)=\left(x^2\right)^2-\left(2y\right)^2=x^4-4y^2\)
30) \(\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x^2\right)^2-\left(y^2\right)^2=x^4-y^4\)
A
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a, Ta có : \(\frac{15}{x-1}\in Z\Leftrightarrow x-1\inƯ_{15}=\left\{-15;-5;-3;-1;1;3;5;15\right\}.\)
\(\Rightarrow x=\left\{-14;-4;-2;0;2;4;6;16\right\}\)
b, Ta có : \(\frac{x+6}{x-1}=\frac{x-1+7}{x-1}=\frac{x-1}{x-1}+\frac{7}{x-1}=1+\frac{7}{x-1}\)
Mà 1 thuộc Z => \(\frac{7}{x-1}\in Z\Leftrightarrow x-1\inƯ_7=\left\{-7;-1;1;7\right\}\)
\(\Rightarrow x=\left\{-6;0;2;8\right\}\)
C, tự làm nha
Vì NBCNN(a;b)=900
nên: a=900.m
b=900.n
=>(m;n)=1
nên: 900.m+900.n=9000
900.(m+n) =9000
m+n =9000:900
m+n = 10
vì (m;n)=1
nên:
cảm ơn bạn nhìu nha