Cách 1: X x 3 = 6

X =6:3( tìm thừa số)

X =2

Vậy số cận tìm là 2.

Cách 2:X x 3=6

X x 3=3 x 2

Ta thấy cả 2 vế đầu có thừa số 3.

Mà thừa 2 thừa số X và 2.

Vậy X=2

16N

Gọi A= 5/1.6+5/6.11+5/11.16+....+5/101.106

Ta cóA= 5-5/6 +5/6-5/11+5/11-5/16+....+5/101-5/106

Đơn giản còn lại

A= 5-5/106

=525/106

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Thấy đúng hông?

A=(\(\dfrac{\left(x+4\right)}{3\left(x+2\right)}-\dfrac{1}{\left(x+2\right)^2}\))(\(\dfrac{x+5+x-1}{x+5}\))
A=\(\dfrac{\left(x+4\right)\left(x+2\right)-3}{3\left(x+2\right)^2}\cdot\dfrac{2x+2}{x+5}\)
A=\(\dfrac{x^2+2x+4x+8-3}{3\left(x-2\right)}\cdot\dfrac{2}{x+5}\)
A=\(\dfrac{x^2+6x+5}{3\left(x-2\right)}\cdot\dfrac{2}{x+5}\)
A=\(\dfrac{x^2+6x+9-4}{3\left(x-2\right)}\cdot\dfrac{2}{x+5}\)
A=\(\dfrac{\left(x+3\right)^2-4}{3\left(x-2\right)}\cdot\dfrac{2}{x+5}\)
A=\(\dfrac{2\left(x+3-2\right)\left(x+3+2\right)}{3\left(x-2\right)\left(x+5\right)}\)
A=\(\dfrac{2\left(x+1\right)}{3\left(x-2\right)}\)

B3(rút gọn)

=\(|4-\sqrt{15}|+\sqrt{15}-3\)
\(=4-\)\(\sqrt{15}+\sqrt{15}-3\) (vì \(4-\sqrt{15}>0\)nên \(|4-\sqrt{15}|=4-\sqrt{15}\))
=1

B1
A= \(7+6\cdot2\sqrt{3}-7\cdot9\sqrt{3}+\dfrac{1}{3}\cdot3\sqrt{3}\)
A=\(7+\sqrt{3}\left(12-63+1\right)\)
A=\(7-50\sqrt{3}\)

B2

a) để A có nghĩa thì 5x-15>=0
=>5x>=15
=>x>=3

b) để B có nghĩa thì 12-3x>=0
=>12>=3x
=>x<=4

B4

<=>\(\left(\dfrac{1}{\sqrt{a}\left(1+\sqrt{a}\right)}-\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}+1\right)}\right)\cdot\dfrac{\sqrt{a}+1}{\sqrt{a}}\)
<=>\(\dfrac{1-\sqrt{a}}{\sqrt{a}\left(1+\sqrt{a}\right)}\cdot\dfrac{\sqrt{a}+1}{\sqrt{a}}\)
<=>\(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}{\sqrt{a}\cdot\sqrt{a}}\)
<=>\(\dfrac{1-a}{a}\)

a) cóA= \(\sqrt{-|x+5|}>=0\)

=> \(-|x+5|>=0\)

mà \(|x+5|>=0\)

nên để A có nghĩa thì x+5=0

=>x=-5

b)B<=>\(\sqrt{\left(1-x\right)\left(1+x\right)}+\sqrt{x+1}\)=0(đk -1<=x<=1

<=>\(\sqrt{x+1}\left(\sqrt{1-x}+1\right)\)=0

có \(\sqrt{1-x}+1\)\(>=\)1 nên để B=0 thì \(\sqrt{x+1}=0\)

<=> x+1=0

=>x=-1

a)A<=>\(\sqrt{\left(x-2\right)\left(x+2\right)}+\sqrt{\left(x+2\right)^2}\)=0(đk -2<=x)

<=>\(\sqrt{x+2}\left(1+\sqrt{x+2}\right)\)=0

vì 1+\(\sqrt{x+2}\) >=1 nên để A=0 thì \(\sqrt{x+2}\)=0

=>x+2=0

=>x=-2