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6 tháng 3 2016

4 số liên tiếp nên chia hết cho 2.3.4=24

giá trị 9luôn có các chữ số tận cùng là 9;1 nên 2 số 9x+1 hoặc 9x+4 sẽ cố số chia hết cho 5 

nên nó chia hết cho 24.5=120 

25 tháng 11 2017

=>21 chia hết 49 h minh nhé

13 tháng 10 2020

Ta có: \(x^3;y^3\equiv1;-1\left(mod9\right)\Rightarrow x^6\equiv y^6\equiv1\left(mod9\right)\Rightarrow x^6-y^6⋮9\)

a, \(16x^2-5=0\)

\(\Rightarrow16x^2=5\)

\(\Rightarrow x^2=\frac{5}{16}\)

\(\Rightarrow x=\sqrt{\frac{5}{16}}\Rightarrow x=\frac{\sqrt{5}}{4}\)

b, \(2\sqrt{x-3}=4\)

\(\Rightarrow\sqrt{x-3}=4:2\)

\(\Rightarrow\sqrt{x-3}=2\)

\(\Rightarrow x-3=4\)

\(\Rightarrow x=4+3\)

\(\Rightarrow x=7\)

c, \(\sqrt{4x^2-4x+1}=3\)

\(\Rightarrow\sqrt{\left(2x-1\right)^2}=3\)

\(\Rightarrow2x-1=3\)

\(\Rightarrow2x=4\)

\(\Rightarrow x=2\)

d, \(\sqrt{x+3}\ge5\)

\(\Rightarrow x+3\ge25\)

\(\Rightarrow x\ge22\)

e, \(\sqrt{3x-1}< 2\)

\(\Rightarrow3x-1< 4\)

\(\Rightarrow3x< 5\)

\(\Rightarrow x< \frac{5}{3}\)

g, \(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\)

\(\Rightarrow\sqrt{\left(x-3\right)\left(x+3\right)}+\sqrt{\left(x-3\right)^2}=0\)

\(\Rightarrow\sqrt{x-3}\left(\sqrt{x+3}+\sqrt{x-3}\right)=0\)

\(\left(\sqrt{x+3}+\sqrt{x-3}\right)>0\)

\(\Rightarrow\sqrt{x-3}=0\)

\(\Rightarrow x-3=0\)

\(\Rightarrow x=3\)

7 tháng 7 2019

a) \(16x^2-5=0\)

\(\Leftrightarrow16x^2=5\)

\(\Leftrightarrow x^2=\frac{5}{16}\)

\(\Leftrightarrow x=\pm\sqrt{\frac{5}{16}}\)

b) \(2\sqrt{x-3}=4\)

\(\Leftrightarrow\sqrt{x-3}=2\)

\(\Leftrightarrow x-3=4\)

\(\Leftrightarrow x=7\)

c) \(\sqrt{4x^2-4x+1}=3\)

\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=3\)

\(\Leftrightarrow2x-1=3\)

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=2\)

d) \(\sqrt{x+3}\ge5\)

\(\Leftrightarrow x+3\ge25\)

\(\Leftrightarrow x\ge22\)

e) \(\sqrt{3x-1}< 2\)

\(\Leftrightarrow3x-1< 4\)

\(\Leftrightarrow3x< 5\)

\(\Leftrightarrow x< \frac{5}{3}\)

g) \(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\)

\(\Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}+\sqrt{\left(x-3\right)^2}=0\)

\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}+\sqrt{x-3}\right)=0\)

Vì \(\left(\sqrt{x+3}+\sqrt{x-3}\right)>0\)

\(\Leftrightarrow\sqrt{x-3}=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

1 tháng 8 2019

)1) \(x^2-9=x^2-3^2=\left(x-3\right)\left(x+3\right)\)

2) \(9x^2-16=\left(3x\right)^2-4^2=\left(3x-4\right)\left(3x+4\right)\)

3) \(x^2-5=x^2-\left(\sqrt{5}\right)^2=\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\)

4) \(x-9=\left(\sqrt{x}\right)^2-3^2=\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\)(ĐK: \(x\ge0\))

5) \(x-3=\left(\sqrt{x}\right)^2-\left(\sqrt{3}\right)^2=\left(\sqrt{x}-\sqrt{3}\right)\left(\sqrt{x}+\sqrt{3}\right)\)(ĐK: nt)

6) \(x+2\sqrt{x}+1=\left(\sqrt{x}\right)^2+2\cdot\sqrt{x}\cdot1+1=\left(\sqrt{x}+1\right)^2\)(ĐK: nt)

7) \(x-4\sqrt{x}+4=\left(\sqrt{x}\right)^2-2\cdot\sqrt{x}\cdot2+2^2=\left(\sqrt{x}-2\right)^2\)(ĐK: nt)

8) \(4x+4\sqrt{x}+1=\left(2\sqrt{x}\right)^2+2\cdot2\sqrt{x}\cdot1+1=\left(2\sqrt{x}+1\right)^2\)(ĐK:nt

9)

\(x+2\sqrt{x}-35\\ =x-5\sqrt{x}+7\sqrt{x}-35\\ =\sqrt{x}\left(\sqrt{x}-5\right)+7\left(\sqrt{x}-5\right)\\=\left(\sqrt{x}-5\right)\left(\sqrt{x}+7\right)\)(ĐK: nt)

22 tháng 8 2018

a. Ta có: x2-11=0

⇌ x2=11

\(\left[{}\begin{matrix}x=\sqrt{11}\\x=-\sqrt{11}\end{matrix}\right.\)

b.Ta có: x2-2\(\sqrt{13}\)x+\(\sqrt{13}\)=0

⇌(x-\(\sqrt{13}\))2=0

⇌ x-\(\sqrt{13}\)=0

⇌ x=\(\sqrt{13}\)

c. Ta có : x2-9x+14=0

⇌ (x-7)(x-2)=0

\(\left[{}\begin{matrix}x-7=0\\z-2=0\end{matrix}\right.\)\(\left[{}\begin{matrix}x=7\\x=2\end{matrix}\right.\)

d.Ta có \(\sqrt{x}\)-6=13

\(\sqrt{x}\)=19

⇌x = 361

e.Ta có: \(\sqrt{x}\)+9=3

\(\sqrt{x}\)≥0∀x⇒\(\sqrt{x}\)+9≥9

⇒ ptvn

f.Ta có:\(\sqrt{x^2}\)-2x+4=x-1

⇌ |x|-3x-5=0(*)

TH1: x≥0

⇒ pt(*) ⇌ x-3x+5=0⇌-2x-5=0⇒x=\(\dfrac{5}{2}\)(t/m)

TH2: x<0

⇒ pt(*) ⇌ -x-3x+5=0⇌-4x+5=0⇒x=\(\dfrac{5}{4}\)(l)

Vậy x=\(\dfrac{5}{2}\)là nghiệm của phương trình

a: =>|x-3|=4-x

\(\Leftrightarrow\left\{{}\begin{matrix}x< =4\\\left(4-x-x+3\right)\left(4-x+x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =4\\\left(7-2x\right)=0\end{matrix}\right.\Leftrightarrow x=\dfrac{7}{2}\)

b: =>|x-5|=3-19x

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{19}\\\left(x-5-3+19x\right)\left(x-5+3-19x\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{19}\\\left(20x-8\right)\left(-18x-2\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{1}{9}\right\}\)

c: =>\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}+\sqrt{x-3}\right)=0\)

=>căn x-3=0

=>x=3