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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, \(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\)
\(\Rightarrow\sqrt{\left(x-3\right)\left(x+3\right)}+\sqrt{\left(x-3\right)^2}=0\)
\(\Rightarrow\sqrt{x-3}\left(\sqrt{x+3}+\sqrt{x-3}\right)=0\)
\(\left(\sqrt{x+3}+\sqrt{x-3}\right)>0\)
\(\Rightarrow\sqrt{x-3}=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
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) \(x^2-9=x^2-3^2=\left(x-3\right)\left(x+3\right)\)
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)
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
Vì \(\sqrt{x}\)≥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
4 số liên tiếp nên chia hết cho 2.3.4=24
giá trị 9x luôn 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