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1: \(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}+\sqrt{x-3}\right)=0\)
=>căn x-3=0
=>x-3=0
=>x=3
2: =>\(\sqrt{2x-3+2\sqrt{2x-3}+1}+\sqrt{2x-3+2\cdot\sqrt{2x-3}\cdot4+16}=5\)
=>\(\left|\sqrt{2x-3}+1\right|+\left|\sqrt{2x-3}+4\right|=5\)
=>2*căn 2x-3+5=5
=>2x-3=0
=>x=3/2
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bài 1) ta có : \(\dfrac{\left(5-x\right)\sqrt{5-x}+\left(3-x\right)\sqrt{3-x}}{\sqrt{5-x}+\sqrt{3-x}}=2\) đk : \(x\le3\)
\(\Leftrightarrow\left(5-x\right)\sqrt{5-x}+\left(3-x\right)\sqrt{3-x}=2\left(\sqrt{5-x}+\sqrt{3-x}\right)\)
\(\Leftrightarrow\left(\sqrt{5-x}+\sqrt{3-x}\right)\left(5-x+\sqrt{\left(5-x\right)\left(3-x\right)}+3-x\right)=2\left(\sqrt{5-x}+\sqrt{3-x}\right)\)
\(\Leftrightarrow5-x+\sqrt{\left(5-x\right)\left(3-x\right)}+3-x=2\)
\(\Leftrightarrow\sqrt{\left(5-x\right)\left(3-x\right)}=2x-6\) đk : \(x\ge3\) \(\Rightarrow x=3\)
thử lại ta thấy \(x=3\) là nghiệm
vậy \(x=3\)
bài 2) ta có : \(\sqrt{x-5}-\dfrac{x-14}{3+\sqrt{x-5}}=3\) đk : \(x\ge5\)
\(\Leftrightarrow\dfrac{3\sqrt{x-5}+x-5-x+14}{3+\sqrt{x-5}}=3\)
\(\Leftrightarrow3\sqrt{x-5}+9=9+3\sqrt{x-5}\) (luôn đúng)
vậy \(x\ge5\)
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a. 2x\(^2\)-8=0
2x\(^2\)=8
x\(^2\)=4
x=2
b.3x\(^3\)-5x=0
x(3x\(^2\)-5)=0
\(\left[{}\begin{matrix}x=0\\x^2-5=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=0\\x^2=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=^+_-\sqrt{5}\end{matrix}\right.\)
c.x\(^4\)+3x\(^2\)-4=0\(^{\left(\cdot\right)}\)
đặt t=x\(^2\) (t>0)
ta có pt: t\(^2\)+3t-4=0 \(^{\left(1\right)}\)
thấy có a+b+c=1+3+(-4)=0 nên pt\(^{\left(1\right)}\) có 2 nghiệm
t\(_1\)=1; t\(_2\)=\(\dfrac{c}{a}\)=-4
khi t\(_1\)=1 thì x\(^2\)=1 ⇒x=\(^+_-\)1
khi t\(_2\)=-4 thì x\(^2\)=-4 ⇒ x=\(^+_-\)2
vậy pt đã cho có 4 nghiệm x=\(^+_-\)1; x=\(^+_-\)2
d)3x\(^2\)+6x-9=0
thấy có a+b+c= 3+6+(-9)=0 nên pt có 2 nghiệm
x\(_1\)=1; x\(_2\)=\(\dfrac{c}{a}=\dfrac{-9}{3}=-3\)
e. \(\dfrac{x+2}{x-5}+3=\dfrac{6}{2-x}\) (ĐK: x#5; x#2 )
⇔\(\dfrac{\left(x+2\right)\left(2-x\right)}{\left(x-5\right)\left(2-x\right)}+\dfrac{3\left(x+2\right)\left(2-x\right)}{\left(x-5\right)\left(2-x\right)}\)=\(\dfrac{6\left(x-5\right)}{\left(x-5\right)\left(2-x\right)}\)
⇒2x - x\(^2\) + 4 - 2x + 6x - 6x\(^2\) + 12 - 6x - 6x +30 = 0
⇔-7x\(^2\) - 6x + 46=0
Δ'=b'\(^2\)-ac = (-3)\(^2\) - (-7)\(\times\)46= 9+53 = 62>0
\(\sqrt{\Delta'}=\sqrt{62}\)
vậy pt có 2 nghiệm phân biệt
x\(_1\)=\(\dfrac{-b'+\sqrt{\Delta'}}{a}=\dfrac{3+\sqrt{62}}{-7}\)
x\(_2\)=\(\dfrac{-b'-\sqrt{\Delta'}}{a}=\dfrac{3-\sqrt{62}}{-7}\)
vậy pt đã cho có 2 nghiệm x\(_1\)=.....;x\(_2\)=......
câu g làm tương tự câu c
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\(x^{11}+3x^{10}+x^9+3x^8+x^7-3x^6-17x^5+3x^4+x^3+3x^2+x+3=0\)
\(\Leftrightarrow\left(x^{11}+2x^{10}+4x^9+6x^8+9x^7+6x^6+4x^5+2x^4+x^3\right)+\left(x^{10}+2x^9+4x^8+6x^7+9x^6+6x^5+4x^4+2x^3+x^2\right)-\left(5x^9+10x^8+20x^7+30x^6+45x^5+30x^4+20x^3+10x^2+5x\right)+\left(3x^8+6x^7+12x^6+18x^5+27x^4+18x^3+12x^2+6x+3\right)=0\)
\(\Leftrightarrow x^3\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)+x^2\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)-5\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)+3\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x^3+x^2-5x+3\right)\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)\left(x+3\right)\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x+3\right)\left(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1\right)=0\)
Dễ thấy: \(x^8+2x^7+4x^6+6x^5+9x^4+6x^3+4x^2+2x+1>0\forall x\)
Nên \(\left[{}\begin{matrix}\left(x-1\right)^2=0\\x+3=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)