Toán Lớp 9: giải các pt sau 1.√(x-3)^2 =3-x
2.√1-12x+36x^2 =5
3. √x+2 √x-1 =2
4.√4x^2-20x+25+2x =5
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Toán Lớp 9: giải các pt sau 1.√(x-3)^2 =3-x
2.√1-12x+36x^2 =5
3. √x+2 √x-1 =2
4.√4x^2-20x+25+2x =5
Comments ( 2 )
<=>|x-3|=3-x
text(TH1):x>=3
x-3=3-x
<=>x-3-3+x=0
<=>2x-6=0
<=>2x=6
<=>x=3(loại)
text(TH2):x<3
x-3=-(3-x)
<=>x-3=-3+x
<=>x-3+3-x=0
<=>0x=0(loại)
Vậy phương trình vô nghiệm.
<=>sqrt((1-6x))^2=5
text(TH1):x>=1/6
1-6x=5
<=>-6x=4
<=>x=-2/3(TM)
text(TH2):x<1/6
1-6x=-5
<=>-6x=-6
<=>x=1(TM)
Vậy x=1
<=>sqrt((sqrtsx−1)^2)=2
<=>|sqrtx−1|=2
text(TH1):x>=1
<=>sqrtx−1=2
<=>sqrtx=3
<=>x=9(TM)
text(TH2): 0<x<1
<=>1−sqrtx=2
<=>sqrtx=-1(loại)
<=>sqrt(4x^2-18x+25)=5
<=>4x^2-18x+25=25
<=>4x^2-18x=0
<=>2x(2x-9)=0
<=>\(\left[ \begin{array}{l}2x=0\\2x-9=0\end{array} \right.⇔\left[ \begin{array}{l}x=0\\x=\dfrac92\end{array} \right.\)