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\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)
\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)
\(2.\left(x-2\right)\left(2x-1\right)=x^2-2x.\Leftrightarrow\left(x-2\right)\left(2x-1\right)-x\left(x-2\right)=0.\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1-x\right)=0.\Leftrightarrow\left(x-2\right)\left(x-1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=1.\end{matrix}\right.\)
\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)
\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)
\(5.\left|3x+2\right|=4.\Leftrightarrow\left[{}\begin{matrix}3x+2=4.\\3x+2=-4.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}.\\x=-2.\end{matrix}\right.\)
\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)
\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)
\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)
\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
\(6,\left|2x-5\right|=\left|-x+2\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=-x+2\\2x-5=x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)
Đặt \(u=x^2-x\)
Phương trình trở thành \(u^2-4u+4=0\)
\(\Leftrightarrow\left(u-2\right)^2=0\)
\(\Leftrightarrow u-2=0\)
\(\Rightarrow x^2-x=2\)
\(\Rightarrow x^2-x-2=0\)
Ta có \(\Delta=1^2+4.2=9,\sqrt{\Delta}=3\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1+3}{2}=2\\x=\frac{1-3}{2}=-1\end{cases}}\)
Đặt \(2x+1=w\)
Phương trình trở thành \(w^2-w=2\)
\(\Rightarrow\orbr{\begin{cases}w=2\\w=-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=2\\2x+1=-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-1\end{cases}}\)
a: Ta có: \(3x-\left(3x+2\right)=x+3\)
\(\Leftrightarrow x+3=-2\)
hay x=-5
b: Ta có: \(\dfrac{5x-1}{4}+\dfrac{2x-1}{3}=\dfrac{3x}{2}\)
\(\Leftrightarrow15x-3+8x-4=18x\)
\(\Leftrightarrow5x=7\)
hay \(x=\dfrac{7}{5}\)
Áp dụng công thức: \(A\left(x\right).B\left(x\right)=0\Leftrightarrow\left[{}\begin{matrix}A\left(x\right)=0\\B\left(x\right)=0\end{matrix}\right.\)
a) \(PT\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{2}{3};-\dfrac{5}{4}\right\}\)
b) \(PT\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-20\end{matrix}\right.\)
Vậy: \(S=\left\{3;20\right\}\)
c) Vì \(x^2+1\ge1>0\forall x\)
\(\Rightarrow4x+2=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)
d) \(PT\Leftrightarrow\left[{}\begin{matrix}2x+7=0\\x-5=0\\5x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\\x=-\dfrac{1}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{7}{2};5;-\dfrac{1}{5}\right\}\)
a: =>3x-2=0 hoặc 4x+5=0
=>x=2/3 hoặc x=-5/4
b: =>(x-3)(x+20)=0
=>x=3 hoặc x=-20
c: =>4x+2=0
hay x=-1/2
d: =>2x+7=0 hoặc x-5=0 hoặc 5x+1=0
=>x=-7/2 hoặc x=5 hoặc x=-1/5
a: =>|x-7|=3-2x
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(-2x+3\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(2x-3-x+7\right)\left(2x-3+x-7\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(x+4\right)\left(3x-10\right)=0\end{matrix}\right.\Leftrightarrow x=-4\)
b: =>|2x-3|=4x+9
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(4x+9-2x+3\right)\left(4x+9+2x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(2x+12\right)\left(6x+6\right)=0\end{matrix}\right.\Leftrightarrow x=-1\)
c: =>3x+5=2-5x hoặc 3x+5=5x-2
=>8x=-3 hoặc -2x=-7
=>x=-3/8 hoặc x=7/2
chẳng ai giải, thôi mình giải vậy!
a) Đặt \(y=x^2+4x+8\),phương trình có dạng:
\(t^2+3x\cdot t+2x^2=0\)
\(\Leftrightarrow t^2+xt+2xt+2x^2=0\)
\(\Leftrightarrow t\left(t+x\right)+2x\left(t+x\right)=0\)
\(\Leftrightarrow\left(2x+t\right)\left(t+x\right)=0\)
\(\Leftrightarrow\left(2x+x^2+4x+8\right)\left(x^2+4x+8+x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-4\end{cases}}\)vậy tập nghiệm của phương trình là:S={-2;-4}
b) nhân 2 vế của phương trình với 12 ta được:
\(\left(6x+7\right)^2\left(6x+8\right)\left(6x+6\right)=72\)
Đặt y=6x+7, ta được:\(y^2\left(y+1\right)\left(y-1\right)=72\)
giải tiếp ra ta sẽ được S={-2/3;-5/3}
c) \(\left(x-2\right)^4+\left(x-6\right)^4=82\)
S={3;5}
d)s={1}
e) S={1;-2;-1/2}
f) phương trình vô nghiệm
a) ĐKXĐ: \(x\ne0\)
Ta có: \(\dfrac{3x^2+7x-10}{x}=0\)
Suy ra: \(3x^2+7x-10=0\)
\(\Leftrightarrow3x^2-3x+10x-10=0\)
\(\Leftrightarrow3x\left(x-1\right)+10\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\3x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{10}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{1;-\dfrac{10}{3}\right\}\)
a/ \(\dfrac{3x^2+7x-10}{x}=0\)
\(< =>3x^2+7x-10=0\)
\(< =>3x^2+10x-3x-10=0\)
\(< =>\left(3x^2+10x\right)-\left(3x+10\right)=0\)
\(< =>x\left(3x+10\right)-\left(3x+10\right)=0\)
\(< =>\left(3x+10\right)\left(x-1\right)=0\)
\(=>\left\{{}\begin{matrix}3x+10=0=>x=-\dfrac{10}{3}\\x-1=0=>x=1\end{matrix}\right.\)
Vậy tập nghiệm của .....
1: \(\Leftrightarrow6\left(3x-1\right)+3\left(6x-2\right)=4\left(1-3x\right)\)
=>18x-6+18x-6=4-12x
=>36x-12=4-12x
=>48x=16
hay x=1/3
2: \(\Leftrightarrow\left(2x-1\right)\left(2x-1+x-3\right)=0\)
=>(2x-1)(3x-4)=0
=>x=1/2 hoặc x=4/3
a) (x-1)(5x+3)=(3x-8)(x-1)
= (x-1)(5x+3)-(3x-8)(x-1)=0
=(x-1)[(5x+3)-(3x-8)]=0
=(x-1)(5x+3-3x+8)=0
=(x-1)(2x+11)=0
\(\Leftrightarrow\) x-1=0 hoặc 2x+11=0
\(\Leftrightarrow\) x=1 hoặc x=\(\dfrac{-11}{2}\)
Vậy S={1;\(\dfrac{-11}{2}\)}
b) 3x(25x+15)-35(5x+3)=0
=3x.5(5x+3)-35(5x+3)=0
=15x(5x+3)-35(5x+3)=0
=(5x+3)(15x-35)=0
\(\Leftrightarrow\) 5x+3=0 hoặc 15x-35=0
\(\Leftrightarrow\) x=\(\dfrac{-3}{5}\) hoặc x=\(\dfrac{7}{3}\)
Vậy S={\(\dfrac{-3}{5};\dfrac{7}{3}\)}
c) (2-3x)(x+11)=(3x-2)(2-5x)
=(2-3x)(x+11)-(3x-2)(2-5x)=0
=(3x-2)[(x+11)-(2-5x)]=0
=(3x-2)(x+11-2+5x)=0
=(3x-2)(6x+9)=0
\(\Leftrightarrow\) 3x-2=0 hoặc 6x+9=0
\(\Leftrightarrow\) x=\(\dfrac{2}{3}\) hoặc x=\(\dfrac{-3}{2}\)
Vậy S={\(\dfrac{2}{3};\dfrac{-3}{2}\)}
d) (2x2+1)(4x-3)=(2x2+1)(x-12)
=(2x2+1)(4x-3)-(2x2+1)(x-12)=0
=(2x2+1)[(4x-3)-(x-12)=0
=(2x2+1)(4x-3-x+12)=0
=(2x2+1)(3x+9)=0
\(\Leftrightarrow\)2x2+1=0 hoặc 3x+9=0
\(\Leftrightarrow\)x=\(\dfrac{1}{2}\)hoặc x=\(\dfrac{-1}{2}\) hoặc x=-3
Vậy S={\(\dfrac{1}{2};\dfrac{-1}{2};-3\)}
e) (2x-1)2+(2-x)(2x-1)=0
=(2x-1)[(2x-1)+(2-x)=0
=(2x-1)(2x-1+2-x)=0
=(2x-1)(x+1)=0
\(\Leftrightarrow\) 2x-1=0 hoặc x+1=0
\(\Leftrightarrow\) x=\(\dfrac{-1}{2}\) hoặc x=-1
Vậy S={\(\dfrac{-1}{2}\);-1}
f)(x+2)(3-4x)=x2+4x+4
=(x+2)(3-4x)=(x+2)2
=(x+2)(3-4x)-(x+2)2=0
=(x+2)[(3-4x)-(x+2)]=0
=(x+2)(3-4x-x-2)=0
=(x+2)(-5x+1)=0
\(\Leftrightarrow\) x+2=0 hoặc -5x+1=0
\(\Leftrightarrow\) x=-2 hoặc x=\(\dfrac{1}{5}\)
Vậy S={-2;\(\dfrac{1}{5}\)}
\(a,\left(x+2\right)^2\left(4x+6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+2\right)^2=0\\4x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x+2=0\\4x=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{3}{2}\end{cases}}\)
Vậy .............
\(b,3x^2-2x-1=0\)
\(\Leftrightarrow3x^2-3x+x-1=0\)
\(\Leftrightarrow\left(3x^2-3x\right)+\left(x-1\right)=0\)
\(\Leftrightarrow3x\left(x-1\right)+\left(x-1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x+1=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x=-1\\x=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{3}\\x=1\end{cases}}\)
Vậy ............
\(\left(x+2\right)^2\left(4x+6\right)=0\)
\(< =>\orbr{\begin{cases}\left(x+2\right)^2=0\\4x+6=0\end{cases}< =>\orbr{\begin{cases}x+2=0\\4x=-6\end{cases}< =>\orbr{\begin{cases}x=-2\\x=-\frac{3}{2}\end{cases}}}}\)
Vậy phương trình đã cho có nghiệm \(x=-2;x=-\frac{3}{2}\)
\(3x^2-2x-1=0\)
\(< =>3x^2-3x+x-1=0\)
\(< =>3x\left(x-1\right)+\left(x-1\right)=0\)
\(< =>\left(3x+1\right)\left(x-1\right)=0\)
\(< =>\orbr{\begin{cases}3x+1=0\\x-1=0\end{cases}}< =>\orbr{\begin{cases}3x=-1\\x=1\end{cases}}< =>\orbr{\begin{cases}x=-\frac{1}{3}\\x=1\end{cases}}\)
Vậy phương trình đã cho có nghiệm \(x=-\frac{1}{3};x=1\)