Giải phương trình:
a) |||x| - 2| - 1| =5
b) |x-3| = (x-3)^2
c) |x+2| + |x| + |x-2| = 4
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1:
a: =>(|x|+4)(|x|-1)=0
=>|x|-1=0
=>x=1; x=-1
b: =>x^2-4>=0
=>x>=2 hoặc x<=-2
d: =>|2x+5|=2x-5
=>x>=5/2 và (2x+5-2x+5)(2x+5+2x-5)=0
=>x=0(loại)
\(a,\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)
\(\Leftrightarrow x^2-4x+4-\left(x^2-9\right)=6\)
\(\Leftrightarrow-4x+13=6\)
\(\Leftrightarrow-4x=-7\)
\(\Leftrightarrow x=\dfrac{7}{4}\)
\(b,\left(x+3\right)^2+\left(4+x\right)\left(4-x\right)=10\)
\(\Leftrightarrow x^2+6x+9+16-x^2=10\)
\(\Leftrightarrow6x+25=10\)
\(\Leftrightarrow6x=-15\)
\(\Leftrightarrow x=-\dfrac{5}{2}\)
\(c,\left(x+4\right)^2+\left(1-x\right)\left(1+x\right)=7\)
\(\Leftrightarrow x^2+8x+16+1-x^2=7\)
\(\Leftrightarrow8x+17=7\)
\(\Leftrightarrow8x=-10\)
\(\Leftrightarrow x=-\dfrac{5}{4}\)
\(d,\left(x-4\right)^2-\left(x-2\right)\left(x+2\right)=6\)
\(\Leftrightarrow x^2-8x+16-\left(x^2-4\right)=6\)
\(\Leftrightarrow-8x+20=6\)
\(\Leftrightarrow-8x=-14\)
\(\Leftrightarrow x=\dfrac{7}{4}\)
#\(Urushi\)
`2/(4-x^2)+1/(x^2-2x)=(x-4)/(x^2+2x)(x ne 0,+-2)`
`<=>(2x)/(4x-x^3)+(x+2)/(x^3-4x)=(x^2-6x+8)/(x^3-4x)`
`<=>-2x+x+2=x^2-6x+8`
`<=>x^2-7x+10=0`
`<=>x^2-2x-5x+10=0`
`<=>x(x-2)-5(x-2)=0`
`<=>(x-2)(x-5)=0`
Vì `x ne 2=>x-2 ne 0`
`=>x-5=0`
`=>x=5`
Vậy `S={5}`
b) ĐKXĐ: \(x\ne1\)
Ta có: \(\dfrac{2}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{x}{x^2+x+1}\)
\(\Leftrightarrow\dfrac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
Suy ra: \(2x^2+2x+1-3x^2-x^2+x=0\)
\(\Leftrightarrow-2x^2+x+1=0\)
\(\Leftrightarrow-2x^2+2x-x+1=0\)
\(\Leftrightarrow-2x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\-2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\-2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(loại\right)\\x=-\dfrac{1}{2}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)
a, \(x^2+7x+10-12x+9=x^2-10x+25\)
\(\Leftrightarrow5x=6\Leftrightarrow x=\dfrac{6}{5}\)
b, bạn ktra lại đề nhé
c, \(x^2-4+3x+3=3+x^2-x-2\)
\(\Leftrightarrow x^2+3x-1=x^2-x+1\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)
Giải phương trình:
a)\(\sqrt{x^2+2x\sqrt{3}+3}=\sqrt{3}+x\)
b)\(\sqrt{x-3+2\sqrt{x-4}}=2\sqrt{x-4}+1\)
a)Pt\(\Leftrightarrow\sqrt{\left(x+\sqrt{3}\right)^2}=x+\sqrt{3}\)
\(\Leftrightarrow\left|x+\sqrt{3}\right|=x+\sqrt{3}\)
\(\Leftrightarrow x+\sqrt{3}\ge0\)\(\Leftrightarrow x\ge-\sqrt{3}\)
Vậy...
b)Đk:\(x\ge4\)
Pt\(\Leftrightarrow\sqrt{\left(x-4\right)+2\sqrt{x-4}+1}=2\sqrt{x-4}+1\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-4}+1\right)^2}=1+2\sqrt{x-4}\)
\(\Leftrightarrow\sqrt{x-4}+1=2\sqrt{x-4}+1\)
\(\Leftrightarrow\sqrt{x-4}=0\)
\(\Leftrightarrow x=4\) (tm)
Vậy...
a) Ta có: \(\sqrt{x^2+2x\sqrt{3}+3}=x+\sqrt{3}\)
\(\Leftrightarrow\left|x+\sqrt{3}\right|=x+\sqrt{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\sqrt{3}=x+\sqrt{3}\left(x\ge-\sqrt{3}\right)\\x+\sqrt{3}=-x-\sqrt{3}\left(x< -\sqrt{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge-\sqrt{3}\\x=-\sqrt{3}\left(loại\right)\end{matrix}\right.\Leftrightarrow x\ge-\sqrt{3}\)
a, \(\left(x-5\right)\left(x-5+3\right)=0\Leftrightarrow x=5;x=2\)
b, \(-4x=\dfrac{274}{21}\Leftrightarrow x=-\dfrac{137}{42}\)
c, đk x khác - 2 ; 2
\(x^2-3x+2-x^2-2x=6-7x\Leftrightarrow-5x+2=6-7x\)
\(\Leftrightarrow2x-4=0\Leftrightarrow x=2\left(ktm\right)\)
Vậy pt vô nghiệm
\(\dfrac{2x}{x-1}+\dfrac{4}{x^2+2x-3}=\dfrac{2x-5}{x+3}\)
\(\Leftrightarrow\dfrac{2x}{x-1}+\dfrac{4}{\left(x-1\right)\left(x+3\right)}=\dfrac{2x-5}{x+3}\)
\(ĐK:x\ne1;-3\)
\(\Leftrightarrow\dfrac{2x\left(x+3\right)+4}{\left(x-1\right)\left(x+3\right)}=\dfrac{\left(2x-5\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow2x\left(x+3\right)+4=\left(2x-5\right)\left(x-1\right)\)
\(\Leftrightarrow2x^2+6x+4=2x^2-2x-5x+5\)
\(\Leftrightarrow13x=1\)
\(\Leftrightarrow x=\dfrac{1}{13}\left(tm\right)\)
a: \(\Leftrightarrow2x\left(x^2+2x+5\right)=0\)
=>x=0
b: \(\Leftrightarrow\dfrac{x}{x-1}-\dfrac{x+1}{x-3}=\dfrac{1}{2}\)
\(\Leftrightarrow x^2-4x+3=2x\left(x-3\right)-2\left(x^2-1\right)\)
\(\Leftrightarrow x^2-4x+3=2x^2-6x-2x^2+2=-6x+2\)
\(\Leftrightarrow x^2+2x+1=0\)
=>x=-1(nhận)
\(a,2x^3+4x^2+10x=0\\ \Leftrightarrow2x\left(x^2+2x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x=0\\x^2+2x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x^2+2x+1\right)+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x+1\right)^2+4=0\left(vô..lí\right)\end{matrix}\right.\)
\(b,ĐKXĐ:\left\{{}\begin{matrix}x\ne1\\x\ne3\\x\ne4\end{matrix}\right.\\ \dfrac{x^2-4x}{x^2-5x+4}-\dfrac{1}{2}=\dfrac{x+1}{x-3}\\ \Leftrightarrow\dfrac{x\left(x-4\right)}{\left(x-1\right)\left(x-4\right)}-\dfrac{1}{2}=\dfrac{x+1}{x-3}\\ \Leftrightarrow\dfrac{x}{x-1}-\dfrac{1}{2}-\dfrac{x+1}{x-3}=0\\ \Leftrightarrow\dfrac{2x\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}-\dfrac{\left(x-1\right)\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}-\dfrac{2\left(x+1\right)\left(x-1\right)}{2\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\dfrac{2x^2-6x}{2\left(x-1\right)\left(x-3\right)}-\dfrac{x^2-4x+3}{2\left(x-1\right)\left(x-3\right)}-\dfrac{2x^2-2}{\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\dfrac{2x^2-6x-x^2+4x-3-2x^2+2}{2\left(x-1\right)\left(x-3\right)}=0\)
\(\Rightarrow-x^2-2x-1=0\)
\(\Leftrightarrow x^2+2x+1=0\\ \Leftrightarrow\left(x+1\right)^2=0\\ \Leftrightarrow x+1=0\\ \Leftrightarrow x=-1\left(tm\right)\)
a) |||x| - 2| - 1| = 5
=> ||x| - 2| - 1 = 5
=> ||x| - 2| = 6
=> |x| - 2 = 6
=> |x| = 8
=> x = 8 hoặc x = -8
Vậy phương trình có tập nghiệm S = {8; -8}
b) |x - 3| = (x - 3)^2
=> x - 3 = (x - 3)^2
=> x - 3 = x^2 - 6x + 9
=> x^2 - 6x + 9 - x + 3 = 0
=> x^2 - 7x + 12 = 0
=> x^2 - 3x - 4x + 12 = 0
=> x(x - 3) - 4(x - 3) = 0
=> (x - 3)(x - 4) = 0
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=4\end{cases}}\)
Vậy phương trình có tập nghiệm S = {3; 4}
Chẳng biết có đúng không nữa.