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`(2x-6)(4x+16)=0`
`@TH1:`
`2x-6=0`
`<=>2x=6`
`<=>x=3`
`@TH2:`
`4x+16=0`
`<=>4x=-16`
`<=>x=-4`
\(\left(2x-6\right).\left(4x+16\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\4x+16=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=6\\4x=-16\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)
\(S=\left\{3;-4\right\}\)
\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)
\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)
\(\Leftrightarrow x^2-9-x^2+3x=0\)
\(\Leftrightarrow3x-9=0\)
\(\Leftrightarrow3x=9\)
\(\Leftrightarrow x=3\left(n\right)\)
Vậy \(S=\left\{3\right\}\)
\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)
\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)
\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)
\(\Leftrightarrow12x-9-12x+20+2x-7>0\)
\(\Leftrightarrow2x+4>0\)
\(\Leftrightarrow2x>-4\)
\(\Leftrightarrow x>-2\)
\(a,\Leftrightarrow\left(4-5x\right)\left(4+5x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+1-2\right)\left(x+1+2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(3x+1-2x\right)\left(3x+1+2x\right)=0\\ \Leftrightarrow\left(x+1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{1}{5}\end{matrix}\right.\\ d,Sửa:\left(4x+1\right)^2-\left(x-2\right)^2=0\\ \Leftrightarrow\left(4x+1-x+2\right)\left(4x+1+x-2\right)=0\\ \Leftrightarrow\left(3x+3\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{5}\end{matrix}\right.\\ e,\Leftrightarrow\left(2x+1-x-3\right)\left(2x+1+x+3\right)=0\\ \Leftrightarrow\left(x-2\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
a, pt <=> (x^3+x^2)-(4x^2-4) = 0
<=> (x+1).(x^2-4x+4) = 0
<=> (x+1).(x-2)^2 = 0
<=> x+1=0 hoặc x-2=0
<=> x=-1 hoặc x=2
b, pt <=> (x^4-x^3)+(2x^3-2x^2)-(2x^2-2x)+(3x-3) = 0
<=> (x-1).(x^3+2x^2-2x+3) = 0
<=> (x-1).[(x^3+3x^2)-(x^2+3x)+(3x+3)] = 0
<=> (x-1).(x+3).(x^2-3x+3) = 0
<=> x-1=0 hoặc x+3=0 ( vì x^2-3x+3 > 0 )
<=> x=1 hoặc x=-3
c, pt <=> (4^x-10.2^x+25)-9 =0
<=> (2^x-5)^2-9 = 0
<=> (2^x-5-3).(2^x-5+3) = 0
<=> (2^x-8).(2^x-2) = 0
<=> 2^x-8=0 hoặc 2^x-2=0
<=> x=3 hoặc x=1
Tk mk nha
a) \(x^3-3x^2+4=0\)
\(\Leftrightarrow\)\(x^3+x^2-4x^2+4=0\)
\(\Leftrightarrow\)\(x^2\left(x+1\right)-4\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\)\(\left(x+1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\)\(\left(x+1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
Vậy....
=>(2x-3)(2x+3)(x-4)-(2x-3)(x-4)(x+4)=0
=>(2x-3)(x-4)(2x+3-x-4)=0
=>(2x-3)(x-4)(x-1)=0
=>\(x\in\left\{1;4;\dfrac{3}{2}\right\}\)
\(\Leftrightarrow4x^3+8x^2-12x^2-24x+9x+18=0\)
\(\Leftrightarrow\left(x+2\right)\left(4x^2-12x+9\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x-3\right)^2=0\)
=>x=3/2 hoặc x=-2
4x^3-4x^2-15x+18=0
<=> 4x^3 +8x^2-12x^2-24x+9x+18=0
<=> 4x^2(x+2)-12x(x+2)+9(x+2)=0
<=> (x+2).(4x^2-12x+9)=0
<=> (x+2).(2x-3)^2=0
<=> x+2=0
(2x-3)^2=0
<=> x=-2
x=3/2
Trả lời:
\(\frac{x-1}{2x^2-4x}-\frac{7}{8x}=\frac{5-x}{4x^2-8x}-\frac{1}{8x-16}\)\(\left(đkxđ:x\ne0;x\ne2\right)\)
\(\Leftrightarrow\frac{x-1}{2x\left(x-2\right)}-\frac{7}{8x}=\frac{5-x}{4x\left(x-2\right)}-\frac{1}{8\left(x-2\right)}\)
\(\Leftrightarrow\frac{4\left(x-1\right)}{8x\left(x-2\right)}-\frac{7\left(x-2\right)}{8x\left(x-2\right)}=\frac{2\left(5-x\right)}{8x\left(x-2\right)}-\frac{x}{8x\left(x-2\right)}\)
\(\Rightarrow4\left(x-1\right)-7\left(x-2\right)=2\left(5-x\right)-x\)
\(\Leftrightarrow4x-4-7x+14=10-2x-x\)
\(\Leftrightarrow10-3x=10-3x\)
\(\Leftrightarrow-3x+3x=10-10\)
\(\Leftrightarrow0x=0\)( luôn thỏa mãn )
Vậy S = R với \(x\ne0;x\ne2\)
Phương trình \(\Leftrightarrow\left(2^x\right)^2+10.2^x+16=0.\)
Đăt \(y=2^x>0\)
\(\Rightarrow y^2+10y+16=0\)
Giải phương trình bậc 2 tìm y từ đó suy ra x
Ta có 4x-10.2x+16=0
<=> (2x)2-10.2x+25-9=0
<=> (2x-5)2-9=0
<=> (2x-5+3)(2x-5-3)=0
<=> (2x-2)(2x-8)=0
=> 2x-2=0 hoặc 2x-8=0
• 2x-2=0 => 2x=2 => x=1
• 2x-8=0 => 2x=8 => x=3
Vậy ...