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\(5x^2+3x-26=0\)
\(\Leftrightarrow5x^2+13x-10x-26=0\)
\(\Leftrightarrow x\left(5x+13\right)-2\left(5x+13\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(5x+13\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\5x+13=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-\frac{13}{5}\end{array}\right.\)
1) <=> x2 - 4x - x2 + 8 = 0 <=> x2 - 4x + 8 = 0
Dễ thấy phương trình vô nghiệm vì x2 - 4x + 8 = ( x - 2 )2 + 4 > 0
2) <=> ( x - 1 )3 = 0 <=> x = 1
3) <=> ( x - 2 )3 = 0 <=> x = 2
4) <=> ( 2x - 1 )3 = 0 <=> x = 1/2
1 ) 2x2 - 5x + 4x - 10 = 0
=> 2x2 + 4x - 5x - 10 = 0
=> 2x ( x + 2 ) - 5. ( x + 2 ) = 0
=> ( x + 2 ) . ( 2x - 5 ) = 0
=> \(\orbr{\begin{cases}x+2=0\\2x-5=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-2\\x=\frac{5}{2}\end{cases}}\)
Vậy \(x\in\left\{-2;\frac{5}{2}\right\}\)
2 ) x2 ( 2x - 3 ) + 3 - 2x = 0
=> x2 ( 2x - 3 ) - ( 2x - 3 ) = 0
=> ( 2x - 3 ) . ( x2 - 1 ) = 0
=> \(\orbr{\begin{cases}2x-3=0\\x^2-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}2x=3\\x^2=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{2}\\x=\pm1\end{cases}}\)
Vậy \(x\in\left\{\frac{3}{2};\pm1\right\}\)
Sửa để\(\frac{x-10}{1994}+\frac{x-8}{1996}+\frac{x-6}{1998}+\frac{x-4}{2000}+\frac{x-2}{2002}=5\)
\(\Leftrightarrow\frac{x-10}{1994}-1+\frac{x-8}{1996}-1+\frac{x-6}{1998}-1+\frac{x-4}{2000}-1+\frac{x-2}{2002}-1=0\)
\(\Leftrightarrow\frac{x-2004}{1994}+\frac{x-2004}{1996}+\frac{x-2004}{1998}+\frac{x-2004}{2000}+\frac{x-2004}{2002}=0\)
\(\Leftrightarrow\left(x-2004\right)\left(\frac{1}{1994}+\frac{1}{1996}+...+\frac{1}{2002}\right)=0\)
|_____________A__________________|
Vì A > 0 nên x - 2004 = 0
=> x = 2004
Vậy ..........
\(A=x^2+2x+2=x^2+2x+1+1=\left(x+1\right)^2+1\ge1>0\)
Vậy \(A_{min}=1\Leftrightarrow x=-1\)
\(B=x^2+4x=6=x^2+4x+4+2=\left(x+2\right)^2+2\ge2>0\)
Vậy \(B_{min}=2\Leftrightarrow x=-2\)
\(x^2-5x+6=\left(x-3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}}\)
1, <=>x^2-x-2 = x^2-4
<=>x^2-4-x^2+x+2 = 0
<=> x-2 = 0
<=> x=2
2, <=> (x-2).(x-3)=0
<=> x-2 = 0 hoặc x-3 = 0
<=> x=2 hoặc x=3
Tìm x
1) 3(x−1)2−3x(x−5)=1
2) (6x−2)2+(5x−2)2−4(3x−1)(5x−2)=0
3) (2x−5)(2x+5)−1=0
4) 5x2−20=0
Giusp mk vs
\(a,3(x-1)^2-3x(x-5)=1\)
\(\Leftrightarrow3x^2-6x+3-3x^2-15x=1\)
\(\Leftrightarrow\left[3x^2-3x^2\right]+3-\left[15x-6x\right]=1\)
\(\Leftrightarrow3-9x=1\)
\(\Leftrightarrow9x=2\Leftrightarrow x=\frac{2}{9}\)
=>x(x-2)=0
=>x=0 hoặc x=2
\(3x^4-24x=0\\ \Rightarrow3x\left(x^3-8\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x=0\\x^3-8=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^3=8\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)