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a, \(x^2-25-\left(x+5\right)=0\)
\(\Rightarrow x^2-5^2-\left(x+5\right)=0\)
\(\Rightarrow\left(x-5\right)\times\left(x+5\right)-\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\times\left(x-5-1\right)=0\)
\(\Rightarrow\left(x+5\right)\times\left(x-6\right)=0\)
\(\Rightarrow\hept{\begin{cases}x+5=0\\x-6=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=0-5=\left(-5\right)\\x=0+6=6\end{cases}}\)
b, \(\left(2x-1\right)^2-\left(4x^2-1\right)=0\)
\(\Rightarrow\left(2x-1\right)^2-\left(\left(2x\right)^2-1^2\right)=0\)
\(\Rightarrow\left(2x-1\right)^2-\left(2x-1\right)\times\left(2x+1\right)=0\)
\(\Rightarrow\left(2x-1\right)\times\left(2x-1-\left(2x+1\right)\right)=0\)
\(\Rightarrow\left(2x-1\right)\times\left(2x-1-2x-1\right)=0\)
\(\Rightarrow\left(2x-1\right)\times\left(-2\right)=0\)\(\Rightarrow\left(-4x\right)+2=0\)
\(\Rightarrow\left(-4x\right)=0-2=-2\)
\(\Rightarrow x=\frac{-2}{-4}=\frac{1}{2}\)
c, \(x^2\times\left(x^2+4\right)-x^2-4=0\)
\(\Rightarrow x^2\times\left(x^2+4\right)-\left(x^2+4\right)=0\)
\(\Rightarrow\left(x^2-1\right)\times\left(x^2+4\right)=0\)
\(\Rightarrow\hept{\begin{cases}x^2-1=0\\x^2+4=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x^2=1\\x^2=\left(-4\right)\end{cases}}\)
\(\Rightarrow x=1\)
a ) \(9x^2-49=9\)
\(\Leftrightarrow9x^2=58\)
\(\Leftrightarrow x^2=29\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=29\\x=-29\end{array}\right.\)
Vậy ......................
b ) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-1\right)\left(x+1\right)-27=0\)
\(\Leftrightarrow\left(x^3+3^3\right)-x.\left(x^2-1^2\right)-27=0\)
\(\Leftrightarrow x^3+27-x^3+x-27=0\)
\(\Leftrightarrow x=0\)
c ) \(\left(x-1\right)\left(x+2\right)-x-2=0\)
\(\Leftrightarrow x^2+2x-x-2-x-2=0\)
\(\Leftrightarrow x^2-4=0\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-2\end{array}\right.\)
Vây .....................
\(a)2\left(x+1\right)=3+2x\\ \Leftrightarrow2x+2=3+2x\\ \Leftrightarrow2x-2x=3-1\\ \Leftrightarrow0x=2\left(VN\right)\)
Vậy phương trình vô nghiệm
\(b)4x\left(1-x\right)-8=1-\left(4x^2+3\right)\\ \Leftrightarrow4x-4x^2-8=1-4x^2-3\\ \Leftrightarrow4x-8=-2\\ \Leftrightarrow4x=6\\ \Leftrightarrow x=\dfrac{3}{2}\)
Vậy \(S=\left\{\dfrac{3}{2}\right\}\)
\(c)x^3+1=x\left(x+1\right)\\ \Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)=x\left(x+1\right)\\ \Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)-x\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x^2-x+1-x\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x^2-2x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\x^2-2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
Vậy \(S=\left\{-1;1\right\}\)
\(d)\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\)
\(\Leftrightarrow 12\left(\dfrac{3x-2}{6}-5\right)=12.\dfrac{3-2\left(x+7\right)}{4}\)
\(\Leftrightarrow 6x-4-60=9-6\left(x+7\right)\)
\(\Leftrightarrow 6x-64=9-6x-42\)
\(\Leftrightarrow 6x-64=-6x-33\)
\(\Leftrightarrow 6x+6x=-33+64\\\Leftrightarrow 12x=31\\\Leftrightarrow x=\dfrac{31}{12}\)
Vậy \(S=\left\{\dfrac{31}{12}\right\}\)
a) 3x(4x - 3) - 2x(5 - 6x) = 0
=> 6x2 - 9x - 10x + 12x2 = 0
=> 18x2 - 19x = 0
=> x(18x - 19) = 0
=> \(\orbr{\begin{cases}x=0\\18x-19=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=\frac{19}{18}\end{cases}}\)
b) 5(2x - 3) + 4x(x - 2) + 2x(3 - 2x) = 0
=> 10x - 15 + 4x2 - 8x + 6x - 4x2 = 0
=> 8x - 15 = 0
=> 8x = 15
=> x = 15 : 8 = 15/8
c) 3x(2 - x) + 2x(x - 1) = 5x(x + 3)
=> 6x - 3x2 + 2x2 - 2x = 5x2 + 15x
=> 4x - x2 - 5x2 - 15x = 0
=> -6x2 - 11x = 0
=> -x(6x - 11) = 0
=> \(\orbr{\begin{cases}-x=0\\6x-11=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=\frac{11}{6}\end{cases}}\)
a) \(3x\left(4x-3\right)-2x\left(5-6x\right)=0\)
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow-19x=0\Leftrightarrow x=0\)
b) \(5\left(2x-3\right)+4x\left(x-2\right)+2x\left(3-2x\right)=0\)
\(\Leftrightarrow10x-15+4x^2-8x+6x-4x^2=0\)
\(\Leftrightarrow8x-15=0\Leftrightarrow x=\frac{15}{8}\)
1) \(\left(5x-4\right)\left(4x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-4=0\\4x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=4\\4x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{4}{5};\dfrac{3}{2}\right\}\)
2) \(\left(4x-10\right)\left(24+5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-10=0\\24+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=10\\5x=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-24}{5}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{5}{2};\dfrac{-24}{5}\right\}\)
3) \(\left(x-3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{3;\dfrac{-1}{2}\right\}\)
a) Ta có: \(x\left(x+3\right)-2x-6=0\)
\(\Leftrightarrow x\left(x+3\right)-\left(2x+6\right)=0\)
\(\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{-3;2\right\}\)
b) Ta có: \(4x^2-1+x\left(2x-1\right)=0\)
\(\Rightarrow\left(2x\right)^2-1^2+x\left(2x-1\right)=0\)
\(\Rightarrow\left[\left(2x\right)^2-1^2\right]+x\left(2x-1\right)=0\)
\(\Rightarrow\left(2x-1\right)\left(2x+1\right)+x\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1+x\right)=0\)
\(\Rightarrow\left(2x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\3x=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{-1}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{2};\frac{-1}{3}\right\}\)
a, x(x+3)-2x-6=0
⇔x(x+3)-2(x+3)=0
⇔(x+3)(x-2)=0
⇔x+3=0 hoặc x-2=0
⇔x= -3 hoặc x=2
b, 4x2-1+x(2x-1)=0
⇔(2x-1)(2x+1)+x(2x-1)=0
⇔(2x-1)(2x+1+x)=0
⇔(2x-1)(3x+1)=0
⇔2x-1=0 hoặc 3x+1=0
⇔x=1/2 hoặc x= -1/3