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a) \(\left(x^2+5x-6\right):\left(x-1\right)\)
\(=\left[x\left(x+6\right)-\left(x+6\right)\right]:\left(x-1\right)\)
\(=\left(x-1\right)\left(x+6\right):\left(x-1\right)\)
\(=x+6\)
b) \(\left(x^3-x^2-5x+21\right):\left(x^2-4x+7\right)\)
\(=\left(x+3\right)\left(x^2-4x+7\right):\left(x^2-4x+7\right)\)
\(=x+3\)
f(5) =0 <=> 5^2 -5^2 +b =0 => b = 0
b =0 ; f(x) =x^2 -5x =x(x-5) => nghiệm thứ 2 x2 =0
Thay x=5 vào phương trình, ta có:
52-5.5+b=0
\(\Rightarrow\)b=0
Ta có phương trình:
x2-5x=0
=> x2=0
Cho\(\frac{2x-1}{5x-10}=0\)(điều kiện là x khác 2)
\(\Rightarrow2x-1=0\Rightarrow x=\frac{1}{2}\)(thỏa mãn điều kiện)
Vậy nghiệm của phân thức trên là \(\frac{1}{2}\)
a) (x - 2)(x - 3). b) 3(x - 2)(x + 5).
c) (x - 2)(3x + 1). d) (x-2y)(x - 5y).
e) (x + l)(x + 2)(x - 3). g) (x-1)(x + 3)( x 2 + 3).
h) (x + y - 3)(x - y + 1).
a) Ta có: \(\left(x^2-5x\right)^2+10\left(x^2-5x\right)+24=0\)
\(\Leftrightarrow\left(x^2-5x\right)^2+4\left(x^2-5x\right)+6\left(x^2-5x\right)+24=0\)
\(\Leftrightarrow\left(x^2-5x\right)\left(x^2-5x+4\right)+6\left(x^2-5x+4\right)=0\)
\(\Leftrightarrow\left(x^2-5x+6\right)\left(x^2-5x+4\right)=0\)
\(\Leftrightarrow\left(x^2-2x-3x+6\right)\left(x^2-x-4x+4\right)=0\)
\(\Leftrightarrow\left[x\left(x-2\right)-3\left(x-2\right)\right]\left[x\left(x-1\right)-4\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=3\\x=4\end{matrix}\right.\)
Vậy: S={1;2;3;4}
b) Ta có: \(\left(2x+1\right)^2-2x-1=2\)
\(\Leftrightarrow\left(2x+1\right)^2-\left(2x+1\right)-2=0\)
\(\Leftrightarrow\left(2x+1\right)^2-2\left(2x+1\right)+\left(2x+1\right)-2=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x+1-2\right)+\left(2x+1-2\right)=0\)
\(\Leftrightarrow\left(2x+1+1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x+2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+2=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-2\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{-1;\dfrac{1}{2}\right\}\)
c) Ta có: \(x\left(x-1\right)\left(x^2-x+1\right)-6=0\)
\(\Leftrightarrow x\left(x^3-x^2+x-x^2+x-1\right)-6=0\)
\(\Leftrightarrow x\left(x^3-2x^2+2x-1\right)-6=0\)
\(\Leftrightarrow x^4-2x^3+2x^2-x-6=0\)
\(\Leftrightarrow x^4-2x^3+2x^2-4x+3x-6=0\)
\(\Leftrightarrow x^3\left(x-2\right)+2x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+2x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-x+3x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x\left(x^2-1\right)+3\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left[x\left(x-1\right)\left(x+1\right)+3\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(x^2-x+3\right)=0\)
mà \(x^2-x+3>0\forall x\)
nên (x-2)(x+1)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Vậy: S={2;-1}
d) Ta có: \(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0\)
\(\Leftrightarrow\left(x^2+1\right)^2+2x\left(x^2+1\right)+x\left(x^2+1\right)+2x^2=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x^2+1+2x\right)+x\left(x^2+1+2x\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x^2+x+1\right)=0\)
mà \(x^2+x+1>0\forall x\)
nên x+1=0
hay x=-1
Vậy: S={-1}
1) Ta có: \(x^2-4x+4=0\)
\(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
Vậy: S={2}
\(\left(x^2+5x+6\right)\left(x^2-11x+3x\right)=180\\ \Leftrightarrow\left(x+2\right)\left(x+3\right)\left(x-5\right)\left(x-6\right)=180\\ \Leftrightarrow\left[\left(x+2\right)\left(x-5\right)\right]\left[\left(x+3\right)\left(x-6\right)\right]\\ \Leftrightarrow\left(x^2-3x-10\right)\left(x^2-3x-18\right)=0\left(1\right)\)
Đặt \(x^2-3x-14=a\)
\(\left(1\right)\Leftrightarrow\left(a+4\right)\left(a-4\right)=180\\ \Leftrightarrow a^2-16=180\\ \Leftrightarrow a^2=196\\ \Leftrightarrow\left[{}\begin{matrix}a=14\\a=-14\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2-3x-14=14\\x^2-3x-14=-14\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2-3x=0\left(vô.lí\right)\\x^2-3x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
\(x^2-5x+6=0\)
\(\Leftrightarrow x^2-2x-3x+6=0\)
\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}\)
Vậy ....
x2 - 5x + 6 =0
<=> x2 - 2x - 3x + 6 = 0
<=> x( x - 2) - 3(x - 2)
<=>(x - 2)(x - 3)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}}\)
Vậy ......