Ta có
\(-x^2+6x-8=-\left(x^2-6x+8\right)=-\left(x^2-2x-4x+8\right)=-\left[x\left(x-2\right)-4\left(x-2\right)\right]\)
\(=-\left(x-2\right)\left(x-4\right)\)
MTC:\(\left(x-2\right)\left(x-4\right)\)
\(\frac{\left(x-1\right)\left(x-4\right)+\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=\frac{-2}{\left(x-2\right)\left(x-4\right)}\)
\(\frac{x^2-x-4x+4+x^2+3x-2x-6}{\left(x-2\right)\left(x-4\right)}=-\frac{2}{\left(x-2\right)\left(x-4\right)}\)
\(\frac{2x^2-4x-2}{\left(x-2\right)\left(x-4\right)}=-\frac{2}{\left(x-2\right)\left(x-4\right)}\)
\(2x^2-4x-2+2=0\Rightarrow2x^2-4x=0\Rightarrow2x\left(x-4\right)=0\Rightarrow\orbr{\begin{cases}2x=0\Rightarrow x=0\\x-4=0\Rightarrow x=4\end{cases}}\)
a) trước nha
T i c k cho mình nha bạn cảm ơn sẽ làm típ câu b)
b) x3 - 8 - (x - 2) (x + 2) = 0
=> x3 - (2)3 - x2 - 22 = 0
mk làm bậy thui!!! mới lên lớp 8 mà!! 6756886787696969768658585685685685858978467
a/ ĐKXĐ: \(x\ne2;x\ne4\)
\(\frac{x-1}{x-2}+\frac{x+3}{x-4}=\frac{2}{\left(2-x\right)\left(x-4\right)}\)
\(\Rightarrow\frac{x-1}{x-2}+\frac{x-3}{x-4}=-\frac{2}{\left(x-2\right)\left(x-4\right)}\)
=> (x - 1)(x - 4) + (x - 2)(x - 3) = -2
=> x2 - 5x + 4 + x2 - 5x + 6 + 2 = 0
=> 2x2 - 10x + 12 = 0
=> x2 - 5x + 6 = 0 => (x - 2)(x - 3) = 0 => x = 2 (loại) hoặc x = 3
Vậy x = 3
b/ x3 - 8 - (x - 2)(x + 2) = 0
=> (x - 2)(x2 + 2x + 4) - (x - 2)(x + 2) = 0
=> (x - 2)(x2 + x + 2) = 0
=> x - 2 = 0 => x = 2
hoặc x2 + x + 2 = 0 , mà x2 + x + 2 > 0 => vô nghiệm
Vậy x = 2
\(\frac{x-1}{x-2}+\frac{x+3}{x-4}=\frac{2}{-x^2+6x-8}\)
\(\frac{\left(x-1\right)\left(x-4\right)+\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=\frac{2}{-\left(x^2-6x+8\right)}\)
\(\frac{\left(x^2-4x-x+4+x^2-2x+3x-6\right)}{\left(x-2\right)\left(x-4\right)}=\frac{-2}{x^2-2x-4x+8}\)
\(\frac{-4x-2}{\left(x-2\right)\left(x-4\right)}=\frac{-2}{\left(x-2\right)\left(x-4\right)}\)
\(-4x-2=-2\)
\(x=0\)
\(x^3-8-\left(x-2\right)\left(x+2\right)=0\Rightarrow\left(x-2\right)\left(x^2+2x+4\right)-\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x^2+2x+4-x-2\right)=0\Rightarrow\left(x-2\right)\left(x^2-x+2\right)=0\Rightarrow\left(x-2\right)\left(x^2-2.\frac{1}{2}.x+\frac{1}{4}+\frac{7}{4}\right)=0\)
\(\Rightarrow\left(x-2\right)\left[\left(x-\frac{1}{2}\right)^2+\frac{7}{4}\right]=0\Rightarrow\orbr{\begin{cases}x-2=0\\\left(x-\frac{1}{2}\right)^2+\frac{7}{4}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\\left(x-\frac{1}{2}\right)^2=-\frac{7}{4}\end{cases}}}\)
loại trường hợp\(\left(x-\frac{1}{2}\right)^2=-\frac{7}{4}\) do bình phương luôn là số dương