Lớp 6 ? Bạn xem lại lớp nhé đây là lớp 8.
`a)(x^2-x+1)(x^2-x+3)=0`
Đặt `t=x^2-x+2`
`pt<=>(t-1)(t+1)=0`
`<=>t^2=1`
`<=>[(t=1),(t=-1):}`
`=>[(x^2-x+1=0),(x^2-x+3=0):}`
`=>[((x-1/2)^2+3/4=0(vô lý)),((x-1/2)^2+11/4=0(vô lý)):}`
`=>` pt vô nghiệm
`b)(x+1/x)^2-4,5(x+1/x)+5=0(x ne 0)`
Đặt `t=x+1/x`
`pt<=>t^2-4,5t+5=0`
`<=>t^2-2t-2,5t+5=0`
`<=>(t-2)(t-2,5)=0`
`**t-2=0<=>x+1/x=2<=>x^2+1=2x<=>(x-1)^2=0<=>x=1(tm)`
`**t-2,5=0<=>2x+2/x=5<=>2x^2-5x+2=0<=>(2x-1)(x-2)=0<=>[(x=2),(x=1/2):}(TM)`
Vậy `x in {1;2;1/2}.`
a: Ta có: \(\left(x^2-x+1\right)\left(x^2-x+3\right)=0\)
mà \(x^2-x+1>0\forall x\)
và \(x^2-x+3>0\forall x\)
nên \(x\in\varnothing\)
b: Ta có: \(\left(x+\dfrac{1}{x}\right)^2-4.5\left(x+\dfrac{1}{2}\right)+5=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{x}-2\right)\left(x+\dfrac{1}{x}-\dfrac{5}{2}\right)=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)\left(2x^2-5x+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left(2x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)
