(X^2-x)(x^2-x+1)=6 giải pt
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22 tháng 4 2021
a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)
\(\Leftrightarrow x^2-2x+12-8-x^2=0\)
\(\Leftrightarrow-2x+4=0\)
\(\Leftrightarrow-2x=-4\)
hay x=2(loại)
Vậy: \(S=\varnothing\)
22 tháng 4 2021
b) Ta có: \(\left|2x+6\right|-x=3\)
\(\Leftrightarrow\left|2x+6\right|=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)
Vậy: S={-3}
12 tháng 4 2022
\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)=15x^2\)
\(\Leftrightarrow\left(x^2-7x+6\right)\left(x^2-5x+6\right)-15x^2=0\) (*)
-Đặt \(t=x^2-5x+6\)
(*) \(\Leftrightarrow t\left(t-2x\right)-15x^2=0\)
\(\Leftrightarrow t^2-2xt-15x^2=0\)
\(\Leftrightarrow t^2-5xt+3xt-15x^2=0\)
\(\Leftrightarrow t\left(t-5x\right)+3x\left(t-5x\right)=0\)
\(\Leftrightarrow\left(t-5x\right)\left(t+3x\right)=0\)
\(\Leftrightarrow t-5x=0\) hay \(t+3x=0\)
\(\Leftrightarrow x^2-5x+6-5x=0\) hay \(x^2-5x+6+3x=0\)
\(\Leftrightarrow x^2-10x+6=0\) hay \(x^2-2x+6=0\)
\(\Leftrightarrow x^2-2.5x+25-19=0\) hay \(\left(x-1\right)^2+5=0\) (pt vô nghiệm)
\(\Leftrightarrow\left(x-5\right)^2-19=0\)
\(\Leftrightarrow\left(x-5-\sqrt{19}\right)\left(x-5+\sqrt{19}\right)=0\)
\(\Leftrightarrow x=5+\sqrt{19}\) hay \(x=5-\sqrt{19}\)
-Vậy \(S=\left\{5+\sqrt{19};5-\sqrt{19}\right\}\)
NH
1
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 1
ĐKXĐ: \(x\ge-1\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x+1}+1\right)^2}+\sqrt{\left(\sqrt{x+1}-3\right)^2}=2\sqrt{\left(\sqrt{x+1}-1\right)^2}\)
\(\Leftrightarrow\left|\sqrt{x+1}+1\right|+\left|\sqrt{x+1}-3\right|=\left|2\sqrt{x+1}-2\right|\)
Áp dụng BĐT trị tuyệt đối:
\(\left|\sqrt{x+1}+1\right|+\left|\sqrt{x+1}-3\right|\ge\left|\sqrt{x+1}+1+\sqrt{x+1}-3\right|=\left|2\sqrt{x+1}-2\right|\)
Dấu "=" xảy ra khi và chỉ khi \(\left(\sqrt{x+1}+1\right)\left(\sqrt{x+1}-3\right)\ge0\)
\(\Leftrightarrow\sqrt{x+1}-3\ge0\)
\(\Leftrightarrow x+1\ge9\)
\(\Leftrightarrow x\ge8\)
26 tháng 8 2022
\(\dfrac{1}{x-3}=\dfrac{x^2-3x+5}{x^2-x-6}\)
Suy ra: \(x^2-3x+5=x+2\)
=>x2-4x+3=0
=>(x-3)*(x-1)=0
=>x=1(nhận) hoặc x=3(loại)
GK
1
14 tháng 9 2023
\(a,\dfrac{2x-1}{3}< \dfrac{x+6}{2}\)
\(\Leftrightarrow\dfrac{4x-2}{6}< \dfrac{3x+18}{6}\)
\(\Leftrightarrow4x-2< 3x+18\)
\(\Leftrightarrow4x-3x< 2+18\)
\(\Leftrightarrow x< 20\)
\(b,\dfrac{5\left(x-1\right)}{6}-1>\dfrac{2\left(x+1\right)}{3}\)
\(\Leftrightarrow\dfrac{5x-11}{6}>\dfrac{4x+4}{6}\)
\(\Leftrightarrow5x-11>4x+4\)
\(\Leftrightarrow5x-4x>11+4\)
\(\Leftrightarrow x>15\)
`(x^2-x)(x^2-x+1)=6`
Đặt `a=x^2-x(x>=-1/4)`
`pt<=>a(a+1)=6`
`<=>a^2+a-6=0`
`Delta=1+24=25`
`=>a_1=-2(l),a_2=1(tm)`
`<=>x^2-x=1`
`<=>x^2-x-1=0`
`Delta=1+4=5`
`=>x_{12}=(+-sqrt5+1)/2`
\(\left(x^2-x\right)\left(x^2-x+1\right)=6\)
\(\Leftrightarrow\left(x^2-x\right)^2+\left(x^2-x\right)-6=0\)
\(\Leftrightarrow\left[\left(x^2-x\right)-2\right]\left[\left(x^2-x\right)+3\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x=2\\x^2-x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-2=0\\x^2-x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-2\right)\left(x+1\right)=0\\\left(x-\dfrac{1}{2}\right)^2+\dfrac{11}{4}=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)