(x^2+3x)^2+4(x^2+3x)=12
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NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 3 2023
a.
ĐKXĐ: \(x\ge-1\)
\(7+12\sqrt{x+1}=x+4\sqrt{x^2+3x+2}\)
\(\Leftrightarrow4\sqrt{\left(x+1\right)\left(x+2\right)}-12\sqrt{x+1}+x-7=0\)
\(\Leftrightarrow4\sqrt{x+1}\left(\sqrt{x+2}-3\right)+x-7=0\)
\(\Leftrightarrow4\sqrt{x+1}\left(\dfrac{x-7}{\sqrt{x+2}+3}\right)+x-7=0\)
\(\Leftrightarrow\left(x-7\right)\left(\dfrac{4\sqrt{x+1}}{\sqrt{x+2}+3}+1\right)=0\)
\(\Leftrightarrow x-7=0\) (do \(\dfrac{4\sqrt{x+1}}{\sqrt{x+2}+3}+1>0;\forall x\ge-1\))
\(\Rightarrow x=7\)
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 3 2023
b.
ĐKXĐ: \(x\ne-\dfrac{1}{3}\)
\(\Rightarrow3x^2+3x+2=\left(3x+1\right)\sqrt{x^2+x+2}\)
\(\Leftrightarrow x^2+x+2-\left(3x+1\right)\sqrt{x^2+x+2}+2x^2+2x=0\)
Đặt \(\sqrt{x^2+x+2}=t\)
\(\Rightarrow t^2-\left(3x+1\right)t+2x^2+2x=0\)
\(\Delta=\left(3x+1\right)^2-4\left(2x^2+2x\right)=\left(x-1\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{3x+1+x-1}{2}=2x\\t=\dfrac{3x+1-\left(x-1\right)}{2}=x+1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2+x+2}=2x\left(x\ge0\right)\\\sqrt{x^2+x+2}=x+1\left(x\ge-1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+2=4x^2\left(x\ge0\right)\\x^2+x+2=x^2+2x+1\left(x\ge-1\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{3}\\\end{matrix}\right.\)
22 tháng 3 2021
a) Ta có: \(\dfrac{x+5}{3x-6}-\dfrac{1}{2}=\dfrac{2x-3}{2x-4}\)
\(\Leftrightarrow\dfrac{2\left(x+5\right)}{6\left(x-2\right)}-\dfrac{3\left(x-2\right)}{6\left(x-2\right)}=\dfrac{3\left(2x-3\right)}{6\left(x-2\right)}\)
Suy ra: \(2x+5-3x+6=6x-9\)
\(\Leftrightarrow-x+11-6x+9=0\)
\(\Leftrightarrow20-7x=0\)
\(\Leftrightarrow7x=20\)
hay \(x=\dfrac{20}{7}\)(thỏa ĐK)
Vậy: \(S=\left\{\dfrac{20}{7}\right\}\)
23 tháng 3 2021
a) ĐKXĐ: \(x\notin\left\{\dfrac{1}{3};-\dfrac{1}{3}\right\}\)
Ta có: \(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\)
\(\Leftrightarrow\dfrac{\left(1-3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}-\dfrac{\left(1+3x\right)^2}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}\)
Suy ra: \(9x^2-6x+1-9x^2-6x-1=12\)
\(\Leftrightarrow-12x=12\)
hay x=-1(thỏa ĐK)
Vậy: S={-1}
19 tháng 2 2022
b: \(\Leftrightarrow9x^2+12x+4-18x+12=9x^2\)
=>-6x+16=0
=>-6x=-16
hay x=8/3(nhận)
c: \(\Leftrightarrow\dfrac{x+1+x-1}{\left(x-1\right)\left(x+1\right)}=\dfrac{2}{x+2}\)
\(\Leftrightarrow2x\left(x+2\right)=2\left(x^2-1\right)\)
\(\Leftrightarrow2x^2+4x-2x^2+2=0\)
=>4x+2=0
hay x=-1/2(nhận)