a) \(x^2-4-\left(x-2\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm \(x\in\left\{2;\dfrac{1}{3}\right\}\).
b) \(\dfrac{4x}{x^2+4x+3}-1=6\cdot\left(\dfrac{1}{x+3}-\dfrac{1}{2x+2}\right)\) (ĐKXĐ: \(x\ne-1;x\ne-3\))
\(\Leftrightarrow\dfrac{4x-x^2-4x-3}{x^2+4x+3}=6\cdot\left[\dfrac{2\left(x+1\right)}{2\left(x+1\right)\left(x+3\right)}-\dfrac{x+3}{2\left(x+1\right)\left(x+3\right)}\right]\)
\(\Leftrightarrow\dfrac{-x^2-3}{\left(x+1\right)\left(x+3\right)}=\dfrac{6\cdot\left(2x+2-x-3\right)}{2\left(x+1\right)\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{-x^2-3}{\left(x+1\right)\left(x+3\right)}=\dfrac{3\cdot\left(x-1\right)}{\left(x+1\right)\left(x+3\right)}\)
\(\Rightarrow-x^2-3=3x-3\)
\(\Leftrightarrow-x^2-3x-3+3=0\)
\(\Leftrightarrow-x\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(\text{tm ĐKXĐ}\right)\\x=-3\left(\text{ktm ĐKXĐ}\right)\end{matrix}\right.\)
Vậy phương trình đã cho có 1 nghiệm duy nhất là \(x=0\).
c) \(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left[x\left(x+1\right)\right]\cdot\left[\left(x-1\right)\left(x+2\right)\right]=24\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\) (1)
Đặt \(x^2+x=y\), khi đó pt (1) trở thành:
\(y\left(y-2\right)=24\)
\(\Leftrightarrow y^2-2y+1=25\)
\(\Leftrightarrow\left(y-1\right)^2-25=0\)
\(\Leftrightarrow\left(y-1-5\right)\left(y-1+5\right)=0\)
\(\Leftrightarrow\left(y-6\right)\left(y+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y-6=0\\y+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x^2+x-6=0\\x^2+x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x+3x-6=0\\\left[x^2+2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]-\dfrac{1}{4}+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\left(x-2\right)+3\left(x-2\right)=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\left(\text{vô lí vì }\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}>0\forall x\right)\end{matrix}\right.\)
\(\Rightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm là: \(x\in\left\{2;-3\right\}\).
\(\text{#}Toru\)
d) \(2x^4-5x^3+6x^2-5x+2=0\)
\(\Leftrightarrow2x^4-2x^3-3x^3+3x^2+3x^2-3x-2x+2=0\)
\(\Leftrightarrow2x^3\left(x-1\right)-3x^2\left(x-1\right)+3x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x^3-3x^2+3x-2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[2\left(x^3-1\right)-3x\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left[2\left(x-1\right)\left(x^2+x+1\right)-3x\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(2x^2+2x+1-3x\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(2x^2-x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x^2-x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\2\left(x^2-\dfrac{1}{2}x\right)+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\2\left[x^2-2\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2\right]-2\cdot\left(\dfrac{1}{4}\right)^2+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\2\left(x-\dfrac{1}{4}\right)^2+\dfrac{7}{8}=0\left(\text{vô lí vì }2\left(x-\dfrac{1}{4}\right)^2+\dfrac{7}{8}>0\forall x\right)\end{matrix}\right.\)
\(\Rightarrow x=1\)
Vậy phương trình đã cho có 1 nghiệm duy nhất là \(x=1\).
e) \(\dfrac{\left(x+1\right)^2}{x^2+2x+2}-\dfrac{x^2+2x}{\left(x+1\right)^2}=\dfrac{1}{90}\) (ĐKXĐ: \(x\ne-1\))
\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2+1}-\dfrac{\left(x+1\right)^2-1}{\left(x+1\right)^2}=\dfrac{1}{90}\) (1)
Đặt \(\left(x+1\right)^2=y\), khi đó pt (1) trở thành:
\(\dfrac{y}{y+1}-\dfrac{y-1}{y}=\dfrac{1}{90}\) (ĐKXĐ: \(y\ne0;y\ne-1\))
\(\Leftrightarrow\dfrac{90y^2}{90y\left(y+1\right)}-\dfrac{90\left(y-1\right)\left(y+1\right)}{90y\left(y+1\right)}=\dfrac{y\left(y+1\right)}{90y\left(y+1\right)}\)
\(\Rightarrow90y^2-90\left(y^2-1\right)=y^2+y\)
\(\Leftrightarrow90y^2-90y^2+90=y^2+y\)
\(\Leftrightarrow y^2+y-90=0\)
\(\Leftrightarrow y^2-9y+10y-90=0\)
\(\Leftrightarrow y\left(y-9\right)+10\left(y-9\right)=0\)
\(\Leftrightarrow\left(y-9\right)\left(y+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y-9=0\\y+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=9\left(\text{tm ĐKXĐ}\right)\\y=-10\left(\text{tm ĐKXĐ}\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left(x+1\right)^2=9\\\left(x+1\right)^2=-10\left(\text{vô lí vì }\left(x+1\right)^2\ge0\forall x\right)\end{matrix}\right.\)
\(\Rightarrow\left(x+1\right)^2=9\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(\text{tm ĐKXĐ}\right)\\x=-4\left(\text{tm ĐKXĐ}\right)\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm là: \(x\in\left\{2;-4\right\}\).
f) \(\left(x-6\right)^4+\left(x-8\right)^4=16\) (1)
Đặt \(x-7=a\), khi đó pt (1) trở thành:
\(\left(a+1\right)^4+\left(a-1\right)^4=16\)
\(\Leftrightarrow a^4+4a^3+6a^2+4a+1+a^4-4a^3+6a^2-4a+1=16\)
\(\Leftrightarrow2a^4+12a^2+2=16\)
\(\Leftrightarrow a^4+6a^2+1=8\)
\(\Leftrightarrow a^4+6a^2-7=0\)
\(\Leftrightarrow a^4-a^2+7a^2-7=0\)
\(\Leftrightarrow a^2\left(a^2-1\right)+7\left(a^2-1\right)=0\)
\(\Leftrightarrow\left(a^2+7\right)\left(a^2-1\right)=0\)
\(\Leftrightarrow a^2-1=0\left(\text{vì }a^2+7>0\forall x\right)\)
\(\Leftrightarrow a^2=1\Leftrightarrow\left[{}\begin{matrix}a=1\\a=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-7=1\\x-7=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=6\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm là: \(x\in\left\{6;8\right\}\).
\(\text{#}Toru\)
a: \(x^2-4-\left(x-2\right)\left(3-2x\right)=0\)
=>\(\left(x-2\right)\left(x+2\right)+\left(x-2\right)\left(2x-3\right)=0\)
=>(x-2)(x+2+2x-3)=0
=>(x-2)(3x-1)=0
=>\(\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
c: \(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)
=>\(\left(x^2+x\right)\left(x^2+x-2\right)=24\)
=>\(\left(x^2+x\right)^2-2\left(x^2+x\right)-24=0\)
=>\(\left(x^2+x-6\right)\left(x^2+x+4\right)=0\)
mà \(x^2+x+4=\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}>=0\forall x\)
nên \(x^2+x-6=0\)
=>(x+3)(x-2)=0
=>\(\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
d: \(2x^4-5x^3+6x^2-5x+2=0\)
=>\(2x^4-2x^3-3x^3+3x^2+3x^2-3x-2x+2=0\)
=>\(\left(x-1\right)\left(2x^3-3x^2+3x-2\right)=0\)
=>\(\left(x-1\right)\left(2x^3-2x^2-x^2+x+2x-2\right)=0\)
=>\(\left(x-1\right)^2\cdot\left(2x^2-x+2\right)=0\)
=>\(\left(x-1\right)^2=0\)
=>x-1=0
=>x=1