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<=> (x2 +x +4)2 + 2 . 4x(x2+ x + 4) + (4x)2 = 0
<=> ( x2 + x+ 4 +4x )2 = 0
<=> [(x2 + x) + (4 +4x)] =0
<=> [x(x+1) + 4(1+x)] =0
<=> (x+1) + (x+4) =0
- x+1 = 0 <=> x= -1
- x+4 = 0 <=> x= -4
*vn:vô nghiệm.
a. \(\left(x^2-2\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2=0\\x^2+x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\pm\sqrt{2}\)
-Vậy \(S=\left\{\pm\sqrt{2}\right\}\).
b. \(16x^2-8x+5=0\)
\(\Leftrightarrow16x^2-8x+1+4=0\)
\(\Leftrightarrow\left(4x-1\right)^2+4=0\) (vô lí)
-Vậy S=∅.
c. \(2x^3-x^2-8x+4=0\)
\(\Leftrightarrow x^2\left(2x-1\right)-4\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\pm2\end{matrix}\right.\)
-Vậy \(S=\left\{\dfrac{1}{2};\pm2\right\}\).
d. \(3x^3+6x^2-75x-150=0\)
\(\Leftrightarrow3x^2\left(x+2\right)-75\left(x+2\right)=0\)
\(\Leftrightarrow3\left(x+2\right)\left(x^2-25\right)=0\)
\(\Leftrightarrow3\left(x+2\right)\left(x+5\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\pm5\end{matrix}\right.\)
-Vậy \(S=\left\{-2;\pm5\right\}\)
anh ơi, vậy là sai đề hả anh, chứ đề kêu chứng minh phương trình vô nghiệm mà em thấy anh ghi x=2
\(\Leftrightarrow1+\dfrac{2}{x+2}+1+\dfrac{8}{x+8}=1+\dfrac{4}{x+4}+1+\dfrac{6}{x+6}\)
\(\Leftrightarrow\dfrac{1}{x+2}+\dfrac{4}{x+8}=\dfrac{2}{x+4}+\dfrac{3}{x+6}\)
\(\Leftrightarrow\dfrac{4}{x+8}-\dfrac{3}{x+6}=\dfrac{2}{x+4}-\dfrac{1}{x+2}\)
\(\Leftrightarrow\dfrac{4x+24-3\left(x+8\right)}{\left(x+8\right)\left(x+6\right)}=\dfrac{2x+4-\left(x+4\right)}{\left(x+4\right)\left(x+2\right)}\)
\(\dfrac{x}{\left(x+8\right)\left(x+6\right)}=\dfrac{x}{\left(x+4\right)\left(x+2\right)}\)
x=0 là nghiệm
x khác 0
\(\left\{{}\begin{matrix}x\ne\left\{-8;-6;-4;-2\right\}\\\left(x+4\right)\left(x+2\right)=\left(x+8\right)\left(x+6\right)\end{matrix}\right.\)<=>x^2 +6x+8 =x^2 +14x+48
-40 =8x=> x =-5 nhận
x={-5;0}
b: \(\Leftrightarrow32x^5+1-32x^5+1=2\)
=>2=2(luôn đúng)
a: \(\Leftrightarrow\left[\left(x-3\right)^2-\left(x+3\right)^2\right]\left[\left(x-3\right)^2+\left(x+3\right)^2\right]+24x^3=216\)
\(\Leftrightarrow-12x\left(2x^2+18\right)+24x^3=216\)
=>-216x=216
hay x=-1
=>\(\frac{\left(x+2\right)^2+2}{x+2}+\frac{\left(x+8\right)^2+8}{x+8}\)=\(\frac{\left(x+4\right)+4}{x+4}+\frac{\left(x+6\right)^2+6}{x+6}\)
=>2x+10+\(\frac{2}{x+2}+\frac{8}{x+8}\)=2x+10+\(\frac{4}{x+4}+\frac{6}{x+6}\)
=>-x\(\left(\frac{1}{x+2}-\frac{1}{x+4}-\frac{1}{x+6}+\frac{1}{x+8}\right)\)=0
=>\(\orbr{\begin{cases}x=0\\\frac{1}{x+2}-.....+\frac{1}{x+8}=0\end{cases}}\)
Voi \(\frac{1}{x+2}-....\)=0 ta co
Dat x+5=t
=>\(\frac{1}{t-3}-\frac{1}{t-1}-\frac{1}{t+1}+\frac{1}{t+3}\)=0
=> \(2t\left(\frac{1}{t^2-1}+\frac{1}{t^2-9}\right)=0\)
=>t=0
=>x=-5
Vay phuong trinh co nghiem x=0;-5
\(=\left(x^2+x+4\right)^2+2\cdot\left(x^2+x+4\right)\cdot4x+\left(4x\right)^2\)
\(=\left(x^2+5x+4\right)^2\)
\(=\left(x+1\right)^2\cdot\left(x+4\right)^2\)
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