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\(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]+15=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15=0\)\(Dat:x^2+8x+7=a\Rightarrow a\left(a+8\right)+15=0\Leftrightarrow a^2+8a+15=0\Leftrightarrow\left(a+3\right)\left(a+5\right)=0\Leftrightarrow\left[{}\begin{matrix}a=-3\\a=-5\end{matrix}\right.\)\(+,a=-5\Rightarrow x^2+8x+7=-5\Leftrightarrow x^2+8x+16=4\Leftrightarrow\left(x+4\right)^2=4\Rightarrow\left[{}\begin{matrix}x+4=-2\\x+4=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\left(thoaman\right)\\x=2\left(loai\right)\end{matrix}\right.\)\(+,a=-3\Rightarrow x^2+8x+7=-3\Leftrightarrow x^2+8x+16=6\Leftrightarrow\left(x+4\right)^2=6\Leftrightarrow\left[{}\begin{matrix}x+4=-\sqrt{6}\\x+4=\sqrt{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\left(\sqrt{6}+4\right)\left(thoaman\right)\\x=\sqrt{6}-4\left(thoaman\right)\end{matrix}\right.\) \(\Rightarrow x\in\left\{\sqrt{6}-4;-\sqrt{6}-4;-6\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(x - 1) + 2(1 - x) = 0
<=> x - 1 + 2 - 2x = 0
<=> -x + 1 = 0
<=> - x = -1
<=> x = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2-3x+2+\left|x-1\right|=0\)
\(\Leftrightarrow x^2-2x-x+2+\left|x-1\right|=0\)
\(\Leftrightarrow x\left(x-2\right)-\left(x-2\right)+\left|x-1\right|=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)+\left|x-1\right|=0\)
\(\Leftrightarrow\left|x-1\right|=\left(x-1\right)\left(2-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=\left(x-1\right)\left(2-x\right)\left(x\ge1\right)\\x-1=\left(x-1\right)\left(x-2\right)\left(x< 1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(2-x-1\right)=0\\\left(x-1\right)\left(x-2-1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left[{}\begin{matrix}x=1\left(loai\right)\\x=3\left(loai\right)\end{matrix}\right.\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì \(\left(x+1\right)^4\ge0\forall x\); \(\left(x-3\right)^4\ge0\forall x\)
\(\Rightarrow\left(x+1\right)^4+\left(x-3\right)^4\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^4=0\\\left(x-3\right)^4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}\left(ktm\right)}\)
=> Pt vô nghiệm
a) ( x + 1 ) 4 + ( x - 3 ) 4 = 0
Vì \(\left(x+1\right)^4\ge0\forall x\inℤ\)
\(\left(x-3\right)^4\ge0\forall x\inℤ\)
Nên \(\left(x+1\right)^4+\left(x-3\right)^4=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(x+1\right)^4=0\\\left(x-3\right)^4=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+1=0\\x-3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-1\\x=3\end{cases}}}\)
Vậy .....
![](https://rs.olm.vn/images/avt/0.png?1311)
\(bpt\Leftrightarrow\left[\left(x+1\right)^2+3\right]\left(x-1\right)< 0\)
\(\left(x+1\right)^2+3>0\Leftrightarrow x-1< 0\Leftrightarrow x< 1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\left(x+1\right)^4\ge0\forall x\)
\(\left(x+3\right)^4\ge0\forall x\)
\(\Rightarrow\left(x+1\right)^4+\left(x+3\right)^4\ge0\forall x\)
Dấu = xảy ra khi : \(\left(x+1\right)^4+\left(x+3\right)^4=0\)
\(\Rightarrow\hept{\begin{cases}\left(x+1\right)^4=0\\\left(x+3\right)^4=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-1\\x=-3\end{cases}\left(ktm\right)}\)
\(\Rightarrow\)phương trình vô ngiệm
Ta có :
\(\left(x+1\right)^4\ge0\forall x\)
\(\left(x+3\right)^4\ge0\forall x\)
Phương trình = 0 \(\Leftrightarrow\hept{\begin{cases}\left(x+1\right)^4=0\\\left(x+3\right)^4=0\end{cases}}\)
\(\hept{\begin{cases}x+1=0\\x+3=0\end{cases}}\)
\(\hept{\begin{cases}x=-1\\x=-3\end{cases}}\)
\(x\in\varnothing\)
x=1
Ai kb vs mink ko mink k cho
x(x-1)+(1-x)=0\(\Leftrightarrow x^2-x+1-x=0\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)
Vậy phương trình có 1 nghiệm duy nhất là x=1