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Question

giải hộ mk vs

1/2x^4+3x^3-x^2+3x+2=0

2/x^4-5x^3+7x^2-5x-16=0

3/(x+2)^4+(x+4)^4=16

Phong Thần · Accepted Answer

1) $2x^4+3x^3-x^2+3x+2=0$

$\Rightarrow2x^4+x^3+2x^3+x^2-2x^2-x+4x+2=0$

$\Rightarrow x^3\left(2x+1\right)+x^2\left(2x+1\right)-x\left(2x+1\right)+2\left(2x+1\right)=0$

$\Rightarrow\left(2x+1\right)\left(x^3+x^2-x+2\right)=0$

$\Rightarrow\left(2x+1\right)\left(x^3+2x^2-x^2-2x+x+2\right)=0$

$\Rightarrow\left(2x+1\right)\left[x^2\left(x+2\right)-x\left(x+2\right)+\left(x+2\right)\right]=0$

$\Rightarrow\left(2x+1\right)\left(x+2\right)\left(x^2-x+1\right)=0$

$\Rightarrow\left[{}\begin{matrix}2x+1=0\x+2=0\x^2-x+1=0\end{matrix}\right.$

Ta có:

$x^2-x+1$

$=x^2-2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+1$

$=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}$

Vì $\left(x-\dfrac{1}{2}\right)^2\ge0$ với mọi x

$\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}$ với mọi x

$\Rightarrow x^2-x+1$ vô nghiệm

$\Rightarrow\left[{}\begin{matrix}2x+1=0\x+2=0\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\x=-2\end{matrix}\right.$Câu trả lời: 1) $2x^4+3x^3-x^2+3x+2=0$

$\Rightarrow2x^4+x^3+2x^3+x^2-2x^2-x+4x+2=0$

$\Rightarrow x^3\left(2x+1\right)+x^2\left(2x+1\right)-x\left(2x+1\right)+2\left(2x+1\right)=0$

$\Rightarrow\left(2x+1\right)\left(x^3+x^2-x+2\right)=0$

$\Rightarrow\left(2x+1\right)\left(x^3+2x^2-x^2-2x+x+2\right)=0$

$\Rightarrow\left(2x+1\right)\left[x^2\left(x+2\right)-x\left(x+2\right)+\left(x+2\right)\right]=0$

$\Rightarrow\left(2x+1\right)\left(x+2\right)\left(x^2-x+1\right)=0$

$\Rightarrow\left[{}\begin{matrix}2x+1=0\x+2=0\x^2-x+1=0\end{matrix}\right.$

Ta có:

$x^2-x+1$

$=x^2-2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+1$

$=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}$

Vì $\left(x-\dfrac{1}{2}\right)^2\ge0$ với mọi x

$\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}$ với mọi x

$\Rightarrow x^2-x+1$ vô nghiệm

$\Rightarrow\left[{}\begin{matrix}2x+1=0\x+2=0\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\x=-2\end{matrix}\right.$

Phong Thần · Answer

3) $\left(x+2\right)^4+\left(x+4\right)^4=16$

Đặt x + 3 = a, ta được

$\left(a-1\right)^4+\left(a+1\right)^4=16$

$\Rightarrow\left[\left(a-1\right)^2\right]^2+\left[\left(a+1\right)^2\right]^2=16$

$\Rightarrow\left(a^2-2a+1\right)^2+\left(a^2+2a+1\right)^2=16$

$\Rightarrow a^4+4a^2+1+2a^2-4a^3-4a+a^4+4a^2+1+2a^2+4a^3+4a=16$

$\Rightarrow2a^4+2.4a^2+2+2.2a^2=16$

$\Rightarrow2a^4+8a^2+4a^2+2=16$

$\Rightarrow2a^4+12a^2+2-16=0$

$\Rightarrow2a^4+12a^2-14=0$

$\Rightarrow2a^4-2a^2+14a^2-14=0$

$\Rightarrow2a^2\left(a^2-1\right)+14\left(a^2-1\right)=0$

$\Rightarrow\left(a^2-1\right)\left(2a^2+14\right)=0$

$\Rightarrow\left(a-1\right)\left(a+1\right).2\left(a^2+7\right)=0$

$\Rightarrow\left(a-1\right)\left(a+1\right)\left(a^2+7\right)=0$

$\Rightarrow\left[{}\begin{matrix}a-1=0\a+1=0\a^2+7=0\end{matrix}\right.$

Vì $a^2\ge0$ với mọi a

$\Rightarrow a^2+7\ge7$ với mọi a

$\Rightarrow a^2+7$ vô nghiệm

$\Rightarrow\left[{}\begin{matrix}a-1=0\a+1=0\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x+3-1=0\x+3+1=0\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x+2=0\x+4=0\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=-2\x=-4\end{matrix}\right.$Câu trả lời: 3) $\left(x+2\right)^4+\left(x+4\right)^4=16$