x² - 16x² + 32x - 16 ???????????????

tham khảo tại đây nhé:

https://hoc24.vn/hoi-dap/question/578694.html

ĐKXĐ:\(x\ne\pm\dfrac{1}{2}\)

\(\dfrac{1+8x}{4+8x}-\dfrac{4x}{12x-6}+\dfrac{32x^2}{3\left(4-16x^2\right)}=0\)

\(\Leftrightarrow\dfrac{1+8x}{4\left(2x+1\right)}-\dfrac{4x}{6\left(2x-1\right)}+\dfrac{32x^2}{-6\cdot\left(2x-1\right)\left(2x+1\right)}=0\)

\(\Leftrightarrow\dfrac{6\cdot\left(1+8x\right)\left(2x-1\right)}{24\left(2x-1\right)\left(2x+1\right)}-\dfrac{4\cdot4x\left(2x+1\right)}{24\left(2x-1\right)\left(2x+1\right)}-\dfrac{32x^2\cdot4}{24\left(2x-1\right)\left(2x+1\right)}=0\)

\(\Leftrightarrow96x^2-36x-6-36x^2-16x-144x^2=0\)

\(\Leftrightarrow-84x^2-52x-6=0\)

\(\Leftrightarrow\Delta=688\)

\(\Leftrightarrow\left[{}\begin{matrix}x_1=\dfrac{52-\sqrt{688}}{-168}=\dfrac{-13+\sqrt{43}}{42}\\x_2=\dfrac{52+\sqrt{688}}{-168}=\dfrac{-13-\sqrt{43}}{43}\end{matrix}\right.\)

Vậy pt có 2 nghiệm phân biệt............

\(x^4-6x^3+16x^2-22x+16=0\)

\(\Rightarrow x^4-2x^3+3x^2-4x^3+8x^2-12x+5x^2-10x+15+1=0\)

\(\Rightarrow x^2\left(x^2-2x+3\right)-4x\left(x^2-2x+3\right)+5\left(x^2-2x+3\right)x^2+1=0\)

\(\Rightarrow\left(x^2-2x+3\right)\left(x^2-4x+5\right)=-1\)

\(\Rightarrow\left(x^2-2x+1+2\right)\left(x^2-4x+4+1\right)=-1\)

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

mà \(\left\{{}\begin{matrix}\left(x-1\right)^2+2>0,\forall x\\\left(x-2\right)^2+1>0,\forall x\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left[\left(x-1\right)^2+2\right]\left[\left(x-2\right)^2+1\right]>0,\forall x\\\left[\left(x-1\right)^2+2\right]\left[\left(x-2\right)^2+1\right]=-1\end{matrix}\right.\) (vô lí)

Vậy phương trình trên vô nghiệm (dpcm)

\(\Leftrightarrow\left(x^2-12x-6\right)\left(x^2-4x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-12x-6=0\\x^2-4x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(x-6\right)^2=42\\\left(x-2\right)^2=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6\in\left\{\sqrt{42};-\sqrt{42}\right\}\\x-2\in\left\{\sqrt{2};-\sqrt{2}\right\}\end{matrix}\right.\Leftrightarrow x\in\left\{\sqrt{42}+6;-\sqrt{42}+6;\sqrt{2}+2;2-\sqrt{2}\right\}\)

Đây đích thực có phải là lớp 1 ko bn?

Lên lớp trên mà gửi đi gửi ở lớp 1 làm gì vậy bạn .

Ta có: \(\sqrt{9x+9}+\sqrt{4x+4}-2\sqrt{16x+16}=\sqrt{x+1}-8\)

\(\Leftrightarrow3\sqrt{x+1}+2\sqrt{x+1}-8\sqrt{x+1}-\sqrt{x+1}=-8\)

\(\Leftrightarrow\sqrt{x+1}=2\)

\(\Leftrightarrow x+1=4\)

hay x=3

hết cứu đi mà làm

\(\Delta=32^2+4\cdot900=4624\)

\(\Leftrightarrow\sqrt{\Delta}=\sqrt{4624}=68\)

\(\Leftrightarrow\left[{}\begin{matrix}x_1=\dfrac{-68-32}{2}\\x_2=\dfrac{-32+68}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x_1=-50\\x_2=18\end{matrix}\right.\)

Vậy phương trình đã cho có 2 nghiệm \(S=\left\{-50;18\right\}\)

\(a,x^4-16x^2+32x-16=0\)

\(\Leftrightarrow\left(x^4-16\right)-16x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x^4+4\right)\left(x-2\right)\left(x+2\right)-16x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3+2x^2-12x+8\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3-2x^2+4x^2-8x-4x+8\right)=0\)\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x-2\right)+4x\left(x-2\right)-4\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-2\right)\left(x^2+4x-4\right)=0\)

\(\Leftrightarrow\left(x-2\right)^2\left[\left(x+2\right)^2-8\right]=0\Rightarrow\left[{}\begin{matrix}\left(x-2\right)^2=0\\\left(x+2\right)^2-8=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-2=0\\\left(x+2\right)^2=8\Rightarrow\left[{}\begin{matrix}x+2=\sqrt{8}\\x+2=-\sqrt{8}\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{8}-2\\x=-\sqrt{8}-2\end{matrix}\right.\)

câu nào dễ xơi trước

g) \(x^3+3x^2-2x-6=0\Leftrightarrow x^2\left(x+3\right)-2\left(x+3\right)=0\)

\(\Leftrightarrow\left(x^2-2\right)\left(x+3\right)=0\Leftrightarrow\left\{{}\begin{matrix}x=\pm\sqrt{2}\\x=-3\end{matrix}\right.\)

kl: ...........