\(x^4+x^2+6x-8=0\)

\(\Leftrightarrow x^4-x^3+x^3-x^2+2x^2-2x+8x-8=0\)

\(\Leftrightarrow x^3\left(x-1\right)+x^2\left(x-1\right)+2x\left(x-1\right)+8\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2+2x+8\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[\left(x+2\right)\left(x^2-2x+4\right)+x\left(x+2\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-x+4\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x-1=0\\x+2=0\\x^2-x+4=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=1\\x=-2\\x\in\varnothing\end{cases}}\)

Bạn đổi ngoặc nhọn thành ngoặc vuông giúp mình nhé

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

\(=6x^5-8x^4+10x^3-6x^2\)

\(x^3-8x^2+21x-18=0\)

\(\Leftrightarrow x^3-2x^2-6x^2+12x+9x-18=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2-6x+9\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

Kham khảo 

\(\frac{x+1}{x-3}-\frac{1}{x-1}=\frac{2}{\left(x-1\right)\left(x-3\right)}\)

\(\Leftrightarrow x^2-1-x+3-2=0\)

\(\Leftrightarrow x^2-x=0\)

\(\Leftrightarrow x\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

\(\frac{x+1}{x-3}-\frac{1}{x-1}=\frac{2}{\left(x-1\right)\left(x-3\right)}\)

điều kiện: \(x\ne3;1\)

quy đồng mẫu hai phân số

\(\frac{\left(x+1\right)\left(x-1\right)-\left(x-3\right)}{\left(x-3\right)\left(x-1\right)}=\frac{2}{\left(x-1\right)\left(x-3\right)}\)

\(\frac{x^2-1-x+3}{\left(x-3\right)\left(x-1\right)}-\frac{2}{\left(x-1\right)\left(x-3\right)}=0\)

\(\frac{x^2-1-x+3-2}{\left(x-1\right)\left(x-3\right)}=0\)

\(\frac{x^2-x}{\left(x-1\right)\left(x-3\right)}=0\)

\(\Rightarrow x^2-x=0\)

\(x\left(x-1\right)=0\)

vây x = 0 

hoặc x = 1 (không thỏa điều kiện)

vậy x = 0

(2x + 7)² = 9(x + 2)²

(2x + 7)² - 9(x + 2)² = 0

áp dụng hằng đẳng thức hiệu 2 bình phương:

[2x + 7 - 3(x + 2)] . [2x + 7 + 3(x + 2)] = 0

(2x + 7 - 3x - 6)(2x + 7 + 3x + 6) = 0

(1 - x)(5x + 13) = 0

vậy 1 - x = 0 hoặc 5x + 13 = 0

hay x = 1 hoặc x = -13/5

\(\left(2x+7\right)^2=9\left(x+2\right)^2\)

<=>\(\left(2x+7\right)^2=\left(3x+6\right)^2\)

,<=>\(2x+7=3x+6\)

<=>\(2x-3x=6-7\)

<=>\(-x=-1\)

<=>\(x=1\)

a,\(2x^2-6x+1=0\)

\(=>x.\left(2x-6\right)=1\)

\(th1:\orbr{\begin{cases}x=1\\2x-6=1\end{cases}=>\orbr{\begin{cases}x=1\\x=\frac{7}{2}\end{cases}}}\)

\(th2:\orbr{\begin{cases}x=-1\\2x-6=-1\end{cases}=>\orbr{\begin{cases}x=-1\\x=\frac{5}{2}\end{cases}}}\)

b,\(4x^2-12x+5=0\)

\(=>x.\left(4x-12\right)=-5\)

\(th1:\orbr{\begin{cases}x=1\\4x-12=-5\end{cases}=>\orbr{\begin{cases}x=1\\x=\frac{7}{4}\end{cases}}}\)

\(th2:\orbr{\begin{cases}x=-1\\4x-12=5\end{cases}=>\orbr{\begin{cases}x=-1\\x=\frac{17}{4}\end{cases}}}\)

\(th3:\orbr{\begin{cases}x=5\\4x-12=-1\end{cases}=>\orbr{\begin{cases}x=5\\x=\frac{11}{4}\end{cases}}}\)

\(th4:\orbr{\begin{cases}x=-5\\4x-12=1\end{cases}=>\orbr{\begin{cases}x=-5\\x=\frac{13}{4}\end{cases}}}\)

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

Ta có \(\Delta=6^2+4.16=100,\sqrt{\Delta}=10\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-6+10}{2}=2\\x=\frac{-6-10}{2}=-8\end{cases}}\)

\(3x^2+2x-1=0\)

\(3x.3x+2x-1=0\)

\(9x+2x-1=0\)

\(11x=1\)

\(x=\frac{1}{11}\)

\(3x^2+2x-1=0\)

\(\Leftrightarrow3x^2+3x-x-1=0\)

\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-1\end{cases}}\)

Ta có: \(\left|x-4\right|+\left|x-9\right|=\left|x-4\right|+\left|9-x\right|\)

\(\ge\left|\left(x-4\right)+\left(9-x\right)\right|=5\)

Dấu "=" khi \(\left(x-4\right)\left(9-x\right)\ge0\)

\(\Rightarrow4\le x\le9\)