ĐKXĐ : \(x\ne\pm2\)

\(A=\left[\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\div\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)

\(=\left[\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\right]\div\left(\frac{x^2-4+10-x^2}{x+2}\right)\)

\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}\times\frac{x+2}{6}=-\frac{1}{x-2}\)

\(\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{7x-3}{9-x^2}\)ĐK : \(x\ne\pm3\)

\(\Leftrightarrow\frac{x-1}{x+3}+\frac{x}{3-x}=\frac{7x-3}{9-x^2}\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(3-x\right)+x\left(x+3\right)}{\left(x+3\right)\left(3-x\right)}=\frac{7x-3}{\left(3-x\right)\left(x+3\right)}\)

\(\Rightarrow3x-x^2-3+x+x^2+3x=7x-3\)

\(\Leftrightarrow7x-3=7x-3\Leftrightarrow0x=0\)

Vậy phương trình có vô số nghiệm 

Trả lời:

\(\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{7x-3}{9-x^2}\)\(\left(ĐKXĐ:x\ne\pm3\right)\)

\(\Leftrightarrow\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{3-7x}{x^2-9}\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(x-3\right)}{x^2-9}-\frac{x\left(x+3\right)}{x^2-9}=\frac{3-7x}{x^2-9}\)

\(\Rightarrow x^2-3x-x+3-\left(x^2+3x\right)=3-7x\)

\(\Leftrightarrow x^2-4x+3-x^2-3x=3-7x\)

\(\Leftrightarrow3-7x=3-7x\)

\(\Leftrightarrow-7x+7x=3-3\)

\(\Leftrightarrow0x=0\)( luôn thỏa mãn )

Vậy \(S=ℝ\)với \(x\ne\pm3\)

Bài này dễ mà cứ áp dụng đúng công thức là ra kết quả!

20,8 nha! lấy 18,72:9x10 là raaaaaaaaaaaa

\(x^3-6x^2+10x-8=0\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x^2-2x+2=0\\x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^2=-1\left(vn\right)\\x=4\end{cases}}\Leftrightarrow x=4\)(vn : vô nghiệm).

Vậy phương trình có nghiệm duy nhất : \(x=4\)

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A B C D E N

a) Xét \(\Delta BAC\)có phân giác BD (giả thiết).

\(\Rightarrow\frac{BA}{BC}=\frac{AD}{CD}\)(tính chất).

\(\Rightarrow\frac{BA}{BC+BA}=\frac{AD}{CD+AD}=\frac{AD}{AC}\)(tính chất của tỉ lệ thức).

\(\Rightarrow\frac{6}{10+6}=\frac{AD}{8}\)(thay số).

\(\Rightarrow\frac{6}{16}=\frac{AD}{8}\)

\(\Rightarrow AD=\frac{6}{16}.8=\frac{3}{8}.8=3\left(cm\right)\)

Do đó \(CD=AC-AD=8-3=5\left(cm\right)\)

Vậy \(AD=3cm,CD=5cm\)

ta có

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

áp dụng bất đẳng thức AM-GM với các số thực dương ta có

\(\left(x-1\right)+\frac{1}{x-1}\ge2\sqrt{\left(x-1\right)\frac{1}{x-1}=2}\)

dấu "=" xảy ra khi

\(\Leftrightarrow x-1=\frac{1}{x-1}\)

\(\left(x-1\right)^2=1\)

\(\Leftrightarrow x=2\)

\(\Rightarrow p\ge2+2=4\)

VẬY MINP là 

\(4\Leftrightarrow x=1\)

cảm ơn nhé nhưng còn cách khác không vì mình cũng làm giống như này :P

nhìn khó ghê nha

khó vãi nhỉ

\(5+\frac{8}{x^2-4}=\frac{2x-1}{x+2}-\frac{3x-1}{2-x}\left(ĐKXĐ:x\ne\pm2\right)\)

\(\Leftrightarrow5+\frac{8}{\left(x-2\right)\left(x+2\right)}=\frac{2x-1}{x+2}+\frac{3x-1}{x-2}\)

\(\Leftrightarrow\frac{5\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{8}{\left(x-2\right)\left(x+2\right)}\)\(=\frac{\left(2x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{\left(3x-1\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}\)

\(\Rightarrow5\left(x-2\right)\left(x+2\right)+8=\)\(\left(2x-1\right)\left(x-2\right)+\left(3x-1\right)\left(x+2\right)\)

\(\Leftrightarrow5\left(x^2-4\right)+8=2x^2-4x-x+2\)\(+3x^2+6x-x-2\)

\(\Leftrightarrow5x^2-20+8=\)\(5x^2\)

\(\Leftrightarrow-12=5x^2-5x^2\)

\(\Leftrightarrow0=-12\)(vô nghiệm).

Vậy phương trình đã cho vô nghiệm.

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