Ta có bảng xét dấu :

+) Nếu \(x< -3\Leftrightarrow|x+3|=-x-3\)

\(pt\Leftrightarrow3\left(-x-3\right)-3x=-1\)

\(\Leftrightarrow-3x-9-3x=-1\)

\(\Leftrightarrow-6x=8\)

\(x=\frac{-4}{3}\) ( loại )

+) Nếu  \(x\ge-3\Leftrightarrow|x+3|=x+3\)

\(pt\Leftrightarrow3\left(x+3\right)-3x=-1\)

\(\Leftrightarrow3x+9-3x=-1\)

\(\Leftrightarrow9=-1\) ( vô lí )

Vậy phương trình vô nghiệm

\(2x^3-8x^2+8x-11=0\)

??? cái j z

A B C x D E y

Xét  tam giác \(ABE\) \(\&ADC\)

\(BAE=ADC\)(góc chung)

\(\frac{AB}{CD}=\frac{8}{10}=\frac{4}{5};\frac{AE}{AC}=\frac{12}{15}=\frac{4}{5}\)

\(\Rightarrow tamgiácABE~tamgiacADC\left(C.G.C\right)\)

b) Từ tam giác \(ABE\) \(~\)tam giác \(ADC\)\(\Rightarrow\frac{AB}{CD}=\frac{BE}{DC}\Rightarrow DC=\frac{AD\cdot BE}{AB}=\frac{10\cdot10}{8}=12,5\)

c) Từ tam giác \(ABE~\)tam giác \(ADC\left(cmt\right)\)

\(\Rightarrow\frac{S_{ABE}}{S_{ADC}}=\left(\frac{AB}{AD}\right)^2=\left(\frac{8}{10}\right)^2\left(\frac{4}{5}\right)^2=\frac{16}{25}\)

a)  \(ĐKXĐ:\) \(x\ne\pm1\)

\(A=\left(\frac{3x^2-4}{x^2-1}-\frac{2}{1-x}-\frac{2}{x+1}\right):\left(\frac{1-x}{x+1}\right)\)

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

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

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

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

Nhanh , mọi người , mình đang cần gấp lắm

        \(\frac{3x-4}{5}+\frac{x}{6}=\frac{x+1}{2}\)

\(\Leftrightarrow\)\(\frac{6\left(3x-4\right)}{30}+\frac{5x}{30}=\frac{15\left(x+1\right)}{30}\)

\(\Rightarrow\)\(6\left(3x-4\right)+5x=15\left(x+1\right)\)

\(\Leftrightarrow\)\(18x-24+5x=15x+15\)

\(\Leftrightarrow\)\(8x=39\)

\(\Leftrightarrow\)\(x=\frac{39}{8}\)

Vậy...

a)   Ta có:   \(\frac{4}{8}=\frac{5}{10}=\frac{6}{12}\left(=\frac{1}{2}\right)\)

hay   \(\frac{AB}{A'B'}=\frac{AC}{A'C'}=\frac{BC}{B'C'}\)

\(\Rightarrow\)\(\Delta A'B'C'~\Delta ABC\) 

b)   \(\Delta A'B'C'~\Delta ABC\)

\(\Rightarrow\)\(\frac{P_{A'B'C'}}{P_{ABC}}=\frac{A'B'}{AB}=\frac{8}{4}=2\)

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đừng đăng linh tinh nha bạn

\(a)\) Ta có : 

\(M=\frac{2\left|x-3\right|}{x^2+2x-15}=\frac{2\left|x-3\right|}{\left(x^2+2x+1\right)-16}=\frac{2\left|x-3\right|}{\left(x+1\right)^2-16}=\frac{2\left|x-3\right|}{\left(x+1\right)^2-4^2}=\frac{2\left|x-3\right|}{\left(x+5\right)\left(x-3\right)}\)

+) Nếu \(x-3\ge0\) \(\Rightarrow\) \(x\ge3\) ta có : 

\(M=\frac{2\left|x-3\right|}{\left(x+5\right)\left(x-3\right)}=\frac{2\left(x-3\right)}{\left(x+5\right)\left(x-3\right)}=\frac{2}{x+5}\)

+) Nếu \(x-3< 0\)\(\Rightarrow\)\(x< 3\) ta có : 

\(M=\frac{2\left|x-3\right|}{\left(x+5\right)\left(x-3\right)}=\frac{-2\left(x-3\right)}{\left(x+5\right)\left(x-3\right)}=\frac{-2}{x+5}\)

Vậy : +) Nếu \(x\ge3\) thì \(M=\frac{2}{x+5}\) 

         +) Nếu \(x< 3\) thì \(M=\frac{-2}{x+5}\)

Chúc bạn học tốt ~