x²+4

=x²+4+4x-4x

=(x²+2.x.2+2²)-4x

=(x+2)²-(2√x)²

=(x+2-2√x)(x+2+2√x)

Trả lời:

a, \(\frac{4x-5}{x-1}=2+\frac{x}{x-1}\left(đkxđ:x\ne1\right)\)

\(\Leftrightarrow\frac{4x-5}{x-1}=\frac{2\left(x-1\right)+x}{x-1}\)

\(\Rightarrow4x-5=2x-2+x\)

\(\Leftrightarrow4x-5=3x-2\)

\(\Leftrightarrow4x-3x=-2+5\)

\(\Leftrightarrow x=3\left(tm\right)\)

Vậy \(S=\left\{3\right\}\)

b, \(\frac{7}{x+2}=\frac{3}{x-5}\left(đkxđ:x\ne-2;x\ne5\right)\)

\(\Leftrightarrow\frac{7\left(x-5\right)}{\left(x+2\right)\left(x-5\right)}=\frac{3\left(x+2\right)}{\left(x+2\right)\left(x-5\right)}\)

\(\Rightarrow7x-35=3x+6\)

\(\Leftrightarrow7x-3x=6+35\)

\(\Leftrightarrow4x=41\)

\(\Leftrightarrow x=\frac{41}{4}\left(tm\right)\)

Vậy \(S=\left\{\frac{41}{4}\right\}\)

c, \(\frac{2x+5}{2x}-\frac{x}{x+5}=0\left(đkxđ:x\ne0;x\ne-5\right)\)

\(\Leftrightarrow\frac{2x+5}{2x}=\frac{x}{x+5}\)

\(\Leftrightarrow\left(2x+5\right)\left(x+5\right)=2x^2\)

\(\Leftrightarrow2x^2+15x+25=2x^2\)

\(\Leftrightarrow2x^2+15x-2x^2=-25\)

\(\Leftrightarrow15x=-25\)

\(\Leftrightarrow x=\frac{-5}{3}\left(tm\right)\)

Vậy \(S=\left\{\frac{-5}{3}\right\}\)

d, \(\frac{12x+1}{11x-4}+\frac{10x-4}{9}=\frac{20x+17}{18}\left(đkxđ:x\ne\frac{4}{11}\right)\)

\(\Leftrightarrow\frac{18\left(12x+1\right)+2\left(10x-4\right)\left(11x-4\right)}{18\left(11x-4\right)}=\frac{\left(20x+17\right)\left(11x-4\right)}{18\left(11x-4\right)}\)

\(\Rightarrow216x+18+\left(20x-8\right)\left(11x-4\right)=220x^2+107x-68\)

\(\Leftrightarrow216x+18+220x^2-168x+32=220x^2+107x-68\)

\(\Leftrightarrow220x^2+48x+50=220x^2+107x-68\)

\(\Leftrightarrow220x^2+48x-220x^2-107x=-68-50\)

\(\Leftrightarrow-59x=-118\)

\(\Leftrightarrow x=2\left(tm\right)\)

Vậy \(S=\left\{2\right\}\)

ĐKXĐ: \(x\ne\pm2\)

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

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

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

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

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

\(=\frac{x^2+1}{x^2-4}\)

Vì \(x^2+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)

\(\Rightarrow\)Để A không âm thì \(x^2-4>0\)(do \(x\ne\pm2\)nên \(x^2-4\ne0\))

\(\Leftrightarrow x^2>4\)

\(\Leftrightarrow\orbr{\begin{cases}x>2\\x< -2\end{cases}}\)

Vậy để A không âm thì \(x>2\)hoặc \(x< -2\)

tự kẻ hình ná

trong tam giác AHC có 

AK=KH

HN=CN

=> KN là đtb=> KN//AC và KN=AC/2

tương tự, ta có MK//AB và MK=AB/2

MN//BC và MN=BC/2

Xét tam giác ABC và tam giác KMN có

KN/AC=MN/BC=MK/AB(=1/2) (cũng là tỉ số đồng dạng của 2 tam giác)

=> tam giác ABC đồng dạng với tam giác KMN(ccc)

\(\frac{x+3}{x-3}-\frac{x+5}{x+6}=\frac{47}{x^2+3x-18}\) (ĐK: \(x\ne3,x\ne-6\)) 

\(\Leftrightarrow\frac{\left(x+3\right)\left(x+6\right)-\left(x+5\right)\left(x-3\right)}{\left(x-3\right)\left(x+6\right)}=\frac{47}{\left(x-3\right)\left(x+6\right)}\)

\(\Rightarrow7x+33=47\)

\(\Leftrightarrow x=2\)(tm).

Trả lời:

\(\frac{x+3}{x-3}-\frac{x+5}{x+6}=\frac{47}{x^2+3x-18}\left(đkxđ:x\ne3;x\ne-6\right)\)

\(\Leftrightarrow\frac{x+3}{x-3}-\frac{x+5}{x+6}=\frac{47}{x^2-3x+6x-18}\)

\(\Leftrightarrow\frac{x+3}{x-3}-\frac{x+5}{x+6}=\frac{47}{x\left(x-3\right)+6\left(x-3\right)}\)

\(\Leftrightarrow\frac{x+3}{x-3}-\frac{x+5}{x+6}=\frac{47}{\left(x-3\right)\left(x+6\right)}\)

\(\Leftrightarrow\frac{\left(x+3\right)\left(x+6\right)-\left(x+5\right)\left(x-3\right)}{\left(x-3\right)\left(x+6\right)}=\frac{47}{\left(x-3\right)\left(x+6\right)}\)

\(\Rightarrow x^2+9x+18-\left(x^2+2x-15\right)=47\)

\(\Leftrightarrow x^2+9x+18-x^2-2x+15=47\)

\(\Leftrightarrow7x+33=47\)

\(\Leftrightarrow7x=14\)

\(\Leftrightarrow x=2\left(tm\right)\)

Vậy phương trình trên có một nghiệm là x = 2 

mày bị hấy à

Có mà bạn ấy. Ngồi đó mà chửi người khác 

xinh mà học dốt Toán thì cũng dở thôi :(

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

ĐKXĐ : \(x\ne\pm1\)

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

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

\(\Rightarrow x^2-x-2x^2+x+3=2x+3\)

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

\(\Leftrightarrow2x+3+x^2-3=0\)

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

\(\Leftrightarrow x\left(x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x=-2\left(tm\right)\end{cases}}\)

Vậy S = { 0 ; -2 }

Trả lời:

\(\frac{x}{x+1}-\frac{2x-3}{x-1}=\frac{2x+3}{x^2-1}\left(đkxđ:x\ne\pm1\right)\)

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

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

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

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

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

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

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

\(\Leftrightarrow x=0\left(tm\right)\)  hoặc \(x+2=0\)

                                          \(\Leftrightarrow x=-2\left(tm\right)\)

Vậy \(S=\left\{0;-2\right\}\)