a,\(\left(x-1\right)^2-\left(2x\right)^2=0< =>\left(x-1-2x\right)\left(x-1+2x\right)=0\)

\(< =>\left(-x-1\right)\left(3x-1\right)=0< =>\orbr{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}\)

b,\(\left(3x-5\right)^2-x\left(3x-5\right)=0< =>\left(3x-5\right)\left(3x-5-x\right)=0\)

\(< =>\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{5}{2}\end{cases}}\)

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

\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\Leftrightarrow x=-1;x=\frac{1}{3}\)

b, \(\left(3x-5\right)^2-x\left(3x-5\right)=0\)

\(\Leftrightarrow\left(3x-5\right)\left(3x-5-x\right)=0\Leftrightarrow\left(3x-5\right)\left(2x-5\right)=0\Leftrightarrow x=\frac{5}{3};x=\frac{5}{2}\)

a;b tìm nhân tử chung ở mẫu bạn tự làm nhé 

c, \(\frac{3}{1-4x}=\frac{2}{4x+1}-\frac{8+6x}{16x^2-1}\)ĐK : \(x\ne\pm\frac{1}{4}\)

\(\Leftrightarrow-\frac{3}{4x-1}=\frac{2}{4x+1}-\frac{8+6x}{\left(4x-1\right)\left(4x+1\right)}\)

\(\Leftrightarrow-\frac{3\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}=\frac{2\left(4x-1\right)-8-6x}{\left(4x+1\right)\left(4x-1\right)}\)

\(\Rightarrow-12x-3=8x-2-8-6x\Leftrightarrow-14x=-7\Leftrightarrow x=\frac{1}{2}\)

i, \(\frac{x+2}{x+3}-\frac{x+1}{x-1}=\frac{4}{x^2+2x-3}\)ĐK : \(x\ne-3;1\)

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

\(\Rightarrow x^2+x-2-x^2-4x-3=4\Leftrightarrow-3x-5=4\Leftrightarrow x=-3\)(ktm)

Vậy pt vô nghiệm 

g, \(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)ĐK : \(x\ne1\)

\(\Leftrightarrow\frac{x^2+x+1+2x^2-5}{x^3-1}=\frac{4\left(x-1\right)}{x^3-1}\)

\(\Rightarrow3x^2+x-4=4x-4\Leftrightarrow3x^2-3x=0\Leftrightarrow x=0\left(tm\right);x=1\left(ktm\right)\)

h, \(\frac{3}{5x-1}+\frac{2}{3-5x}=\frac{4}{\left(1-5x\right)\left(5x-3\right)}\)ĐK : \(x\ne\frac{1}{5};\frac{3}{5}\)

\(\Leftrightarrow\frac{3\left(5x-3\right)-2\left(5x-1\right)}{\left(5x-1\right)\left(5x-3\right)}=\frac{-4}{\left(5x-1\right)\left(5x-3\right)}\)

\(\Rightarrow15x-9-10x+2=-4\Leftrightarrow5x=3\Leftrightarrow x=\frac{3}{5}\)(ktm)

Vậy pt vô nghiệm 

đề là ptđt thành nhân tử hả bạn ? 

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Đặt \(x^2+7x+10=t\)

\(t\left(t+2\right)-24=t^2+2t-24=\left(t-4\right)\left(t+6\right)\)

Theo cách đặt : \(\left(x^2+7x+6\right)\left(x^2+7x+16\right)=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Đặt \(x^2+7x+11=t\)

\(\Rightarrow\left(t+1\right)\left(t-1\right)-24\)

\(=t^2-1-24\)

\(=t^2-25\)

\(=\left(t-5\right)\left(t+5\right)\)

\(\Rightarrow\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

Ta có : \(VT=3+\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\ge3+\frac{\left(1+1+2\right)^2}{a+b+c}=3+16=19\)

Dấu "=" tự tìm nha b yeuuuuu

\(\frac{a+1}{a}+\frac{b+1}{b}+\frac{c+4}{c}=1+\frac{1}{a}+1+\frac{1}{b}+1+\frac{4}{c}\)

Theo BĐT Cauchy Schwarz dạng Engel ta có : 

\(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\ge\frac{\left(1+1+2\right)^2}{a+b+c}=16\)

\(\Rightarrow\frac{a+1}{a}+\frac{b+1}{b}+\frac{c+4}{c}\ge16+3=19\)

Dấu ''='' xảy ra khi \(a=b=c=\frac{1}{3}\)

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

\(< =>x^2+6x+9+x^2-4x+4=2x^2\)

\(< =>2x+13=0< =>x=-\frac{13}{2}\)

b,\(5x\left(x-2\right)=x-2< =>\left(x-2\right)\left(5x-1\right)=0< =>\hept{\begin{cases}x=2\\x=\frac{1}{5}\end{cases}}\)

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

\(\Leftrightarrow x^2+6x+9+x^2-4x+4-2x^2=0\)

\(\Leftrightarrow2x+13=0\)

\(\Leftrightarrow2x=-13\)

\(\Leftrightarrow x=-\frac{13}{2}\)
Vậy \(S=\left\{-\frac{13}{2}\right\}\)

b) \(5x\left(x-2\right)=x-2\)

\(\Leftrightarrow5x\left(x-2\right)-\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\5x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{5}\end{cases}}\)

Vậy \(S=\left\{2;\frac{1}{5}\right\}\)

a, \(5y^2-5x^2+6x+6y=5\left(y-x\right)\left(x+y\right)+6\left(x+y\right)\)

\(=\left(x+y\right)\left(5y-5x+6\right)\)

b, \(12x^2+19x+7=12x^2+12x+7x+7\)

\(=12x\left(x+1\right)+7\left(x+1\right)=\left(12x+7\right)\left(x+1\right)\)

5y² - 5x² + 6x + 6y 

= 5(y² - x²) + 6(x + y) 

= 5(y - x)(x + y) + 6(x + y) 

= (x + y)(5y - 5x + 6) 

b) 12x² + 19x + 7 

= 12x² + 12x + 7x + 7 

= 12x(x + 1) + 7(x + 1) 

= (x + 1)(12x + 7) 

mik ko nhìn thấy đa thức đâu bạn ơiiiiiiiiiiiiiiiiiiiiiiii

\(\left(x-2\right)^2-x\left(x-1\right)\left(x+1\right)+x\left(7x-6\right)=0\)

\(\Leftrightarrow x^2-4x+4-x\left(x^2-1\right)+7x^2-6x=0\)

\(\Leftrightarrow8x^2-10x+4-x^3+x=0\Leftrightarrow-x^3+8x^2-9x+4=0\Leftrightarrow x=6,7...\)

\(\left(x-2\right)^2-x\left(x-1\right)\left(x+1\right)+x\left(7x-6\right)=0\)

\(\Leftrightarrow x^2-4x+4-x^3-x^2+x^2+x+7x^2-6x=0\)

\(\Leftrightarrow-x^3-8x^2-9x+4=0\)

Làm tiếp nhé =))

mik cần gấp , ai làm nhanh mik k cho .

\(a,\)\(x^2+2x-15=0\Rightarrow\left(x+1\right)^2-16=0\Rightarrow\left(x+1\right)^2=16\)

\(\hept{\begin{cases}x+1=4\\x+1=-4\end{cases}}\Rightarrow\hept{\begin{cases}x=3\\x=-5\end{cases}}\)\(\)vậy \(S=\left\{3;-5\right\}\)

\(b,\)\(2x^2-7x+6=0\Rightarrow2x\left(x-1\right)-5\left(x-1\right)+1=0\)

\(\Rightarrow\left(x-1\right)\left(2x-5\right)=-1=1.-1=-1.1\)

\(x-1=1;2x-5=-1\)và \(x-1=-1;2x-5=1\)

\(\Rightarrow x=2\)và\(x=0;x=3\)vậy \(S=\left\{2;0;3\right\}\)

Mình làm nốt nhé

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

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

Vì \(\left(x-2\right)^2+1>0\forall x\Leftrightarrow x=0\)

d) \(x^3+5x^2+3x-9=0\Leftrightarrow x^3+3x^2+2x^2+6x-3x-9=0\)

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

\(\Leftrightarrow\left(x+3\right)\left[\left(x+1\right)^2-4\right]=\left(x+3\right)\left(x+1-2\right)\left(x+1+2\right)=0\)

\(\Leftrightarrow\left(x+3\right)^2\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}\left(x+3\right)^2=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=1\end{cases}}}\)

Tự KL