1) Nhìn cái pt hết ham, nhưng bấm nghiệm đẹp v~`~

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

\(\Leftrightarrow\left(\sqrt{2}+2\right)\left(x\sqrt{2}-1\right)-2x\sqrt{2}+\sqrt{2}=0\)

\(\Leftrightarrow2x-\sqrt{2}+2x\sqrt{2}-2-2x\sqrt{2}+\sqrt{2}=0\)

\(\Leftrightarrow2x-2=0\Leftrightarrow2x=2\Rightarrow x=1\)

Mấy bài kia sao cái phương trình dài thê,s giải sao nổi

Đề phải vậy chứ nhỉ?

\(\frac{1}{x-1}+\frac{3x^2}{1-x^3}=\frac{2x}{x^2+x+1}\left(Đkxđ:x\ne1\right)\)

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

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

\(\Leftrightarrow4x^2-3x-1=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(ktmđk\right)\\x=-\frac{1}{4}\left(tmđk\right)\end{matrix}\right.\)

Vậy ...........

1.\(\frac{x+1}{2013}\)+\(\frac{x+2}{2012}\)=\(\frac{x+3}{2011}\)+\(\frac{x+4}{2010}\)

⇔\(\frac{x+1}{2013}\)+1+\(\frac{x+2}{2012}\)+1=\(\frac{x+3}{2011}\)+1+\(\frac{x+4}{2010}\)+1

⇔\(\frac{x+2014}{2013}\)+\(\frac{x+2014}{2012}\)=\(\frac{x+2014}{2011}\)+\(\frac{x+2014}{2010}\)

⇔\(\frac{x+2014}{2013}\)+\(\frac{x+2014}{2012}\)-\(\frac{x+2014}{2011}\)-\(\frac{x+2014}{2010}\)=0

⇔(x+2014)(\(\frac{1}{2013}\)+\(\frac{1}{2012}\)-\(\frac{1}{2011}\)-\(\frac{1}{2010}\))=0

Mà \(\frac{1}{2013}\)+\(\frac{1}{2012}\)-\(\frac{1}{2011}\)-\(\frac{1}{2010}\)≠0

⇔x+2014=0

⇔x=-2014

Vậy tập nghiệm của phương trình đã cho là:S={-2014}

2.\(\frac{3x+2}{4}\)+\(\frac{x+3}{2}\)=\(\frac{x-1}{3}\)-\(\frac{-x-1}{12}\)

⇔\(\frac{3\left(3x+2\right)}{12}\)+\(\frac{6\left(x+3\right)}{12}\)=\(\frac{4\left(x-1\right)}{12}\)+\(\frac{x+1}{12}\)

⇒9x+6+6x+18=4x-4+x+1

⇒15x+24=5x-3

⇒15x-5x=-3-24

⇒10x=-27

⇒ x=-\(\frac{27}{10}\)

Vậy tập nghiệm của phương trình đã cho là S={-\(\frac{27}{10}\)}

\(3.\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2+3}{1-x^2}=0ĐKXĐ:x\ne\pm1\)

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

\(-3+7x-5x^2+x^3=0\)

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

\(\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)Theo ĐKXĐ => x=3

Đề chỗ mấu thức của phân thức cuối ý cho mình hỏi là \(x^2\) hay \(x^3\)

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)

Bài 1:

a, \(\frac{1}{x+1}+\frac{2}{x-1}=\frac{1+x^2}{x^2-1}\) (ĐKXĐ: x \(\ne\) \(\pm\) 1)

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

\(\Rightarrow\) x - 1 + 2(x + 1) = 1 + x²

\(\Leftrightarrow\) x - 1 + 2x + 2 - 1 - x² = 0

\(\Leftrightarrow\) -x² + 3x = 0

\(\Leftrightarrow\) x(3 - x) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TMĐKXĐ\right)\\x=3\left(TMĐKXĐ\right)\end{matrix}\right.\)

Vậy S = {0; 3}

b, \(\frac{x-2}{x+2}-\frac{x}{x-2}=\frac{8}{x^2-4}\) (ĐKXĐ: x \(\ne\) \(\pm\) 2)

\(\Leftrightarrow\) \(\frac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}-\frac{x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{8}{\left(x+2\right)\left(x-2\right)}\)

\(\Rightarrow\) (x - 2)² - x(x + 2) = 8

\(\Leftrightarrow\) (x - 2)² - x(x + 2) - 8 = 0

\(\Leftrightarrow\) x² - 4x + 4 - x² - 2x - 8 = 0

\(\Leftrightarrow\) -6x - 4 = 0

\(\Leftrightarrow\) x = \(\frac{-2}{3}\) (TMĐKXĐ)

Vậy S = {\(\frac{-2}{3}\)}

c, \(\frac{1}{x}\) + \(\frac{2}{x-3}\) = \(\frac{1-5x}{x^2-3x}\) (ĐKXĐ: x \(\ne\) 0; x \(\ne\) 3)

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

\(\Rightarrow\) x - 3 + 2x = 1 - 5x

\(\Leftrightarrow\) 3x - 3 = 1 - 5x

\(\Leftrightarrow\) 3x + 5x = 1 + 3

\(\Leftrightarrow\) 8x = 4

\(\Leftrightarrow\) x = \(\frac{1}{2}\) (TMĐKXĐ)

Vậy S = {\(\frac{1}{2}\)}

Bài 2:

a, \(\frac{1}{x+2}=\frac{5}{2-x}+\frac{12+x}{x^2-4}\)

\(\Leftrightarrow\) \(\frac{1}{x+2}=\frac{-5}{x-2}+\frac{12+x}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow\) \(\frac{x-2}{\left(x+2\right)\left(x-2\right)}=\frac{-5\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{12+x}{\left(x+2\right)\left(x-2\right)}\)

\(\Rightarrow\) x - 2 = -5(x + 2) + 12 + x

\(\Leftrightarrow\) x - 2 = -5x - 10 + 12 + x

\(\Leftrightarrow\) x - 2 = -4x + 2

\(\Leftrightarrow\) x + 4x = 2 + 2

\(\Leftrightarrow\) 5x = 4

\(\Leftrightarrow\) x = \(\frac{4}{5}\)

Vậy S = {\(\frac{4}{5}\)}

Chúc bn học tốt!! (Phần b hình như không có gì thì phải)

\(\frac{3x+2}{x+4}+\frac{2x+1}{x-2}=5-\frac{x-32}{x^2+2x-8}\)

\(\Leftrightarrow\) \(\frac{\left(3x+2\right)\left(x-2\right)}{\left(x+4\right)\left(x-2\right)}+\frac{\left(2x+1\right)\left(x+4\right)}{\left(x+4\right)\left(x-2\right)}=\frac{5\left(x+4\right)\left(x-2\right)}{\left(x+4\right)\left(x-2\right)}-\frac{x-32}{\left(x+4\right)\left(x-2\right)}\)

\(\Rightarrow\) (3x + 2)(x - 2) + (2x + 1)(x + 4) = 5(x + 4)(x - 2) - x + 32

\(\Leftrightarrow\) 3x² - 6x + 2x - 4 + 2x² + 8x + x + 4 = 5x² - 10x + 20x - 40 - x + 32

\(\Leftrightarrow\) 5x² + 5x = 5x² + 9x - 8

\(\Leftrightarrow\) 5x² + 5x - 5x² - 9x + 8 = 0

\(\Leftrightarrow\) -4x + 8 = 0

\(\Leftrightarrow\) x - 2 = 0

\(\Leftrightarrow\) x = 2

Vậy S = {2}

\(\frac{x+2m}{x+3}+\frac{x-m}{x-3}=\frac{mx\left(x+1\right)}{x^2-9}\) (đkxđ: x \(\ne\) \(\pm\) 3)

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

\(\Rightarrow\) (x + 2m)(x - 3) + (x - m)(x + 3) = mx(x + 1)

\(\Leftrightarrow\) x² - 3x + 2mx - 6m + x² + 3x - mx - 3m - mx² - mx = 0

\(\Leftrightarrow\) (2 - m)x² - 9m = 0

Thay m = 1 ta được:

(2 - 1)x² - 9 . 1 = 0

\(\Leftrightarrow\) x² - 9 = 0

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

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(KTM\right)\\x=-3\left(KTM\right)\end{matrix}\right.\)

Vậy S = \(\varnothing\)

Thay m = 2 ta được:

(2 - 2)x² - 9 . 2 = 0

\(\Leftrightarrow\) -18 = 0

\(\Rightarrow\) Pt vô nghiệm

Vậy S = \(\varnothing\)

Chúc bn học tốt!!

x=-7206932,631

tính = máy tính đó 

nhớ tink cho mình nữa nha

giải ra đi