hỏi j v

a)\(\frac{1}{x-1}\)-\(\frac{3x2}{x3-1}\)=\(\frac{2x}{x2+x+1}\)

<=> \(\frac{1}{x-1}\)-\(\frac{3x2}{\left(x-1\right)\left(x2+x+1\right)}\)=\(\frac{2x}{x2+x+1}\) ĐKXĐ: x khác 1

<=> x²+x+1 - 3x² = 2x(x-1)

<=>x²+x+1 - 3x² = 2x²-2x

<=>x²-3x-1=0( đoạn này làm nhanh nhé)

<=>x²-2*\(\frac{3}{2}\)x +\(\frac{9}{4}\)-\(\frac{9}{4}\)-1=0

<=>(x-\(\frac{3}{2}\))²-\(\frac{13}{4}\)=0

<=>(x-\(\frac{3-\sqrt{13}}{2}\))(x-\(\frac{3+\sqrt{13}}{2}\))=0

\(\begin{cases}x=\frac{3+\sqrt{13}}{2}\\x=\frac{3-\sqrt{13}}{2}\end{cases}\)

b) pt... đkxđ x khác 1;2;3

<=>  3(x-3) +2(x-2)=x-1

<=>  3x-9 +2x-4 = x-1

<=> 4x= 12

<=>  x=3 ( ko thỏa đk)

vậy pt vô nghiệm

c) 1+\(\frac{1}{x+2}\)=\(\frac{12}{\left(x+2\right)\left(x2+2x+4\right)}\)đkxđ : x khác -2

<=> x³+8 + x²+2x+4 = 12

<=>  x³+x²+2x=0

<=> x²+x+2=0( chia cả 2 vế cho x)

pt này chắc chắn vô nghiệm nhé bạn

a) ĐKXĐ: \(x\ne-1;x\ne2\)

Ta có: \(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)

⇔\(\frac{1}{x+1}-\frac{5}{x-2}+\frac{15}{\left(x+1\right)\left(x-2\right)}=0\)

⇔\(\frac{x-2}{\left(x+1\right)\left(x-2\right)}-\frac{5\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}+\frac{15}{\left(x+1\right)\left(x-2\right)}=0\)

⇔\(x-2-5x-5+15=0\)

⇔\(-4x+8=0\)

⇔\(-4x=-8\)

⇔\(x=\frac{-8}{-4}=2\)(loại)

Vậy: x không có giá trị

b) ĐKXĐ: \(x\ne0;x\ne\frac{3}{2}\)

Ta có: \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)

⇔\(\frac{x}{\left(2x-3\right)\cdot x}-\frac{3}{x\left(2x-3\right)}-\frac{5\left(2x-3\right)}{x\left(2x-3\right)}=0\)

⇔\(x-3-10x+15=0\)

⇔\(-9x+12=0\)

⇔\(-9x=-12\)

⇔\(x=\frac{-12}{-9}=\frac{4}{3}\)

Vậy: \(x=\frac{4}{3}\)

c) ĐKXĐ:\(x\ne3;x\ne1\)

Ta có: \(\frac{6}{x-1}-\frac{4}{x-3}=\frac{8}{2x-6}\)

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

⇔\(\frac{6}{x-1}-\frac{4}{x-3}=\frac{4}{x-3}\)

⇔\(\frac{6}{x-1}-\frac{4}{x-3}-\frac{4}{x-3}=0\)

⇔\(\frac{6}{x-1}-\frac{8}{x-3}=0\)

⇔\(\frac{6\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}-\frac{8\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}=0\)

⇔\(6\left(x-3\right)-8\left(x-1\right)=0\)

⇔6x-18-8x+8=0

⇔-2x-10=0

⇔-2(x+5)=0

Vì 2≠0 nên x+5=0

hay x=-5

Vậy: x=-5

a) => 4x + 2/3 = 0 hoặc 2/3x - 1 =0 

4x= -2/3 hoặc 2/3x= 1

x = -2/3 . 1/4 hoặc x = 1.3/2

x = -1/6 hoặc x = 3/2 

b) x+2 / x -1 = 5/2 

=> 2(x+2) = 5(x-1)

2x + 4 = 5x - 5

5x - 2x= 4+5

3x = 9

=> x= 3

a) (4x+\(\frac{2}{3}\)) . ( \(\frac{2}{3}\)x-1)=0

\(\Rightarrow\)\(\orbr{\begin{cases}4x+\frac{2}{3}=0\\\frac{2}{3}x-1=0\end{cases}}\)

\(\Rightarrow\)\(\orbr{\begin{cases}x=\\x=\end{cases}}\)........

Tới đây bn tự giải nha

còn câu c nữa các bạn ơi ,giúp mình với

Cái này là phương trình chứa ẩn ở mẫu đó nha, mình cần sớm

mann nào trả lời đc thui k hết 5 cái nick lun :D

\(B=\left[\left(\frac{x}{y}-\frac{y}{x}\right):\left(x-y\right)-2.\left(\frac{1}{y}-\frac{1}{x}\right)\right]:\frac{x-y}{y}\)

\(=\left[\frac{x^2-y^2}{xy}.\frac{1}{x-y}-2.\frac{x-y}{xy}\right].\frac{y}{x-y}\)

\(=\left(\frac{\left(x-y\right)\left(x+y\right)}{xy.\left(x-y\right)}-\frac{2.\left(x-y\right)}{xy}\right).\frac{y}{x-y}\)

\(=\left(\frac{x+y}{xy}-\frac{2x-2y}{xy}\right).\frac{y}{x-y}=\frac{x+y-2x+2y}{xy}.\frac{y}{x-y}=\frac{y.\left(3y-x\right)}{xy.\left(x-y\right)}=\frac{3y-x}{x.\left(x-y\right)}\)

\(C=\left(\frac{x+y}{2x-2y}-\frac{x-y}{2x+2y}-\frac{2y^2}{y-x}\right):\frac{2y}{x-y}\)

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

\(=\frac{\left(x+y\right)^2-\left(x-y\right)^2+2.2y^2.\left(x+y\right)}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}\)

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

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

\(D=3x:\left\{\frac{x^2-y^2}{x^3+y^3}.\left[\left(x-\frac{x^2+y^2}{y}\right):\left(\frac{1}{x}-\frac{1}{y}\right)\right]\right\}\)

\(=3x:\left\{\frac{\left(x+y\right)\left(x-y\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}.\left[\frac{xy-x^2-y^2}{y}:\frac{y-x}{xy}\right]\right\}\)

\(=3x:\left[\frac{x-y}{x^2-xy+y^2}.\left(\frac{xy-x^2-y^2}{y}.\frac{xy}{y-x}\right)\right]\)

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

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

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

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

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

\(=2.\frac{x^2+2x+3x+6+x^2+3x+x^2+x}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)

\(=2.\frac{3x^2+9x+6}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=2.\frac{3.\left(x^2+3x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)

\(=\frac{6.\left(x^2+x+2x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\frac{6.\left[x.\left(x+1\right)+2.\left(x+1\right)\right]}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)

\(=\frac{6.\left(x+1\right)\left(x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\frac{6}{x.\left(x+3\right)}\)

bÀI LÀM

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)

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

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

\(=>-2x^2+x+1=2x^2-2x\)

\(=>-4x^2+3x+1=0\)

\(=>\left(x-1\right)\left(x+\frac{1}{4}\right)=0\)'

\(=>\orbr{\begin{cases}x-1=0\\x+\frac{1}{4}\end{cases}=>\orbr{\begin{cases}x=1\\x=-\frac{1}{4}\end{cases}}}\)