13(x+3)+(x+3)(x-3)=6(2x+7)

13x+39+x^2-9-12x-42=0

x^2+x-12=0

x=3 và x=-4

**** cho mk nha!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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

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

\(=>\frac{2}{x+1}=\frac{1}{x+1}\)

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

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

\(=>x=-1\)

vậy nghiệm của phương trình trên là {-1}

À quên ĐKXĐ của câu a là \(x\ne-1\)

Nên \(x\in\varnothing\)nhé :v

\(\left(2x^2+1\right)\left(4x-2\right)=\left(2x^2+1\right)\left(x-12\right)\)

\(\Leftrightarrow8x^3-6x^2+4x-3=2x^3-24x^2+x-12\)

\(\Leftrightarrow8x^3-6x^2+4x-3-2x^3+24x^4-x+12=0\)

\(\Leftrightarrow6x^3+18x^2+3x+9=0\)

\(\Leftrightarrow3\left(x+3\right)\left(2x^2+1\right)=0\)

\(\Leftrightarrow x+3=0\)

\(\Leftrightarrow x=-3\)

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

\(\Leftrightarrow4x^2+7x-13=0\)(pt vô nghiệm)

\(b.\Leftrightarrow x^3+3x^2+3x+1-x^2+2x+8=x^3-8+2x^2\)

\(\Leftrightarrow5x=-17\Rightarrow x=\frac{-17}{5}\)

Đặt \(t=x^2+2x+2\left(t\ge1\right)\)

\(c.\Leftrightarrow\frac{t-1}{t}+\frac{t}{t+1}=\frac{7}{6}\)\(\Leftrightarrow\frac{t^2-1+t^2}{t^2+t}=\frac{7}{6}\)\(\Leftrightarrow12t^2-6=7t^2+7t\)

\(\Leftrightarrow5t^2-7t-6=0\Rightarrow\orbr{\begin{cases}t=2\left(tm\right)\\t=\frac{-3}{5}\left(l\right)\end{cases}}\)

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

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

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

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

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

\(\Rightarrow\frac{x^2-7x+12+x^2-4}{\left(x-2\right)\left(x-4\right)}=-1\)

\(\Rightarrow\frac{2x^2-7x+8}{\left(x-2\right)\left(x-4\right)}=-1\)

\(\Rightarrow\frac{2x^2-7x+8}{\left(x-2\right)\left(x-4\right)}=-1\)

.................

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

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

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

\(\Rightarrow\left(x^3-1\right)\left[2\left(x^2+x+1\right)+\left(2x+3\right)\left(x-1\right)\right]=\left(x^3-1\right)\left(2x-1\right)\left(2x+1\right)\)

\(\Rightarrow2\left(x^2+x+1\right)+\left(2x+3\right)\left(x-1\right)=\left(2x-1\right)\left(2x+1\right)\)

\(\Rightarrow2\left(x^2+x+1\right)+\left(2x+3\right)\left(x-1\right)-\left(2x-1\right)\left(2x+1\right)=0\)

\(\Rightarrow2x^2+2x+2+2x^2-2x+3x-3-\left(4x^2-1\right)=0\)

\(\Rightarrow2x^2+2x+2+2x^2-2x+3x-3-4x^2+1=0\)

\(\Rightarrow3x=0\)

\(\Rightarrow luon-dung-voi-moi-x\)

nhầm phải là

3x=0

=>không có giá trị x thỏa mãn yêu cầu

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

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

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

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

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

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

a, \(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right).\left(2x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}=0\)

\(\Leftrightarrow\frac{x^2-4x+4}{3}+\frac{9-4x^2}{8}+\frac{x^2-8x+16}{6}=0\)

\(\Leftrightarrow\frac{8\left(x^2-4x+4\right)+3\left(9-4x^2\right)+4\left(x^2-8x+16\right)}{24}=0\)

\(\Leftrightarrow\frac{8x^2-32x+32+27-12x^2+4x^2-32x+64}{24}=0\)

\(\Leftrightarrow\frac{123-64x}{24}=0\Leftrightarrow123-64x=0\Leftrightarrow x=\frac{123}{64}\)

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

\(\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}-\frac{2.2x}{2\left(x-3\right)\left(x+1\right)}=0\)

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

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

\(\frac{2x^2-6x}{2\left(x-3\right)\left(x+1\right)}=0\)

=>\(2x^2-6x=0\)

\(2x\left(x-3\right)=0\)

=>\(x=0\)

\(x=3\)

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