\(3\left(x-2\right)^2+9\left(x-1\right)=3\left(x^2+x-3\right)\\ \Leftrightarrow3\left(x^2-4x+4\right)+9x-9=3x^2+3x-9\\ \Leftrightarrow3x^2-12x+12+9x-9-3x^2-3x+9=0\\ \Leftrightarrow-6x+12=0\\ \Leftrightarrow x=2\)

\(3\left(x-2\right)^2+9\left(x-1\right)=3\left(x^2+x-3\right)\)

\(\Leftrightarrow3\left(x^2-4x+4\right)+9x-9=3x^2+3x-9\)

\(\Leftrightarrow3x^2-12x+12+9x-9-3x^2-2x+9=0\)

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

\(\Leftrightarrow-6\left(x+1\right)=0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

Vậy phương trình có nghiệm là \(-1\)

ĐKXĐ: x<>0; x<>-1

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

=>x^2-1-x=2x-1

=>x^2-x-2x=0

=>x(x-3)=0

=>x=0(loại) hoặc x=3(nhận)

\(ĐK: x\ne 0; x\ne-1\)

Khi đó: 

\(\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\\ \Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}-\dfrac{1.x}{\left(x+1\right).x}-\dfrac{2x-1}{x\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{x^2-1}{x\left(x+1\right)}-\dfrac{x}{x\left(x+1\right)}-\dfrac{2x-1}{x\left(x+1\right)}=0\\ \Leftrightarrow x^2-1-x-2x+1=0\\ \Leftrightarrow x^2-3x=0\\ \Leftrightarrow x\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=3\left(nhận\right)\end{matrix}\right.\)

Vậy PT có nghiệm duy nhất \(S=\left\{3\right\}\)

\(ĐKXĐ:x\ne0;x\ne-1\)

\(\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(x-1\right)}{x\left(x+1\right)}-\dfrac{1.x}{x\left(x+1\right)}=\dfrac{2x-1}{x\left(x+1\right)}\\ \Rightarrow x^2-x+x-1-x=2x-1\\ \Leftrightarrow x^2-x+x-x-2x=-1+1\\ \Leftrightarrow x^2-3x=0\\ \Leftrightarrow x\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=3\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{3\right\}\)

a: =>3x+3=4x-4

=>-x=-7

hay x=7(nhận)

b: (x-1)(x-3)=0

=>x-1=0 hoặc x-3=0

=>x=1 hoặc x=3

c: 2(x-1)+x=0

=>2x-2+x=0

=>3x-2=0

hay x=2/3

a, ĐKXĐ : x ≠ 1 ; x ≠ -1

\(\Rightarrow3\left(x+1\right)=4\left(x-1\right)\)

\(\Leftrightarrow3x+3=4x-4\)

\(\Leftrightarrow-x=-7\)

\(\Leftrightarrow x=7\left(N\right)\)

b,

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

c,

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

\(\Leftrightarrow3x=2\)

\(\Leftrightarrow x=\dfrac{2}{3}\)

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

\(ĐKXĐ:x\ne2,x\ne4\)

\(MC:\left(x-2\right)\left(x-4\right)\)

\(PT\Leftrightarrow\left(x-1\right)\left(x-4\right)+\left(x+3\right)\left(x-2\right)=2\)

\(\Leftrightarrow x^2-5x+4+x^2+x-6=2\)

\(\Leftrightarrow2x^2-4x-4=0\)

\(\Leftrightarrow2\left(x^2-2x-2\right)=0\)

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

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

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

Vậy giờ mình kết luận x=? hả bạn? Mình dở toán lắm. (T-T)

ĐKXĐ:\(x\ne-2\)

\(\dfrac{1}{x+2}-1=\dfrac{5x+7}{x+2}\\ \Leftrightarrow\dfrac{1}{x+2}-\dfrac{5x+7}{x+2}=1\\ \Leftrightarrow\dfrac{1-5x-7}{x+2}=1\\ \Leftrightarrow-5x-6=x+2\\ \Leftrightarrow x+2+5x+6=0\\ \Leftrightarrow6x+8=0\\ \Leftrightarrow x=-\dfrac{4}{3}\left(tm\right)\)

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

\(x^4+x^3+2x^2+x^3+x^2+2x+x^2+x+2=12\)

\(x^4+2x^3+4x^2+3x+2=12\)

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

\(x^4+2x^3+4x^2+3x-10=0\)

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

TH1 : \(x^2+x+5=0\)

\(\Delta=1^2-4.1.5=1-20=-19< 0\)

Nên phương trình vô nghiệm.

TH2 : \(x+2=0\Leftrightarrow x=-2\)  

TH3 : \(x-1=0\Leftrightarrow x=1\)

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

Đặt \(x^2+x+1=t\)

\(\Rightarrow t\left(t+1\right)=12\)\(\Leftrightarrow t^2+t=12\)

\(\Leftrightarrow t^2+t-12=0\)\(\Leftrightarrow\left(t^2-3t\right)+\left(4t-12\right)=0\)

\(\Leftrightarrow t\left(t-3\right)+4\left(t-3\right)=0\)\(\Leftrightarrow\left(t-3\right)\left(t+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t-3=0\\t+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}t=3\\t=-4\end{cases}}\)

Ta thấy: \(x^2+x+1=x^2+2.\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

\(\Rightarrow t>0\)\(\Rightarrow t=3\)thoả mãn

\(\Rightarrow x^2+x+1=3\)\(\Leftrightarrow x^2+x+1-3=0\)\(\Leftrightarrow x^2+x-2=0\)

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

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

Vậy tập nghiệm của phương trình là \(S=\left\{-2;1\right\}\)

Đặt : \(x^2+x+1=u\)

Ta có pt mới sau : \(u\left(u+1\right)-12=0\)

\(< =>u^2+u-12=0\)

\(< =>\orbr{\begin{cases}u=3\\u=-4\end{cases}}\)(giải delta)

Xét : \(x^2+x-1=x^2+1x+\frac{1}{2}^2+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0< =>u>0\)

Nên chỉ có \(u=3\)thỏa mãn

Ta có : \(u=3< =>x^2+x+1=3\)

\(< =>x^2+x-2=0\)

\(< =>\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)(giải delta)

Vậy tập nghiệm của pt trên là {...}

\(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}=\dfrac{1}{x}\left(đk:x\ne0,x\ne2\right)\)

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

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

\(\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{1}{2}\)

Cho mình sửa lại nhé:

\(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}=\dfrac{1}{x}\left(đk:x\ne0,x\ne2\right)\)

\(\Leftrightarrow\dfrac{\left(x+2\right)x-2}{x\left(x-2\right)}=\dfrac{x-2}{x\left(x-2\right)}\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

ĐKXĐ: \(x^2-4x+1\ge0\)

\(2x+2+2\sqrt{x^2-4x+1}=6\sqrt{x}\)

\(\Leftrightarrow2x+2-5\sqrt{x}+2\sqrt{x^2-4x+1}-\sqrt{x}=0\)

\(\Leftrightarrow\dfrac{4x^2-17x+4}{2x+2+5\sqrt{x}}+\dfrac{4x^2-17x+4}{2\sqrt{x^2-4x+1}+\sqrt{x}}=0\)

\(\Leftrightarrow\left(4x^2-17x+4\right)\left(\dfrac{1}{2x+2+5\sqrt{x}}+\dfrac{1}{2\sqrt{x^2-4x+1}+\sqrt{x}}\right)=0\)

\(\Leftrightarrow4x^2-17x+4=0\)

\(\Leftrightarrow...\)

đặt   \(x^2+x+2\)  là a  ; đặt  \(x+1\)là b

\(\Rightarrow a+b=x^2+x+2+x+1\)\(=x^2+2x+3\)

\(\Rightarrow a^3+b^3=\left(a+b\right)^3\)

\(\Rightarrow a^3+b^3=a^3+3a^2b+3ab^2+b^3\)

\(\Rightarrow3a^2b+3ab^2=0\)\(\Rightarrow3ab\left(a+b\right)=0\)\(\Rightarrow\)\(a=0\)hoặc \(b=0\)hoặc \(a+b=0\)

* nếu a = 0  \(\Rightarrow\) \(x^2+x+2=0\)( vô lí vì luôn dương, cái này dễ chứng minh nha)

* nếu b = 0   \(\Rightarrow x+1=0\Rightarrow x=-1\)

* nếu a + b = 0 \(\Rightarrow x^2+2x+3=0\)(cái này cũng luôn dương nhé)

Vậy phương trình có 1 nghiệm là x = -1 

chúc bạn học tốt nha <3

Thanks bạn nhìu