a: =>4x-3x=1-2

=>x=-1

b: =>3x=12

=>x=4

c: =>2(x^2-6)=x(x+3)

=>2x^2-12=x^2+3x

=>x^2-3x-12=0

=>\(x=\dfrac{3\pm\sqrt{57}}{2}\)

\(a,\)( sửa lại xíu đề cho đúng nhé )

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

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

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

\(\Rightarrow x=1\)

\(g,\)\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)=-16\)

\(\Rightarrow\left(x^2+10x+16\right)\left(x^2+10x+24\right)=-16\)

Đặt \(x^2+10x+16=a\)

\(\Rightarrow a\left(a+8\right)=-16\)

\(\Rightarrow a^2+8a+16=0\)

\(\Rightarrow\left(a+4\right)^2=0\)

\(\Leftrightarrow\left(x^2+10x+20\right)^2=0\)

\(\Rightarrow x^2+10x+25-25=0\)

\(\Rightarrow\left(x+5\right)^2-\left(\sqrt{5}\right)^2=0\)

\(\Rightarrow\left(x+5-\sqrt{5}\right)\left(x+5+\sqrt{5}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=-5+\sqrt{5}\\x=-5-\sqrt{5}\end{cases}}\)

\(h,\)\(x^3+3x^2+3x+2=0\)

\(\Rightarrow x^3+2x^2+x^2+2x+x+2=0\)

\(\Rightarrow x^2\left(x+2\right)+x\left(x+2\right)+\left(x+2\right)=0\)

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

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

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

<=>4x-8=0 

<=>4x=8 

=.x=2(nhan)

\(a, x(x+3)-(2x-1)(x+3)=0\)

\(⇔(x+3)(1-x)=0\)

\(⇔\left[\begin{array}{} x+3=0\\ 1-x=0 \end{array}\right.\)

\(⇔\left[\begin{array}{} x=-3\\ x=1 \end{array}\right.\)

Vậy phương trình có tập nghiệm là S={\(-3; 1\)}

\(b, 3x-5(x+2)=3(4-2x)\)

\(⇔3x-5x-10=12-6x\)

\(⇔3x-5x+6x=12+10\)

\(⇔4x=22\)

\(⇔x=\dfrac{22}{4}\)

Vậy pt có 1 nghiệm là \(x=\dfrac{22}{4}\)

\(c, (4x-3)(5x-6)=(4x-3)(2x-3)\)

\(⇔5x-6=2x-3\)

\(⇔5x-2x=-3+6\)

\(⇔3x=3\)

\(⇔x=1\)

Vậy pt có 1 nghiệm là \(x=1\)

c.

ĐKXĐ: \(\left[{}\begin{matrix}x\le-5\\x\ge6\end{matrix}\right.\)

\(\sqrt{\left(x-3\right)\left(x-5\right)}+\sqrt{\left(x-3\right)\left(x+5\right)}=\sqrt{\left(x-3\right)\left(x-6\right)}\)

- Với \(x\ge6\) , do \(x-3>0\) pt trở thành:

\(\sqrt{x-5}+\sqrt{x+5}=\sqrt{x-6}\)

Do \(\left\{{}\begin{matrix}\sqrt{x-5}>\sqrt{x-6}\\\sqrt{x+5}>0\end{matrix}\right.\) \(\Rightarrow\sqrt{x-5}+\sqrt{x+5}>\sqrt{x-6}\) pt vô nghiệm

- Với \(x\le-5\) pt tương đương:

\(\sqrt{\left(3-x\right)\left(5-x\right)}+\sqrt{\left(3-x\right)\left(-x-5\right)}=\sqrt{\left(3-x\right)\left(6-x\right)}\)

Do \(3-x>0\) pt trở thành:

\(\sqrt{5-x}+\sqrt{-x-5}=\sqrt{6-x}\)

\(\Leftrightarrow-2x+2\sqrt{x^2-25}=6-x\)

\(\Leftrightarrow2\sqrt{x^2-25}=x+6\) (\(x\ge-6\))

\(\Leftrightarrow4\left(x^2-25\right)=x^2+12x+36\)

\(\Leftrightarrow3x^2-12x-136=0\Rightarrow x=\dfrac{6-2\sqrt{111}}{3}\)

a.

Kiểm tra lại đề, pt này không giải được

b.

ĐKXĐ: \(x\ge0\)

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

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

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

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

\(\Leftrightarrow...\)

a. \(3-4x\left(25-2x\right)-8x^2+x-300=0\)

\(\Leftrightarrow3-100x+8x^2-8x^2+x-300=0\)

\(\Leftrightarrow-297-99x=0\)

\(\Leftrightarrow x=3\)

Vậy \(n_0\) của PT là: x=3

b. \(\Leftrightarrow\frac{\left(2-6x\right)}{5}-2+\frac{3x}{10}=7-\frac{3x+3}{4}\)

\(\Leftrightarrow\frac{\left(4-12x\right)}{5}-\frac{20}{10}+\frac{3x}{10}=\frac{\left(28-3x-3\right)}{4}\)

\(\Leftrightarrow\frac{\left(-16-9x\right)}{10}=\frac{\left(25-3x\right)}{4}\)

\(\Leftrightarrow-64-36x=250-30x\)

\(\Leftrightarrow-6x=314\)

\(\Leftrightarrow x=-\frac{157}{3}\)

Vậy -\(n_0\) của PT là: \(x=\frac{-157}{3}\)

c. \(5x+\frac{2}{6}-8x-\frac{1}{3}=4x+\frac{2}{5}-5\)

\(\Leftrightarrow-3x=4x-\frac{23}{5}\)

\(\Leftrightarrow7x=\frac{23}{5}\)

\(\Leftrightarrow x=\frac{23}{35}\)

Vậy \(n_0\) của PT là: \(x=\frac{23}{35}\)

d. \(3x+\frac{2}{3}-3x+\frac{1}{6}=2x+\frac{5}{3}\)

\(\Leftrightarrow\frac{5}{6}=2x+\frac{5}{3}\)

\(\Leftrightarrow x=-\frac{5}{12}\)

Vậy \(n_0\) của Pt là: \(x=-\frac{5}{12}\)

a, \(\sqrt{2x^2-3}=\sqrt{4x-3}\) (x \(\ge\) \(\sqrt{\dfrac{3}{2}}\))

Vì hai vế ko âm, bp 2 vế ta được:

2x² - 3 = 4x - 3

\(\Leftrightarrow\) 2x² = 4x

\(\Leftrightarrow\) x² = 2x

\(\Leftrightarrow\) x² - 2x = 0

\(\Leftrightarrow\) x(x - 2) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(KTM\right)\\x=2\left(TM\right)\end{matrix}\right.\)

Vậy S = {2}

b, \(\sqrt{2x-1}=\sqrt{x-1}\) (x \(\ge\) 1)

Vì hai vế ko âm, bp 2 vế ta được:

2x - 1 = x - 1

\(\Leftrightarrow\) x = 0 (KTM)

Vậy x = \(\varnothing\)

c, \(\sqrt{x^2-x-6}=\sqrt{x-3}\) (x \(\ge\) 3)

Vì hai vế ko âm, bp 2 vế ta được:

x² - x - 6 = x - 3

\(\Leftrightarrow\) x² - 2x - 3 = 0

\(\Leftrightarrow\) x² - 3x + x - 3 = 0

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

\(\Leftrightarrow\) (x - 3)(x + 1) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(TM\right)\\x=-1\left(KTM\right)\end{matrix}\right.\)

Vậy S = {3}

d, \(\sqrt{x^2-x}=\sqrt{3x-5}\) (x \(\ge\) \(\dfrac{5}{3}\))

Vì hai vế ko âm, bp 2 vế ta được:

x² - x = 3x - 5

\(\Leftrightarrow\) x² - 4x + 5 = 0

\(\Leftrightarrow\) x² - 4x + 4 + 1 = 0

\(\Leftrightarrow\) (x - 2)² + 1 = 0

Vì (x - 2)² \(\ge\) 0 với mọi x \(\ge\) \(\dfrac{5}{3}\) \(\Rightarrow\) (x - 2)² + 1 > 0 với mọi x \(\ge\) \(\dfrac{5}{3}\)

\(\Rightarrow\) Pt vô nghiệm

Vậy S = \(\varnothing\)

Chúc bn học tốt!

Nguyễn Lê Phước Thịnh nhờ anh xíu ạ

a: \(\dfrac{3x-7}{2}+\dfrac{x-1}{3}=-16\)

\(\Leftrightarrow3\left(3x-7\right)+2\left(x-1\right)=-96\)

\(\Leftrightarrow9x-21+2x-2=-96\)

=>11x=-73

hay x=-73/11

b: \(x-\dfrac{x-1}{3}=\dfrac{2x+1}{5}\)

=>15x-5(x-1)=3(2x+1)

=>15x-5x+5=6x+3

=>10x+5=6x+3

=>4x=-2

hay x=-1/2

c: \(\dfrac{2x-1}{3}-\dfrac{5x+2}{7}=x+13\)

=>14x-7-15x-6=21(x+13)

=>21x+273=-x-13

=>22x=-286

hay x=13