\(1.x-\dfrac{2}{3}\times\left(x+9\right)=1\)

\(x-\dfrac{2}{3}\times x-6=1\)

\(x\times\left(1-\dfrac{2}{3}\right)=7\)

\(x\times\dfrac{1}{3}=7\)

\(x=21\)

\(2.x-\dfrac{11}{15}=\dfrac{3+x}{5}\)

\(\dfrac{15x}{15}-\dfrac{11}{15}=\dfrac{9+3x}{15}\)

\(15x-11=9+3x\)

\(12x=20\)

\(x=\dfrac{5}{3}\)

a: Ta có: \(\left(x+1\right)^3-\left(x+2\right)\left(x-1\right)^2-3\left(x-3\right)\left(x+3\right)=5\)

\(\Leftrightarrow x^3+3x^2+3x+1-\left(x+2\right)\left(x^2-2x+1\right)-3\left(x^2-9\right)=5\)

\(\Leftrightarrow x^3+3x^2+3x+1-\left(x^3-2x^2+x+2x^2-4x+2\right)-3\left(x^2-9\right)=5\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x-2-3x^2+9=5\)

\(\Leftrightarrow6x=-3\)

hay \(x=-\dfrac{1}{2}\)

b: Ta có: \(\left(x+1\right)^3+\left(x-1\right)^3=\left(x+2\right)^3+\left(x-2\right)^3\)

\(\Leftrightarrow x^3+3x^2+3x+1+x^3-3x^2+3x-1=x^3+6x^2+12x+8+x^3-6x^2+12x-8\)

\(\Leftrightarrow2x^3+6x=2x^3+24x\)

\(\Leftrightarrow x=0\)

c: Ta có: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-1=-10\)

\(\Leftrightarrow12x=-11\)

hay \(x=-\dfrac{11}{12}\)

\(3\left(x-2\right)+4\left(x-1\right)=25\) 

\(\Leftrightarrow3x-6+4x-4=25\) 

\(\Leftrightarrow7x=35\) 

\(\Leftrightarrow x=5\)

\(\left(5x-3\right)\left(x-2\right)=\left(x-1\right)\left(x-2\right)\) 

\(\Leftrightarrow\left(5x-3\right)\left(x-2\right)-\left(x-1\right)\left(x-2\right)=0\) 

\(\Leftrightarrow\left(x-2\right)\left(5x-3-x+1\right)=0\) 

\(\Leftrightarrow\left(x-2\right)\left(4x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow\left(x-2\right)^2-4\left(x-1\right)^2=0\) 

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

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

\(\Leftrightarrow\left(-x\right)\left(3x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\3x-4=0\end{matrix}\right.\) 

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)

1C

2A

1C        2A

1 

\(\left(x-2\right):2.3=6\)

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

\(\Leftrightarrow\left(x-2\right)=4\)

\(\Leftrightarrow x=4+2=6\)

c) ta có

\(\left[\left(2x+1\right)+1\right]m:2=625\)

\(\Leftrightarrow\left[\left(2x+1\right)+1\right]\left\{\left[\left(2x+1\right)-1\right]:2+1\right\}=1250\)

\(\Leftrightarrow\left(2x+1\right)^2+1-1:2+1=1250\)

\(\Leftrightarrow\left(2x+1\right)^2+1-2+1=1250\)

\(\Leftrightarrow\left(2x+1\right)^2+1-2=1249\)

\(\Leftrightarrow\left(2x+1\right)^2+1=1251\)

\(\Leftrightarrow\left(2x+1\right)^2=1250\)

...

2

\(\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{7}{4}-\frac{1}{2}\)

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

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

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

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

\(\Leftrightarrow x=\frac{3}{4}+\frac{1}{2}=\frac{5}{4}\)

 \(C=\left(\dfrac{2x^2+1}{x^3-1}-\dfrac{1}{x-1}\right)\div\left(1-\dfrac{x^2-2}{x^2+x+1}\right)\)

ĐKXĐ: \(x\ne1\)

\(C=[\left(\dfrac{2x^2+1}{(x-1)\left(x^2+x+1\right)}-\dfrac{1}{x-1}\right)]\div\left(1-\dfrac{x^2-2}{x^2+x+1}\right)\)

\(\Leftrightarrow C=[\left(\dfrac{2x^2+1}{(x-1)\left(x^2+x+1\right)}-\dfrac{1\left(x^2+x+1\right)}{(x-1)\left(x^2+x+1\right)}\right)]\div[\dfrac{(x-1)\left(x^2+x+1\right)}{(x-1)\left(x^2+x+1\right)}-\dfrac{(x^2-2)(x-1)}{(x^2+x+1)\left(x-1\right)}]\)

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

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

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

\(a,\Leftrightarrow\left(x+2\right)\left(x+2-x+3\right)=0\\ \Leftrightarrow5\left(x+2\right)=0\Leftrightarrow x=-2\\ b,\Leftrightarrow2x\left(x-1\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\ c,\Leftrightarrow\left(x-1-2x-1\right)\left(x-1+2x+1\right)=0\\ \Leftrightarrow3x\left(-x-2\right)=0\Leftrightarrow-3x\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

d. (x - 3)(x² + 3x + 9) + x(x + 2)(2 - x) = 1

<=> x³ - 9 + (x² + 2x)(2 - x) = 1

<=> x³ - 9 + 2x² - x³ + 4x - 2x² = 1

<=> 4x = 10

<=> x = \(\dfrac{10}{4}=\dfrac{5}{2}\)

d)(x - 3)(x^2 + 3x + 9) + x(x + 2)(2 - x) = 1

\(<=> x^3-27-x(x^2-4)=1\)

\(<=> x^3-27-x^3-4x=1<=>-4x=28<=> x=-7\)

=> ptrình có tập nghiệm S={-7}

e) (x + 1)^3 - (x - 1)^3 - 6(x - 1)^2 = -19

\(<=> x^3+3x^2+3x+1-(x^3-3x^2+3x-1)-6(x^2-2x+1)+19=0\)

\(<=>x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+19=0\)

\(<=>12x=15<=>x=12/15 \)

=> ptrình có tập nghiệm S={12/15}

1/a)Ta có -5=-1*5=-5*1

​Sau đó bn lập bảng rồi dùng ước bội để tính nhé​

​ Tk mình nha bn!