a) \(3\left(x-2\right)+2\left(x-3\right)=5\)

\(\Rightarrow3x-6+2x-6=5\)

\(\Rightarrow5x=17\Rightarrow x=\dfrac{17}{5}\)

b) \(\left(2x-8\right)^2-16=0\)

\(\Rightarrow\left(2x-8-4\right)\left(2x-8+4\right)=0\)

\(\Rightarrow\left(2x-12\right)\left(2x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x=12\\2x=4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=6\\x=2\end{matrix}\right.\)

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

\(\Rightarrow4x^2-4x+1-4x^2+12x-x+3=3\)

\(\Rightarrow7x=-1\Rightarrow x=-\dfrac{1}{7}\)

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

\(\Leftrightarrow3x-6+2x-6=5\)

\(\Leftrightarrow5x=17\)

hay \(x=\dfrac{17}{5}\)

b: Ta có: \(\left(2x-8\right)^2-16=0\)

\(\Leftrightarrow\left(2x-4\right)\left(2x-12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

a. \(3\left(x-2\right)+2\left(x-3\right)=5\)

\(\Leftrightarrow3x-6+2x-6=5\)

\(\Leftrightarrow5x=17\)

\(\Leftrightarrow x=\dfrac{17}{5}\)

b. \(\left(2x-8\right)^2-16=0\)

\(\Leftrightarrow\left(2x-8-4\right)\left(2x-8+4\right)=0\)

\(\Leftrightarrow4\left(x-6\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=2\end{matrix}\right.\)

c. \(\left(2x-1\right)^2-\left(4x+1\right)\left(x-3\right)=3\)

\(\Leftrightarrow4x^2-4x+1-4x^2+11x+3-3=0\)

\(\Leftrightarrow7x+1=0\)

\(\Leftrightarrow x=-\dfrac{1}{7}\)

a) x³-1-(x²+2x)(x-2)=5

⇔ x³-1-x³+4x=5

⇔ 4x=6

⇔ \(x=\dfrac{3}{2}\)

a: (x-2)(y-3)=5

=>\(\left(x-2\right)\cdot\left(y-3\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)

=>\(\left(x-2;y-3\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(3;8\right);\left(7;4\right);\left(1;-2\right);\left(-3;2\right)\right\}\)

b: (2x-1)*(y-4)=-11

=>\(\left(2x-1\right)\cdot\left(y-4\right)=1\cdot\left(-11\right)=\left(-11\right)\cdot1=\left(-1\right)\cdot11=11\cdot\left(-1\right)\)

=>\(\left(2x-1;y-4\right)\in\left\{\left(1;-11\right);\left(-11;1\right);\left(-1;11\right);\left(11;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(1;-7\right);\left(-5;5\right);\left(0;15\right);\left(6;3\right)\right\}\)

c: xy-2x+y=3

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

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

=>\(\left(x+1\right)\cdot\left(y-2\right)=1\cdot1=\left(-1\right)\cdot\left(-1\right)\)

=>\(\left(x+1;y-2\right)\in\left\{\left(1;1\right);\left(-1;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(0;3\right);\left(-2;1\right)\right\}\)

a: \(x\left(x-3\right)+2x-6=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

b: \(\left(x+1\right)^2-4\left(x+1\right)=0\)

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

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

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

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

hay x=5

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

\(\Leftrightarrow2x\left(x-1\right)-\left(x-1\right)^2=0\)

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

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

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

c) Ta có: \(\left(x+3\right)\cdot\left(x^2-3x+9\right)-x\left(x-2\right)^2=19\)

\(\Leftrightarrow x^3+27-x\left(x^2-4x+4\right)-19=0\)

\(\Leftrightarrow x^3+8-x^3+4x^2-4x=0\)

\(\Leftrightarrow4x^2-4x+8=0\)(Vô lý)

a) PT \(\Leftrightarrow x^2-x-x^2+2x=5\) \(\Rightarrow x=5\)

  Vậy ...

b) PT \(\Leftrightarrow8x=16\) \(\Rightarrow x=2\)

  Vậy ...

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

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

hay x=5

b: Ta có: \(2x\left(3x+4\right)-6x^2=16\)

\(\Leftrightarrow6x^2+8x-6x^2=16\)

\(\Leftrightarrow8x=16\)

hay x=2

a: Ta có: \(\left(8x^2-4x\right):\left(-4x\right)-\left(x+2\right)=8\)

\(\Leftrightarrow-2x+1-x-2=8\)

\(\Leftrightarrow-3x=9\)

hay x=-3

b: Ta có: \(\left(2x^4-3x^3+x^2\right):\left(-\dfrac{1}{2}x^2\right)+4\left(x-1\right)^2=0\)

\(\Leftrightarrow-4x^2+6x-2+4x^2-8x+4=0\)

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

hay x=1

e: ta có: \(4x^2+4x-6=2\)

\(\Leftrightarrow4x^2+4x-8=0\)

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

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

f: Ta có: \(2x^2+7x+3=0\)

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

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

a, \(x,y\in Z\Rightarrow\left\{{}\begin{matrix}x-3,2y-6\in Z\\x-3,2y-6\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\end{matrix}\right.\)

Ta có bảng:

Vậy không có x,y thỏa mãn đề bài 

b, tương tự câu a

 \(c,xy-5x+2y=7\\ \Rightarrow x\left(y-5\right)+2y-10=-3\\ \Rightarrow x\left(y-5\right)+2\left(y-5\right)=-3\\ \Rightarrow\left(x+2\right)\left(y-5\right)=-3\)

Rồi làm tương tự câu a

\(d,xy-3x-4y=5\\ \Rightarrow x\left(y-3\right)-4y+12=17\\ \Rightarrow x\left(y-3\right)-4\left(y-3\right)=17\\ \Rightarrow\left(x-4\right)\left(y-3\right)=17\)

Rồi làm tương tự câu a

a: \(\Leftrightarrow2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28\)

=>\(13\sqrt{2x}=28\)

=>căn 2x=28/13

=>2x=784/169

=>x=392/169

b: \(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

=>2*căn x-5=4

=>căn x-5=2

=>x-5=4

=>x=9

c: =>\(\sqrt{x-2}\left(\sqrt{x+2}-1\right)=0\)

=>x-2=0 hoặc x+2=1

=>x=-1 hoặc x=2