1,a, \(\left(2x+1\right)\left(4x^2-2x+1\right)-8x\left(x^2+2\right)=17\)

\(\Leftrightarrow8x^3+1-8x^3-16x=17\)

\(\Leftrightarrow-16x=16\)

\(\Leftrightarrow x=-1\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}}\)

2,\(M=x^2+2x+6=x^2+2x+1+5=\left(x+1\right)^2+5\ge5\)

Dấu "=" xảy ra <=> x +  1 = 0

                        <=> x = -1

Vậy \(M_{min}=5\Leftrightarrow x=-1\)

Bài 1:

a) \(A=-\left(2x-5\right)^2+6\left|2x-5\right|+4=-\left[\left(2x-5\right)^2-6\left|2x-5\right|+9\right]+13=-\left(\left|2x-5\right|-3\right)^2+13\le13\)

\(maxA=13\Leftrightarrow\) \(\left[{}\begin{matrix}2x-5=3\\2x-5=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)

b) \(B=-x^2-y^2+2x-6y+9=-\left(x^2-2x+1\right)-\left(y^2+6y+9\right)+19=-\left(x-1\right)^2-\left(y+3\right)^2+19\le19\)

\(maxC=19\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)

Bài 2:

\(A=2\left(x^3-y^3\right)-3\left(x+y\right)^2=2\left(x-y\right)\left(x^2+xy+y^2\right)-3\left(x^2+2xy+y^2\right)=4\left(x^2+xy+y^2\right)-3\left(x^2+2xy+y^2\right)=x^2-2xy+y^2=\left(x-y\right)^2=2^2=4\)

bài 2
\(A=2\left(x-y\right)\left(x^2+xy+y^2\right)-3\left(x^2+2xy+y^2\right)\)
\(A=2.2\left(x^2+xy+y^2\right)-3\left(x^2+2xy+y^2\right)\)
\(A=\left(4x^2+4xy+4y^2\right)+\left(-3x^2-6xy-3y^2\right)\)
\(A=x^2-2xy+y^2=\left(x-y\right)^2=2^2=4\)

Câu 2: 

a: (x+3)(y+2)=1

\(\Leftrightarrow\left(x+3;y+2\right)\in\left\{\left(-1;-1\right);\left(1;1\right)\right\}\)

hay \(\left(x,y\right)\in\left\{\left(-4;-3\right);\left(-2;-3\right)\right\}\)

b: (2x-5)(y-6)=17

\(\Leftrightarrow\left(2x-5;y-6\right)\in\left\{\left(1;17\right);\left(17;1\right);\left(-1;-17\right);\left(-17;-1\right)\right\}\)

hay \(\left(x,y\right)\in\left\{\left(3;23\right);\left(11;7\right);\left(2;-11\right);\left(-6;5\right)\right\}\)

c: \(\left(x-1\right)\left(x+y\right)=33\)

\(\Leftrightarrow\left(x-1;x+y\right)\in\left\{\left(1;33\right);\left(33;1\right);\left(-1;-33\right);\left(-33;-1\right);\left(3;11\right);\left(11;3\right);\left(-11;-3\right);\left(-3;-11\right)\right\}\)

hay \(\Leftrightarrow\left(x;x+y\right)\in\left\{\left(2;33\right);\left(34;1\right);\left(0;-33\right);\left(-32;-1\right);\left(4;11\right);\left(12;3\right);\left(-10;-3\right);\left(-2;-11\right)\right\}\)

hay \(\left(x,y\right)\in\left\{\left(2;31\right);\left(34;-33\right);\left(0;-33\right);\left(-32;31\right);\left(4;7\right);\left(12;-9\right);\left(-10;7\right);\left(-2;-9\right)\right\}\)

(x+1)(y+1) = 5

=> x+1 và y + 1 thuộc Ư(5) = {-1;1;-5;5}

ta có bảng :

a.

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

b.

\(\Leftrightarrow2x^2+2x-x-1-2x^2-3x+1=0\)

\(\Leftrightarrow-2x=0\)

\(\Leftrightarrow x=0\)

phân tích các số sau ra thừa số nguyên tố 48;56;84;105;360

bộ bạn tự nghĩ ra đề hả, ko có điều kiện của x,y thì làm sao tìm được

câu b thì ko có kết quả, tìm kiểu gì

a: \(\Leftrightarrow6x^2-2x=6x^2-13\)

=>-2x=-13

hay x=13/2

b: \(\dfrac{x}{3}-\dfrac{2x+1}{6}=\dfrac{x}{6}-x\)

=>2x-2x-1=x-6x

=>-5x=-1

hay x=1/5

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

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

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

=>x=3

\(x+22⋮x+1\)

\(\Rightarrow x+1+21⋮x+1\)

     \(x+1⋮x+1\)

\(\Rightarrow21⋮x+1\)

\(\Rightarrow x+1\inƯ\left(21\right)\)

\(\Rightarrow x+1\in\left\{-1;1;-3;3;-7;7;-21;21\right\}\)

\(\Rightarrow x\in\left\{-2;0;-4;2;-7;6;-22;20\right\}\)