Bài 1:tìm x ,biết:

a) (2x - 1)(3x + 2) - 6x(x + 1) = 0

\(\Leftrightarrow6x^2+x-2-6x^2-6x=0\)

\(\Leftrightarrow-5x=2\)

\(\Leftrightarrow x=\frac{-2}{5}\)

b) \(\left(4x-1\right)^2-\left(2x+1\right)\left(8x-3\right)=0\)

\(\Leftrightarrow16x^2-8x+1-16x^2-2x+3=0\)

\(\Leftrightarrow-10x=-4\)

\(\Leftrightarrow x=\frac{2}{5}\)

c) \(4x^2-1=2\left(2x+1\right)\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{2}\end{cases}}\)

2a) \(4x^2-9y^2-6y-1=4x^2-\left(3y+1\right)^2\)

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

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

\(=1.\left(2x-1\right)\)

c) \(x^2-8x-4y^2+16=\left(x-4\right)^2-4y^2\)

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

d) \(9x^2-12x-y^2+4=\left(3x-2\right)^2-y^2\)

\(=\left(3x-2-y\right)\left(3x-2+y\right)\)

e) \(4x^2+10x-5=4x^2+2.2.\frac{5}{2}x+\frac{25}{4}-\frac{25}{4}-5\)

\(=\left(2x+\frac{5}{2}\right)^2-\frac{45}{4}\)

\(=\left(2x+\frac{5+3\sqrt{5}}{2}\right)\left(2x+\frac{5-3\sqrt{5}}{2}\right)\)

\(x^2\left(x-2\right)^2-\left(x-2\right)^2-x^2+1\)

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

​\(=\left[\left(x-2\right)^2-1\right]\left(x^2-1\right)\)​

Làm tiếp cho chắc nhé

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

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

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

\(S=\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\)

\(S=\left(a^2+ab+bc+ac\right)\left(b^2+ab+bc+ac\right)\left(c^2+ab+bc+ac\right)\)

\(S=\left(a+b\right)\left(a+c\right)\left(b+c\right)\left(b+a\right)\left(c+a\right)\left(c+b\right)=\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2\)

là số chính phương (đpcm)

b, Mk đặt số đó là B nhé để làm cái đề thôi !!!( và viết dưới dạng chia hết nhé ngại viết bằng phân số :))thay dấu chia hết thahf phân số nhé 

Để B \(\in Z\)

\(2a+9⋮a+3\)+\(5a+17⋮a+3\)-\(3a⋮a+3\)

\(=2a+9+5a+17-3a⋮a+3\)

\(=4a+26⋮a+3\)

\(=4a+12+14⋮a+3\)

\(=4a+12⋮3+14⋮a+3\)

\(=4\left(a+3\right)⋮a+3+14⋮a+3\)

\(=4+14⋮a+3\in Z\)

\(=\Rightarrow14⋮a+3\in Z\)

\(\Rightarrow14⋮a+3\)

\(\Rightarrow a+3\inƯ\left(14\right)=\left\{\mp1;\mp2;\mp7;\mp14\right\}\)

Ta có bảng 

\(A=\left(2x+5y\right)^2+\left|3x-9\right|+200\)

\(\left(2x+5y\right)^2\ge0;\left|3x-9\right|\ge0\)

\(\Rightarrow\left(2x+5y\right)^2+\left|3x-9\right|\ge0\)

\(\Rightarrow\left(2x+5y\right)^2+\left|3x-9\right|+200\ge200\)

\(\Rightarrow A\ge200\)

dấu "=" xảy ra khi : 

\(\hept{\begin{cases}\left(2x+5y\right)^2=0\\\left|3x-9\right|=0\end{cases}\Rightarrow\hept{\begin{cases}2x+5y=0\\3x-9=0\end{cases}\Rightarrow}\hept{\begin{cases}2x=-5y\\x=3\end{cases}}}\)

=> 2.3 = -5.y

=> -5y = 6

=> y = -6/5

vậy Min A = 200 khi x = 3 và y = -6/5

Ta có: (2x + 5y)² \(\ge\)0 \(\forall\)x; y

|3x - 9| \(\ge\)0 \(\forall\)x

=> (2x + 5y) + |3x - 9| + 200 \(\ge\)200 \(\forall\)x;y

Hay A \(\ge\)200 \(\forall\)x; y

Dấu "=" xảy ra khi : \(\hept{\begin{cases}2x+5y=0\\3x-9=0\end{cases}}\) <=> \(\hept{\begin{cases}5y=-2x\\3x=9\end{cases}}\) <=> \(\hept{\begin{cases}y=-\frac{2}{5}x\\x=3\end{cases}}\) <=> \(\hept{\begin{cases}y=-\frac{6}{5}\\x=3\end{cases}}\)

Vậy A_min = 200 tại x = 3 và y = -6/5