Ta có : \(\left(3x-2\right)\left(4x+3\right)=\left(2-3x\right)\left(x-1\right)\)

\(\Leftrightarrow12x^2-8x+9x-6=2x-3x^2-2+3x\)

\(\Leftrightarrow12x^2-8x+9x-6-2x+3x^2+2-3x=0\)

\(\Leftrightarrow15x^2-4x-4=0\)​

\(\Leftrightarrow15x^2-10x+6x-4=0\)

Lỗi :vvvv

\(\Leftrightarrow10x\left(\dfrac{3}{2}x-1\right)+4\left(\dfrac{3}{2}x-1\right)=0\)

\(\Leftrightarrow\left(10x+4\right)\left(\dfrac{3}{2}x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{5}\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy ...

\(\frac{2}{x-2}-\frac{3}{x+2}=\frac{x+1}{x^2-4}\left(x\ne\pm2\right)\)

\(\Leftrightarrow\frac{2}{x-2}-\frac{3}{x+2}-\frac{x+1}{x^2-4}=0\)

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

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

\(\Leftrightarrow\frac{2x+4-3x+6-x-1}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{-2x-9}{\left(x-2\right)\left(x+2\right)}=0\)

=> -2x-9=0

<=> -2x=9

<=> \(x=\frac{-9}{2}\left(tmđk\right)\)

\(x^2-4x-1=0\)

\(\left(x^2-2\cdot x\cdot2+4\right)-5=0\)

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

\(\Rightarrow x-2=\pm\sqrt{5}\)

Tự giải tiếp nha ...

\(x^2-4x-1=0\)

\(\Delta=\left(-4\right)^2-4.\left(-1\right)=20\)

pt có 2 nghiệm

\(x_1=\frac{-4-\sqrt{20}}{2}=-2-\sqrt{5}\)

\(x_2=\frac{-4+\sqrt{20}}{2}=-2+\sqrt{5}\)

Em nghĩ là như vầy ạ:

\(B=\frac{4-x+x+1}{\left(4-x\right)\left(x+1\right)}=\frac{5}{-x^2+3x+4}\) (-1 < x < 4)

Ta có: \(-x^2+3x+4=-\left(x-\frac{3}{2}\right)^2+\frac{25}{4}\le\frac{25}{4}\)

Do đó: \(B=\frac{5}{-x^2+3x+4}\ge\frac{5}{\frac{25}{4}}=\frac{20}{25}=\frac{4}{5}\)

Vậy min B = 4/5 khi x = 3/2 (TMĐK)

1/(x + 1) + 1/(4 - x) ≥ (1 + 1)^2/(x + 1 + 4 - x) = 4/5

\(\left(-3x-2\right)^2+\left(3x+5\right)\left(5-3x\right)=-7\)

\(\Leftrightarrow9x^2+12x+4+15x-9x^2+25-15x=-7\)

\(\Leftrightarrow12x+36=0\Leftrightarrow x=-3\)

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

\(\Leftrightarrow x^3+2x^2+2x+2x^2+4x+4-x\left(x^2-16x+64\right)=16x^2-9\)

\(\Leftrightarrow x^3+4x^2+6x+4-x^3+16x^2-64=16x^2-9\)

\(\Leftrightarrow4x^2+6x-51=0\)

\(\cdot\Delta=6^2-4.4.\left(-51\right)=852\)

Vậy pt có 2 nghiệm phân biệt

\(x_1=\frac{-6+\sqrt{852}}{8}\);\(x_2=\frac{-6-\sqrt{852}}{8}\)

Vì xy = 1 

Suy ra : x , y thuộc Ư(1) = {-1;1}

+ x = -1 và y = -1 thì GTNN của |x + y| = 0 

+ x = 1 và y = 1 thì GTNN của |x + y| = 2

Vậy GTNN của |x + y| = 0 

Banj lam sai roi. Maf minhf cungx biet lam roi

Để B là số nguyên thì x chia hết cho 2x-1

=>2x chia hết cho 2x-1

=>2x-1+1 chia hết cho 2x-1

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

=>\(x\in\left\{1;0\right\}\)

bạn ơi cho mình hỏi tại sao là 2x vậy

(x;y)=(0;0);(2;2)