không có bài

\(\text{a) }3x+6=8x+3\)

\(\Leftrightarrow3x-8x=3-6\)

\(\Leftrightarrow-5x=-3\)

\(\Leftrightarrow x=\frac{-3}{-5}=\frac{3}{5}\)

\(\text{Câu b và câu c bạn ghi rõ lại giùm}\)

b) x+2/5 + x+7/3 = 2x+8/15

giúp mình với nhé

Câu b) trên là 8 1/5 chứ đâu phải là 8/15 đâu

1) 3(x + 2) = 5x + 8

<=> 3x + 6 = 5x + 8

<=> 3x + 6 - 5x - 8 = 0

<=> -2x - 2 = 0

<=> -2x = 0 + 2

<=> -2x = 2

<=> x = -1

2) 2(x - 1) = 3(3 + x) + 3

<=> 2x - 2 = 9 + x + 3

<=> 2x - 2 = 12 + x

<=> 2x - 2 - 12 - x = 0

<=> x - 14 = 0

<=> x = 0 + 14

<=> x = 14

3) 5 - (x - 6) = 4(3 - 2x)

<=> 5 - x + 6 = 12 - 8x

<=> 11 - x = 12 - 8x

<=> 11 - x - 12 + 8x = 0

<=> -1 + 7x = 0

<=> 7x = 0 + 1

<=> 7x = 1

<=> x = 1/7

ĐKXĐ; ...

a/ \(P=\frac{x^2}{x+4}\left[\frac{\left(x+4\right)^2}{x}\right]+9=x\left(x+4\right)+9=\left(x+2\right)^2+5\ge5\)

\(P_{min}=5\) khi \(x=-2\)

b/ \(Q=\left(\frac{\left(x+2\right)\left(x^2-2x+4\right).4\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)\left(x-2\right)\left(x+2\right)}-\frac{4x}{x-2}\right).\frac{x\left(x-2\right)^3}{-16}\)

\(=\left(\frac{4\left(x^2-2x+4\right)-4x\left(x-2\right)}{\left(x-2\right)^2}\right).\frac{-x\left(x-2\right)^3}{16}\)

\(=\frac{16}{\left(x-2\right)^2}.\frac{-x\left(x-2\right)^3}{16}=-x\left(x-2\right)=-x^2+2x\)

\(=1-\left(x-1\right)^2\le1\)

\(Q_{max}=1\) khi \(x=1\)

1)

ĐK: \(x,y\neq 0\); \(x+y\neq 0\)

\(\frac{x^2-y^2}{6x^2y^2}: \frac{x+y}{12xy}\)

\(=\frac{x^2-y^2}{6x^2y^2}. \frac{12xy}{x+y}=\frac{(x-y)(x+y).12xy}{6x^2y^2(x+y)}=\frac{2(x-y)}{xy}\)

2) ĐK: \(x\neq \frac{\pm 1}{2}; 0; 1\)

\(\frac{5x}{2x+1}: \frac{3x(x-1)}{4x^2-1}=\frac{5x}{2x+1}.\frac{4x^2-1}{3x(x-1)}\)

\(=\frac{5x(2x-1)(2x+1)}{(2x+1).3x(x-1)}=\frac{5(2x-1)}{3(x-1)}\)

3) ĐK: \(x\neq \frac{\pm 1}{2}; 0\)

\(\left(\frac{2x-1}{2x+1}-\frac{2x-1}{2x+1}\right): \frac{4x}{10x-5}=0: \frac{4x}{10x-5}=0\)

4) ĐK: \(x\neq \frac{\pm 1}{3}\)

\(\frac{2}{9x^2+6x+1}-\frac{3x}{9x^2-1}=\frac{2}{(3x+1)^2}-\frac{3x}{(3x-1)(3x+1)}\)

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

\(=\frac{6x-2-9x^2-3x}{(3x+1)^2(3x-1)}=\frac{-9x^2+3x-2}{(3x-1)(3x+1)^2}\)

5) ĐK: \(x\neq \pm 1; \frac{-7\pm \sqrt{89}}{4}\)

\(\left(\frac{5}{x^2+2x+1}+\frac{2x}{x^2-1}\right): \frac{2x^2+7x-5}{3x-3}\)

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

\(=\frac{5(x-1)+2x(x+1)}{(x-1)(x+1)^2}. \frac{3(x-1)}{2x^2+7x-5}=\frac{2x^2+7x-5}{(x+1)^2(x-1)}.\frac{3(x-1)}{2x^2+7x-5}\)

\(=\frac{3}{(x+1)^2}\)

6) ĐK: \(x\neq \pm 3\); 0

\(\left(\frac{3}{x-3}+\frac{2x}{x^2-9}+\frac{x}{x+3}\right): \frac{2x}{x+3}\)

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

\(=\frac{3(x+3)+2x+x(x-3)}{(x-3)(x+3)}.\frac{x+3}{2x}\)

\(\frac{(x^2+2x+9)(x+3)}{(x-3)(x+3).2x}=\frac{x^2+2x+9}{2x(x-3)}\)

7) ĐK: \(x\neq 2; \pm 3;0\)

\(\left(\frac{3}{x^2-9}+\frac{1}{x^2+3x}-\frac{1}{x^2-3x}\right): \frac{x-2}{2x^2+6x}\)

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

\(=\frac{3x+x-3-(x+3)}{x(x-3)(x+3)}.\frac{2x(x+3)}{x-2}\)

\(=\frac{3x-6}{x(x-3)(x+3)}.\frac{2x(x+3)}{x-2}=\frac{3(x-2).2x(x+3)}{x(x-3)(x+3)(x-2)}=\frac{6}{x-3}\)