a, (x-1)³ - x(x-1)² = 5(2-x) - 11(x+2)

b, (x-2)³ + (3x-1)(3x+1) = (x+1)³

c, $\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{5}$$\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{5}$

d, $\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}=\frac{13x+4}{21}$$\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}=\frac{13x+4}{21}$

e, $\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}$

_Giúp với_~

Tìm x :

a)$5\frac{8}{17}:x+\left(-\frac{4}{17}\right):x\; +3\frac{1}{7}:17\frac{1}{3}=\frac{4}{11}$$5\frac{8}{17}:x+\left(-\frac{4}{7}\right)+3\frac{1}{7}:17\frac{1}{3}=\frac{4}{11}$

b) ${\left(x+\frac{1}{5}\right)}^2+\frac{17}{25}=\frac{26}{25}$

giúp mik với

Cho a,b,c khác nhau đôi một và $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$ .Rút gọn các biểu thức sau:

a, $M$$=\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}$

$\mathrm{b,\; N}=\frac{\mathrm{bc}}{a^2+2bc}+\frac{\mathrm{ac}}{b^2+2ac}+\frac{\mathrm{ab}}{c^2+2ab}$

Giải các phương trình sau:

a) $\frac{10}{\left(x+5\right)\left(x-1\right)}+\frac{3}{1-x}=\frac{5}{x+5}$$\frac{10}{\left(x+5\right)\left(x-1\right)}+\frac{3}{1-x}=\frac{5}{x+5}$

b) $\frac{x-1}{x+2}+\frac{x+3}{x-4}=\frac{2}{\left(x-2\right)\left(4-x\right)}$$\frac{x-1}{x+2}+\frac{x+3}{x-4}=\frac{2}{\left(x-2\right)\left(4-x\right)}$

c) $\frac{7x-3}{x-x^3}=\frac{1}{x-1}-\frac{5}{\left(x-1\right)}$$\frac{7x-3}{x-x^3}=\frac{1}{x-1}-\frac{5}{\left(x-1\right)}$

d) $\frac{1}{x+2}+\frac{1}{x+3}=\frac{1}{\left(x+2\right)\left(x+3\right)}$

Cho a,b,c là độ dài 3 cạnh tam giác. Chứng minh rằng

$\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\le \frac{a+b+c}{2abc}$

Bài 1: Giải các phương trình sau:

a. (x - 1)² (3x - 1)= 0

b.$\left(\frac{2x}{3}+4\right)\left(2x-3\right)\left(\frac{x}{2}-1\right)=0$$\left(\frac{2x}{3}+4\right)\left(2x-3\right)\left(\frac{x}{2}-1\right)=0$

c.$\left(x^2-4\right)\left(\frac{x}{4}-\frac{4x+5}{3}\right)=0$

Câu 1: Cho biểu thức: A=$\left(\frac{7}{x-2}+\frac{8}{x+2}\frac{14x}{x^2-4}\right):\left(\frac{1}{x-2}-\frac{1}{x+2}\right)$$\left(\frac{7}{x-2}+\frac{8}{x+2}\frac{14x}{x^2-4}\right):\left(\frac{1}{x-2}-\frac{1}{x+2}\right)$

a) Rút gọn biểu thức A

b) Tìm x để biểu thức A nhận giá trị âm

Giải các phương trình sau:

a. $\frac{9x-0,7}{4}-\frac{5x-1,5}{7}=\frac{7x-1,1}{3}-\frac{5\left(0,4-2x\right)}{6}$$\frac{9x-0,7}{4}-\frac{5x-1,5}{7}=\frac{7x-1,1}{3}-\frac{5\left(0,4-2x\right)}{6}$

b. $\frac{3x-1}{x-1}-\frac{2x+5}{x+3}=1-\frac{4}{\left(x-1\right)\left(x+3\right)}$$\frac{3x-1}{x-1}-\frac{2x+5}{x+3}=1-\frac{4}{\left(x-1\right)\left(x+3\right)}$

c. $\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=-\frac{7}{6\left(x+5\right)}$$\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=-\frac{7}{6\left(x+5\right)}$

d. $\frac{8x^2}{3\left(1-4x^2\right)}=\frac{2x}{6x-3}-\frac{1+8x}{4+8x}$

rút gọn biểu thức

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