\(bpt\Leftrightarrow2x^2-x-8< 5-2x+2x^2\\ \Leftrightarrow x< 13\)
Vậy ...
Ta có: x(2x-1)-8<5-2x(1-x)
\(\Leftrightarrow2x^2-x-8< 5-2x+2x^2\)
\(\Leftrightarrow-x+2x< 5+8\)
hay x<13
Vậy: S={x|x<13}
P=\(\left(\dfrac{3\left(x+2\right)}{2x^2+8}-\dfrac{2x^2-x-10}{\left(x+1\right)\left[\left(x+1\right)^2-2x\right]}\right):\left(\dfrac{5}{x^2+1}+\dfrac{3}{2\left(x+1\right)}-\dfrac{3}{x-1}\right)\cdot\dfrac{2}{x-1}\)
a) rút gọn P
b)tìm tất cả các giá trị nguyên của x để P có giá trị là bội của 4
BÀI 6 tìm x
1,\(2x\left(x-5\right)-\left(3x+2x^2\right)=0\) 2,\(x\left(5-2x\right)+2x\left(x-1\right)=13\)
3,\(2x^3\left(2x-3\right)-x^2\left(4x^2-6x+2\right)=0\) 4,\(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
5,\(6x^2-\left(2x-3\right)\left(3x+2\right)=1\) 6,\(2x\left(1-x\right)+5=9-2x^2\)
1\()\)\((\)x-1\()\)\(\left(5x+3\right)\)\(=\)\(\left(3x-8\right)\)\(\left(x-1\right)\)
2/\(\left(2x-1\right)^2\)\(+\)\(\left(2-x\right)\)\(\left(2x-1\right)\)\(=\)0
Tính nhanh :
a) \(\left(x+2\right)^3-x\left(x^2+6x-5\right)-8\)
b) \(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)+3x^2-2x\)
c) \(\left(2x+1\right)\left(4x^2+2x+1\right)-8x\left(x^2+1\right)-5\)
Nhanh nhé ! Ai nhanh và đúng nhất mình tick cho nhé ! Thank you !
Tìm x
a, \(\dfrac{\left(x+2\right)^2}{2}\) + \(\dfrac{\left(1+2x\right)^2}{4}\) + \(\dfrac{\left(1-2x\right)^2}{8}\) – (1 + x)2 = 0
b, \(\dfrac{\left(x+1\right)^2}{2}\) - \(\dfrac{\left(1-2x\right)^2}{3}\) + \(\dfrac{\left(1+2x\right)^2}{4}\) - \(\dfrac{\left(5-x\right)^2}{6}\)= 0
c, (3 + x)3 – 3x2(x + 4) + (x + 2)3 = (1 – x)3 – 8
\(\frac{2x+9}{\left(x+1\right)\left(x+8\right)}-\frac{2x+15}{\left(x+8\right)\left(x+7\right)}+\frac{2x+10}{\left(x+7\right)\left(x+3\right)}=\frac{4}{3}\)
Giải các bất phương trình:
\(a,\left(x-1\right)\left(x+2\right)>\left(x-1\right)^2+3\)
\(b,x\left(2x-1\right)-8< 5-2x\left(1-x\right)\)
Tìm x
a)\(3x\left(2x+1\right)=5\left(2x+1\right)\)
b)\(\left(3x-8\right)^2=\left(2x-7\right)^2\)
c)\(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2=0\)
d)\(\left(9x^2-16\right)^2-4\left(3x+4\right)^2\)
e)\(\left(2x-1\right)\left(4x^2+2x+1\right)=x\left(x-8\right)\)
\(\dfrac{3}{x-5}-\dfrac{x+1}{x\left(x-5\right)}\)
\(\dfrac{8\left(y+2\right)}{3x^2}.\dfrac{15x^5}{4\left(y+2\right)^2}\)
\(\dfrac{8\left(y-1\right)}{3x^2-3}:\dfrac{4\left(y-1\right)^3}{x^2-2x+1}\)
