xét hiệu
\(\dfrac{x^2+y^2+z^2}{3}-\dfrac{\left(x+y+z\right)^2}{9}\ge0\)
<=> \(\dfrac{3\left(x^2+y^2+z^2\right)}{9}-\dfrac{x^2+y^2+z^2+2xy+2yz+2zx}{9}\ge0\)
=> \(3x^2+3y^2+3z^2-x^2-y^2-z^2-2yx-2yz-2xz\ge0\)
<=> \(2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\)
<=>\(\left(x^2-2yx+y^2\right)+\left(y^2-2yz+z^2\right)+\left(x^2-2xz+z^2\right)\ge0\)
<=> (x-y)2 +(y-z)2 +(x-z)2 ≥ 0 (luôn đúng )
=> đpcm
\(\dfrac{\left(3x-2\right)^2}{3}-\dfrac{\left(2x+1\right)^2}{3}\le x\left(x+1\right)\)
bài 1 giải BPT
\(\left(x-5\right)^{2^{ }}-x\left(x-2\right)\le4\)
\(\dfrac{x-3}{4}-\dfrac{x-1}{5}< 2\)
\(3-\dfrac{x-4}{2}>x-3\)
bài 3: giải phương trình
a) \(\dfrac{5x-7
}{3}=\dfrac{5-3x}{2}\)
b) \(\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\)
c) \(\dfrac{7x-1}{6}+2x=\dfrac{16-x}{5}\)
d) \(4\left(0,5-1,5x\right)=-\dfrac{5x-6}{3}\)
Cho a, b, c thuộc R. CM:
1, \(ab\le\left(\dfrac{a+b}{2}\right)^2\le\dfrac{a^2+b^2}{2}\)
2, \(\dfrac{a^3+b^3}{2}\ge\left(\dfrac{a+b}{2}\right)^3\)
3, \(a^4+b^4\ge a^3b+ab^3\)
4, \(a^4+3\ge4a\)
5, \(a^3+b^3+c^3\ge3abc\left(a,b,c>0\right)\)
6, \(a^4+b^4\le\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}\left(a,b\ne0\right)\)
7, \(\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}\ge\dfrac{2}{1+ab}\left(a,b\ge1\right)\)
8, \(\left(a^5+b^5\right)\left(a+b\right)\ge\left(a^4+b^4\right)\left(a^2+b^2\right)\)
Giải phương trình sau:
1) 1+\(\dfrac{2x-5}{x-2}-\dfrac{3x-5}{x-1}=0\)
2)\(\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1\)
3)\(\dfrac{x+1}{x-2}-\dfrac{x-1}{x+2}=\dfrac{2\left(x^2+2\right)}{x^2-4}\)
\(\text{Cho A=}\dfrac{x-2014}{\left(\dfrac{x-2}{x+1}-\dfrac{x+1}{x-2}\right):\left(\dfrac{x-2}{x+1}+\dfrac{x+1}{x-2}\right)}\)
\(\text{Tìm x để A}\ge0\)
1.
\(\dfrac{2x}{x+1}=\dfrac{x^2-x+4}{\left(x-1\right)\left(x-4\right)}\)
Bài 3: cho biểu thức
a) tìm đkxđ, rút gọn p
b) tìm x để p>10
P=\(\dfrac{2\left(x^2+1\right)}{x^2-3x-4}+\dfrac{x}{X+1}-\dfrac{8}{x-4}\)
giải bất phương trình và biểu diễn tập nghiệm:
c) 2x - 8 \(\ge\) 2\(\times\) ( x + 1/2)
d) \(\dfrac{5x^2-3x}{5}+\dfrac{3x+1}{4}< \dfrac{x\left(2x+1\right)}{2}-\dfrac{3}{2}\)
