(x - 2) (x - 1) (x + 1) (x + 2) = 4

x (-2 - 2 + 1 + 2) = 4

x . 0 = 4

0 = 4

\(\rightarrow\)vô nghiệm

ủa nhầm

x ( -2 - 1 + 1 + 2)

\(\dfrac{x+3}{x}-1< 0\Leftrightarrow\dfrac{x+3-x}{x}< 0\Leftrightarrow\dfrac{3}{x}< 0\Rightarrow x< 0\)

x^2+y^2+z^2≤8→x^2+z^2≤8 vì y^2≥0

->2∣xz∣≤8(BĐT cosi)->-4≤xz≤4.(1)

Ta có -xy-yz=-y(x+z)≥-∣y∣.∣(x+z)∣≥-(y^2+(x+z)^2)/2(BĐT cosi)

Ta có -(y^2+(x+z)^2)/2=-(x^2+y^2+z^2+2xz)/2≥-4-xz (2)

Vậy ta có từ (1) và (2):T=507xz-xy-yz≥507xz-4-xz=506xz-4≥-4.506-4=-2028
Dấu = xảy ra chẳng hạn khi x=2,y=0,z=-2

\(\dfrac{x+1}{98}+\dfrac{x+2}{97}+\dfrac{x+90}{9}+\dfrac{x+84}{15}>-4\\ \Leftrightarrow\left(\dfrac{x+1}{98}+1\right)+\left(\dfrac{x+2}{97}+1\right)+\left(\dfrac{x+90}{9}+1\right)+\left(\dfrac{x+84}{15}+1\right)>0\\ \Leftrightarrow\dfrac{x+99}{98}+\dfrac{x+99}{97}+\dfrac{x+99}{9}+\dfrac{x+99}{15}>0\\ \Leftrightarrow\left(x+99\right)\left(\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{9}+\dfrac{1}{15}\right)>0\)

Vì \(\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{9}+\dfrac{1}{15}>0\Rightarrow x+99>0\Rightarrow x>-99\)

cho tớ hỏi ông  

cái chỗ trên cạnh AB và AC là AB và BC đk

minh cung chua hoc bai nay

tt

liuliu

\(\frac{x}{3}-x+1=\frac{x}{4}-\frac{x+1}{3}\)

\(\Leftrightarrow12.\left(\frac{x}{3}-x+1\right)=12.\left(\frac{x}{4}-\frac{x+1}{3}\right)\)

\(\Leftrightarrow4x-12x+12=3x-4.\left(x+1\right)\)

\(\Leftrightarrow4x-12x+12=3x-4x-4\)

\(\Leftrightarrow4x-12x-3x+4x=\left(-12\right)-4\)

\(\Leftrightarrow\left(-7\right)x=-16\)

\(\Leftrightarrow x=\frac{16}{7}\)

Vậy \(x=\frac{16}{7}\)

\(4x-12x+12=3x-4\left(x+1\right)\Leftrightarrow-8x+12=-x-4\)

\(\Leftrightarrow-7x=-16\Leftrightarrow x=\dfrac{16}{7}\)