Tìm giá trị nhỏ nhất: \(\dfrac{2\left|x-1\right|+11}{\left|x-1\right|+7}\)
Ta có: \(B-\dfrac{2}{3}=\dfrac{x^2+1}{x^2-x+1}-\dfrac{2}{3}=\dfrac{\left(x-1\right)^2}{3\left(x^2-x+1\right)}=\dfrac{\left(x-1\right)^2}{3\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]}\ge0\Rightarrow B\ge\dfrac{2}{3}\).
Đẳng thức xảy ra khi x = 1.
Vậy Min B = \(\frac{2}{3}\) khi x = 1.
\(\dfrac{7\left(3x^2-1\right)}{1-3x^2}\)
= \(\dfrac{-7\left(3x^2-1\right)}{3x^2-1}\)
= -7
Ta có: \(\dfrac{7\left(3x^2-1\right)}{1-3x^2}\)
\(=\dfrac{-7\cdot\left(1-3x^2\right)}{1-3x^2}\)
=-7
Ta có: \(\dfrac{1-x^2}{x\left(x-1\right)}\)
\(=\dfrac{-\left(x-1\right)\left(x+1\right)}{x\left(x-1\right)}\)
\(=\dfrac{-x-1}{x}\)
Ta có: \(\dfrac{x-1}{x^3}-\dfrac{x+1}{x^3-x^2}+\dfrac{3}{x^3-2x^2+x}\)
\(=\dfrac{x-1}{x^3}-\dfrac{x+1}{x^2\left(x-1\right)}+\dfrac{3}{x\left(x^2-2x+1\right)}\)
\(=\dfrac{\left(x-1\right)^3}{x^3\cdot\left(x-1\right)^2}-\dfrac{x\left(x+1\right)\left(x-1\right)}{x^3\cdot\left(x-1\right)^2}+\dfrac{3x^2}{x^3\cdot\left(x-1\right)^2}\)
\(=\dfrac{x^3-3x^2+3x-1-x\left(x^2-1\right)+3x^2}{x^3\cdot\left(x-1\right)^2}\)
\(=\dfrac{x^3+3x-1-x^3+x}{x^3\cdot\left(x-1\right)^2}\)
\(=\dfrac{4x-1}{x^3\cdot\left(x-1\right)^2}\)
có A = \(a^4-2a^3+3a^2-4a+5\)
\(\Leftrightarrow A=\left(a^2\right)^2-2a^2.a+a^2+2a^2-4a+2+3\)
\(\Leftrightarrow A=\left(a^2-a\right)^2+\left(\sqrt{2}.a-\sqrt{2}\right)^2+3\)
\(\Rightarrow\) A luôn luôn lớn hơn hoặc bằng 3 với mọi giá trị của x
=> giá trị nhỏ nhất của A = 3 khi
( \(\left(a^2-a\right)^2=0\) \(\Leftrightarrow a^2-a=0\Leftrightarrow a\left(a-1\right)=0\) )
\(\Rightarrow\) a= 0 hoặc a= 1
a) Thay \(x=2\) vào phương trình, ta được:
\(15\left(m+6\right)+12=80\) \(\Rightarrow m=-\dfrac{22}{15}\)
Vậy \(m=-\dfrac{22}{15}\)
b) Thay \(x=1\) vào phương trình, ta được:
\(15\left(2+m\right)-32=43\) \(\Rightarrow m=3\)
Vậy \(m=3\)
\(\left\{{}\begin{matrix}a+b+c=3\\a^2+b^2+c^2=3\end{matrix}\right.\)
=> (a+b+c)2= 9
=> ab + bc + ca =3
=>\(a^2+b^2+c^2=ab+ac+cb\)
=>\(2a^2+2b^2+2c^2=2ab+2bc+2ca\)
=> \(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
=>a-b=0, b-c=0, c-a=0 => a=b=c
mà a+b+c=3 suy ra a=b=c=1
Vậy A=\(\dfrac{2^2.2^2.2^2}{4.4.4}=1\)