`a)`
`A=(x+1)(2x-1)`
`=2x^{2}+x-1`
`=2(x^{2}+(1)/(2)x-(1)/(2))`
`=2(x^{2}+(1)/(2)x+(1)/(16)-(9)/(16))`
`=2(x+(1)/(4))^{2}-(9)/(8)>= -9/8` với mọi `x`
Dấu `=` xảy ra khi :
`x+(1)/(4)=0<=>x=-1/4`
Vậy `min=-9/8<=>x=-1/4`
``
`b)`
`(4x+1)(2x-5)`
`=8x^{2}-18x-5`
`=8(x^{2}-(9)/(4)x-(5)/(8))`
`=8(x^{2}-(9)/(4)x+(81)/(64)-(121)/(64))`
`=8(x-(9)/(8))^{2}-(121)/(8)>= -(121)/(8)` với mọi `x`
Dấu `=` xảy ra khi :
`x-(9)/(8)=0<=>x=9/8`
Vậy `min=-121/8<=>x=9/8`
\(A=2x^2+x-1=2\left(x+\dfrac{1}{4}\right)^2-\dfrac{9}{8}\ge-\dfrac{9}{8}\)
\(A_{min}=-\dfrac{9}{8}\) khi \(x=-\dfrac{1}{4}\)
\(B=8x^2-18x-5=8\left(x-\dfrac{9}{8}\right)^2-\dfrac{121}{8}\ge-\dfrac{121}{8}\)
\(B_{min}=-\dfrac{121}{8}\) khi \(x=\dfrac{9}{8}\)
a) \(A=\left(x+1\right)\left(2x-1\right)=2x^2+x-1=2\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{9}{8}=2\left(x+\dfrac{1}{4}\right)^2-\dfrac{9}{8}\)Vì \(2\left(x+\dfrac{1}{4}\right)^2\ge0\Rightarrow2\left(x+\dfrac{1}{4}\right)^2-\dfrac{9}{8}\ge-\dfrac{9}{8}\)
\(ĐTXR\Leftrightarrow x=-\dfrac{1}{4}\)
b) \(B=\left(4x+1\right)\left(2x-5\right)=8x^2-18x-5=8\left(x^2-\dfrac{9}{4}x+\dfrac{81}{64}\right)-\dfrac{121}{8}=8\left(x-\dfrac{9}{8}\right)^2-\dfrac{121}{8}\)
Vì \(8\left(x-\dfrac{9}{8}\right)^2\ge0\Rightarrow8\left(x-\dfrac{9}{8}\right)^2-\dfrac{121}{8}\ge-\dfrac{121}{8}\)
\(ĐTXR\Leftrightarrow x=\dfrac{9}{8}\)
a: Ta có: \(A=\left(x+1\right)\left(2x-1\right)\)
\(=2x^2-x+2x-1\)
\(=2x^2+x-1\)
\(=2\left(x^2+\dfrac{1}{2}x-\dfrac{1}{2}\right)\)
\(=2\left(x^2+2\cdot x\cdot\dfrac{1}{4}+\dfrac{1}{16}-\dfrac{9}{16}\right)\)
\(=2\left(x+\dfrac{1}{4}\right)^2-\dfrac{9}{8}\ge-\dfrac{9}{8}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{1}{4}\)
b: Ta có: \(B=\left(4x+1\right)\left(2x-5\right)\)
\(=8x^2-20x+2x-5\)
\(=8x^2-18x-5\)
\(=8\left(x^2-\dfrac{9}{4}x-\dfrac{5}{8}\right)\)
\(=8\left(x^2-2\cdot x\cdot\dfrac{9}{8}+\dfrac{81}{64}-\dfrac{121}{64}\right)\)
\(=8\left(x-\dfrac{9}{8}\right)^2-\dfrac{121}{8}\ge-\dfrac{121}{8}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{9}{8}\)