Rút gọn các biểu thức sau:
a, M=|2x-3|+|x-1| với x>1,59
b, N=|2-x| -3|x+1| với x<-1
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a: Ta có: \(\left(x+5\right)^2-4x\left(2x+3\right)^2-\left(2x-1\right)\left(x+3\right)\left(x-3\right)\)
\(=x^2+10x+25-4x\left(4x^2+12x+9\right)-\left(2x-1\right)\left(x^2-9\right)\)
\(=x^2+10x+25-16x^3-48x^2-36x-2x^3+18x+x^2-9\)
\(=-18x^3-46x^2-8x+16\)
a: Ta có: \(3x\left(2x+1\right)+\left(2x-3\right)\left(x+1\right)\)
\(=6x^2+3x+2x^2+2x-3x-3\)
\(=8x^2+2x-3\)
a: M=A:B
\(=\dfrac{x+\sqrt{x}+10-\sqrt{x}-3}{x-9}\cdot\dfrac{\sqrt{x}-3}{1}=\dfrac{x+7}{\sqrt{x}+3}\)
b: \(M=\dfrac{x-9+16}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{16}{\sqrt{x}+3}\)
=>\(M=\sqrt{x}+3+\dfrac{16}{\sqrt{x}+3}-6>=2\sqrt{16}-6=2\)
Dấu = xảy ra khi (căn x+3)^2=16
=>căn x+3=4
=>x=1
`a)(2x-1)^2+(x+3)^2-5(x-7)(x+7)`
`=4x^2-4x+1+x^2+6x+9-5(x^2-49)`
`=5x^2-5x^2-4x+6x+1+9+245`
`=2x+255`
`b)(x-2)(x^2+2x+4)-(25+x^3)`
`=x^3-8-x^3-25=-33`
Lời giải:
a.
$(2x-1)^2+(x+3)^2-5(x-7)(x+7)$
$=4x^2-4x+1+(x^2+6x+9)-5(x^2-49)$
$=5x^2+2x+10-(5x^2-245)=2x+255$
b.
$(x-2)(x^2+2x+4)-(25+x^3)=(x^3-2^3)-(25+x^3)$
$=-8-25=-33$
\(a,\left(x-5\right)\left(2x+1\right)-2x\left(x-3\right)\\ =x.2x-5.2x+x-5-2x.x-2x.\left(-3\right)\\ =2x^2-10x+x-5-2x^2+6x\\ =2x^2-2x^2-10x+x+6x-5\\ =-3x-5\)
\(b,\left(2+3x\right)\left(2-3x\right)+\left(3x+4\right)^2\\ =\left[2^2-\left(3x\right)^2\right]+\left[\left(3x\right)^2+2.3x.4+4^2\right]\\=4-9x^2+\left(9x^2+24x+16\right)\\ =24x+20\)
a) 2x(x+3) – 3x2(x+2) + x(3x2 + 4x – 6)
= (2x . x + 2x . 3) – (3x2 . x + 3x2 . 2) + (x . 3x2 + x . 4x – x . 6)
= 2x2 + 6x – (3x3 + 6x2) + (3x3 + 4x2 - 6x)
= 2x2 + 6x – 3x3 – 6x2 + 3x3 + 4x2 - 6x
= (– 3x3 + 3x3 ) + (2x2 - 6x2 + 4x2 ) + (6x – 6x)
= 0 + 0 + 0
= 0
b) 3x(2x2 – x) – 2x2(3x+1) + 5(x2 – 1)
= [3x . 2x2 + 3x . (-x)] – (2x2 . 3x + 2x2 . 1) + [5x2 + 5 . (-1)]
= 6x3 – 3x2 – (6x3 +2x2) + 5x2 – 5
= 6x3 – 3x2 – 6x3 - 2x2 + 5x2 – 5
= (6x3 – 6x3 ) + (-3x2 – 2x2 + 5x2) – 5
= 0 + 0 – 5
= - 5
a.
\(B=\dfrac{\sqrt{x}+1+\sqrt{x}\left(\sqrt{x}-1\right)+2\sqrt{x}}{1-x}=\dfrac{\sqrt{x}+1+x-\sqrt{x}+2\sqrt{x}}{1-x}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b.
\(P=\dfrac{B}{A}=\dfrac{x+3}{\sqrt{x}+1}:\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{\left(x+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{x+3}{\sqrt{x}-1}=\dfrac{x-1+4}{\sqrt{x}-1}\)
\(=\sqrt{x}+1+\dfrac{4}{\sqrt{x}-1}\)\(=\sqrt{x}-1+\dfrac{4}{\sqrt{x}-1}+2\)
Theo BĐT AM - GM ta có: \(\sqrt{x}-1+\dfrac{4}{\sqrt{x}-1}\ge2\sqrt{\left(\sqrt{x}-1\right)\dfrac{4}{\sqrt{x}-1}}=4\)
\(\Rightarrow\dfrac{1}{P}\ge6\Rightarrow Min_{\dfrac{1}{P}}=6\)
Dấu "=" xảy ra \(\Leftrightarrow\left(\sqrt{x}-1\right)^2=4\Rightarrow x=9\) (loại trường hợp \(\sqrt{x}-1=-2\))
Vậy GTNN của biểu thức \(\dfrac{1}{P}=6\) khi x = 9.