Tìm GTNN, GTLN
L= -4x + 5 - 16x2
Bài 1) a) (2x+3y)2
b) (25x2-10x+1)
c) (x2-2y)2
d) 16x2-9y2
Bài 2) Tìm GTNN của biểu thức
D= x2+2y2-2xy-6y+2x+2020
Q= 2x2-4xy+y2-4x+6y+10
Tìm X:
a) 16x2-24x+9=25
b) x2+10x+9=0
c) x2-4x-12=0
d) x2-5x-6=0
e) 4x2-3x-1=0
f) x4+4x2-5=0
`a)16x^2-24x+9=25`
`<=>(4x-3)^2=25`
`+)4x-3=5`
`<=>4x=8<=>x=2`
`+)4x-3=-5`
`<=>4x=-2`
`<=>x=-1/2`
`b)x^2+10x+9=0`
`<=>x^2+x+9x+9=0`
`<=>x(x+1)+9(x+1)=0`
`<=>(x+1)(x+9)=0`
`<=>` \(\left[ \begin{array}{l}x=-9\\x=-1\end{array} \right.\)
`c)x^2-4x-12=0`
`<=>x^2+2x-6x-12=0`
`<=>x(x+2)-6(x+2)=0`
`<=>(x+2)(x-6)=0`
`<=>` \(\left[ \begin{array}{l}x=-2\\x=6\end{array} \right.\)
`d)x^2-5x-6=0`
`<=>x^2+x-6x-6=0`
`<=>x(x+1)-6(x+1)=0`
`<=>(x+1)(x-6)=0`
`<=>` \(\left[ \begin{array}{l}x=6\\x=-1\end{array} \right.\)
`e)4x^2-3x-1=0`
`<=>4x^2-4x+x-1=0`
`<=>4x(x-1)+(x-1)=0`
`<=>` \(\left[ \begin{array}{l}x=1\\x=-\dfrac14\end{array} \right.\)
`f)x^4+4x^2-5=0`
`<=>x^4-x^2+5x^2-5=0`
`<=>x^2(x^2-1)+5(x^2-1)=0`
`<=>(x^2-1)(x^2+5)=0`
Vì `x^2+5>=5>0`
`=>x^2-1=0<=>x^2=1`
`<=>` \(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\)
Tìm x, biết:
a) 16x2-(4x-5)2=15 b) (2x+1)(1-2x)+(1-2x)2=18
c) (x-5)2-x(x-4)=9 d) (x-5)2+(x-4)(1-x)=0
a) <=> (4x - 4x + 5)(4x + 4x - 5) = 15 <=> 40x = 15 <=> x = 3/8
a) <=> (4x - 4x + 5)(4x + 4x - 5) = 15 <=> 5(8x-5) = 15
<=> 40x = 40 <=> x = 1
Cái này mới chuẩn
b) (2x+1)(1-2x)+(1-2x)2=18 <=> 1 - 4x2 + 4x2 - 4x + 1 = 18
<=> -4x = 16 <=> x = -4
(4x + 2)(4x – 2) bằng:
A. 4 x 2 + 4
B. 4 x 2 - 4
C. 16 x 2 + 4
D. 16 x 2 - 4
Cho các số x,y,z dương thỏa mãn:
x2 +y2 +z2 = 1. Tìm GTNN của M= 1/16x2 +1/4y2 + 1/z2
\(M=\dfrac{\dfrac{1}{16}}{x^2}+\dfrac{\dfrac{1}{4}}{y^2}+\dfrac{1}{z^2}\ge\dfrac{\left(\dfrac{1}{4}+\dfrac{1}{2}+1\right)^2}{x^2+y^2+z^2}=\dfrac{49}{16}\)
\(M_{min}=\dfrac{49}{16}\) khi \(\left(x;y;z\right)=\left(\dfrac{1}{\sqrt{7}};\dfrac{2}{\sqrt{14}};\dfrac{2}{\sqrt{7}}\right)\)
Cho các số x,y,z dương thỏa mãn:
x2 +y2 +z2 = 7/4. Tìm GTNN của M= 1/16x2 +1/4y2 + 1/z2
\(M=\dfrac{\dfrac{1}{16}}{x^2}+\dfrac{\dfrac{1}{4}}{y^2}+\dfrac{1}{z^2}\ge\dfrac{\left(\dfrac{1}{4}+\dfrac{1}{2}+1\right)^2}{x^2+y^2+z^2}=\dfrac{7}{4}\)
\(M_{min}=\dfrac{7}{4}\) khi \(\left(x;y;z\right)=\left(\dfrac{1}{2};\dfrac{1}{\sqrt{2}};1\right)\)
a)4x2-4y2-20x+20y
b)16x2-25+(4x-5)
c)(x+5y)3
e)x2+4x+4-y2
g)x2-3x-4
\(a,4x^2-4y^2-20x+20y=4\left(x^2-y^2\right)-\left(20x-20y\right)=4\left(x-y\right)\left(x+y\right)-20\left(x-y\right)=\left(x-y\right)\left(4x+4y-20\right)=4\left(x-y\right)\left(x+y-5\right)\\ b,16x^2-25+\left(4x-5\right)=\left(4x-5\right)\left(4x+5\right)+\left(4x-5\right)=\left(4x-5\right)\left(4x+5+1\right)=\left(4x-5\right)\left(4x+6\right)=2\left(4x-5\right)\left(2x+3\right)\)
\(c,\left(x+5y\right)^3=x^3+15x^2y+75xy^2+125y^3\\ e,x^2-4x+4-y^2=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\\ g,x^2-3x-4=\left(x^2-4x\right)+\left(x-4\right)=x\left(x-4\right)+\left(x-4\right)=\left(x+1\right)\left(x-4\right)\)
rút gọn biểu thức c,C=(5x+2)3+(5x-2)3-2(x-2)(x+2)
d,D=(4x-3)(16x2+12x+9)-(4x+3)(16x2-12x+9)
c: C=125x^3+150x^2+60x+8+125x^3-150x^2+60x-8-2(x^2-4)
=250x^3+120x-2x^2+8
=250x^3-2x^2+120x+8
d: D=(4x)^3-3^3-(4x)^3-3^3
=64x^3-27-64x^3-27
=-54
c) \(C=\left(5x+2\right)^3+\left(5x-2\right)^3-2\left(x-2\right)\left(x+2\right)\)
\(=\left[\left(5x\right)^3+3\cdot\left(5x\right)^2\cdot2+3\cdot5x\cdot2^2+2^3\right]+\left[\left(5x\right)^3-3\cdot\left(5x\right)^2\cdot2+3\cdot5x\cdot2^2-2^3\right]-2\left(x^2-4\right)\)
\(=125x^3+150x^2+60x+8+125x^3-150x^2+60x-8-2x^2+8\)
\(=\left(125x^3+125x^3\right)+\left(150x^2-150x^2-2x^2\right)+\left(60x+60x\right)+\left(8-8+8\right)\)
\(=250x^3-2x^2+120x+8\)
d) \(D=\left(4x-3\right)\left(16x^2+12x+9\right)-\left(4x+3\right)\left(16x^2-12x+9\right)\)
\(=\left(4x\right)^3-3^3-\left[\left(4x\right)^3+3^3\right]\)
\(=64x^3-27-\left(64x^3+27\right)\)
\(=64x^3-27-64x^3-27\)
\(=-27-27\)
\(=-54\)
Tìm a sao cho biểu thức A chia hết cho B(tìm a sao cho A:B ∈ Z)
1)A=x3-3x2-ax+3;B=x-1
2)A=3x3-16x2+25x+a;B=x2-4x+3
3)A=x4-x3+6x2-x+a;B=x2-x+5
\(1,A⋮B\Leftrightarrow x^3-3x^2-ax+3=\left(x-1\right)\cdot a\left(x\right)\)
Thay \(x=1\)
\(\Leftrightarrow1-3-a+3=0\\ \Leftrightarrow a=1\)
\(2,A⋮B\Leftrightarrow3x^3-16x^2+25x+a=\left(x^2-4x+3\right)\cdot b\left(x\right)\\ \Leftrightarrow3x^3-16x^2+25x+a=\left(x-3\right)\left(x-1\right)\cdot b\left(x\right)\)
Thay \(x=1\)
\(\Leftrightarrow3-16+25+a=0\\ \Leftrightarrow a=-12\)
Thay \(x=3\)
\(\Leftrightarrow3\cdot27-16\cdot9+25\cdot3+a=0\\ \Leftrightarrow81-144+75+a=0\\ \Leftrightarrow12+a=0\Leftrightarrow a=-12\)
Vậy \(a=-12\)