Tìm x biết
a)2x(x-5)-x(3+2x)=26
b)(4x^3-6x^2-6x):(-2x)-(3-2x)(x+1)=18
Mong mọi người giúp
tìm nghiệm của đa thức f(x)=x4+2x2-2x2-6x-x4+2x2-x3+8x-x3-2
mọi người giúp mình nha, cảm ơn
ta có: f(x) = x4 + 2x2 - 2x2 - 6x - x4 + 2x2 - x3 + 8x -x3 - 2
f(x) = (x4 - x4) + (2x2 + 2x2 -2x2) + (8x-6x) - (x3 + x3 ) - 2
f(x) = 2x2 + 2x - 2x3 - 2 = 2x2- 2x3 + 2x - 2
Để f(x) = 0
=> 2x2 - 2x3 + 2x - 2 = 0
2x2.(x-1) + 2.(x-1) = 0
(x-1).(2x2+2) = 0
=> x - 1 = 0 => x = 1
2x2 + 2 = 0 => 2x2 = -2 => x2 = - 1 => không tìm được x
KL:...
Bài 1 : Tìm giá trị nhỏ nhất của các biểu thức sau :
a, A = x2 + 3x + 4 | d, D = 4x2+ 4x - 24 |
b, B = 2x2 - x + 1 | e, E = x2 + 6x - 11 |
c, C = 5x2 + 2x - 3 | g, G = \(\dfrac{1}{4}x^2+x-\dfrac{1}{3}\) |
MONG MỌI NGƯỜI GIÚP VỚI Ạ !!! EM CẦN GẤP !
a) \(A=x^2+3x+4=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)
\(minA=\dfrac{7}{4}\Leftrightarrow x=-\dfrac{3}{2}\)
b) \(B=2x^2-x+1=2\left(x-\dfrac{1}{4}\right)^2+\dfrac{7}{8}\ge\dfrac{7}{8}\)
\(minB=\dfrac{7}{8}\Leftrightarrow x=\dfrac{1}{4}\)
c) \(C=5x^2+2x-3=5\left(x+\dfrac{1}{5}\right)^2-\dfrac{16}{5}\ge-\dfrac{16}{5}\)
\(minC=-\dfrac{16}{5}\Leftrightarrow x=-\dfrac{1}{5}\)
d) \(D=4x^2+4x-24=\left(2x+1\right)^2-25\ge-25\)
\(minD=-25\Leftrightarrow x=-\dfrac{1}{2}\)
e) \(E=x^2+6x-11=\left(x+3\right)^2-20\ge-20\)
\(minE=-20\Leftrightarrow x=-3\)
f) \(G=\dfrac{1}{4}x^2+x-\dfrac{1}{3}=\left(\dfrac{1}{2}x+1\right)^2-\dfrac{4}{3}\ge-\dfrac{4}{3}\)
\(minG=-\dfrac{4}{3}\Leftrightarrow x=-2\)
a: Ta có: \(A=x^2+3x+4\)
\(=x^2+2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{7}{4}\)
\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{3}{2}\)
d: Ta có: \(D=4x^2+4x-24\)
\(=4x^2+4x+1-25\)
\(=\left(2x+1\right)^2-25\ge-25\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{1}{2}\)
e: ta có: \(E=x^2+6x-11\)
\(=x^2+6x+9-20\)
\(=\left(x+3\right)^2-20\ge-20\forall x\)
Dấu '=' xảy ra khi x=-3
Tìm x,biết:
a)3x(12x - 4) - 9x(4x - 3) = 30
b)x(5 - 2x) + 2x(x - 1) =15
mọi người giúp em lẹ nha sắp đi học r
a) \(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)
\(\Rightarrow36x^2-12x-36x^2+27x=30\)
\(\Rightarrow\left(36x^2-36x^2\right)+\left(-12+27\right)=30\)
\(\Rightarrow0+15x=30\Leftrightarrow x=30:15=2\)
b) \(x\left(5-2x\right)+2x\left(x-1\right)=15\)
\(\Rightarrow5x-2x^2+2x^2-2x=15\)
\(\Rightarrow\left(5x-2x\right)+\left(-2x^2+2x^2\right)=15\)
\(\Rightarrow3x+0=15\Leftrightarrow x=15:3=5\)
a, 3x(12x - 4) - 9x(4x-3) = 30
36x2 - 12x - 36x2 + 27x = 30
- 12x + 27x = 30
15x = 30
x = 2
b, x(5 - 2x) + 2x(x - 1) = 15
5x - 2x2 + 2x2 - 2x = 15
5x - 2x = 15
3x = 15
x = 5
1) 5x^2 = 13x
2) (5x^2 + 3x – 2 )^2 = (4x^2 – 3x – 2 )^2
3) x^3 + 27 + (x + 3)(x – 9) = 0
4) 5x(x – 2000) – x + 2000 = 0
5) 5x(x – 2) – x – 2 = 0
6) 4x(x + 1) = 8( x + 1)
7) x(x – 4) + (x – 4)^2 = 0
8) x^2 – 6x + 8 = 0
9) 9x^2 + 6x – 8 = 0
10) x^3 + x^2 + x + 1 = 0
11) x^3 - x^2 - x + 1 = 0
12) (5 – 2x)(2x + 7) = 4x^2 – 25
13) x(2x - 1) + 1/3 . 2/3x = 0
14) 4(2x + 7) – 9(x + 3)^2 = 0
GIÚP TUI ZỚI MỌI NGƯỜI OIWIII!!!
1,\(5x^2=13x\Leftrightarrow5x^2-13x=0\Leftrightarrow x\left(5x-13\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{13}{5}\end{cases}}\)
2,\(\left(5x^2+3x-2\right)^2=\left(4x^2-3x-2\right)^2\Leftrightarrow\orbr{\begin{cases}5x^2+3x-2=4x^2-3x-2\\5x^2+3x-2=-4x+3x+2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+6x=0\\9x^2-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x\left(x+6\right)=0\\\left(3x\right)^2=2^2\end{cases}\Leftrightarrow}}\orbr{\begin{cases}x=0or-6\\x=-\frac{2}{3}or\frac{2}{3}\end{cases}}\)
3,\(x^3+27+\left(x+3\right)\left(x-9\right)=0\Leftrightarrow\left(x+3\right)\left(x^2+3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+3x+9+x-9\right)=0\Leftrightarrow\left(x+3\right)\left(x^2+4x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x^2+4x=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3\\x\left(x+4\right)=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3\\x=0or-4\end{cases}}\)
4,\(5x\left(x-2000\right)-x+2000=0\Leftrightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\)
\(\Leftrightarrow\left(x-2000\right)\left(5x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=2000\\x=\frac{1}{5}\end{cases}}\)
5,\(5x\left(x-2\right)-x+2=0\Leftrightarrow5x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(5x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x-2=0\\5x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=\frac{1}{5}\end{cases}}\)
6,\(4x\left(x+1\right)=8\left(x+1\right)\Leftrightarrow4x\left(x+1\right)-8\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(4x-8\right)=0\Leftrightarrow\orbr{\begin{cases}x+1=0\\4x-8=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
7,\(x\left(x-4\right)+\left(x-4\right)^2=0\Leftrightarrow\left(x-4\right)\left(2x-4\right)=0\Leftrightarrow\orbr{\begin{cases}x-4=0\\2x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=2\end{cases}}\)
tí làm nửa kia
8,\(x^2-6x+8=0\Leftrightarrow x^2-6x+9-1=0\Leftrightarrow\left(x-3\right)^2-1^2=0\)
\(\Leftrightarrow\left(x-3-1\right)\left(x-3+1\right)=0\Leftrightarrow\left(x-4\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\\x=2\end{cases}}\)
9,\(9x^2+6x-8=0\Leftrightarrow9x^2+6x+1-9=0\Leftrightarrow\left(3x+1\right)^2-3^2=0\)
\(\Leftrightarrow\left(3x+1-3\right)\left(3x+1+3\right)=0\Leftrightarrow\left(3x-2\right)\left(3x+4\right)=0\Leftrightarrow\orbr{\begin{cases}3x-2=0\\3x+4=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{4}{3}\end{cases}}\)
10,\(x^3+x^2+x+1=0\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^2+1=0\end{cases}\Leftrightarrow}x=-1\)
11,\(x^3-x^2-x+1=0\Leftrightarrow\left(x-1\right)\left(x^2-1\right)=0\Leftrightarrow\orbr{\begin{cases}x-1=0\\x^2-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
12,\(\left(5-2x\right)\left(2x+7\right)=4x^2-25\Leftrightarrow\left(5-2x\right)\left(2x+7\right)-4x^2+25=0\)
\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)-\left(5-2x\right)\left(5+2x\right)=0\)
\(\Leftrightarrow\left(5-2x\right)\left(2x+7-5-2x\right)=0\Leftrightarrow\left(5-2x\right).2=0\Leftrightarrow5-2x=0\Leftrightarrow x=\frac{5}{2}\)
13,\(x\left(2x-1\right)+\frac{1}{3}.\frac{2}{3}x=0\Leftrightarrow x\left(2x-1\right)+\frac{2}{9}x=0\)
\(\Leftrightarrow x\left(2x-1+\frac{2}{9}\right)=0\Leftrightarrow x\left(2x-\frac{7}{9}\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\2x=\frac{7}{9}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{7}{18}\end{cases}}\)
14,\(4\left(2x+7\right)-9\left(x+3\right)^2=0\Leftrightarrow8x+28-9x^2-54x-81=0\)
\(\Leftrightarrow-9x^2+\left(8x-54x\right)+\left(28-81\right)=0\Leftrightarrow-9x^2-46x-53=0\)
\(\Leftrightarrow9x^2+46x+53=0\)Ta có : \(\Delta'=\frac{2116}{4}-477=529-477=52\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-23+\sqrt{52}}{9}\\x=\frac{-23-\sqrt{52}}{9}\end{cases}}\)
làm phép chia :
a) (x^4 -2x^3 + 2x -1) : (x^2 - 1)
b) (x^3 -8) : (x^2 + 2x +4)
c) (x^6 - 2x^5 + 2x^4 + 6x^3 - 4x^2)n: 6x^2
d) (-2x^5 + 3x^2 - 4x^3) :2x^2
e) (15x^3 - 10x^2 + x - 2) : (x - 2)
f) (2x^4 - 3x^3 - 3x^2 + 6x - 2) : (x^2 - 2)
b: =x-2
d: \(=-x^3+\dfrac{3}{2}-2x\)
Mọi người ơi giúp mình với, mình cần gấp ạ !
Tìm x : (8x-3)(3x+2)-(4x+7)(x+4)=(2x+1)(5x-1)-33
(8x-3)(3x+2)-(4x+7)(x+4) = (2x+1)(5x-1)-33
(24x2-9x+16x-6)-(4x2+7x+16x+28) = (10x2+5x-2x-1)-33
24x2+7x-6-4x2-23x-28 = 10x2+3x-1-33
20x2-16x-34 = 10x2+3x-34
<=> 20x2-16x = 10x2+3x
2x2-19x=0
2x(x-19)=0
=>\(\left[{}\begin{matrix}2x=0\Rightarrow x=0\\x-19=0\Rightarrow x=19\end{matrix}\right.\)
Không chắc lắm :)
Tìm GTNN của bt
A=2x^2-4x+10
B=2x^2+y^3+2xy+6x+2y+2015
C=(x-1)(x+2)+3x+5
D=4x+3/x^2+1
giúp mk nka 5 tk lun !!^_^
A)\(A=2.x^2-4.x+10\)
\(2A=4.x^2-8x+20\)
\(2A=4.x^2-2.2x.2+2^2+16\)
\(2A=\left(2x-2\right)^2+16\ge16\forall x\)
\(A=8\)
DẤU =XẢY RA KHI \(\left(2x-2\right)^2=0\leftrightarrow x=1\)
VẬY GTNN CỦA A LÀ 8 VỚI x=1
C)\(\left(x-1\right)\left(x+2\right)+3x+5\)
\(C=x^2+2x-x-2+3x+5\)
\(C=x^2+4x+3\)
\(4C=4x^2+16x+12\)
\(4C=4x^2+2.2x.4+4^2-4\)
\(4C=\left(2x+4\right)^2-4\ge-4\forall x\)
\(C=-1\)
DẤU = XẢY RA KHI\(\left(2x+4\right)^2=0\leftrightarrow x=-2\)
VẬY GTNN CỦA C LÀ -1 VỚI X=-2
XIN LỖI MÌNH CHỈ BIẾT LÀM 2 CÂU THÔI
1,chứng tỏ
a,x mũ2 -x+1>0 với mọi x
b,25x mũ2 +10x+2>0 với mọi x
c,3x mũ2+2x+14>0 với mọi x
d,2x mũ2+y mũ2+ 2xy- 2x+2>0 với mọi x
2,tìm giá trị nhỏ nhất của
A=3x mũ2-3x
B=4x mũ 2+4x+3
C=x mũ2+5x-2
D=2x mũ2+6x+7
E=x mũ2+y mũ2-x+6y+10
mk ko viết đc dấu mũ,thông cảm nha,giúp mk vs,hii
1,chứng tỏ
a,x mũ2 -x+1>0 với mọi x
b,25x mũ2 +10x+2>0 với mọi x
c,3x mũ2+2x+14>0 với mọi x
d,2x mũ2+y mũ2+ 2xy- 2x+2>0 với mọi x
2,tìm giá trị nhỏ nhất của
A=3x mũ2-3x
B=4x mũ 2+4x+3
C=x mũ2+5x-2
D=2x mũ2+6x+7
E=x mũ2+y mũ2-x+6y+10
mk ko viết đc dấu mũ,thông cảm nha,giúp mk vs,hii
Bài 1:
a) \(x^2-x+1\)
\(=x^2-x+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0;\forall x\)
b) \(25x^2+10x+2\)
\(=25x^2+10x+1+1\)
\(=\left(5x+1\right)^2+1\ge1>0;\forall x\)
c) \(3x^2+2x+14\)
\(=3x^2+2x+\dfrac{1}{3}+\dfrac{41}{3}\)
\(=\left(\sqrt{3}x+\dfrac{\sqrt{3}}{3}\right)^2+\dfrac{41}{3}\ge\dfrac{41}{3}>0;\forall x\)
d) \(2x^2+y^2-2xy-2x+2\)
\(=x^2+y^2-2xy-2x+x^2+1+1\)
\(=\left(x-y\right)^2+\left(x-1\right)^2+1\ge1>0;\forall x\)
Vậy ...