Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,x^2+2x+7\)
\(=x^2+2x+1+6\)
\(=\left(x+1\right)^2+6\)
\(V\text{ì}\left(x+1\right)^2\ge0\)
\(\left(x+1\right)^2+6\ge0+6\)
\(\left(x+1\right)^2+6\ge6\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x+1\right)^2=0\)
\(x+1=0\)
\(x=-1\)
Vậy MinA=6 khi x=-1
b) \(x^2+x+1\)
\(=x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Vì \(\left(x+\dfrac{1}{2}\right)^2\ge0\)
\(\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=0\)
\(x=\dfrac{1}{2}\)
a) <=> \(3x^4-9x^3+9x^2-27x=0\)
<=>\(3x\left(x^3-3x^2+3x-9\right)=0\)
<=>\(3x\left(x-3\right)\left(x^2+3\right)\)=0
<=>x=0 hoặc x=3
b) \(\left(x+3\right)\left(x^2-3x+5\right)-x\left(x+3\right)=0\)
<=>\(\left(x+3\right)\left(x^2-4x+5\right)=0\)
<=>\(\left(x+3\right)\left(\left(x-2\right)^2+1\right)=0\)
=> x=-3
a) 3x4 - 9x3 = -9x2 + 27x
3x4 - 9x3 + 9x2 - 27x = 0
3x(x3 - 3x2 + 3x - 9) = 0
3x[x2(x - 3) + 3(x - 3)] = 0
3x(x - 3)(x2 + 3) = 0
vì x2 + 3 > 0 nên:
3x = 0 hoặc x - 3 = 0
x = 0 : 3 x = 0 + 3
x = 0 x = 3
=> x = 0 hoặc x = 3
b) (x + 3)(x2 - 3x + 5) = x2 + 3x
x3 - 3x2 + 5x + 3x2 - 9x = x2 + 3x
x3 - 4x + 15 = x2 + 3x
x3 - 4x + 15 - x2 - 3x = 0
x3 - 7x + 15 - x2 = 0
(x2 - 4x + 5)(x + 3) = 0
vì x2 - 4x + 5 > 0 nên
x + 3 = 0
=> x = -3
a) 2x (x-5) -(x2-10x +25)=0
\(\Leftrightarrow\)2x(x-5)-(x-5)2=0
\(\Leftrightarrow\)(x-5)(2x-x+5)=0
\(\Leftrightarrow\)(x-5)(x+5)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-5=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
b) x2 - 9 +3x(x+3) = 0
\(\Leftrightarrow\)(x2 - 9) +3x(x+3) =0
\(\Leftrightarrow\)(x-3)(x+3)+3x(x+3)=0
\(\Leftrightarrow\)(x+3)(x-3+3x)=0
\(\Leftrightarrow\)(x+3)(4x-3)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x+3=0\\4x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-3\\4x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{3}{4}\end{matrix}\right.\)
c) x3 - 16x = 0
\(\Leftrightarrow\)x(x2-16)=0
\(\Leftrightarrow\)x(x-4)(x+4)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
d) (2x+3)(x-2) - (x2 -4x+4) = 0
\(\Leftrightarrow\)(2x+3)(x-2) -(x-2)2=0
\(\Leftrightarrow\)(x-2)(2x+3-x+2)=0
\(\Leftrightarrow\)(x-2)(x+5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
e) 9x2 -(x2 -2x +1)=0
\(\Leftrightarrow\)(3x)2-(x-1)2=0
\(\Leftrightarrow\)(3x-x+1)(3x+x-1)=0
\(\Leftrightarrow\)(2x+1)(4x-1)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x+1=0\\4x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x=-1\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{1}{4}\end{matrix}\right.\)
f)x3-4x2 -9x +36 = 0
\(\Leftrightarrow\)(x3-9x)-(4x2-36)=0
\(\Leftrightarrow\)x(x2-9)-4(x2-9)=0
\(\Leftrightarrow\)(x-4)(x2-9)=0
\(\Leftrightarrow\)(x-4)(x-3)(x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=4\\x=3\\x=-3\end{matrix}\right.\)
g) 3x - 6 = (x-1).(x-2)
\(\Leftrightarrow\)3(x-2)=(x-1)(x-2)
\(\Leftrightarrow\)x-1=3
\(\Leftrightarrow\)x=4
i) (x-2).(x+2) +(2x+1)2 =-5x.(x-3) =5 (?? đề sao vậy ??)
k) x2 -1 = (x-1).(2x+3)
\(\Leftrightarrow\)(x-1)(x+1)=(x-1)(2x+3)
\(\Leftrightarrow\)x+1=2x+3
\(\Leftrightarrow\)x-2x=3-1
\(\Leftrightarrow\)-x=2
\(\Leftrightarrow\)x=-2
l) (2x-1)2 +(x+3).(x-3) -5x(x-2)=6
\(\Leftrightarrow\)4x2-4x+1+x2-9-5x2+10x=6
\(\Leftrightarrow\)6x-8=6
\(\Leftrightarrow\)6x=14
\(\Leftrightarrow\)x=\(\frac{7}{3}\)
\(A=x^2+12x+36=x^2+12x+36+3=\left(x+6\right)^2+3\ge3\)
Dấu '=' xảy ra khi x=-6
\(B=9x^2-12x+4-4=\left(3x-2\right)^2-4\ge-4\)
Dấu '=' xảy ra khi x=2/3
\(C=-x^2+4x+1\)
\(=-\left(x^2-4x-1\right)=-\left(x^2-4x+4-5\right)\)
\(=-\left(x-2\right)^2+5\le5\forall x\)
Dấu '=' xảy ra khi x=2
a. 5x.(12x+7)-3x.(20x-5)=-150
x=-3
b. ( 2x-1).(3-x)+(x+4).(2x-5)=20
x=43/10
c. 9x2-1+(3x-1)2=0
x=1/3
d. 3x.(x-2)-(3x+2).(x-1)=7
x=-5/2
e. (2x-1)2-(2x+5).(2x-5)=20
x=3/2
f. 4x2-5=4
x=3/2
~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~
~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~
a) x2 + 5x + 4
= x2 + x + 4x + 4
= x (x+1) + 4 (x+1)
= (x+1) ( x+4)
c) x2 - 7x + 12
= x2 - 3x - 4x +12
= x(x-3) - 4(x-3)
= (x-3)( x-4)
m) \(5x^2+6x+1\)
\(=5x^2+5x+x+1\)
\(=5x\left(x+1\right)+\left(x+1\right)\)
\(=\left(5x+1\right)\left(x+1\right)\)
câu d nè bạn
\(x^3+9x^2+23x+15=x^3+5x^2+4x^2+20x+3x+15\)
=\(x^2\left(x+5\right)+4x\left(x+5\right)+3\left(x+5\right)\)
=\(\left(x^2+4x+3\right)\left(x+5\right)=\left(x+1\right)\left(x+3\right)\left(x+5\right)\)
câu c nè
\(x^3-6x^2-x+30=\left(x^3-5x^2\right)-\left(x^2-5x\right)-\left(6x-30\right)\)
\(=x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)=\left(x^2-x-6\right)\left(x-5\right)\)
=\(\left(x+2\right)\left(x-3\right)\left(x-5\right)\)
tick rui minh làm tiếp cho
D= 2( \(x^2\)+5x-\(\dfrac{1}{2}\))
D= 2( \(x^2\)+ 2. \(\dfrac{5}{2}\)x + \(\dfrac{25}{4}\)-\(\dfrac{27}{4}\))
D= 2( x+\(\dfrac{5}{2}\))\(^2\)+ \(\dfrac{27}{8}\) lớn hơn hoặc bằng \(\dfrac{27}{8}\)
vậy min P = \(\dfrac{27}{8}\) <=> x = -\(\dfrac{5}{2}\)
e)\(E=5x-x^2=-x^2+5x=-x^2+2\cdot x\cdot\dfrac{5}{2}-\dfrac{25}{4}+\dfrac{25}{4}=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\)
(Vì: \(\left(x-\dfrac{5}{2}\right)^2\ge0\Rightarrow-\left(x-\dfrac{5}{2}\right)^2\le0\))
Vậy \(MaxE=\dfrac{25}{4}\) khi \(x=\dfrac{5}{2}\)
\(A=x^2+9x+56=\left(x+\frac{9}{2}\right)^2+\frac{143}{4}\)
Vì \(\left(x+\frac{9}{2}\right)^2\ge0\forall x\)\(\Rightarrow\left(x+\frac{9}{2}\right)^2+\frac{143}{4}\ge\frac{143}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x+\frac{9}{2}\right)^2=0\Leftrightarrow x=-\frac{9}{2}\)
Vậy minA = 143/4 <=> x = - 9/2
\(B=x^2-2x+15=\left(x-1\right)^2+14\)
Vì \(\left(x-1\right)^2\ge0\)\(\Rightarrow\left(x-1\right)^2+14\ge14\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)
Vậy minB = 14 <=> x = 1
\(C=9x^2-12x=9\left(x-\frac{2}{3}\right)^2-4\)
Vì \(\left(x-\frac{2}{3}\right)^2\ge0\forall x\)\(\Rightarrow9\left(x-\frac{2}{3}\right)^2-4\ge-4\)
Dấu "=" xảy ra \(\Leftrightarrow9\left(x-\frac{2}{3}\right)^2=0\Leftrightarrow x-\frac{2}{3}=0\Leftrightarrow x=\frac{2}{3}\)
Vậy minC = - 4 <=> x = 2/3
Bài 1.
A = x2 + 9x + 56
= ( x2 + 9x + 81/4 ) + 143/4
= ( x + 9/2 )2 + 143/4
( x + 9/2 )2 ≥ 0 ∀ x => ( x + 9/2 )2 + 143/4 ≥ 143/4
Đẳng thức xảy ra <=> x + 9/2 = 0 => x = -9/2
=> MinA = 143/4 <=> x = -9/2
B = x2 - 2x + 15
= ( x2 - 2x + 1 ) + 14
= ( x - 1 )2 + 14
( x - 1 )2 ≥ 0 ∀ x => ( x - 1 )2 + 14 ≥ 14
Đẳng thức xảy ra <=> x - 1 = 0 => x = 1
=> MinB = 14 <=> x = 1
C = 9x2 - 12x
= 9( x2 - 4/3x + 4/9 ) - 4
= 9( x - 2/3 )2 - 4
9( x - 2/3 )2 ≥ 0 ∀ x => 9( x - 2/3 )2 - 4 ≥ -4
Đẳng thức xảy ra <=> x - 2/3 = 0 => x = 2/3
=> MinC = -4 <=> x = 2/3
Bài 2.
D = -9x2 + x
= -9( x2 - 1/9x + 1/324 ) + 1/36
= -9( x - 1/18 )2 + 1/36
-9( x - 1/18 )2 ≤ 0 ∀ x => -9( x - 1/18 )2 + 1/36 ≤ 1/36
Đẳng thức xảy ra <=> x - 1/18 = 0 => x = 1/18
=> MaxD = 1/36 <=> x = 1/18
E = -x2 + 3x - 5
= -( x2 - 3x + 9/4 ) - 11/4
= -( x - 3/2 )2 - 11/4
-( x - 3/2 )2 ≤ 0 ∀ x => -( x - 3/2 )2 - 11/4 ≤ -11/4
Đẳng thức xảy ra <=> x - 3/2 = 0 => x = 3/2
=> MaxE = -11/4 <=> x = 3/2
F = -16x2 - 5x
= -16( x2 + 5/16x + 25/1024 ) + 25/64
= -16( x + 5/32 )2 + 25/64
-16( x + 5/32 )2 ≤ 0 ∀ x => -16( x + 5/32 )2 + 25/64 ≤ 25/64
Đẳng thức xảy ra <=> x + 5/32 = 0 => x = -5/32
=> MaxF = 25/64 <=> x = -5/32