X= 1+1*2.Tìm 3X
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Tìm min:
$F=3x^2+x-2=3(x^2+\frac{x}{3})-2$
$=3[x^2+\frac{x}{3}+(\frac{1}{6})^2]-\frac{25}{12}$
$=3(x+\frac{1}{6})^2-\frac{25}{12}\geq \frac{-25}{12}$
Vậy $F_{\min}=\frac{-25}{12}$. Giá trị này đạt tại $x+\frac{1}{6}=0$
$\Leftrightarrow x=\frac{-1}{6}$
Tìm min
$G=4x^2+2x-1=(2x)^2+2.2x.\frac{1}{2}+(\frac{1}{2})^2-\frac{5}{4}$
$=(2x+\frac{1}{2})^2-\frac{5}{4}\geq 0-\frac{5}{4}=\frac{-5}{4}$ (do $(2x+\frac{1}{2})^2\geq 0$ với mọi $x$)
Vậy $G_{\min}=\frac{-5}{4}$. Giá trị này đạt tại $2x+\frac{1}{2}=0$
$\Leftrightarrow x=\frac{-1}{4}$
d: ta có: \(x^2-4x+4=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=11\end{matrix}\right.\)
a)x.(5-2x)-2x.(1-x)=15
x [ 5 - 2x -2.(1-x) ] = 15
x ( 5 - 2x -2 + 2x ) =15
x . 3 =15
x = 5
b)(3x+2)2+(1+3x).(1-3x)=2
9x2+12x+4+1-9x2=2
12x + 5 = 2
12x = -3
x = -1/4
a) \(x\left(5-2x\right)-2x\left(1-x\right)=15\\ \Leftrightarrow5x-2x^2-2x+2x^2=15\\ \Leftrightarrow3x=15\\ \Leftrightarrow x=5\)
Vậy x = 5 là nghiệm của pt.
b) \(\left(3x+2\right)^2+\left(1+3x\right)\left(1-3x\right)=2\\ \Leftrightarrow\left(9x^2+12x+4\right)+1-9x^2=2\\ \Leftrightarrow12x+5=2\\ \Leftrightarrow12x=-3\\ \Leftrightarrow x=\dfrac{-1}{4}\)
Vậy \(x=-\dfrac{1}{4}\) là nghiệm của pt.
a)x+(x+1)+(x+2)+(x+3)+...+(x+99)+(x+100)=5555
=> 101x +5050 = 5555
=> 101x = 505
=> x = 505 : 101 = 5
Vậy, x = 5
b)1+2+3+4+...+x=820
=> ( x+1) x :2 = 820
=> (x+1)x = 1640
Mà 1640 = 40 . 41
=> x = 40 ( vì {x+1} - x = 1)
Vậy, x = 40
c) 3x+1 = 9.27=243
=> 3x+1 = 35
=>x + 1 = 5
=> x = 4
Vậy, x=4
d) x+2x+3x+...+99x+100x=15150
=> [( 100 + 1) x 100 :2 ] x = 15150
=> 5050x = 15150
=> x = 15150:5050 = 3
Vậy, x =3
e)(x+1)+(x+2)+(x+3)+...+(x+100)=205550
=> 100x + 5050 = 205550
=> 100x = 205550 - 5050= 200500
=> x = 200500 : 100 = 2005
Vậy, x = 2005
f)3x+3x+1+3x+2=351
=> 3x + 3x . 3 + 3x x 9 = 351
=> 3x ( 1+3+9) = 351
=> 3x . 13 = 351
=> 3x = 351 :13=27 mà 27 = 33
=> x=3
Vậy, x=3
a) \(\left(2x-3\right)\left(2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=3\\2x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b) \(\left(x-4\right)\left(x-1\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=1\\x=2\end{matrix}\right.\)
c) \(2x\left(3x-1\right)-3x\left(5+2x\right)=0\)
\(\Rightarrow x\left[2\left(3x-1\right)-3\left(5+2x\right)\right]=0\)
\(\Rightarrow x\left(6x-2-15-6x\right)\)
\(\Rightarrow-16x=0\)
\(\Rightarrow x=0\)
d) \(\left(3x-2\right)\left(3x+2\right)-4\left(x-1\right)=0\)
\(\Rightarrow9x^2-4-4x+4=0\)
\(\Rightarrow9x^2-4x=0\)
\(\Rightarrow x\left(9x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\9x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\end{matrix}\right.\)
\(a,\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ b,\left(x-4\right)\left(x-1\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\\x=2\end{matrix}\right.\)
a) \(\Rightarrow9x^2+24x+16-9x^2+1=49\)
\(\Rightarrow24x=32\Rightarrow x=\dfrac{4}{3}\)
b) \(\Rightarrow x^2-13x+22=0\)
\(\Rightarrow\left(x-11\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=11\\x=2\end{matrix}\right.\)
c) \(\Rightarrow x^2-3x-10=0\)
\(\Rightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
X = 1+1x2= 3
3X = 3.3=9