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16 tháng 9 2018
\(a,A=-x^2-6x-10=-\left(x^2+6x+9\right)-1=-\left(x+3\right)^2-1\le-1\)
Dấu = xảy ra ⇔ x +3 =0 ⇔ x = -3
\(Max_A=-1\text{ ⇔}x=-3\)
\(b,B=12x-4x^2+3=-\left(4x^2-12x+9\right)+12=-\left(2x-3\right)^2+12\le12\)
Dấu = xảy ra \(\Leftrightarrow2x-3=0\Leftrightarrow x=\dfrac{3}{2}\)
\(Max_B=12\text{ ⇔}x=\dfrac{3}{2}\)
\(c,8x-8x^2+3=-8\left(x^2-x+\dfrac{1}{4}\right)+5=-8\left(x-\dfrac{1}{2}\right)^2+5\le5\)
\(d,-x^2-8x+2018-y^2+4y\)
\(=-\left(x^2+8x+16\right)-\left(y^2-4y+4\right)+2038\le2038\)
\(e,-4x^4-12x^2+11=-\left(4x^4+12x^2+9\right)+20=-\left(2x^2+3\right)^2+20\le20\)
\(f,C=x-\dfrac{x^2}{4}\Rightarrow4C=4x-x^2\)\(=-\left(x^2-4x+4\right)+4=-\left(x-2\right)^2+4\)
\(\Rightarrow C=-\dfrac{\left(x-2\right)^2}{4}+1\le1\)
\(g,D=x-\dfrac{9x^2}{25}\Rightarrow25D=-\left(9x^2-25x\right)=-\left(9x^2-2.3x.\dfrac{25}{6}+\dfrac{625}{36}\right)+\dfrac{625}{36}=-\left(3x-\dfrac{25}{6}\right)^2+\dfrac{625}{36}\)
\(\Rightarrow D=\dfrac{-\left(3x-\dfrac{25}{6}\right)^2}{25}+\dfrac{25}{36}\le\dfrac{25}{36}\)
S
0
HT
1
14 tháng 9 2015
( x2 + 8x - 34)2 - ( 3x2 - 8x + 2)2
=> x4 + 8x2 - 342 - 3x4 + 8x2 - 22
=> x4( 1 - 4 ) + ( 8x2 - 8x2 ) + ( 342 - 22 )
-3x4 + 0 + 1152
= -3x4 + 1152
Lk t nhá, Michiel na`h, h pải dùng nik ko có tn
21 tháng 5 2022
a: \(=\dfrac{\left(2\cdot547+1\right)\cdot3}{547\cdot211}-\dfrac{546}{547\cdot211}-\dfrac{4}{547\cdot211}\)
\(=\dfrac{2735}{547\cdot211}=\dfrac{5}{211}\)
b: x=7 nên x+1=8
\(x^{15}-8x^{14}+8x^{13}-8x^{12}+...-8x^2+8x-5\)
\(=x^{15}-x^{14}\left(x+1\right)+x^{13}\left(x+1\right)-x^{12}\left(x+1\right)+...-x^2\left(x+1\right)+x\left(x+1\right)-5\)
\(=x^{15}-x^{15}-x^{14}+x^{14}-...-x^3-x^2+x^2+x-5\)
=x-5=7-5=2
DD
1
15 tháng 12 2017
\(Q=\)\(1+\frac{x+3}{x^2+5x+6}:\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)
\(Q=1+\frac{x+3}{x^2+3x+2x+6}:\left[\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}+\frac{x-2}{\left(x+2\right)\left(x-2\right)}\right]\)
\(Q=1+\frac{x+3}{\left(x+3\right)\left(x+2\right)}:\left[\frac{2}{x-2}-\frac{x}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x+2\right)\left(x-2\right)}\right]\)
\(Q=1+\frac{1}{x+2}:\left[\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x+x-2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(Q=1+\frac{1}{x+2}:\left[\frac{2x+4-2x+2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(Q=1+\frac{1}{x+2}:\frac{6}{\left(x-2\right)\left(x+2\right)}\)
\(Q=1+\frac{1}{x+2}.\frac{\left(x-2\right)\left(x+2\right)}{6}\)
\(Q=1+\frac{x-2}{6}\)
\(Q=\frac{6+x-2}{6}\)
\(Q=\frac{x+4}{6}\)
b) khi \(Q=0\)thì \(\frac{x+4}{6}=0\)
\(\Rightarrow x+4=0\)
\(\Rightarrow x=-4\)
vậy \(x=-4\)khi \(Q=0\)
c) khi \(Q>0\)thì \(\frac{x+4}{6}>0\)
\(\Rightarrow x+4>0\)
\(\Leftrightarrow x>-4\)
vậy \(x>-4\)thì \(Q>0\)
NT
0
AH
Akai Haruma
Giáo viên
29 tháng 6 2021
1. ĐKXĐ: $x\neq 1$
Sửa lại đề 1 chút:
$\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}$
$\Leftrightarrow \frac{x^2+x+1}{(x-1)(x^2+x+1)}-\frac{3x^2}{(x-1)(x^2+x+1)}=\frac{2x(x-1)}{(x-1)(x^2+x+1)}$
$\Leftrightarrow x^2+x+1-3x^2=2x(x-1)$
$\Leftrightarrow 4x^2-3x-1=0$
$\Leftrightarrow (4x+1)(x-1)=0$
Vì $x\neq 1$ nên $x=-\frac{1}{4}$
AH
Akai Haruma
Giáo viên
29 tháng 6 2021
2. ĐKXĐ: $x\neq 0;2$
PT \(\Leftrightarrow \frac{7}{8x}+\frac{5-x}{4x(x-2)}=\frac{x-1}{2x(x-2)}+\frac{1}{8(x-2)}\)
\(\Leftrightarrow \frac{7(x-2)}{8x(x-2)}+\frac{2(5-x)}{8x(x-2)}=\frac{4(x-1)}{8x(x-2)}+\frac{x}{8x(x-2)}\)
\(\Leftrightarrow 7(x-2)+2(5-x)=4(x-1)+x\)
\(\Leftrightarrow 5x-4=5x-4\) (luôn đúng)
Vậy pt có nghiệm $x\in\mathbb{R}$ với $x\neq 0;2$
26 tháng 8 2020
a) ĐKXĐ: \(x\notin\left\{\frac{1}{2};\frac{-1}{2}\right\}\)
Ta có: \(\frac{1+8x}{8x+4}=\frac{2x}{6x-3}-\frac{8x^2}{3-12x^2}\)
\(\Leftrightarrow\frac{8x+1}{4\left(2x+1\right)}=\frac{2x}{3\left(2x-1\right)}+\frac{8x^2}{3\left(4x^2-1\right)}\)
\(\Leftrightarrow\frac{3\left(8x+1\right)\left(2x-1\right)}{12\left(2x+1\right)\left(2x-1\right)}=\frac{2x\cdot4\cdot\left(2x+1\right)}{12\left(2x+1\right)\left(2x-1\right)}+\frac{32x^2}{12\left(2x-1\right)\left(2x+1\right)}\)
Suy ra: \(3\left(8x+1\right)\left(2x-1\right)=8x\left(2x+1\right)+32x^2\)
\(\Leftrightarrow3\left(16x^2-8x+2x-1\right)=16x^2+8x+32x^2\)
\(\Leftrightarrow3\left(16x^2-6x-1\right)=48x^2+8x\)
\(\Leftrightarrow48x^2-18x-3-48x^2-8x=0\)
\(\Leftrightarrow-26x-3=0\)
\(\Leftrightarrow-26x=3\)
hay \(x=-\frac{3}{26}\)
Vậy: \(S=\left\{-\frac{3}{26}\right\}\)
b) Ta có: \(\left(x-2\right)\left(x-3\right)< \left(x-4\right)^2-2\left(x+3\right)\)
\(\Leftrightarrow x^2-5x+6< x^2-8x+16-2x-6\)
\(\Leftrightarrow x^2-5x+6< x^2-10x+10\)
\(\Leftrightarrow x^2-5x+6-x^2+10x-10< 0\)
\(\Leftrightarrow5x-4< 0\)
\(\Leftrightarrow5x< 4\)
hay \(x< \frac{4}{5}\)
Vậy: S={x|\(x< \frac{4}{5}\)}
x2 – 8x – 33
= x2 – 11x + 3x - 33
= (x2 – 11x ) + ( 3x - 33)
= x ( x - 11) + 3(x-11)
= ( x+ 3 ) ( x-11)
TL
x2 - 8x - 33
= x2 - 11x + 3x - 33
= ( x2 - 11x ) + ( 3x - 33 )
= x . ( x - 11) + 3 . ( x - 11 )
= ( x - 11 ) . ( x + 3 )