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27 tháng 11 2018
\(A=\frac{x^3-3x^2-7x-15}{x^5-x^4-10x^3-38x^2-51x-45}\)
\(=\frac{x^2\left(x-5\right)+2x\left(x-5\right)+3\left(x-5\right)}{x^4\left(x-5\right)+4x^3\left(x-5\right)+10x^2\left(x-5\right)+12x\left(x-5\right)+9\left(x-5\right)}\)
\(=\frac{\left(x-5\right)\left(x^2+2x+3\right)}{\left(x-5\right)\left(x^4+4x^3+10x^2+12x+9\right)}\)
\(=\frac{x^2+2x+3}{x^4+4x^3+10x^2+12x+9}\)
\(=\frac{x^2+2x+3}{\left(x^2\right)^2+2.x^2.2x+\left(2x\right)^2+6x^2+12x+9}\)
\(=\frac{x^2+2x+3}{\left(x^2+2x\right)^2+2.\left(x^2+2x\right).3+3^2}\)
\(=\frac{\left(x^2+2x+3\right)}{\left(x^2+2x+3\right)^2}=\frac{1}{x^2+2x+3}\)
b, \(A=\frac{1}{x^2+2x+3}=\frac{1}{\left(x+1\right)^2+2}\le\frac{1}{2}\forall x\)
Dấu "=" xảy ra khi: \(x+1=0\Rightarrow x=-1\)
Vậy GTLN của A là \(\frac{1}{2}\) khi x = -1
23 tháng 11 2018
a.x2+5x+4
=>x2+4x+x+4
=>x(x+4)+x+4
=>(x+4)(x+1)
b. x2-7x=6
=>x2-x-6x=6
=>x(x-1)-6(x-1)
=>(x-1)(x-6)
c.x2-6x+8
=>x2-2x-4x+8
=>x(x-2)-4(x-2)
=>(x-2)(x-4)
d.x2-10x+21
=>x2-3x-7x+21
=>x(x-3)-7(x-3)
=>(x-3)(x-7)
e.Bạn xem lại í e đi xem có đúng đề không?
g.x2+2x-15
=>x2+5x-3x-15
=>x(x+5)-3(x+5)
=>(x-5)(x-3)
h.x2-3x-28
=>x2+4x-7x-28
=>x(x+4)-7(x+4)
=>(x+4)(x-7)
Nhớ tick cho mình nhé
TT
23 tháng 11 2018
1, \(x^2+5x+4\)
\(=x^2+x+4x+4\)
\(=x\left(x+1\right)+4\left(x+1\right)\)
\(=\left(x+1\right).\left(x+4\right)\)
2, \(x^2-7x+6\)
\(=x^2-x-6x+6\)
\(=x\left(x-1\right)-6\left(x-1\right)\)
\(=\left(x+1\right)\left(x-6\right)\)
3, \(x^2-6x+8\)
\(=x^2-2x-4x+8\)
\(=x\left(x-2\right)-4\left(x-2\right)\)
\(=\left(x-2\right)\left(x-4\right)\)
4, \(x^2-10x+21\)
\(=x^2-3x-7x+21\)
\(=x\left(x-3\right)-7\left(x-3\right)\)
\(=\left(x-3\right)\left(x-7\right)\)
5, \(x^2-2x+8\)
\(=x^2+2x-4x+8\)
\(=x\left(x+2\right)-4\left(x-2\right)\)
\(=x\left(x+2\right)+4\left(x+2\right)\)
\(=\left(x+2\right)\left(x+4\right)\)
6, \(x^2+2x-15\)
\(=x^2+5x-3x-15\)
\(=x\left(x+5\right)-3\left(x+5\right)\)
\(=\left(x+5\right)\left(x-3\right)\)
7, \(x^2-3x-28\)
\(=x^2+4x-7x-28\)
\(=x\left(x+4\right)-7\left(x+4\right)\)
\(=\left(x+4\right)\left(x-7\right)\)
TN
1
28 tháng 1
\(\dfrac{x}{x^2+x+1}=\dfrac{1}{4}\)
=>\(x^2+x+1=4x\)
=>\(x^2-3x+1=0\)
\(\dfrac{x^5-3x^3-10x+12}{x^4+7x^2+15}\)
\(=\dfrac{x^5-3x^4+x^3+3x^4-9x^3+3x^2+5x^3-15x^2+5x+12x^2-36x+12+21x}{x^4+7x^2+15}\)
\(=\dfrac{x^3\left(x^2-3x+1\right)+3x^2\left(x^2-3x+1\right)+5x\left(x^2-3x+1\right)+12\left(x^2-3x+1\right)+21x}{x^4+7x^2+15}\)
\(=\dfrac{21x}{x^4-3x^3+x^2+3x^3-9x^2+3x+15x^2-45x+15+42x}\)
\(=\dfrac{21x}{x^2\left(x^2-3x+1\right)+3x\left(x^2-3x+1\right)+15\left(x^2-3x+1\right)+42x}\)
\(=\dfrac{21x}{42x}=\dfrac{1}{2}\)
OP
29 tháng 1 2022
1.
<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)
2.
<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
3.
<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)
4.
<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)
5.
<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)
6,7. ko đủ điều kiện tìm
10 tháng 4 2020
a, 15x - 6 = 12x + 3
\(\Leftrightarrow\) 15x - 12x = 3 + 6
\(\Leftrightarrow\) 3x = 9
\(\Leftrightarrow\) x = 3
Vậy S = {3}
b, \(\frac{x+2}{2}-\frac{2x-3}{5}=10x+\frac{13}{10}\)
\(\Leftrightarrow\) \(\frac{5\left(x+2\right)}{10}-\frac{2\left(2x-3\right)}{10}=\frac{100x}{10}+\frac{13}{10}\)
\(\Leftrightarrow\) 5(x + 2) - 2(2x - 3) - 100x - 13 = 0
\(\Leftrightarrow\) 5x + 10 - 4x + 6 - 100x - 13 = 0
\(\Leftrightarrow\) -99x + 3 = 0
\(\Leftrightarrow\) x = \(\frac{1}{33}\)
Vậy S = {\(\frac{1}{33}\)}
d, (3x + 2)(4x - 5) = 0
\(\Leftrightarrow\) 3x + 2 = 0 hoặc 4x - 5 = 0
\(\Leftrightarrow\) x = \(\frac{-2}{3}\) và x = \(\frac{5}{4}\)
Vậy S = {\(\frac{-2}{3}\); \(\frac{5}{4}\)}
Phần c với phần e bạn viết vậy mình ko hiểu, bn viết lại đi!
Chúc bn học tốt!!
TN
1 tháng 7 2017
\(A=x^2-10x+26\)
\(=\left(x^2-10x+25\right)+1\)
\(=\left(x-5\right)^2+1\ge1\)
Vậy \(Min_A=1\) khi \(x-5=0\Rightarrow x=5\)
\(B=x^2+7x+10=\left(x^2+7x+\dfrac{49}{4}\right)-\dfrac{9}{4}=\left(x+\dfrac{7}{2}\right)^2-\dfrac{9}{4}\ge\dfrac{-9}{4}\)Vậy \(Min_B=\dfrac{-9}{4}\) khi \(x+\dfrac{7}{2}=0\Rightarrow x=\dfrac{-7}{2}\)
\(C=4x^2+8x+15=4\left(x^2+2x+1\right)+11=4\left(x+1\right)^2+11\ge11\)Vậy \(Min_C=11\) khi \(x+1=0\Rightarrow x=-1\)
\(D=3x^2-7x+20=3\left(x^2-\dfrac{7}{3}x+\dfrac{49}{36}\right)+\dfrac{191}{12}=3\left(x-\dfrac{7}{6}\right)^2+\dfrac{191}{12}\ge\dfrac{191}{12}\)Vậy \(Min_D=\dfrac{191}{12}\) khi \(x-\dfrac{7}{6}=0\Rightarrow x=\dfrac{7}{6}\)
\(E=x^2-4xy+5y^2-22y+8\)
\(=\left(x^2-4xy+4y^2\right)+\left(y^2-22y+121\right)-113\)\(=\left(x-2y\right)^2+\left(y-11\right)^2-113\ge-113\)
Vậy \(Min_E=-113\) khi \(\left[{}\begin{matrix}x-2y=0\\x-11=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}11-2y=0\\x=11\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2y=11\\x=11\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=11\end{matrix}\right.\)
CS
0
8(3x-2)-10x=2(4-7x)+15
\(\Leftrightarrow24x-16-10x=8-14x+15\)
\(\Leftrightarrow14x-16=-14x+23\)
\(\Leftrightarrow14x+14x=23+16\)
\(\Leftrightarrow28x=39\)
\(\Leftrightarrow x=\frac{39}{28}\)
8(3x-2)-10x = 2(4-7x)+15
24x - 16 - 10x = 8 - 14x + 15
14x - 16 = 23 - 14x
14x + 14x = 23 + 16
28x = 39
x = 39:28
x = 39/28