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19 tháng 7 2018
1, P=5-8x-x^2
= -(x^2+2*4*x+4^2) +21
=-(x+4)^2+21
Vì (x+4)^2> hoặc= 0 nên -(x+4)< hoặc =0=>P< hoặc bằng 21
=>GTLN của P là 21
2,P=4x-x^2+1
=-(x^2-2*2*x+2^2)+5
=-(x-2)^2+5
Tương tự như câu 1, ta có GTLN của P là 5
AH
Akai Haruma
Giáo viên
17 tháng 10 2018
a)
\(25x^2-9(x+y)^2=(5x)^2-(3x+3y)^2\)
\(=(5x-3x-3y)(5x+3x+3y)=(2x-3y)(8x+3y)\)
b)
\(x^2-x-2=x^2+x-2x-2=x(x+1)-2(x+1)=(x-2)(x+1)\)
c)
\(3x^2-11x+6=3x^2-9x-2x+6\)
\(=3x(x-3)-2(x-3)=(x-3)(3x-2)\)
d)
\(x^2+5x+8\): biểu thức không phân tích được thành nhân tử
AH
Akai Haruma
Giáo viên
17 tháng 10 2018
e)
\(x^2+8x+7=x^2+x+7x+7\)
\(=x(x+1)+7(x+1)=(x+1)(x+7)\)
g)
\(x^2-6x-16=x^2-6x+9-25\)
\(=(x-3)^2-5^2=(x-3-5)(x-2+5)=(x-8)(x+2)\)
h)
\(4x^2-8x+3=4(x^2-2x+1)-1\)
\(=4(x-1)^2-1=(2x-2)^2-1^2=(2x-2-1)(2x-2+1)\)
\(=(2x-3)(2x-1)\)
i)
\(3x^2-11x+6=3x^2-9x-2x+6\)
\(=3x(x-3)-2(x-3)=(3x-2)(x-3)\)
B
22 tháng 8 2017
Ta có: x2 – x – 12 = x2 – x – 16 + 4
= (x2 – 16) – (x – 4)
= (x – 4).(x + 4) – (x – 4)
= (x – 4).(x + 4 – 1)
= (x – 4).(x + 3)
9 tháng 6 2019
Bài 1 :
\(a,\)\(x^3+6x^2+11x+6\)
\(=x^2\left(x+1\right)+5x\left(x+1\right)+6\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+5x+6\right)\)
\(=\left(x+1\right)\left(x+2\right)\left(x+3\right)\)
7 tháng 12 2019
Bài làm
a) \(\frac{4x-5}{8xy}+\frac{5-y}{8xy}=\frac{4x-5+5-y}{8xy}=\frac{4x-y}{8xy}\)
b) \(\frac{4x^2}{x-2}+\frac{3}{x-2}+\frac{19}{2-x}=\frac{4x^2}{x-2}+\frac{3}{x-2}-\frac{19}{x-2}=\frac{4x^2+3-19}{x-2}=\frac{4x^2-16}{x-2}=\frac{2\left(x-2\right)\left(2x+4\right)}{x-2}=2\left(2x+4\right)\)
c) \(\frac{2x^3+5}{x^2-x+1}-\frac{x^3+4}{x^2-x+1}=\frac{2x^3+5-x^3-4}{x^2-x+1}=\frac{2x^2-x^3+1}{x^2-x+1}\)
d) \(\frac{6}{5x-20}-\frac{x-5}{x^2-8x+16}=\frac{6}{5\left(x-4\right)}-\frac{x-5}{\left(x-4\right)^2}=\frac{6\left(x-4\right)}{5\left(x-4\right)^2}-\frac{\left(x-5\right)5}{5\left(x-4\right)^2}=\frac{6x-4-5x+25}{5\left(x-4\right)^2}=\frac{x+21}{5\left(x-4\right)^2}\)
# Học tốt #
16 tháng 8 2020
Lần sau đăng 3 - 4 ý/câu hỏi thôi :V
1/ -x2 + 4x - 5 = -( x2 - 4x + 4 ) - 1 = -( x - 2 )2 - 1
\(-\left(x-2\right)^2\le0\forall x\Rightarrow-\left(x-2\right)^2-1\le-1\)
Đẳng thức xảy ra <=> x - 2 = 0 => x = 2
=> GTLN = -1 <=> x = 2
2/ -x2 + 2x - 7 = -( x2 - 2x + 1 ) - 6 = -( x - 1 )2 - 6
\(-\left(x-1\right)^2\le0\forall x\Rightarrow-\left(x-1\right)^2-6\le-6\)
Đẳng thức xảy ra <=> x - 1 = 0 => x = 1
=> GTLN = -6 <=> x = 1
3/ -x2 - 6x - 10 = -( x2 + 6x + 9 ) - 1 = -( x + 3 )2 - 1
\(-\left(x+3\right)^2\le0\forall x\Rightarrow-\left(x+3\right)^2-1\le-1\)
Đẳng thức xảy ra <=> x + 3 = 0 => x = -3
=> GTLN = -1 <=> x = -3
4/ -x2 + 2x - 2 = -( x2 - 2x + 1 ) - 1 = -( x - 1 )2 - 1
\(-\left(x-1\right)^2\le0\forall x\Rightarrow-\left(x-1\right)^2-1\le-1\)
Đẳng thức xảy ra <=> x - 1 = 0 => x = 1
=> GTLN = -1 <=> x = 1
5/ -9x2 + 24x - 18 = -9( x2 - 8/3x + 16/9 ) - 2 = -9( x - 4/3 )2 - 2
\(-9\left(x-\frac{4}{3}\right)^2\le0\forall x\Rightarrow-9\left(x-\frac{4}{3}\right)^2-2\le-2\)
Đẳng thức xảy ra <=> x - 4/3 = 0 => x = 4/3
=> GTLN = -2 <=> x = 4/3
6/ -4x2 + 4x - 7 = -4( x2 - x + 1/4 ) - 6 = -4( x - 1/2 )2 - 6
\(-4\left(x-\frac{1}{2}\right)^2\le0\forall x\Rightarrow-4\left(x-\frac{1}{2}\right)^2-6\le-6\)
Đẳng thức xảy ra <=> x - 1/2 = 0 => x = 1/2
=> GTLN = -6 <=> x = 1/2
7/ -16x2 + 8x - 2 = -16( x2 - 1/2x + 1/16 ) - 1 = -16( x - 1/4 )2 - 1
\(-16\left(x-\frac{1}{4}\right)^2\le0\forall x\Rightarrow-16\left(x-\frac{1}{4}\right)^2-1\le-1\)
Đẳng thức xảy ra <=> x - 1/4 = 0 => x = 1/4
=> GTLN = -1 <=> x = 1/4
16 tháng 8 2020
8/ -5x2 + 20x - 49 = -5( x2 - 4x + 4 ) - 29 = -5( x - 2 )2 - 29
\(-5\left(x-2\right)^2\le0\forall x\Rightarrow-5\left(x-2\right)^2-29\le-29\)
Đẳng thức xảy ra <=> x - 2 = 0 => x = 2
=> GTLN = -29 <=> x = 2
9/ -x2 + x - 1 = -( x2 - x + 1/4 ) - 3/4 = -( x - 1/2 )2 - 3/4
\(-\left(x-\frac{1}{2}\right)^2\le0\forall x\Rightarrow-\left(x-\frac{1}{2}\right)^2-\frac{3}{4}\le-\frac{3}{4}\)
Đẳng thức xảy ra <=> x - 1/2 = 0 => x = 1/2
=> GTLN = -3/4 <=> x = 1/2
10/ -x2 + 3x - 3 = -( x2 - 3x + 9/4 ) - 3/4 = -( x - 3/2 )2 - 3/4
\(-\left(x-\frac{3}{2}\right)^2\le0\forall x\Rightarrow-\left(x-\frac{3}{2}\right)^2-\frac{3}{4}\le-\frac{3}{4}\)
Đẳng thức xảy ra <=> x - 3/2 = 0 => x = 3/2
=> GTLN = -3/4 <=> x = 3/2
11/ -x2 + 5x - 8 = -( x2 - 5x + 25/4 ) - 7/4 = -( x - 5/2 )2 - 7/4
\(-\left(x-\frac{5}{2}\right)^2\le0\forall x\Rightarrow-\left(x-\frac{5}{2}\right)^2-\frac{7}{4}\le-\frac{7}{4}\)
Đẳng thức xảy ra <=> x - 5/2 = 0 => x = 5/2
=> GTLN = -7/4 <=> x = 5/2
12/ -9x2 + 12x - 5 = -9( x2 - 4/3x + 4/9 ) - 1 = -9( x - 2/3 )2 - 1
\(-9\left(x-\frac{2}{3}\right)^2\le0\forall x\Rightarrow-9\left(x-\frac{2}{3}\right)^2-1\le-1\)
Đẳng thức xảy ra <=> x - 2/3 = 0 => x = 2/3
=> GTLN = -1 <=> x = 2/3
13/ -x2 - 8x - 19 = -( x2 + 8x + 16 ) - 3 = -( x + 4 )2 - 3
\(-\left(x+4\right)^2\le0\forall x\Rightarrow-\left(x+4\right)^2-3\le-3\)
Đẳng thức xảy ra <=> x + 4 = 0 => x = -4
=> GTLN = -3 <=> x = -4
14/ -x2 + 2/3x - 1 = -( x2 - 2/3x + 1/9 ) - 8/9 = -( x - 1/3 )2 - 8/9
\(-\left(x-\frac{1}{3}\right)^2\le0\forall x\Rightarrow-\left(x-\frac{1}{3}\right)^2-\frac{8}{9}\le-\frac{8}{9}\)
Đẳng thức xảy ra <=> x - 1/3 = 0 => x = 1/3
=> GTLN = -8/9 <=> x = 1/3
Mệt :)
9 tháng 1 2017
a) A=\(\frac{x+1}{6x^3-6x^2}-\frac{x-2}{8x^3-8x}=\frac{x+1}{6x^2\left(x-1\right)}-\frac{x-2}{8x\left(x-1\right)\left(x+1\right)}=\frac{4\left(x+1\right)^2-3x\left(x-2\right)}{24x^2\left(x-1\right)\left(x+1\right)}=\frac{4x^2+8x+4-3x^2+6x}{24x^2\left(x-1\right)\left(x+1\right)}=\frac{x^2+14x+10}{24x^2\left(x-1\right)\left(x+1\right)}\)
CM
4 tháng 12 2018
a) \(\left(3x-5\right)\left(2x+3\right)-\left(2x-3\right)\left(3x+7\right)-2x\left(x-4\right)\)
\(=\left(6x^2-x-15\right)-\left(6x^2+5x-21\right)-\left(2x^2-8x\right)\)
\(=6x^2-x-15-6x^2-5x+21-2x^2+8x\)
\(=-2x^2+2x+6\)
\(=-2\left(x^2-x-3\right)\)
b) \(\left(x^2+2\right)^2-\left(x+2\right)\left(x-2\right)\left(x^2+4\right)\)
\(=\left(x^2+2\right)^2-\left(x^2-4\right)\left(x^2+4\right)\)
\(=\left(x^2+2\right)^2-\left(x^4-16\right)\)
\(=\left(x^4+4x^2+4\right)-\left(x^4-16\right)\)
\(=x^4+4x^2+4-x^4+16\)
\(=4x^2+20\)
\(=4\left(x^2+5\right)\)
c) \(\left(2x-y\right)^2-2\left(x+3y\right)^2-\left(1+3x\right)\left(3x-1\right)\)
\(=\left(4x^2-4xy+y^2\right)-2\left(x^2+6xy+9y^2\right)-\left(9x^2-1\right)\)
\(=4x^2-4xy+y^2-2x^2-16xy-18y^2-9x^2+1\)
\(=-7x^2-20xy-17y^2+1\)
d) \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)\)
\(=\left(x^6-3x^4+3x^2-1\right)-\left(x^6-1\right)\)
\(=x^6-3x^4+3x^2-1-x^6+1\)
\(=-3x^4+3x^2\)
\(=-3x^2\left(x^2-1\right)\)
\(=-3x^2\left(x-1\right)\left(x+1\right)\)
e) \(\left(2x-1\right)^2-2\left(4x^2-1\right)+\left(2x+1\right)^2\)
\(=\left(2x-1\right)^2-2\left(2x-1\right)\left(2x+1\right)+\left(2x+1\right)^2\)
\(=\left[\left(2x-1\right)-\left(2x+1\right)\right]^2\)
\(=\left(2x-1-2x-1\right)^2\)
\(=\left(-2\right)^2=4\)
g) \(\left(x-y+z\right)^2+\left(y-z\right)^2-2\left(x-y+z\right)\left(z-y\right)\)
\(=\left(x-y+z\right)^2+2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2\)
\(=\left(x-y+z+y+z\right)^2\)
\(=\left(x+2z\right)^2\)
h) \(\left(2x+3\right)^2+\left(2x+5\right)^2-\left(4x+6\right)\left(2x+5\right)\)
\(=\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\)
\(=\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\)
\(=\left(2x+3-2x-5\right)^2\)
\(=\left(-2\right)^2=4\)
i) \(5x^2-\dfrac{10x^3+15x^2-5x}{-5x}-3\left(x+1\right)\)
\(=5x^2-\dfrac{-5x\left(-2x^2-3x+1\right)}{-5x}-3\left(x+1\right)\)
\(=5x^2-\left(-2x^2-3x+1\right)-3\left(x+1\right)\)
\(=5x^2+2x^2+3x-1-3x-3\)
\(=7x^2-4\)
\(x^3-x^2=4x^2-8x+4\)
=> \(x^2\left(x-1\right)=4\left(x^2-2x+1\right)\)
=> \(x^2\left(x-1\right)=4\left(x-1\right)^2\)
=> \(x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
=> \(\left(x-1\right)\left(x^2-4x+4\right)=0\)
=> \(\left(x-1\right)\left(x-2\right)^2=0\)
=> \(x-1=0\) hoặc \(x-2=0\)
=> x = 1 hoặc x = 2
\(x^3-x^2=4x^2-8x+4\)
\(\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\)
\(\Leftrightarrow x^2\left(x-1\right)-\left(4x-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy \(S=\left\{1;2\right\}\)