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TL
2
29 tháng 6 2017
a, \(3\left(2x-1\right)^2+7\left(3y+5\right)^2=0\)
Ta thấy:
\(3\left(2x-1\right)^2\ge0\) với mọi x
\(7\left(3y+5\right)^2\ge0\) với mọi y
=> \(3\left(2x-1\right)^2+7\left(3y+5\right)^2\ge0\) với mọi x, y
Mà \(3\left(2x-1\right)^2+7\left(3y+5\right)^2=0\)
Dấu '' = '' xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\3y+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=1\\3y=-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-5}{3}\end{matrix}\right.\)
Vậy x = \(\dfrac{1}{2}\) ; y = \(\dfrac{-5}{3}\)
9 tháng 10 2020
1.
a) x4 + x3 + x + 1 = x3( x + 1 ) + ( x + 1 ) = ( x + 1 )( x3 + 1 ) = ( x + 1 )( x + 1 )( x2 - x + 1 ) = ( x + 1 )2( x2 - x + 1 )
b) x2y + xy2 - x - y = xy( x + y ) - ( x + y ) = ( x + y )( xy - 1 )
c) x2 - 2xy + y2 - xz + yz = ( x2 - 2xy + y2 ) - ( xz - yz ) = ( x - y )2 - z( x - y ) = ( x - y )( x - y - z )
d) ax - ab + b - x = ( ax - x ) - ( ab - b ) = x( a - 1 ) - b( a - 1 ) = ( a - 1 )( x - b )
2.
( 2x - 1 )2 - 25 = 0
<=> ( 2x - 1 )2 - 52 = 0
<=> ( 2x - 1 - 5 )( 2x - 1 + 5 ) = 0
<=> ( 2x - 6 )( 2x + 4 ) = 0
<=> \(\orbr{\begin{cases}2x-6=0\\2x+4=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
3x( x - 1 ) + x - 1 = 0
<=> 3x( x - 1 ) + ( x - 1 ) = 0
<=> ( x - 1 )( 3x + 1 ) = 0
<=> \(\orbr{\begin{cases}x-1=0\\3x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{3}\end{cases}}\)
9 tháng 10 2020
B1:
a) \(x^4+x^3+x+1\)
\(=x^3\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+1\right)\)
\(=\left(x+1\right)^2\left(x^2-x+1\right)\)
b) \(x^2y+xy^2-x-y\)
\(=xy\left(x+y\right)-\left(x+y\right)\)
\(=\left(xy-1\right)\left(x+y\right)\)
c) \(x^2-2xy+y^2-xz+yz\)
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-z\right)\)
d) \(ax-ab+b-x\)
\(=a\left(x-b\right)-\left(x-b\right)\)
\(=\left(a-1\right)\left(x-b\right)\)
H9
HT.Phong (9A5)
CTVHS
11 tháng 7 2023
a) \(\left(\dfrac{x^2}{2}+y^2\right)^2\)
\(=\left(\dfrac{1}{2}x^2+y^2\right)^2\)
\(=\left(\dfrac{1}{2}x^2\right)^2+2\cdot\dfrac{1}{2}x^2\cdot y^2+\left(y^2\right)^2\)
\(=\dfrac{1}{4}x^4+x^2y^2+y^4\)
b) \(\left(\dfrac{4}{5}x^2-\dfrac{2}{3}y\right)^2\)
\(=\left(\dfrac{4}{5}x^2\right)^2-2\cdot\dfrac{4}{5}x^2\cdot\dfrac{2}{3}y+\left(\dfrac{2}{3}y\right)^2\)
\(=\dfrac{16}{25}x^4-\dfrac{16}{15}x^2y+\dfrac{4}{9}y^2\)
c) \(\left(2x+\dfrac{1}{2}\right)\left(2x-\dfrac{1}{2}\right)\)
\(=\left(2x\right)^2-\left(\dfrac{1}{2}\right)^2\)
\(=4x^2-\dfrac{1}{4}\)
10 tháng 7 2023
a: (1/2x^2+y^2)^2
=(1/2x^2)^2+2*1/2x^2*y^2+y^4
=1/4x^4+x^2y^2+y^4
b: (4/5x^2-2/3y)^2
=(4/5x^2)^2-2*4/5x^2*2/3y+4/9y^2
=16/25x^4-16/15x^2y+4/9y^2
c: =(2x)^2-(1/2)^2
=4x^2-1/4
15 tháng 2 2020
a. (3x - 1).(2x + 7) - (x + 1).(6x - 5) = 16
<=> 6x^2 + 19x - 7 - (6x^2 + x - 5) = 16
<=> 18x - 2 = 16
<=> 18x = 18
<=> x = 1
b. (10x + 9).x - (5x - 1).(2x + 3) = 8
<=> 10x^2 + 9x - (10x^2 + 13x - 3) = 8
<=> -4x + 3 = 8
<=> -4x = 5
<=> x = -5/4
c. (3x - 5).(7 - 5x) + (5x + 2).(3x - 2) - 2 = 0
<=> -15x^2 + 46x - 35 + 15x^2 - 4x - 4 - 2 = 0
<=> 42x - 41 = 0
<=> x = 41/42
H9
HT.Phong (9A5)
CTVHS
11 tháng 7 2023
d) \(\left(x+2\right)\left(x^2-2x+4\right)\)
\(=\left(x+2\right)\left(x^2-2\cdot x+2^2\right)\)
\(=x^3+2^3\)
\(=x^3+8\)
e) \(\left(\dfrac{1}{4}-\dfrac{x}{5}\right)\left(\dfrac{x^2}{25}+\dfrac{x}{20}+\dfrac{1}{16}\right)\)
\(=\left(\dfrac{1}{4}-\dfrac{1}{5}x\right)\left(\dfrac{1}{25}x^2+\dfrac{1}{5}x\cdot\dfrac{1}{4}+\dfrac{1}{16}\right)\)
\(=\left(\dfrac{1}{4}-\dfrac{1}{5}x\right)\left[\left(\dfrac{1}{5}x\right)^2+\dfrac{1}{5}x\cdot\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2\right]\)
\(=\left(\dfrac{1}{4}\right)^3-\left(\dfrac{1}{5}x\right)^3\)
\(=\dfrac{1}{64}-\dfrac{1}{125}x^3\)
\(=\dfrac{1}{64}-\dfrac{x^3}{125}\)
10 tháng 7 2023
d: (x+2)(x^2-2x+4)
=(x+2)(x^2-x*2+2^2)
=x^3+8
e: (1/4-x/5)(1/16+x/20+x^2/25)
=(1/4-x/5)[(1/4)^2+1/4*x/5+(x/5)^2]
=1/64-x^3/125
N
20 tháng 3 2017
Bài 1:
Áp dụng BĐt cauchy dạng phân thức:
\(\dfrac{1}{2x+y}+\dfrac{1}{x+2y}\ge\dfrac{4}{3\left(x+y\right)}\)
\(\Rightarrow\left(3x+3y\right)\left(\dfrac{1}{2x+y}+\dfrac{1}{x+2y}\right)\ge\left(3x+3y\right).\dfrac{4}{3x+3y}=4\)
dấu = xảy ra khi 2x+y=x+2y <=> x=y
N
20 tháng 3 2017
Bài 2:
ta có: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\ge\dfrac{4^2}{a+b+c+d}=\dfrac{16}{a+b+c+d}\)(theo BĐt cauchy-schwarz)
\(\Rightarrow\dfrac{1}{a+b+c+d}\le\dfrac{1}{16}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right)\)
Áp dụng BĐT trên vào bài toán ta có:
\(A=\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}\left(\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{c}\right)\)\(A\le\dfrac{1}{16}.4\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
......
dấu = xảy ra khi a=b=c
Bài 2:
Áp dụng BĐT cauchy cho 2 số dương:
\(a^2+1\ge2a\)
\(\Leftrightarrow\dfrac{a}{a^2+1}\le\dfrac{a}{2a}=\dfrac{1}{2}\)
thiết lập tương tự:\(\dfrac{b}{b^2+1}\le\dfrac{1}{2};\dfrac{c}{c^2+1}\le\dfrac{1}{2}\)
cả 2 vế các BĐT đều dương ,cộng vế với vế,ta có dpcm
dấu = xảy ra khi a=b=c=1
\(-4+2x< 0\)
\(2x< 4\)
\(x< 2\)