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Bài 1:
a: \(\left(2x-1\right)^4=16\)
=>2x-1=2 hoặc 2x-1=-2
=>2x=3 hoặc 2x=-1
=>x=3/2 hoặc x=-1/2
b: \(\left(2x-y+7\right)^{2012}+\left|x-3\right|^{2013}< =0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-y+7=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2x+7=y=2\cdot3+7=13\end{matrix}\right.\)
c: \(10800=2^4\cdot3^3\cdot5^2\)
mà \(2^{x+2}\cdot3^{x+1}\cdot5^x=10800\)
nên \(\left\{{}\begin{matrix}x+2=4\\x+1=3\\x=2\end{matrix}\right.\Leftrightarrow x=2\)
Bài 1 :
A + B = 4x2 - 5xy + 3y2 + 3x2 + 2xy - y2
= ( 4x2 + 3x2 ) - ( 5xy - 2xy ) + ( 3y2 - y2 )
= 7x2 - 3xy + 2y2
A - B = 4x2 - 5xy + 3y2 - ( 3x2 + 2xy - y2 )
= 4x2 - 5xy + 3y2 - 3x2 - 2xy + y2
= ( 4x2 - 3x2 ) - ( 5xy + 2xy ) + ( 3y2 + y2 )
= x2 - 7xy + 4y2
Bài 2 :
a) M + (5x2 - 2xy) = 6x2 + 9xy - y2
M = 6x2 + 9xy - y2 - (5x2 - 2xy)
M = 6x2 + 9xy - y2 - 5x2 + 2xy
M = ( 6x2 - 5x2 ) + ( 9xy + 2xy ) - y2
M = x2 + 11xy - y2
Vậy M = x2 + 11xy - y2
b) (3xy - 4y2) - N = x2 - 7xy + 8y2
N = 3xy - 4y2 - x2 - 7xy + 8y2
N = ( 3xy - 7xy ) - ( 4y2 - 8y2 ) - x2
N = -4xy + 4y2 - x2
Vậy N = -4xy + 4y2 - x2
3, Cho đa thức
A(x)+B(x) = (3x4-\(\dfrac{3}{4}\)x3+2x2-3)+(8x4+\(\dfrac{1}{5}\)x3-9x+\(\dfrac{2}{5}\))
= 3x4-\(\dfrac{3}{4}\)x3+2x2-3+8x4+\(\dfrac{1}{5}\)x3-9x+\(\dfrac{2}{5}\)
= (3x4+8x4)+(-3/4x3+1/5x3)+(-3+2/5)+2x2-9x
= 11x4 -0.55x3-2.6+2x2-9x
A(x)-B(x)=(3x4-\(\dfrac{3}{4}\)x3+2x2-3)-(8x4+\(\dfrac{1}{5}\)x3-9x+\(\dfrac{2}{5}\))
= 3x4-\(\dfrac{3}{4}\)x3+2x2-3-8x4-\(\dfrac{1}{5}\)x3+9x-\(\dfrac{2}{5}\)
= (3x4-8x4)+(-3/4x3-1/5x3)+(-3-2/5)+2x2+9x
= -5x4-0.95x3-3.4+2x2+9x
B(x)-A(x)=(8x4+\(\dfrac{1}{5}\)x3-9x+\(\dfrac{2}{5}\))-(3x4-\(\dfrac{3}{4}\)x3+2x2-3)
=8x4+\(\dfrac{1}{5}\)x3-9x+\(\dfrac{2}{5}\)-3x4+\(\dfrac{3}{4}\)x3-2x2+3
=(8x4-3x4)+(1/5x3+3/4x3)+(2/5+3)-9x-2x2
= 5x4+0.95x3+2.6-9x-2x2
a ) \(N=\left(x+1\right)^2+\left(y-\sqrt{2}^2\right)+2008\ge0+0+2008=2008\)
=> MinN đạt được bằng 2008 khi
\(\left\{{}\begin{matrix}x=-1\\y=\sqrt{2}\end{matrix}\right.\)
Thay vào M ,ta có
\(3x+\dfrac{x^2-y^2}{x^2+1}=-3+\dfrac{9-2}{1+1}=-3+3,5=0,5\)
b) Với x , y dương , ta được ngay ĐPCM
Với x âm , y âm , ta cũng được ĐPCM
Vậy nên xét trường hợp x,y trái dấu
\(2x^4y^2\ge0\)
\(7x^3y^5\le0\)
\(\Rightarrow2x^4y^2-7x^3y^5\ge0\) ( ĐPCM)
c)
\(2^{x+1}+2^{x+4}+2^{x+5}=2^5\cdot5^2\)
\(\Rightarrow2^{x+1}\left(1+2^3+2^4\right)=2^5\cdot5^2\)
\(\Rightarrow2^{x+1}\cdot5^2=2^5\cdot5^2\)
\(\Rightarrow2^{x+1}=2^5\Rightarrow x=4\)
Ta có : \(\frac{3x-y}{x+y}=\frac{3}{4}\)
\(\Rightarrow4\left(3x-y\right)=3\left(x+y\right)\)
\(\Rightarrow12x-4y=3x+3y\)
\(\Rightarrow12x-3x=3y+4y\)
\(\Leftrightarrow9x=7y\)
\(\Rightarrow\frac{x}{y}=\frac{7}{9}\)
\(\dfrac{M}{N}=\dfrac{x^4-x^3+6x^2-x+a}{x^2-x+5}\)
\(=\dfrac{x^4-x^3+5x^2+x^2-x+5+a-5}{x^2-x+5}\)
\(=x^2+1+\dfrac{a-5}{x^2-x+5}\)
Để M chiahết cho N thì a-5=0
=>a=5
Để M chia N dư 3 thì a-5=3
=>a=8