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\(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\ge0\) \(\forall x,y\)
mà \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\) (đề bài ) \(\Rightarrow\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}=0\)
\(\Rightarrow\hept{\begin{cases}2x-5=0\\3y+4=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{2}\\y=-\frac{4}{3}\end{cases}}}\)
Rút gọn biểu thức
\(m+\left(5x^2-2xy\right)=6x^2+9xy-y^2\)
=> \(m=x^2+11xy-y^2\)
Thay x,y, vừa tìm được vào biểu thức đã được rút gọn ta tính được m
1)Tìm đa thức M biết rằng :M+(5x2 -2xy)=6x2+9xy-y2
2)Tìm GTLN của :B=\(\frac{x^2+y^2+3}{x^2+y^2+2}\)
1)
\(M+\left(5x^2-2xy\right)=6x^2+9xy-y^2\)
\(M+5x^2-2xy=6x^2+9xy-y^2\)
\(M=\left(6x^2+9xy-y^2\right)-\left(5x^2+2xy\right)\)
\(M=6x^2+9xy-y^2-5x^2-2xy\)
\(M=\left(6x^2-5x^2\right)+\left(9xy-2xy\right)-y^2\)
\(M=x^2+7xy-y^2.\)
Chúc em học tốt!
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
\(M=8x^2-2xy-y^2-5x^2+2xy+3y^2=3x^2+2y^2>=0\forall x,y\)
a)\(x^2-2x+y^2+4y+5=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y+2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\left(tm\right)\)
Tìm đa thức M biết :
a, M +5 (5x2 - 2xy) = 6x2 +9xy - y2
M + 5. 5x2 - 5. 2xy = 6x2 + 9xy - y2
M + 25x2 - 10xy = 6x2 + 9xy - y2
M = 6x2 + 9xy - y2 + 10xy - 25x2
M = ( 6x2 - 25x2 ) + ( 9xy + 10xy ) - y2
M = -19x2 + 19xy - y2
b, M - ( 3xy - 4y2 ) = x2 - 7xy + 8xy
M - 3xy + 4y2 = x2 - 15xy
M = x2 - 15xy - 4y2 + 3xy
M = x2 + ( 15xy + 3xy ) - 4y2
M = x2 + 18xy - 4y2
c, (25 . x2y - 13xy2+ y3 ) - M = 11x2y - 2y3
25x2y - 13xy2+ y3 - M = 11x2y - 2y3
M = 25x2y - 13xy2+ y3 - 11x2y - 2y3
M = ( 25x2y - 11x2y ) + ( y3 - 2y3 ) - 13xy2
M = 14x2y - y3 - 13xy2
d, M + (5x2 - 2xy )= 6x2 + 9xy -y2
M + 5x2 - 2xy = 6x2 + 9xy -y2
M = 6x2 + 9xy -y2 + 2xy - 5x2
M = ( 6x2 - 5x2 ) + ( 9xy + 2xy ) - y2
M = x2 + 11xy - y2
a) \(\left(x-2\right)^3=-27\)
\(\Rightarrow\left(x-2\right)^3=\left(-3\right)^3\)
\(\Rightarrow x-2=-3\)
\(\Rightarrow x=-1\)
Vậy \(x=-1\)
b) \(\left(2x+1\right)^4=81\)
\(\Rightarrow\left(2x+1\right)^4=3^4=\left(-3\right)^4\)
\(\left\{{}\begin{matrix}\left(2x+1\right)^4=3^4\Rightarrow2x+1=3\Rightarrow x=1\\\left(2x+1\right)^4=\left(-3\right)^4\Rightarrow2x+1=-3\Rightarrow x=-2\end{matrix}\right.\)
Vậy \(x=1;x=-2\)
c) Bạn xem lại đề bài nhé!
d) \(\left(5x-2\right)^{10}=\left(5x-2\right)^{100}\)
\(\Rightarrow\left(5x-2\right)^{10}-\left(5x-2\right)^{100}=0\)
\(\Rightarrow\left(5x-2\right)^{10}.\left[1-\left(5x-2\right)^{90}\right]=0\)
+) TH1: \(\left(5x-2\right)^{10}=0\)
\(\Rightarrow5x-2=0\)
\(\Rightarrow x=\dfrac{2}{5}\)
+) TH2: \(1-\left(5x-2\right)^{90}=0\)
\(\Rightarrow\left(5x-2\right)^{90}=1\)
\(\Rightarrow\left(5x-2\right)^{90}=1^{90}=\left(-1\right)^{90}\)
\(\Rightarrow\left\{{}\begin{matrix}\left(5x-2\right)^{90}=1^{90}\Rightarrow5x-2=1\Rightarrow x=\dfrac{3}{5}\\\left(5x-2\right)^{90}=\left(-1\right)^{90}\Rightarrow5x-2=-1\Rightarrow x=\dfrac{1}{5}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{1}{5};\dfrac{2}{5};\dfrac{3}{5}\right\}\)
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)
\(2^{x+1}+2^{x+4}+2^{x+5}=2^5\cdot5^2\)
\(\Rightarrow2^{x+1}\left(1+2^3+2^4\right)=2^5\cdot5^2\)
\(\Rightarrow2^{x+1}\cdot5^2=2^5\cdot5^2\)
\(\Rightarrow2^{x+1}=2^5\Rightarrow x=4\)