Tìm min: E=5x2+y2-4xy+4x-8y+1
tìm GTLN của biểu thức
C=-x2-4x-y2+8y+2
D=2023-8x+2y+4xy-y2-5x2
\(C=-\left(x^2+4x+4\right)-\left(y^2-8y+16\right)+22\\ =-\left(x^2+2x.2+2^2\right)-\left(y^2-2.y.4+4^2\right)+22\\ =-\left(x+2\right)^2-\left(y-4\right)^2+22\\ Vậy:max_C=22.khi.x=-2.và.y=4\)
tim min 4x^2 5y^2-4xy x-8y-2017
tim min 4x^2 5y^2-4xy x-8y-2017
A= 5x2+y2 -4xy+3x-3
B= x(x-1)(x+1)(x+2)
tìm gtnn
\(A=\left(4x^2-4xy+y^2\right)+\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{21}{4}\\ A=\left(2x-y\right)^2+\left(x+\dfrac{3}{2}\right)^2-\dfrac{21}{4}\ge-\dfrac{21}{4}\\ A_{min}=-\dfrac{21}{4}\Leftrightarrow\left\{{}\begin{matrix}2x=y\\x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-3\end{matrix}\right.\)
\(B=\left[\left(x-1\right)\left(x+2\right)\right]\left[x\left(x+1\right)\right]=\left(x^2+x-2\right)\left(x^2+x\right)\\ B=\left(x^2+x\right)^2-2\left(x^2+x\right)\\ B=\left(x^2+x\right)^2-2\left(x^2+x\right)+1-1=\left(x^2+x-1\right)^2-1\ge-1\\ B_{min}=-1\Leftrightarrow x^2+x-1=0\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2-\dfrac{5}{4}=0\\ \Leftrightarrow\left(x+\dfrac{1-\sqrt{5}}{2}\right)\left(x+\dfrac{1+\sqrt{5}}{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{5}}{2}\\x=\dfrac{1+\sqrt{5}}{2}\end{matrix}\right.\)
Phân tích thành nhân tử:
A = (6x - 3y) + (4x2 - 4xy + y2)
B= 9x2 - (y2 - 4y + 4)
C= -25x2 + y2 - 6y + 9
D= x2 - 4x - y2 - 8y -12
\(A=\left(6x-3y\right)+\left(4x^2-4xy+y^2\right)=3\left(2x-y\right)+\left(2x-y\right)^2=\left(2x-y\right)\left(2+2x-y\right)\)
\(B=9x^2-\left(y^2-4y+4\right)=9x^2-\left(y-2\right)^2=\left(3x-y+2\right)\left(3x+y-2\right)\)
\(C=-25x^2+y^2-6y+9=\left(y^2-6y+9\right)-25x^2=\left(y-3\right)^2-\left(5x\right)^2=\left(y-3-5x\right)\left(y-3+5x\right)\)\(D=x^2-4x-y^2-8y-12=\left(x^2-4x+4\right)-\left(y^2+8y+16\right)=\left(x-2\right)^2-\left(y+4\right)^2=\left(x-2-y-4\right)\left(x-2+y+4\right)=\left(x-y-6\right)\left(x+y+2\right)\)
a: Ta có: \(A=\left(6x-3y\right)+\left(4x^2-4xy+y^2\right)\)
\(=3\left(2x-y\right)+\left(2x-y\right)^2\)
\(=\left(2x-y\right)\left(2x-y+3\right)\)
b: Ta có: \(B=9x^2-\left(y^2-4y+4\right)\)
\(=9x^2-\left(y-2\right)^2\)
\(=\left(3x-y+2\right)\left(3x+y-2\right)\)
tìm GTLN của biểu thức
D=2023-8x+2y+4xy-y2-5x2
\(D=2023-8x+2y+4xy-y^2-5x^2\)
\(=-\left(y^2+5x^2-4xy-2y+8x-2023\right)\)
\(=-\left(y^2-2.y.\left(2x+1\right)+\left(2x+1\right)^2-\left(2x+1\right)^2+5x^2+8x-2023\right)\)
\(=-\left[\left(y-2x-1\right)^2-4x^2-4x-1+5x^2+8x-2023\right]\)
\(=-\left[\left(y-2x-1\right)^2+x^2+4x-2024\right]\)
\(=-\left[\left(y-2x-1\right)^2+\left(x+2\right)^2\right]+2028\)
Vì \(-\left[\left(y-2x-1\right)^2+\left(x+2\right)^2\right]\le0\forall x,y\)
\(MaxD=2028\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-3\end{matrix}\right.\)
1) Tìm x, y, z
a) 9x2 +y2 + 2z2 – 18x +4z – 6y +20 = 0
b) 5x2 +5y2 +8xy+2y – 2x+2 = 0
c) 5x2 +2y2 + 4xy – 2x + 4y +5 = 0
d) x2 + 4y2 + z2 =2x + 12y – 4z – 14
e) x2 +y2 – 6x + 4y +2= 0
Giúp mik vs cần gấp!!!
\(a,\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,\Leftrightarrow\left(4x^2+4xy+y^2\right)+\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
a,9x^2+y^2+2z^2−18x+4z−6y+20=0
⇔9(x−1)^2+(y−3)^2+2(z+1)^2=0
⇔x=1;y=3;z=−1
b,5x^2+5y^2+8xy+2y−2x+2=0
⇔4(x+y)2+(x−1)2+(y+1)2=0
⇔x=−y;x=1y=−1⇔x=1y=−1
c,5x^2+2y^2+4xy−2x+4y+5=0
⇔(2x+y)^2+(x−1)^2+(y+2)^2=0
⇔2x=−y;x=1;y=−2
⇔x=1;y=−2
⇔(x−1)^2+(2y−3)^2+(z+2)^2=0
\(d,\Leftrightarrow\left(x^2-2x+1\right)+\left(4y^2-12y+9\right)+\left(z^2+4z+4\right)=0\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
\(\Rightarrow\)PT vô nghiệm vì 11 không phải là tổng 2 số chính phương
tìm min A= x^2+5y^2-4xy+2x-8y+202
\(A=x^2+5y^2-4xy+2x-8y+202\)
\(=x^2+4y^2+1-4xy-4y+2x+\left(y^2-4y+4\right)+197\)
\(=\left(x-2y+1\right)^2+\left(y-2\right)^2+197\ge197\forall x;y\)
Dâu "=" xảy ra khi:
\(\hept{\begin{cases}x-2y+1=0\\y-2=0\end{cases}\Rightarrow\hept{\begin{cases}x-4+1=0\\y=2\end{cases}\Rightarrow}\hept{\begin{cases}x=3\\y=2\end{cases}}}\)
Vậy min A = 197 khi \(x=3,y=2\)
Chúc bạn học tốt.
Phân tích đa thức thành nhân tử:
a) 81x5-x3
b) 9x2y-12xy+4y
c) (5-x)2-16(x-2)2
d) 9x2-y2-21x-7y
e) -y2+8y+9x2-16
f) 5x2-4x-1
g) (x+7)(x+9)-17
h) x(x+2)(x+4)(x+6)+15
a: \(81x^5-x^3\)
\(=x^3\left(81x^2-1\right)\)
\(=x^3\left(9x-1\right)\left(9x+1\right)\)
b: \(9x^2y-12xy+4y\)
\(=y\left(9x^2-12x+4\right)\)
\(=y\left(3x-2\right)^2\)
c: \(\left(5-x\right)^2-16\left(x-2\right)^2\)
\(=\left(x-5\right)^2-\left(4x-8\right)^2\)
\(=\left(x-5-4x+8\right)\left(x-5+4x-8\right)\)
\(=-3\left(x-1\right)\left(5x-13\right)\)
d: Ta có: \(9x^2-y^2-21x-7y\)
\(=\left(3x-y\right)\left(3x+y\right)-7\left(3x+y\right)\)
\(=\left(3x+y\right)\left(3x-y-7\right)\)
e: Ta có: \(-y^2+8y-16+9x^2\)
\(=-\left(y^2-8y+16-9x^2\right)\)
\(=-\left(y-4-3x\right)\left(y-4+3x\right)\)
f: Ta có: \(5x^2-4x-1\)
\(=5x^2-5x+x-1\)
\(=5x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(5x+1\right)\)