Cho 2x-y=3. Tìm giá trị của 4x/2y.
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Câu 1 :
\(E=4x^2+y^2-4x-2y+3\)
\(E=\left(2x\right)^2-2\cdot2x\cdot1+1^2+y^2-2\cdot y\cdot1+1^2+1\)
\(E=\left(2x-1\right)^2+\left(y-1\right)^2+1\ge1\forall x;y\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}2x-1=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=1\end{cases}}\)
Câu 2 :
\(G=x^2+2y^2+2xy-2y\)
\(G=x^2+2xy+y^2+y^2-2.y\cdot1+1^2-1\)
\(G=\left(x+y\right)^2+\left(y-1\right)^2-1\ge-1\forall x;y\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x+y=0\\y-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+1=0\\y=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-1\\y=1\end{cases}}}\)
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Ta có 2 x + 1 3 − y + 1 4 = 4 x − 2 y + 2 5 2 x − 3 4 − y − 4 3 = − 2 x + 2 y − 2
⇔ 40 x + 20 − 15 y − 15 = 48 x − 24 y + 24 6 x − 9 − 4 y + 16 = − 24 x + 24 y − 24
⇔ 8 x − 9 y = − 19 30 x − 28 y = − 31 ⇔ 120 x − 135 = − 285 120 x − 112 = − 124 ⇔ x = 11 2 y = 7
Thay x = 11 2 ; y = 7 vào phương trình 6mx – 5y = 2m – 66 ta được:
6m. 11 2 − 5.7 = 2m – 66 31m = −31 m = −1
Đáp án: A
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\(N=2x^2+y^2+2xy-4x-2y+3\)
\(N=\left(x^2+2xy+y^2\right)+x^2-4x-2y+3\)
\(N=\left[\left(x+y\right)^2-2\left(x+y\right)+1\right]+\left(x^2-2x+1\right)+1\)
\(N=\left(x+y-1\right)^2+\left(x-1\right)^2+1\)
Mà \(\left(x+y-1\right)\ge0\forall x;y\)
\(\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow N\ge1\)
Dấu "=" xảy ra khi :
\(\hept{\begin{cases}x+y-1=0\\x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=0\\x=1\end{cases}}\)
Vậy \(N_{Min}=1\Leftrightarrow\left(x;y\right)=\left(1;0\right)\)
\(N=2x^2+y^2+2xy-4x-2y\)\(+3\)
\(=\left(x^2+2xy+y^2\right)+x^2-2\left(2x+y\right)+3\)
\(=\left[\left(x+y\right)^2-2\left(2x+y\right)+1\right]+2+x^2\)
\(=\left(x+y+1\right)^2+x^2+2\)
\(Do\)\(\left(x+y+1\right)^2\)\(\ge\)\(0\)\(\forall\)\(x\)\(;\)\(y\)
\(x^2\)\(\ge\)\(0\)\(\forall\)\(x\)
=.>\(\left(x+y+1\right)^2+x^2+2\)\(\ge\)\(2\)\(\forall\)\(x\)\(;\)\(y\)
=>\(N\)\(\ge\)\(2\)\(\forall\)\(x\)\(;\)\(y\)
Dấu = xảy ra khi:
\(\hept{\begin{cases}\left(x+y+1\right)^2=0\\x^2=0\end{cases}}\)
=>\(\hept{\begin{cases}x+y+1=0\\x=0\end{cases}}\)
=>\(\hept{\begin{cases}x+y=-1\\x=0\end{cases}}\)
=>\(\hept{\begin{cases}y=-1\\x=0\end{cases}}\)
Vậy \(N_{min}\)\(=\)\(2\)khi \(y=-1\)\(;\)\(x=0\)
Chúc pạn họk tốt~~~!!! :3
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a) A = x2 + 4x - 2 = x2 + 4x + 4 - 6 = (x + 2)2 - 6
(x + 2)2 ≥ 0 => A ≥ -6 => GTNN của A là -6, xảy ra khi x = 2
`a)A=x^2+4x-2`
`A=x^2+4x+4-6=(x+2)^2-6`
Vì `(x+2)^2 >= 0 AA x`
`<=>(x+2)^2-6 >= -6 AA x`
Hay `A >= -6 AA x`
Dấu "`=`" xảy ra`<=>(x+2)^2=0<=>x=-2`
Vậy `GTN N` của `A` là `-6` khi `x=-2`
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`b)B=2x^2-4x+3`
`B=2(x^2-2x+3/2)`
`B=2(x^2-2x+1)+1=2(x-1)^2+1`
Vì `2(x-1)^2 >= 0 AA x`
`<=>2(x-1)^2+1 >= 1 AA x`
Hay `B >= 1 AA x`
Dấu "`=`" xảy ra `<=>(x-1)^2=0<=>x=1`
Vậy `GTN N` của `B` là `1` khi `x=1`
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`c)C=x^2+y^2-4x+2y+5`
`C=x^2-4x+4+y^2+2y+1`
`C=(x-2)^2+(y+1)^2`
Vì `(x-2)^2 >= 0 AA x` và `(y+1)^2 >= 0 AA y`
`=>(x-2)^2+(y+1)^2 >= 0 AA x,y`
Hay `C >= 0 AA x,y`
Dấu "`=`" xảy ra`<=>{((x-2)^2=0),((y+1)^2=0):}`
`<=>{(x=2),(y=-1):}`
Vậy `GTN N` của `C` là `0` khi `x=2`,y=-1
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a) \(A=-x^2+2x=-\left(x^2-2x+1\right)+1=-\left(x-1\right)^2+1\le1\)
\(maxA=1\Leftrightarrow x=1\)
b) \(B=\left(2-3x\right)\left(3+2x\right)=-6x^2-5x+6=-6\left(x^2+\dfrac{5}{6}x+\dfrac{25}{144}\right)+\dfrac{169}{24}=-6\left(x+\dfrac{5}{12}\right)^2+\dfrac{169}{24}\le\dfrac{169}{24}\)
\(minB=\dfrac{169}{24}\Leftrightarrow x=-\dfrac{5}{12}\)
c) \(C=4xy-4x-2y-4x^2-2y^2-3=-\left[4x^2-4x\left(y-1\right)+\left(y-1\right)^2\right]+\left(y^2-4y+4\right)-6=\left(2x-y+1\right)^2+\left(y-2\right)^2-6\le-6\)
\(minC=-6\Leftrightarrow\)\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=2\end{matrix}\right.\)
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a, \(A=x^4-2x^3+2x^2-2x+3\)
\(=\left(x^4+2x^2+1\right)-\left(2x^3+2x\right)+2\)
\(=\left(x^2+1\right)^2-2x\left(x^2+1\right)+2\)
\(=\left(x^2+1\right)\left(x^2-2x+1\right)+2\)
\(=\left(x^2+1\right)\left(x-1\right)^2+2\)
Vì \(\hept{\begin{cases}x^2\ge0\\\left(x-1\right)^2\ge0\end{cases}\Rightarrow\hept{\begin{cases}x^2+1\ge1\\\left(x-1\right)^2\ge0\end{cases}\Rightarrow}\left(x^2+1\right)\left(x-1\right)^2\ge0}\)
\(\Rightarrow A=\left(x^2+1\right)\left(x-1\right)^2+2\ge2\)
Dấu "=" xảy ra khi x = 1
Vậy Amin = 2 khi x = 1
b, \(B=4x^2-2\left|2x-1\right|-4x+5=\left(4x^2-4x+1\right)-2\left|2x-1\right|+4=\left(2x-1\right)^2-2\left|2x-1\right|+4\)
đề sai ko
c, \(C=4-x^2+2x=-\left(x^2-2x+1\right)+5=-\left(x-1\right)^2+5\)
Vì \(-\left(x-1\right)^2\le0\Rightarrow C=-\left(x-1\right)^2+5\le5\)
Dấu "=" xảy ra khi x=1
Vậy Cmin = 5 khi x = 1
2/
+) \(D=-x^2-y^2+x+y+3=-\left(x^2-x+\frac{1}{4}\right)-\left(y^2-y+\frac{1}{4}\right)+\frac{7}{2}=-\left(x-\frac{1}{2}\right)^2-\left(y-\frac{1}{2}\right)^2+\frac{7}{2}\)
Vì \(\hept{\begin{cases}-\left(x-\frac{1}{2}\right)^2\le0\\-\left(y-\frac{1}{2}\right)^2\le0\end{cases}\Rightarrow-\left(x-\frac{1}{2}\right)^2-\left(y-\frac{1}{2}\right)^2\le0}\Rightarrow D=-\left(x-\frac{1}{2}\right)^2-\left(y-\frac{1}{2}\right)^2+\frac{7}{2}\le\frac{7}{2}\)
Dấu "=" xảy ra khi x=y=1/2
Vậy Dmax=7/2 khi x=y=1/2
+) Đề sai
+)bài này là tìm min
\(G=x^2-3x+5=\left(x^2-3x+\frac{9}{4}\right)+\frac{11}{4}=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}\)
Dấu "=" xảy ra khi x=3/2
Vậy Gmin=11/4 khi x=3//2
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Anh/ chị viết rõ đề bằng công thức toán được không ạ?
Vd : 1/2(2x+2y+z)^2 là \(\frac{1}{2\left(2x+2y+z\right)^2}\) hay sao?
\(P=8x^3+8y^3+\frac{z^3}{\left(2x+2y+2z\right)\left(4xy+2yz+2zx\right)}\) đúng ko ạ?
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a: C=A-B
\(=5x^3+y^3-3x^2y+4xy^2-4x^3+6x^2y-xy^2\)
\(=x^3+3x^2y+3xy^2+y^3\)
D=A+B
\(=5x^3+y^3-3x^2y+4xy^2+4x^3-6x^2y+xy^2\)
\(=9x^3-9x^2y+5xy^2+y^3\)
bậc của C là 3
bậc của D là 3
b: Thay x=0 và y=-2 vào D, ta được:
\(D=9\cdot0^3-9\cdot0^2\left(-2\right)+5\cdot0\cdot\left(-2\right)^2+\left(-2\right)^3\)
\(=0-0+0-8=-8\)
c: Thay x=-1 và y=-1 vào C, ta được:
\(C=\left(-1\right)^3+3\cdot\left(-1\right)^2\cdot\left(-1\right)+3\cdot\left(-1\right)\cdot\left(-1\right)^2+\left(-1\right)^3\)
=-8
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1) ADTCDTSBN, ta có:
\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)= \(\frac{2x^2+2y^2-3z^2}{18+32-75}=\frac{-100}{-25}\)= 4
* \(\frac{x}{3}=4\)=> x = 3 . 4 = 12
- \(\frac{y}{4}=4\)=> y = 4 . 4 = 16
* \(\frac{z}{5}=4\)=> z = 5 . 4 = 20
Vậy x = 12
y = 16
z = 20
2x-y=3 => 2x=3
Ta có:
4x/2y=22x/2y=2y+3/2y=8