cho x+4y=1 chứng minh rằng x2+4y2 >=1/5
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8 tháng 12 2023
1: \(x^2-x-y^2-y\)
\(=\left(x^2-y^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)
2: \(x^2-y^2+x-y\)
\(=\left(x^2-y^2\right)+\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y\right)+\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y+1\right)\)
3: \(3x-3y+x^2-y^2\)
\(=\left(3x-3y\right)+\left(x^2-y^2\right)\)
\(=3\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y+3\right)\)
4: \(5x-5y+x^2-y^2\)
\(=\left(5x-5y\right)+\left(x^2-y^2\right)\)
\(=5\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left(5+x+y\right)\)
5: \(x^2-5x-y^2-5y\)
\(=\left(x^2-y^2\right)-\left(5x+5y\right)\)
\(=\left(x-y\right)\left(x+y\right)-5\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-5\right)\)
6: \(x^2-y^2+2x-2y\)
\(=\left(x^2-y^2\right)+\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)+2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y+2\right)\)
7: \(x^2-4y^2+x+2y\)
\(=\left(x^2-4y^2\right)+\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y\right)+\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y+1\right)\)
8: \(x^2-y^2-2x-2y\)
\(=\left(x^2-y^2\right)-\left(2x+2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-2\right)\)
9: \(x^2-4y^2+2x+4y\)
\(=\left(x^2-4y^2\right)+\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)+2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y+2\right)\)
ND
Nguyễn Đức Trí
VIP
10 tháng 9 2023
a) \(x^2+xy+y^2+1\)
\(=x^2+xy+\dfrac{y^2}{4}-\dfrac{y^2}{4}+y^2+1\)
\(=\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}+1\)
mà \(\left\{{}\begin{matrix}\left(x+\dfrac{y}{2}\right)^2\ge0,\forall x;y\\\dfrac{3y^2}{4}\ge0,\forall x;y\end{matrix}\right.\)
\(\Rightarrow\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}+1>0,\forall x;y\)
\(\Rightarrow dpcm\)
ND
Nguyễn Đức Trí
VIP
10 tháng 9 2023
b) \(...=x^2-2x+1+4\left(y^2+2y+1\right)+z^2-6z+9+1\)
\(=\left(x-1\right)^2+4\left(y^{ }+1\right)^2+\left(z-3\right)^2+1>0,\forall x.y\)
\(\Rightarrow dpcm\)
1 tháng 1 2022
\(a,x^2-6x+5=0\\ \Rightarrow\left(x^2-5x\right)-\left(x-5\right)=0\\ \Rightarrow x\left(x-5\right)-\left(x-5\right)=0\\ \Rightarrow\left(x-1\right)\left(x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
\(b,2x^2+4x-8=0\\ \Rightarrow x^2+2x-4=0\\ \Rightarrow\left(x^2+2x+1\right)-5=0\\ \Rightarrow\left(x+1\right)^2-\sqrt{5^2}=0\\ \Rightarrow\left(x+1+\sqrt{5}\right)\left(x+1-\sqrt{5}\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1-\sqrt{5}\\x=-1+\sqrt{5}\end{matrix}\right.\)
\(c,4y^2-4y+1=0\\ \Rightarrow\left(2y-1\right)^2=0\\ \Rightarrow2y-1=0\\ \Rightarrow y=\dfrac{1}{2}\)
\(d,5x^2-x+2=0\)
Ta có:\(\Delta=\left(-1\right)^2-4.5.2=1-40=-39\)
Vì \(\Delta< 0\Rightarrow\) pt vô nghiệm
8 tháng 9 2021
\(a,\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,\Leftrightarrow\left(4x^2+4xy+y^2\right)+\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
MH
8 tháng 9 2021
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
d,x^2+4y^2+z^2=2x+12y−4z−14
⇔(x−1)^2+(2y−3)^2+(z+2)^2=0
⇔x=1;y=3/2;z=−2
e: Ta có: x^2−6x+y2+4y+2=0
⇔x^2−6x+9+y^2+4y+4−11=0
⇔(x−3)^2+(y+2)^2=11
Dấu '=' xảy ra khi x=3 và y=-2
8 tháng 10 2023
a: \(x^2+x+1=x^2+x+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}>0\forall x\)
b: \(4y^2+2y+1\)
\(=4\left(y^2+\dfrac{1}{2}y+\dfrac{1}{4}\right)\)
\(=4\left(y^2+2\cdot y\cdot\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{3}{16}\right)\)
\(=4\left(y+\dfrac{1}{4}\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}>0\forall y\)
c: \(-2x^2+6x-10\)
\(=-2\left(x^2-3x+5\right)\)
\(=-2\left(x^2-3x+\dfrac{9}{4}+\dfrac{11}{4}\right)\)
\(=-2\left(x-\dfrac{3}{2}\right)^2-\dfrac{11}{2}< =-\dfrac{11}{2}< 0\forall x\)
8 tháng 10 2023
`#3107.101107`
a)
`x^2 + x + 1`
`= (x^2 + 2*x*1/2 + 1/4) + 3/4`
`= (x + 1/2)^2 + 3/4`
Vì `(x + 1/2)^2 \ge 0` `AA` `x`
`=> (x + 1/2)^2 + 3/4 \ge 3/4` `AA` `x`
Vậy, `x^2 + x + 1 > 0` `AA` `x`
b)
`4y^2 + 2y + 1`
`= [(2y)^2 + 2*2y*1/2 + 1/4] + 3/4`
`= (2y + 1/2)^2 + 3/4`
Vì `(2y + 1/2)^2 \ge 0` `AA` `y`
`=> (2y + 1/2)^2 + 3/4 \ge 3/4` `AA` `y`
Vậy, `4y^2 + 2y + 1 > 0` `AA` `y`
c)
`-2x^2 + 6x - 10`
`= -(2x^2 - 6x + 10)`
`= -2(x^2 - 3x + 5)`
`= -2[ (x^2 - 2*x*3/2 + 9/4) + 11/4]`
`= -2[ (x - 3/2)^2 + 11/4]`
`= -2(x - 3/2)^2 - 11/2`
Vì `-2(x - 3/2)^2 \le 0` `AA` `x`
`=> -2(x - 3/2)^2 - 11/2 \le 11/2` `AA` `x`
Vậy, `-2x^2 + 6x - 10 < 0` `AA `x.`
TT
2
H9
HT.Phong (9A5)
CTVHS
21 tháng 7 2023
a) \(y^2-x^2+6y+9\)
\(=\left(y^2+6y+9\right)-x^2\)
\(=\left(y+3\right)^2-x^2\)
\(=\left[\left(y+3\right)-x\right]\left[\left(y+3\right)+x\right]\)
\(=\left(y-x+3\right)\left(y+x+3\right)\)
b) \(4y^2-x^2-4y+1\)
\(=\left(4y^2-4x+1\right)-x^2\)
\(=\left(2y-1\right)^2-x^2\)
\(=\left[\left(2y-1\right)+x\right]\left[\left(2y-1\right)-x\right]\)
\(=\left(2y+x-1\right)\left(2y-x-1\right)\)
c) \(\left(x-y\right)^2-x^2+y^2\)
\(=\left(x-y\right)^2-\left(x^2-y^2\right)\)
\(=\left(x-y\right)^2-\left(x+y\right)\left(x-y\right)\)
\(=\left(x-y\right)\left[\left(x-y\right)-\left(x+y\right)\right]\)
\(=\left(x-y\right)\left(x-y-x-y\right)\)
\(=-2y\left(x-y\right)\)
d) \(x^6-y^6\)
\(=\left(x^3\right)^2-\left(y^3\right)^2\)
\(=\left(x^3+y^3\right)\left(x^3-y^3\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)\left(x-y\right)\left(x^2+xy+y^2\right)\)
21 tháng 7 2023
a: =(y+3)^2-x^2
=(y+3+x)(y+3-x)
b: =(2y-1)^2-x^2
=(2y-1-x)(2y-1+x)
c: =x^2-2xy+y^2-x^2+y^2
=2y^2-2xy
=2y(y-x)
d: =(x^3-y^3)(x^3+y^3)
=(x-y)(x+y)(x^2+xy+y^2)(x^2-xy+y^2)
vì x+4y=1 nên x=1-4y (1)
ta có : x^2+4y^2≥1/5
=> x^2+4y^2-1/5 ≥0 (2)
thay (1) vào (2) ta có:(1-4y)^2+4y^2-1/5 ≥ 0
<=>1-8y +16y^2 + 4y^2 - 1/5 ≥ 0
<=>20y^2 - 8y + 4/5 ≥ 0
<=>5(4y^2 - 8/5y + 4/25) ≥ 0
<=>5(2y-8/20)^2 ≥ 0 (luôn đúng)
Vậy với x+4y=1 thì x^2+4y^2≥1/5 ;dấu = xảy ra khi x=y=1/5
Làm gọn thôi bạn ơi! Dùng bất đẳng thức Bunyakovsky