Giúp mình câu này vs mình cần gấp
1. 2+4y=x(2y+1)
2. 3+5x=12y+20xy+20
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15 tháng 5 2021
b) Ta có: \(9x^4+8x^2-1=0\)
\(\Leftrightarrow9x^4+9x^2-x^2-1=0\)
\(\Leftrightarrow9x^2\left(x^2+1\right)-\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(9x^2-1\right)=0\)
mà \(x^2+1>0\forall x\)
nên \(9x^2-1=0\)
\(\Leftrightarrow9x^2=1\)
\(\Leftrightarrow x^2=\dfrac{1}{9}\)
hay \(x\in\left\{\dfrac{1}{3};-\dfrac{1}{3}\right\}\)
Vậy: \(S=\left\{\dfrac{1}{3};-\dfrac{1}{3}\right\}\)
27 tháng 8 2021
\(\dfrac{1}{2}\left(6x-2y\right)\left(3x+y\right)=\dfrac{1}{2}.2\left(3x-y\right)\left(3x+y\right)=9x^2-y^2\)
\(\left(\dfrac{2}{3}z-\dfrac{2}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right).\dfrac{1}{2}=\left(\dfrac{1}{3}z-\dfrac{1}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}z\right).2.\dfrac{1}{2}=\dfrac{1}{9}z^2-\dfrac{1}{25}x^2\)
\(\left(5y-3x\right).\dfrac{1}{4}\left(12x+20y\right)=\left(5y-3x\right)\left(5y+3x\right).4.\dfrac{1}{4}=25y^2-9x^2\)
\(\left(\dfrac{3}{4}y-\dfrac{1}{2}x\right)\left(x+\dfrac{3}{2}y\right)=\left(\dfrac{3}{2}y-x\right)\left(\dfrac{3}{2}y+x\right)=\dfrac{9}{4}y^2-x^2\)
\(\left(a+b+c\right)\left(a+b+c\right)=\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
\(\left(x-y+z\right)\left(x+y-z\right)=x^2-\left(y-z\right)^2=x^2-y^2-z^2+2yz\)
NT
1
DP
2
26 tháng 10 2021
a) \(=2x\left(x-25\right)\)
b) \(=x\left(x-4\right)-\left(x-4\right)=\left(x-4\right)\left(x-1\right)\)
c) \(=x^2-\left(y^2-12y+36\right)=x^2-\left(y-6\right)^2=\left(x-y+6\right)\left(x+y-6\right)\)
d) \(=y\left(x^2+4xz+4yz\right)\)
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
H9
HT.Phong (9A5)
CTVHS
6 tháng 12 2023
\(5x^2+2y^2+6xy-8x-4y+4=0\)
\(\Leftrightarrow4x^2+x^2+y^2+y^2+2xy+4xy-8x-4y+4=0\)
\(\Leftrightarrow\left(4x^2+y^2+4+4xy-8x-4y\right)+\left(x^2+2xy+y^2\right)=0\)
\(\Leftrightarrow\left[\left(2x\right)^2+4xy+y^2-4\left(2x+y\right)+2^2\right]+\left(x+y\right)^2=0\)
\(\Leftrightarrow\left[\left(2x+y\right)^2-2\cdot\left(2x+y\right)\cdot2+2^2\right]+\left(x+y\right)^2=0\)
\(\Leftrightarrow\left(2x+y-2\right)^2+\left(x+y\right)^2=0\)
Ta có: \(\left\{{}\begin{matrix}\left(2x+y-2\right)^2\ge0\forall x,y\\\left(x+y\right)^2\ge0\forall x,y\end{matrix}\right.\)
\(\Rightarrow\left(2x+y-2\right)^2+\left(x+y\right)^2\ge0\forall x,y\)
Mặt khác: \(\left(2x+y-2\right)^2+\left(x+y\right)^2=0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}2x+y-2=0\\x+y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\cdot\left(-y\right)+y-2=0\\x=-y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2y+y-2=0\\x=-y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-y=2\\x=-y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-2\\x=2\end{matrix}\right.\)
Thay x,y vào P ta có:
\(P=2^{2023}+\left(-2\right)^{2023}=2^{2023}-2^{2023}=0\)
Vậy: ...
1. ta có
\(2\left(1+2y\right)=x\left(1+2y\right)\Leftrightarrow\left(1+2y\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}2y+1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}y=-\frac{1}{2}\\x=2\end{cases}}\)
2. đề có phải là \(3+5x=12y+20xy\) không nhỉ, vì nếu như bạn viết ở trên thì sẽ có vô số nghiệm
khi đó \(\left(3+5x\right)=4y\left(3+5x\right)\Leftrightarrow\left(3+5x\right)\left(4y-1\right)=0\Leftrightarrow\orbr{\begin{cases}4y-1=0\\\left(3+5x\right)=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}y=\frac{1}{4}\\x=-\frac{3}{5}\end{cases}}\)