Tìm x biết
a 25x2+4y2-10x+12x+10=0
b 3x2+2y2-12x+12y+30=0
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9 tháng 10 2021
1)
a) \(=15x^3-20x^2+10x\)
b) \(=3x^4-x^3+4x^2-9x^3+3x-12x=3x^4-10x^3+4x^2-9x\)
2)
a) \(\Rightarrow x\left(x^2-6x+12\right)=0\)
\(\Rightarrow x=0\)(do \(x^2-6x+12=\left(x^2-6x+\dfrac{36}{4}\right)+3=\left(x-\dfrac{6}{2}\right)^2+3\ge3>0\))
b) \(\Rightarrow\left(x+3\right)^3=0\Rightarrow x=-3\)
23 tháng 3 2022
(3x²-5x+2)+(3x²+5x)= bao nhiêu ạ
Giúp em vs ạ . Em cảm ơn
5 tháng 9 2021
\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11
5 tháng 9 2021
e: Ta có: \(x^2-6x+y^2+4y+2=0\)
\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Dấu '=' xảy ra khi x=3 và y=-2
2 tháng 11 2021
Bài 1:
a) \(\Rightarrow3x^2+3x-2x^2-4x+x+1=0\)
\(\Rightarrow x^2=-1\left(VLý\right)\Rightarrow S=\varnothing\)
b) \(\Rightarrow\left(x-2020\right)\left(2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2}\end{matrix}\right.\)
c) \(\Rightarrow\left(x-10\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)
d) \(\Rightarrow\left(x+4\right)^2=0\Rightarrow x=-4\)
e) \(\Rightarrow\left(x+6\right)\left(x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)
f) \(\Rightarrow\left(5x-4\right)\left(5x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Bài 2:
a) \(\Rightarrow3x\left(x^2-4\right)=0\Rightarrow3x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow x\left(x-2\right)+5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
23 tháng 9 2021
a: \(x^2+3y^2-4x+6y+7=0\)
\(\Leftrightarrow x^2-4x+4+3y^2+6y+3=0\)
\(\Leftrightarrow\left(x-2\right)^2+3\left(y+1\right)^2=0\)
\(\Leftrightarrow\left(x,y\right)=\left(-2;1\right)\)
a, sửa đề : \(25x^2+4y^2-10x+12y+10=0\)
\(\Leftrightarrow25x^2-10x+1+4y^2+12y+9=0\)
\(\Leftrightarrow\left(5x-1\right)^2+\left(2y+3\right)^2=0\)
Đẳng thức xảy ra khi x = 1/5 ; y = -3/2
b, \(3x^2+2y^2-12x+12y+30=0\)
\(\Leftrightarrow3\left(x^2-4x+4\right)+2\left(y^2+6y+9\right)=0\)
\(\Leftrightarrow3\left(x-2\right)^2+2\left(y+3\right)^2=0\)
Đẳng thức xảy ra khi x = 2 ; y = -3
\(a)\)
\(25x^2+4y^2-10x+12x+10=0\)
\(\Leftrightarrow\left(5x\right)^2-10x+1+\left(2y\right)^2+12y+9=0\)
\(\Leftrightarrow[\left(5x\right)^2-10x+1+\left(2y\right)^2+12y+9=0\)
\(\Leftrightarrow[\left(5x\right)^2-2.5x.1-1^2]+[\left(2y\right)^2+2.2y.3+3^{20}]=0\)
\(\Leftrightarrow\left(5x-1\right)^2+\left(2y+3\right)^2=0\)
\(\Leftrightarrow\left(5x-1\right)^2=0\Leftrightarrow5x-1=0\Leftrightarrow x=\frac{1}{5}\)
\(\Leftrightarrow\left(2y+3\right)^2=0\Leftrightarrow2y+3=0\Leftrightarrow2y=-3\Leftrightarrow y=\frac{-3}{2}\)
\(b)\)
\(3x^2+2y^2-12x+12y+30=0\)
\(\Leftrightarrow3x^2-12x+12+2y^2+12y+18=0\)
\(\Leftrightarrow3\left(x-2\right)^2+2\left(y+3\right)^2=0\)
Mà: \(3\left(x-2\right)^2\ge0\forall x;2\left(y+3\right)^2\ge0\forall y\)
\(\Leftrightarrow3\left(x-2\right)^2+2\left(y+3\right)^2=0\)chỉ khi: \(x-2=y+3=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\y=-3\end{cases}}\)