a) 9x2-12xy+16y2>=0
b) 16a2-30ab+25b2>=0
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GV Nguyễn Trần Thành Đạt
Giáo viên
20 tháng 9 2023
\(a,4x^2+28x+49=\left(2x\right)^2+2.2x.7+7^2=\left(2x+7\right)^2\\ b,16y^2-8y+1=\left(4y\right)^2-2.4y.1+1^2=\left(4y-1\right)^2=\left(1-4y\right)^2\\ 4a^2+20ab+25b^2=\left(2a\right)^2+2.2a.5b+\left(5b\right)^2=\left(2a+5b\right)^2\\ d,9x^2-6xy+y^2=\left(3x\right)^2-2.3x.y+y^2=\left(3x-y\right)^2=\left(y-3x\right)^2\)
14 tháng 10 2021
\(a,\Leftrightarrow x^2\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\\ b,\Leftrightarrow\left(3x-5\right)^2=0\Leftrightarrow x=\dfrac{5}{3}\)
18 tháng 8 2021
a) \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\Rightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\Rightarrow\left(2x-3\right)\left(7x-2x+3\right)=0\Rightarrow\left[{}\begin{matrix}2x-3=0\\5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
b) \(\left(2x-7\right).\left(x-2\right)\left(x^2-4\right)=0\Rightarrow\left(2x-7\right)\left(x-2\right)^2\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}2x-7=0\\\left(x-2\right)^2=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)
c)\(\left(9x^2-25\right)-\left(6x-10\right)=0\Rightarrow\left(3x-5\right)\left(3x+5\right)-2\left(3x-5\right)=0\Rightarrow\left(3x-5\right)\left(3x+5-2\right)=0\Rightarrow\left[{}\begin{matrix}3x-5=0\\3x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=1\end{matrix}\right.\)
18 tháng 8 2021
a: Ta có: \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\)
\(\Leftrightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{5}\end{matrix}\right.\)
b: Ta có: \(\left(2x-7\right)\left(x-2\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(x-2\right)^2\cdot\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)
c: Ta có: \(\left(9x^2-25\right)-\left(6x-10\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+5-2\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)
5 tháng 1 2021
a) Ta có: \(\left(x^2-2x\right)^2-6x^2+12x+9=0\)
\(\Leftrightarrow\left(x^2-2x\right)^2-6\left(x^2-2x\right)+9=0\)
\(\Leftrightarrow\left(x^2-2x-3\right)^2=0\)
\(\Leftrightarrow x^2-2x-3=0\)
\(\Leftrightarrow x^2-3x+x-3=0\)
\(\Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Vậy: S={3;-1}
b) Ta có: \(\left(x^2+x+1\right)\left(x^2+x+2\right)=12\)
\(\Leftrightarrow\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12=0\)
\(\Leftrightarrow\left(x^2+x\right)^2+5\left(x^2+x\right)-2\left(x^2+x\right)-10=0\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x+5\right)-2\left(x^2+x+5\right)=0\)
\(\Leftrightarrow\left(x^2+x+5\right)\left(x^2+x-2\right)=0\)
\(\Leftrightarrow x^2+x-2=0\)(Vì \(x^2+x+5>0\forall x\))
\(\Leftrightarrow x^2+2x-x-2=0\)
\(\Leftrightarrow x\left(x+2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
Vậy: S={-2;1}
5 tháng 1 2021
2 ý a và b anh CTV nãy đã làm rồi nha, còn câu c này thì làm dài dòng+không chắc :VVV
c)\(\left(2x^2-3x+1\right)\left(2x^2+5x+1\right)-9x^2=0\)
\(\Leftrightarrow\left(2x^2-3x+1\right)\left(2x^2-3x+1+8x\right)-9x^2=0\)
\(\Leftrightarrow\left(2x^2-3x+1\right)^2+8x\left(2x^2-3x+1\right)+16x^2-25x^2=0\)
\(\Leftrightarrow\left(2x^2-3x+1+4x\right)^2-25x^2=0\)
\(\Leftrightarrow\left(2x^2+x+1\right)^2-25x^2=0\)
\(\Leftrightarrow\left(2x^2+x+1-5x\right)\left(2x^2+x+1+5x\right)=0\)
\(\Leftrightarrow\left(2x^2-4x+1\right)\left(2x^2+6x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(2x^2-4x+1\right)=0\\\left(2x^2+6x+1\right)=0\end{matrix}\right.\)
Rồi đến đây tự giải nhé, không phân tích được thì bấm máy tính là ra nha:vv
VT
1
14 tháng 6 2021
a) \(\left(x+2\right)\left(x^2-2x+4\right)+\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+4+x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x^2-x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\\left[x^2-2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]+\dfrac{23}{4}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(N\right)\\\left(x-\dfrac{1}{2}\right)^2+\dfrac{23}{4}\ge\dfrac{23}{4}>0\left(L\right)\end{matrix}\right.\)
Vậy \(S=\left\{-2\right\}\)
b) \(9x^2-4-\left(3x-2\right)^2=0\)
\(\Leftrightarrow\left(3x-2\right)\left(3x+2\right)-\left(3x-2\right)^2=0\)
\(\Leftrightarrow\left(3x-2\right)\left[\left(3x+2\right)-\left(3x-2\right)\right]=0\)
\(\Leftrightarrow\left(3x-2\right)\left(3x+2-3x+2\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\cdot4=0\)
\(\Leftrightarrow3x-2=0\)
\(\Leftrightarrow x=\dfrac{2}{3}\)
Vậy \(S=\left\{\dfrac{2}{3}\right\}\)
31 tháng 12 2022
b:=y^2+2y+1+9x^2-12x+4
=(y+1)^2+(3x-2)^2
a:
SỬa đề: 5y^2
=y^2-10y+25+9x^2+4y^2-12xy
=(y-5)^2+(3x-2y)^2
a: \(9x^2-12xy+16y^2=\left(3x-4y\right)^2>=0\)(luôn đúng)
b: \(16a^2-30ab+25b^2=\left(4a-5b\right)^2>=0\)(luôn đúng)