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1.\(3x^2+12x-66=0\)
\(\Rightarrow\)\(3\left(x^2+4x+4\right)-78=0\)
\(\Rightarrow3\left(x+2\right)^2=78\)
\(\Rightarrow\left(x+2\right)^2=26\)
\(\Rightarrow x+2=\sqrt{26}\)hoặc \(x+2=-\sqrt{26}\)
\(\Rightarrow x=\sqrt{26}-2\)hoặc \(x=-\sqrt{26}-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
9x2 +4x2 =20xy =>\(\int^{9x^2-12xy+4y^2=8xy}_{9x^2+12xy+4y^2=32xy}\Leftrightarrow\int^{\left(3x-2y\right)^2=8xy}_{\left(3x+2y\right)^2=32xy}\Leftrightarrow\frac{\left(3x-2y\right)^2}{\left(3x+2y\right)^2}=\frac{1}{4}=A^2\)
A>0 => A =1/2
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Answer:
\(B=\left(2x+1\right)^2+\left(3x-1\right)^2+2.\left(2x+1\right).\left(3x-1\right)+5\)
\(=[\left(2x+1\right)^2+\left(3x-1\right)^2+2.\left(2x+1\right).\left(3x-1\right)]+5\)
\(=[\left(2x+1\right)+\left(3x-1\right)]^2+5\)
\(=\left(2x+1+3x-1\right)^2+5\)
\(=\left(5x\right)^2+5\)
\(=25x^2+5\)
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\(x\left(3x-5\right)-9x+15=0\)
\(\Leftrightarrow x\left(3x-5\right)-3\left(3x-5\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(3x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\3x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{5}{3}\end{cases}}\)
\(3x\left(x-5\right)-2\left(5-x\right)=0\)
\(\Leftrightarrow3x\left(x-5\right)+2\left(x-5\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x+2=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-2}{3}\\x=5\end{cases}}\)
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3x2 + 3x - 5( x + 1 ) = 0
<=> ( 3x2 + 3x ) - 5( x + 1 ) = 0
<=> 3x( x + 1 ) - 5( x + 1 ) = 0
<=> ( x + 1 )( 3x - 5 ) = 0
<=> x + 1 = 0 hoặc 3x - 5 = 0
<=> x = -1 hoặc x = 5/3
\(3x\ge0\)
\(\Leftrightarrow x\ge\dfrac{0}{3}\)
\(\Leftrightarrow x\ge0\)
Vậy: ...