a) x²+8x+16-x²+1=16

8x+17=16

8x=-1

x=-1/8

b) 4x²-4x+1+x²+6x+9-5x²+245=0

2x+255=0

x=-255/2

1: \(=x^2+x+5=x^2+x+\dfrac{1}{4}+\dfrac{19}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}>=\dfrac{19}{4}\)

Dấu '=' xảy ra khi x=-1/2

2: \(=-\left(x^2+4x-9\right)\)

\(=-\left(x^2+4x+4-13\right)\)

\(=-\left(x+2\right)^2+13\le13\)

Dấu '=' xảy ra khi x=-2

3: \(=x^2-4x+4+y^2+2y+1+2\)

\(=\left(x-2\right)^2+\left(y+1\right)^2+2\ge2\)

Dấu '=' xảy ra khi x=2 và y=-1

a) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x\right)^2-5^2-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(-2\right).\left(2x-5\right)=0\)

\(\Leftrightarrow2x-5=0\)

\(\Leftrightarrow x=\dfrac{5}{2}\)

a,\(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Rightarrow\left(4x^2-25\right)-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Rightarrow\left(2x-5\right)^2-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Rightarrow\left(2x-5\right)\left(2x-5-2x-7\right)=0\)

\(\Rightarrow\left(2x-5\right)\left(-12\right)=0\)

\(\Rightarrow2x-5=0\)

\(\Rightarrow2x=5\)

\(\Rightarrow x=\dfrac{5}{2}\)

\(b,2x^3+3x^2+2x+3=0\)

\(\Rightarrow\left(2x^3+2x\right)+\left(3x^2+3\right)=0\)

\(\Rightarrow2x\left(x^2+1\right)+3\left(x^2+1\right)=0\)

\(\Rightarrow\left(2x+3\right)\left(x^2+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x+3=0\\x^2+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=-3\\x^2=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=1\end{matrix}\right.\)

\(c,x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Rightarrow\left(x^3+27\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Rightarrow\left(x+3\right)^3+\left(x+3\right)\left(x-9\right)=0\)

\(\Rightarrow\left(x+3\right)\left(x^2+9+x-9\right)=0\)

\(\Rightarrow\left(x+3\right).x^3=0\)

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x^3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=0\end{matrix}\right.\)

\(d,x^2\left(x+7\right)-4\left(x+7\right)=0\)

\(\Rightarrow\left(x^2-4\right)\left(x+7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2-4=0\\x+7=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2=4\\x=-7\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\)

cái này có vội không? nếu không thì sáng mai mình giải cho bạn?

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