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\(P=\left(x^2-4x+4\right)+\left(y^2+8y+16\right)+2021\\ P=\left(x-2\right)^2+\left(y+4\right)^2+2021\ge2021\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-4\end{matrix}\right.\)

Lời giải:

$P(x)=x^2+y^2-4x+8y+2041=(x^2-4x+4)+(y^2+8y+16)+2021$

$=(x-2)^2+(y+4)^2+2021\geq 0+0+2021=2021$

Vậy $P(x)$ min = $2021$ khi $x-2=y+4=0$

$\Leftrightarrow x=2; y=-4$

a) Ta có: \(x^2-8x+7=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=7\end{matrix}\right.\)

b) Ta có: \(x^2+x-20=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=4\end{matrix}\right.\)

c) Ta có: \(3x^2+4x-4=0\)

\(\Leftrightarrow3x^2+6x-2x-4=0\)

\(\Leftrightarrow3x\left(x+2\right)-2\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(3x-2\right)=0\)

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

d) Ta có: \(3x^2-4x-7=0\)

\(\Leftrightarrow3x^2-7x+3x-7=0\)

\(\Leftrightarrow\left(3x-7\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-1\end{matrix}\right.\)

e) Ta có: \(5x^2-16x+3=0\)

\(\Leftrightarrow5x^2-15x-x+3=0\)

\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

f) Ta có: \(x^2+3x-10=0\)

\(\Leftrightarrow x^2+5x-2x-10=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

a)

\(x^2-8x+7=0\text{⇔}\text{⇔}x^2-7x-x-7=\left(x-7\right)\left(x-1\right)=0\text{⇔}\left[{}\begin{matrix}x=1\\x=7\end{matrix}\right.\)

Vậy nghiệm của đa thức : \(S=\left\{1;7\right\}\)

c)

\(3x^2+4x-4=0\text{⇔}3x^2+6x-2x-4=\left(3x-2\right)\left(x+2\right)=0\text{⇔}\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)

Vậy nghiệm của đa thức : \(S=\left\{\dfrac{2}{3};-2\right\}\)

b)

\(x^2+x-20=0⇔\left(x-4\right)\left(x+5\right)=0\text{⇔}\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)

d)

\(3x^2-4x-7=0\text{⇔}\left(3x-7\right)\left(x+1\right)=0\text{⇔}\left[{}\begin{matrix}x=-1\\x=\dfrac{7}{3}\end{matrix}\right.\)

e)

\(5x^2-16x+3\text{⇔}\left(x-3\right)\left(5x-1\right)=0\text{⇔}\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

f)

\(x^2+3x-10=0\text{⇔}\left(x-2\right)\left(x+5\right)=0\text{⇔}\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

\(\)

giúp mình nha

a)14𝑥^3𝑦∶10𝑥^2=1,4𝑥𝑦

b)(𝑥^3-27):(3-𝑥)=(𝑥-3)(𝑥^2+3𝑥+9):(3-𝑥)=-(𝑥^2+3𝑥+9)=-𝑥^2-3𝑥-9

c)8𝑥^3𝑦^3𝑧∶6𝑥𝑦^3=4/3𝑥^2𝑧

d)(𝑥^2−9𝑦^2+4𝑥+4)∶(𝑥+3𝑦+2)=((𝑥^2+4𝑥+2^2)-(3𝑦)^2):(𝑥+3𝑦+2)=((𝑥+2)^2-(3𝑦)^2):(𝑥+3𝑦+2)=(𝑥+2-3𝑦)(𝑥+2+3𝑦):(𝑥+3𝑦+2)=𝑥+2-3𝑦

a) \(14x^3y:10x^2=\dfrac{7}{5}xy\)

b) \(\left(x^3-27\right):\left(3-x\right)\)

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

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

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

\(=-x^2-3x-9\)

bạn nào giải giúp mình nha

c) \(8x^3y^3z:6xy^3=\dfrac{4}{3}x^2z\)

d) \(\left(x^2-9y^2+4x+4\right):\left(x^2+3y+2\right)\)

\(=\left[\left(x+2\right)^2-\left(3y\right)^2\right]:\left(x^2+3y+2\right)\)

\(=\left(x+3y+2\right)\left(x-3y+2\right):\left(x^2+3y+2\right)\)

\(=x-3y+2\)

a) 

8x³y³z:6xy³=\(\dfrac{4}{3}\)x²z

d) (x²−9y²+4x+4):(x+3y+2)

=[(x+2⁾²−(3y)²]:(x+3y+2)

=(x+3y+2)(x−3y+2):(x+3y+2)

=x-3y+2

Bài 4:

a) x = -3. Ta có: -3a + 5 = 0 -> -3a = -5 -> a = \(\frac{-5}{-3}\)--> a = \(\frac{5}{3}\)

b) x = \(\frac{1}{2}\). Ta có: \(\frac{1}{2}\).2 + 4a\(\frac{1}{2}\) - 5 = 0 -->  \(\frac{1}{2}\). (2 + 4a) = 5 --> 2 +4a = 5:\(\frac{1}{2}\)--> 2+ 4a = 10 --> 4a = 10-2 = 8 --> a = 2

c) x = -1. Ta có: 5.-1.3 + -1.2 - -1 + a = 0 --> -1 (15 + 2 - 1) + a = 0 --> -1. 16 + a = 0 --> -16 + a = 0 --> a = 16

d) x = 1. Ta có: a.1.4 - 2.1.3 + 1- 1 = 0 --> 1. (4a - 2.3 +1) - 1 = 0 --> 1.( 4a - 6 +1) = 1 --> 1.(4a-5) = 1 --> 4a = 6 --> a = \(\frac{3}{2}\)