\(\Rightarrow y^3+5y^2-6y^2-30y+9y+45=0\)

\(\Rightarrow y^2\left(y+5\right)-6y\left(y+5\right)+9\left(y+5\right)=0\)

\(\Rightarrow\left(y^2-6y+9\right)\left(y+5\right)=0\)

\(\Rightarrow\left(y-3\right)^2\left(y+5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\left(y-3\right)^2=0\Rightarrow y=3\\y+5=0\Rightarrow y=-5\end{cases}}\)

Vậy ........................

Ta có : \(y^3-y^2-21y+45=0\)

\(\Leftrightarrow y^3+5y^2-6y^2-30y+9y+45=0\)

\(\Leftrightarrow y^2\left(y+5\right)-6y\left(y+5\right)+9\left(y+5\right)=0\)

\(\Leftrightarrow\left(y+5\right)\left(y^2-6y+9\right)=0\)

\(\Leftrightarrow\left(y+5\right)\left(y-3\right)^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}y+5=0\\y-3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}y=-5\\y=3\end{cases}}\)

Vậy tập nghiệm của phương trình là : \(S=\left\{-5;3\right\}\)

y-21y +45 = 0

-20y + 5 = 0

-20 y = -5

y=0,25

a. y³ - y² - 21y +45 = 0

⇔y³+5y²-6y²-30y+9y+45=0

⇔(y³+5y²)-(6y²+30y)+(9y+45)=0

⇔y²(y+5)-6y(y+5)+9(y+5)=0

⇔(y+5)(y²-6y+9)=0

⇔(y+5)(y-3)²=0

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

vậy s={-5;-3}

d) Ta có: \(\left(y+3\right)^2\ge0\forall y\)

\(\left(y+5\right)^2\ge0\forall y\)

Do đó: \(\left(y+3\right)^2+\left(y+5\right)^2\ge0\forall y\)

mà \(\left(y+3\right)^2+\left(y+5\right)^2=0\)

nên \(\left\{{}\begin{matrix}\left(y+3\right)^2=0\\\left(y+5\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+3=0\\y+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-3\\y=-5\end{matrix}\right.\)

Vậy: y=-3 và y=-5

a, y² + 7y + 2 = 0

y(y + 7) + 2 = 0

y = \(-\frac{7-\sqrt{41}}{2}\) ; \(-\frac{7+\sqrt{41}}{2}\)

*(Nâng cao): - 0.29843788... ; - 6.70156211...

Vậy y = \(-\frac{7-\sqrt{41}}{2}\) ; \(-\frac{7+\sqrt{41}}{2}\).

Chúc bạn học tốt!

Câu x ) là bằng - 5 nhé mấy bạn. Làm giúp mình tất cả nhé ! Mình cảm ơn nhiều lắm !

sai đề rồi bạn ơi

a) ta có : \(N=-21x^{99}-21x^{98}-...-21x^2-21x\)

\(\Rightarrow xN=-21x^{100}-21x^{99}-...-21x^2-21x^2\)

\(\Rightarrow xN-N=-21x^{100}+21x\)

\(\Leftrightarrow\left(x-1\right)N=-21x^{100}+21x\Leftrightarrow N=\dfrac{21x-21x^{100}}{x-1}\)

\(\Rightarrow A=x^{100}-21x^{99}-21x^{98}-...-21x^2-21x+2010\)

\(=x^{100}+\dfrac{21x-21x^{100}}{x-1}+2010\)

\(=\dfrac{21x-21x^{100}+x^{101}-x^{100}+2010x-2010}{x-1}\)

\(=\dfrac{x^{101}-22x^{100}+2031x-2010}{x-1}\)

thay \(x=22\) ta có : \(A=\dfrac{22^{101}-22.22^{100}+2031.22-2010}{22-1}\)

\(=\dfrac{22^{101}-22^{101}+2031.22-2010}{21}=\dfrac{2031.22-2010}{21}=2032\)

vậy ............................................................................................................

câu b lm tương tự .

c.

\(4y^2+1=4y\)

\(\Leftrightarrow4y^2-4y+1=0\)

\(\Leftrightarrow4y^2-2y-2y+1=0\)

\(\Leftrightarrow2y\left(2y-1\right)-\left(2y-1\right)=0\)

\(\Leftrightarrow\left(2y-1\right)^2=0\)

\(\Leftrightarrow y=0\)

d.

\(y^2-2y=80\)

\(\Leftrightarrow y^2-2y-80=0\)

\(\Leftrightarrow y^2-10y+8y-80=0\)

\(\Leftrightarrow y\left(y-10\right)+8\left(y-10\right)=0\)

\(\Leftrightarrow\left(y+8\right)\left(y-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y+8=0\\y-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=-8\\y=10\end{matrix}\right.\)