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bài 1 câu a bỏ x= nhé !

\(ĐK:x\ge-2\)

\(\Leftrightarrow x^3+6x^2+12x+8+2\sqrt{\left(x+2\right)^3}+1-9x^2-18x-9=0\)

\(\Leftrightarrow\left(x+2\right)^3+2\sqrt{\left(x+2\right)^3}+1-\left(9x^2+18x+9\right)=0\)

\(\Leftrightarrow\left[\left(x+2\right)^3+1\right]^2-9\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow\left[\left(x+2\right)^3+1\right]^2-9\left(x+1\right)^2=0\)

ta có: ( 2 trường hợp xảy ra )

TH1: \(\left[\left(x+2\right)^3+1\right]^2=9\left(x+1\right)^2\)

\(\Leftrightarrow\left(x+2\right)^3+1=\left(9x+9\right)\)

\(\Leftrightarrow\left(x+2\right)^3-9x=8\)

\(\Leftrightarrow x^3+6x^2+12x+8-9x-8=0\)

\(\Leftrightarrow x^3+6x^2+3x=0\)

\(\Leftrightarrow x\left(x^2+6x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+6x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(n\right)\\\left[{}\begin{matrix}x=-3+\sqrt{6}\left(n\right)\\-3-\sqrt{6}\left(l\right)\end{matrix}\right.\end{matrix}\right.\)

TH2:\(\left[{}\begin{matrix}\left(x+3\right)^3+1=0\\9\left(x+1\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x+3\right)^3=-1\\\left(9x+9\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=-1\\9x=-9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-4\left(l\right)\\x=-1\left(n\right)\end{matrix}\right.\)

Vậy \(S=\left\{0;-1;-3+\sqrt{6}\right\}\)

( ko bít đúng ko nha bạn ơi )

a) <=> \(6x^2-5x+3-2x+3x\left(3-2x\right)=0\)

<=> \(6x^2-5x+3-2x+9x-6x^2=0\)

<=> \(2x+3=0\)

<=> \(x=\frac{-3}{2}\)

b) <=> \(10\left(x-4\right)-2\left(3+2x\right)=20x+4\left(1-x\right)\)

<=> \(10x-40-6-4x=20x+4-4x\)

<=> \(6x-46-16x-4=0\)

<=> \(-10x-50=0\)

<=> \(-10\left(x+5\right)=0\)

<=> \(x+5=0\)

<=> \(x=-5\)

c) <=> \(8x+3\left(3x-5\right)=18\left(2x-1\right)-14\)

<=> \(8x+9x-15=36x-18-14\)

<=> \(8x+9x-36x=+15-18-14\)

<=> \(-19x=-14\)

<=> \(x=\frac{14}{19}\)

d) <=>\(2\left(6x+5\right)-10x-3=8x+2\left(2x+1\right)\)

<=> \(12x+10-10x-3=8x+4x+2\)

<=> \(2x-7=12x+2\)

<=> \(2x-12x=7+2\)

<=> \(-10x=9\)

<=> \(x=\frac{-9}{10}\)

e) <=> \(x^2-16-6x+4=\left(x-4\right)^2\)

<=> \(x^2-6x-12-\left(x-4^2\right)=0\)

<=> \(x^2-6x-12-\left(x^2-8x+16\right)=0\)

<=> \(x^2-6x-12-x^2+8x-16=0\)

<=> \(2x-28=0\)

<=> \(2\left(x-14\right)=0\)

<=> x-14=0

<=> x=14

Luffy , cậu sai câu c nhé , kia là -17 ạ => x=17/19

\(\frac{x^2}{\left(x+2\right)^2}+3=3x^2-6x\left(đkxđ:x\ne-2\right)\)

\(< =>\frac{x^2}{\left(x+2\right)^2}=3x^2-6x-3\)

\(< =>x^2=\left(3x^2-6x-3\right)\left(x^2+4x+4\right)\)

\(< =>x^2=3x^4+12x^3+12x^2-6x^3-24x^2-14x-3x^2-12x-12\)

\(< =>3x^4+6x^3-16x^2-26x-12=0\)

Đến đây dễ rồi ha !

\(\frac{3x-1}{2}-\frac{2-6x}{5}=\frac{1}{2}+\left(3x-1\right)\)

\(\Leftrightarrow\frac{3x-1}{2}+\frac{2\left(3x-1\right)}{5}-\left(3x-1\right)=\frac{1}{2}\)

\(\Leftrightarrow\left(3x-1\right)\left(\frac{1}{2}+\frac{2}{5}-1\right)=\frac{1}{2}\)

\(\Leftrightarrow\frac{-1}{10}\left(3x-1\right)=\frac{1}{2}\)

\(\Leftrightarrow3x-1=-5\)

\(\Leftrightarrow3x=-4\Leftrightarrow x=\frac{-4}{3}\)

Vậy nghiệm duy nhất của phương trình là\(x=\frac{-4}{3}\)

\(\left(x^2+2x+1\right)-\frac{x+1}{3}=\frac{6\left(x+1\right)^2-5x-5}{6}\)

\(\Leftrightarrow\left(x+1\right)^2-\frac{x+1}{3}=\frac{6\left(x+1\right)^2-5\left(x+1\right)}{6}\)

\(\Leftrightarrow\left(x+1\right)^2-\frac{x+1}{3}=\frac{\left(x+1\right)\left(6x+6-5\right)}{6}\)

\(\Leftrightarrow\left(x+1\right)^2-\frac{x+1}{3}=\frac{\left(x+1\right)\left(6x+1\right)}{6}\)

\(\Leftrightarrow\left(x+1\right)^2-\frac{x+1}{3}-\frac{\left(x+1\right)\left(6x+1\right)}{6}=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+1-\frac{1}{3}-\frac{6x+1}{6}\right)=0\)

\(\Leftrightarrow\frac{1}{2}\left(x+1\right)=0\)

\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

Vậy nghiệm duy nhất của phương trình là\(x=-1\)

\(\frac{x^2}{\left(x+2\right)^2}=3x^2-6x-3\)

\(\frac{x^2}{x^2+4x+4}=3x^2-6x-3\)

\(x^2=\left(3x^2-6x-3\right)\left(x^2+4x+4\right)\)

\(x^2=3x^4+12x^3+12x^2-6x^3-24x^2-24x-3x^2-12x-12\)

\(x^2=3x^4+6x^3-15x^2-36x-12\)

\(x^2=3\left(x^4+2x^3-5x^2-12x-4\right)\)

Đến đây bí rồi

Cmr

A1^3 +a2^3 +....+an^3 chia hết cho 3

Biết rằng a1,a2,a3......,an là các chữa số của 2019^2018

nhân chéo ta có 3x⁴+6x³-16x²-36x-12=0\(\Leftrightarrow\)

3x⁴-18x² +6x³-36x +2x²-12=0\(\Leftrightarrow\)
3x²(x²-6)+6x(x²-6)+2(x²-6)=0

đến đây là ok

\(\frac{x^2}{\left(x+2\right)^2}+3=3x^2-6xĐK:x\ne-2\)

\(\Leftrightarrow\frac{x^2}{\left(x+2\right)^2}+\frac{3\left(x+2\right)^2}{\left(x+2\right)^2}=\frac{\left(3x^2-6\right)\left(x+2\right)^2}{\left(x+2\right)^2}\)

Khử mẫu và rút gọn ta đc : \(-3x^4-12x^3-2x^2+36x+36=0\)

Mời nhân tài chứ e chịu r