a/ \(2x^3=8x\)

\(2.8=2x^3\)

\(16=2x^3\)

\(x^3=16:2\)

\(x^3=8\)

\(x=2\)

phần b mk chưa nghiên cứu dc

a. 12xy² - 8x²y = 4xy . (3y - 2x)

b. 3x + 3y - x² - xy = (3x + 3y) - (x² + xy) = 3 . (x + y) - x . (x + y) = (x + y)(3 - x)

GIUSP MIK VS MN ƠI

Bài 2:

a: =>2x^2-8+3x-2x^2=15

=>3x=23

=>x=23/3

b: \(\Leftrightarrow\left(x+3\right)\left(x-5\right)=0\)

=>x=5 hoặc x=-3

c: =>x^2+7x-30=0

=>(x+10)(x-3)=0

=>x=3 hoặc x=-10

Viết rõ hơn chút mình làm cho !!!

1) \(x^2-2x+5+y^2-4y=0\)

\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2-4y+4\right)=0\)

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

Vì \(\left(x-1\right)^2\ge0;\left(y-2\right)^2\ge0\)

\(\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2\ge0\)

Để PT bằng 0 thì:

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

\(\Rightarrow x=1\)và \(y=2\)

2) \(y^2+2y+5-12x+9x^2=0\)

\(\Leftrightarrow\left(y^2+2y+1\right)+\left(9x^2-12x+4\right)=0\)

\(\Leftrightarrow\left(y+1\right)^2+\left(3x-2\right)^2=0\)

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..............<Giải thích như câu đầu>......................

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\(\left(y+1\right)^2=0\)và \(\left(3x-2\right)^2=0\)

\(\Rightarrow y=-1\)và \(x=\frac{2}{3}\)

3) \(x^2+20+9y^2+8x-12y=0\)

\(\Leftrightarrow\left(x^2+8x+16\right)+\left(9y^2-12y+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)^2+\left(3y-2\right)^2=0\)

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...............<Giải thích như câu đầu>..............

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\(\left(x+4\right)^2=0\)và \(\left(3y-2\right)^2=0\)

\(\Rightarrow x=-4\)và \(y=\frac{2}{3}\)

1) \(x^2-2x+5+y^2-4y=0\)

\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2-4y+4\right)=0\)

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

Vì \(\left(x-1\right)^2\ge0;\left(y-2\right)^2\ge0\)

\(\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2\ge0\)

Để PT bằng 0 thì:

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

\(\Rightarrow x=1\)và \(y=2\)

2) \(y^2+2y+5-12x+9x^2=0\)

\(\Leftrightarrow\left(y^2+2y+1\right)+\left(9x^2-12x+4\right)=0\)

\(\Leftrightarrow\left(y+1\right)^2+\left(3x-2\right)^2=0\)

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..............<Giải thích như câu đầu>......................

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\(\left(y+1\right)^2=0\)và \(\left(3x-2\right)^2=0\)

\(\Rightarrow y=-1\)và \(x=\frac{2}{3}\)

3) \(x^2+20+9y^2+8x-12y=0\)

\(\Leftrightarrow\left(x^2+8x+16\right)+\left(9y^2-12y+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)^2+\left(3y-2\right)^2=0\)

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...............<Giải thích như câu đầu>..............

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\(\left(x+4\right)^2=0\)và \(\left(3y-2\right)^2=0\)

\(\Rightarrow x=-4\)và \(y=\frac{2}{3}\)

\(1,x^2-2x+5+y^2-4y=0\)

\(\Rightarrow\left(x^2-2x+1\right)+\left(y^2-4y+4\right)=0\)

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

\(\Leftrightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y-2\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}}\)

\(2,y^2+2y+5-12x+9x^2=0\)

\(\Rightarrow\left(y^2+2y+1\right)+\left(9x^2-12x+4\right)=0\)

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

\(\Leftrightarrow\hept{\begin{cases}\left(y+1\right)^2=0\\\left(3x-2\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}y=-1\\x=\frac{2}{3}\end{cases}}}\)

\(3,x^2+20+9y^2+8x-12y=0\)

\(\Rightarrow\left(x^2+8x+16\right)+\left(9y^2-12y+4\right)=0\)

\(\Rightarrow\left(x+4\right)^2+\left(3y-2\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}\left(x+4\right)^2=0\\\left(3y-2\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=-4\\y=\frac{2}{3}\end{cases}}}\)

\(\frac{121212}{161616}-\left(\frac{151515}{323232}-x\right)=2\)

=> \(\frac{3}{4}-\left(\frac{15}{32}-x\right)=2\)

=> \(\frac{15}{32}-x=\frac{3}{4}-2\)

=> \(\frac{15}{32}-x=-\frac{5}{4}\)

=> \(x=\frac{15}{32}-\frac{-5}{4}=\frac{15}{32}+\frac{5}{4}=\frac{55}{32}\)

b) \(\frac{x}{2}+\frac{x}{6}+\frac{x}{12}+\frac{x}{20}+\frac{x}{30}+\frac{x}{42}+\frac{x}{56}+\frac{x}{72}+\frac{x}{90}=\frac{9}{5}\)

=> \(\frac{x}{1\cdot2}+\frac{x}{2\cdot3}+\frac{x}{3\cdot4}+\frac{x}{4\cdot5}+\frac{x}{5\cdot6}+\frac{x}{6\cdot7}+\frac{x}{7\cdot8}+\frac{x}{8\cdot9}+\frac{x}{9\cdot10}=\frac{9}{5}\)

=> \(\frac{x}{1}-\frac{x}{2}+\frac{x}{2}-\frac{x}{3}+...+\frac{x}{9}-\frac{x}{10}=\frac{9}{5}\)

=> \(\frac{x}{1}-\frac{x}{10}=\frac{9}{5}\)

=> \(\frac{10x-x}{10}=\frac{9}{5}\)

=> \(\frac{9x}{10}=\frac{9}{5}\)

=> \(\frac{9x}{10}=\frac{9\cdot2}{5\cdot2}=\frac{18}{10}\)

=> x = 2

4a) \(\left(a+b\right)^2=a^2+2ab+b^2\)

\(\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab=a^2+b^2+2ab\)

=> (a+b)^2=(a-b)^2+4ab

(x – 3)(2x + 1) = 0

x = 3 hay x = -1/2

\(5,2.x+\left(3,2.x-1\right)=2\)

\(5,2.x+3,2x-1=2\)

\(5,2x+3,2x=3\)

\(8,4x=3\)

\(x=2,8\)

5,2*x+(3,2*x-1)=2

5,2*x+3,2*x-1=2

5,2*x+3,2*x=3

8.4*x=3

x=2.8

Bài 1:

a) \(x^2-6x+15=\left(x^2-6x+9\right)+6=\left(x-3\right)^2+6\ge6\)

Dấu "=" xảy ra \(\Leftrightarrow x=3\)

b) \(3x^2-15x+4=3\left(x^2-5x+\dfrac{25}{4}\right)-\dfrac{59}{4}=3\left(x-\dfrac{5}{2}\right)^2-\dfrac{59}{4}\ge-\dfrac{59}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{5}{2}\)

Bài 2:

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

\(\Rightarrow\left(x+5\right)\left(x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)

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

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

\(\Rightarrow x=2\left(do.x^2+7\ge7>0\right)\)