What???

Nà ní!!!!!!!!!

1)\(2x^2+9y^2-6xy-6x-12y+2004\)

\(=x^2+x^2-6xy+9y^2-6x-12y+2004\)

\(=x^2+\left(x-3y\right)^2-10x+4x-12y+2004\)

\(=\left(x-3y\right)^2+4\left(x-3y\right)+x^2-10x+2004\)

\(=\left(x-3y\right)^2+4\left(x-3y\right)+x^2-10x+4+25+1975\)

\(=\left[\left(x-3y\right)^2+4\left(x-3y\right)+4\right]+\left(x^2-10x+25\right)+1975\)

\(=\left(x-3y+2\right)^2+\left(x-5\right)^2+1975\ge1975\)

Dấu "=" khi \(\begin{cases}\left(x-5\right)^2=0\\\left(x-3y+2\right)^2=0\end{cases}\)\(\Leftrightarrow\begin{cases}x=5\\y=\frac{7}{3}\end{cases}\)

Vậy Min=1975 khi \(\begin{cases}x=5\\y=\frac{7}{3}\end{cases}\)

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

Đặt \(t=x^2+x\) ta có:

\(t\left(t-4\right)=t^2-4t+4-4\)

\(=\left(t-2\right)^2-4\ge-4\)

Dấu "=" khi \(t-2=0\Leftrightarrow t=2\Leftrightarrow x^2+x=2\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\\x=1\end{array}\right.\)

Vậy Min=-4 khi \(\left[\begin{array}{nghiempt}x=-2\\x=1\end{array}\right.\)

3)\(\left(x^2+5x+5\right)\left[\left(x+2\right)\left(x+3\right)+1\right]\)

\(=\left(x^2+5x+5\right)\left[x^2+5x+6+1\right]\)

Đặt \(t=x^2+5x+5\) ta có:

\(t\left(t+1\right)=t^2+t+\frac{1}{4}-\frac{1}{4}=\left(t+\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)

Dấu "=" khi \(t+\frac{1}{2}=0\Leftrightarrow t=-\frac{1}{2}\Leftrightarrow x^2+5x+5=-\frac{1}{2}\)\(\Leftrightarrow x_{1,2}=\frac{-10\pm\sqrt{12}}{4}\)

Vậy Min=\(-\frac{1}{4}\) khi \(x_{1,2}=\frac{-10\pm\sqrt{12}}{4}\)

4)\(\left(x-1\right)\left(x-3\right)\left(x^2-4x+5\right)\)

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

Đặt \(t=x^2-4x+3\) ta có:

\(t\left(t+2\right)=t^2+2t+1-1=\left(t+1\right)^2-1\ge-1\)

Dấu "=" khi \(t+1=0\Leftrightarrow t=-1\Leftrightarrow x^2-4x+3=-1\Leftrightarrow x=2\)

Vậy Min=-1 khi x=2

Thank you !

mk biết làm

dap an la0 chac ban thi violympic de ot tring ga qua

Theo bài ra , ta có : 

\(\frac{x}{4}+\frac{x}{8}+\frac{x}{16}=\frac{x}{9}+\frac{x}{27}+\frac{x}{81}\)

\(\Rightarrow\frac{x}{4}+\frac{x}{8}+\frac{x}{16}-\frac{x}{9}-\frac{x}{27}-\frac{x}{81}=0\)

\(\Rightarrow x=0\)

Vậy \(x=0\)

Ta có : \(\frac{\left(4^x\right)^2}{2^x}=8\)

\(\Rightarrow4^{2x}=8.2^x\)

\(\Rightarrow4^{2x}=2^3.2^x\)

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

\(\Rightarrow2^{4x}=2^{x+3}\)

=> 4x = x + 3

=> 3x = 3

=> x = 1

Vậy x = 1. 

a) ta có : \(5^5-5^4+5^3=5^3.\left(5^2-5+1\right)=5^3.\left(25-5+1\right)\)

\(5^3.21=5^3.3.7⋮7\) (đpcm)

b) ta có : \(7^6+7^5-7^4=7^4.\left(7^2+7-1\right)=7^4.\left(49+7-1\right)\)

\(=7^4.55=7^4.5.11⋮11\) (đpcm)

c) ta có : \(3^{x+2}-2^{x+3}+3^x-2^{x+1}=3^{x+2}+3^x-2^{x+3}-2^{x+1}\)

\(=3^x\left(3^2+1\right)-2^x\left(2^3+2\right)=3^x.\left(9+1\right)-2^x.\left(8+2\right)\)

\(=3^x.10-2^x.10=10\left(3^x-2^x\right)⋮10\) (đpcm)

d) \(3^{x+3}+3^{x+1}+2^{x+3}+2^{x+2}=3^x.\left(3^3+3\right)+2^x.\left(2^3+2^2\right)\)

\(=3^x.\left(27+3\right)+2^x\left(8+4\right)=3^x.30+2^x.12=6.\left(3^x.5+2^x.2\right)⋮6\) (đpcm)

a)Ta có:\(5^5-5^4+5^3=5^3\left(5^2-5+1\right)=5^3.21\)(vì 21 chia hết cho 7)

\(\)\(\RightarrowĐPCM\)

b)Ta có: \(7^6+7^5-7^4⋮11=7^4\left(7^2+7-1\right)=7^4.55⋮11\)

\(\Rightarrowđpcm\)

\(a,\frac{-1}{2}+\left(x-3\right):\frac{-1}{2}=-1\frac{2}{3}.\)

\(\Rightarrow\left(x-3\right):\frac{-1}{2}=-1\frac{2}{3}-\frac{-1}{2}=\frac{-7}{6}\)

\(\Rightarrow x-3=\frac{-7}{6}\cdot\frac{-1}{2}=\frac{7}{12}\)

\(\Rightarrow x=\frac{7}{12}+3=3\frac{7}{12}\)

\(b.2,25+\frac{3}{2}:\left(x-5\right)=2\frac{1}{2}\)

\(\Rightarrow\frac{3}{2}:\left(x-5\right)=2\frac{1}{2}-2,25=\frac{1}{4}\)

\(\Rightarrow x-5=\frac{3}{2}:\frac{1}{4}=6\)

\(\Rightarrow x=6+5=11\)

\(c,\left(\frac{1}{3}-x\right)^2=\frac{1}{4}=\left(\frac{1}{2}\right)^2=\left(-\frac{1}{2}\right)^2\)

\(\Rightarrow\orbr{\begin{cases}\frac{1}{3}-x=\frac{1}{2}\\\frac{1}{3}-x=-\frac{1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}-\frac{1}{2}=-\frac{1}{6}\\x=\frac{1}{3}-\frac{-1}{2}=\frac{5}{6}\end{cases}}\)

\(d,\frac{3}{2}+\frac{x-1}{3}=1\)

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

\(\Rightarrow x-1=-\frac{1}{2}\cdot3=-\frac{3}{2}\)

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

\(e,-\frac{6}{8}+\frac{x}{12}=\frac{5}{6}\)

\(\Rightarrow\frac{x}{12}=\frac{5}{6}-\frac{-6}{8}=\frac{19}{12}\)

\(\Rightarrow x=19\)

\(g,\frac{1}{2}-\frac{1}{3}\left(x-2\right)=-\frac{2}{3}\)

\(\Rightarrow-\frac{1}{3}\left(x-2\right)=-\frac{2}{3}-\frac{1}{2}=-\frac{7}{6}\)

\(\Rightarrow x-2=\frac{-7}{6}:\frac{-1}{3}=\frac{7}{2}\)

\(\Rightarrow x=\frac{7}{2}+2=2\frac{7}{2}\)

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

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

\(\Rightarrow x+1=3:\frac{5}{2}=\frac{6}{5}\)

\(\Rightarrow x=\frac{6}{5}-1=\frac{1}{5}\)

\(k,\frac{x}{3}-\frac{1}{2}=-2\left(x+1\right)+3\)

\(\Rightarrow x\cdot\frac{1}{3}-\frac{1}{2}=-2x-2+3\)

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

\(\Rightarrow\frac{7}{3}x=\frac{3}{2}\Rightarrow x=\frac{3}{2}:\frac{7}{2}=\frac{3}{7}\)

đề bài yêu cầu gì vậy em

tìm x biết :

a, (x+1).3 = 2.2

=>3 x+3 =4

=> 3x=1

=> x=1/3

b, (x-2) .4 =(x+1).3

=>4x-8=3x+3

=>4x-3x=8+3

=>x=11

c, lam tg tu cau b

d, (x-1)(x+3)=(x+2)(x-2)

\(x^2\)+3x-x-3=\(x^2\)-2x+2x-4

x^2 +2x-3=x^2-4

x^2-x^2+2x=3-4

2x=-1

x=-0,5

\(\frac{x+1}{2}=\frac{2}{3}\)

\(\Rightarrow3.\left(x+1\right)=2.2\)

\(\Rightarrow3x+3=4\)

\(\Rightarrow3x=4-3\)

\(\Rightarrow3x=1\)

\(\Rightarrow x=\frac{1}{3}\)

\(b,\frac{x-2}{3}=\frac{x+1}{4}\)

\(\Rightarrow4.\left(x-2\right)=3.\left(x+1\right)\)

\(\Rightarrow4x-8=3x+3\)

\(\Rightarrow4x-3x=3+8\)

\(\Rightarrow x=11\)

\(c,\frac{x-3}{x+5}=\frac{5}{7}\)

\(\Rightarrow7.\left(x-3\right)=5.\left(x+5\right)\)

\(\Rightarrow7x-21=5x+25\)

\(\Rightarrow7x-5x=25+21\)

\(\Rightarrow2x=46\)

\(\Rightarrow x=23\)

\(d,\frac{x-1}{x+2}=\frac{x-2}{x+3}\)

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

\(\Rightarrow x^2+2x-3=x^2-4\)

\(\Rightarrow2x=-1\)

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