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 !

a: \(\Leftrightarrow\dfrac{7}{2}x-\dfrac{3}{4}=\dfrac{1}{2}x+\dfrac{5}{2}\)

\(\Leftrightarrow3x=\dfrac{5}{2}+\dfrac{3}{4}=\dfrac{10}{4}+\dfrac{3}{4}=\dfrac{13}{4}\)

=>x=13/12

b: \(\Leftrightarrow x\cdot\left(\dfrac{2}{3}-\dfrac{1}{2}\right)=-\dfrac{1}{3}+\dfrac{2}{5}\)

\(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{-5+6}{15}=\dfrac{1}{15}\)

\(\Leftrightarrow x=\dfrac{1}{15}:\dfrac{1}{6}=\dfrac{2}{5}\)

c: \(\Leftrightarrow x\cdot\dfrac{1}{3}+x\cdot\dfrac{2}{5}+\dfrac{2}{5}=0\)

\(\Leftrightarrow x\cdot\dfrac{11}{15}=-\dfrac{2}{5}\)

\(\Leftrightarrow x=-\dfrac{2}{5}:\dfrac{11}{15}=\dfrac{-2}{5}\cdot\dfrac{15}{11}=\dfrac{-30}{55}=\dfrac{-6}{11}\)

d: \(\Leftrightarrow-\dfrac{1}{3}x+\dfrac{1}{2}+\dfrac{2}{3}-x-\dfrac{1}{2}=5\)

\(\Leftrightarrow-\dfrac{4}{3}x+\dfrac{2}{3}=5\)

\(\Leftrightarrow-\dfrac{4}{3}x=5-\dfrac{2}{3}=\dfrac{13}{3}\)

\(\Leftrightarrow x=\dfrac{13}{3}:\dfrac{-4}{3}=\dfrac{-13}{4}\)

e: \(\Leftrightarrow\left(\dfrac{x+2015}{5}+1\right)+\left(\dfrac{x+2016}{4}+1\right)=\left(\dfrac{x+2017}{3}+1\right)+\left(\dfrac{x+2018}{2}+1\right)\)

=>x+2020=0

hay x=-2020

a) \(\left|x-2\right|+\left|x-5\right|=\left|x-2\right|+\left|5-x\right|\ge\left|x-2+5-x\right|=3\)

Dấu''='' xảy ra \(\Leftrightarrow\begin{cases}x-2\ge0\\5-x\le0\end{cases}\)\(\Leftrightarrow\begin{cases}x\ge2\\x\le5\end{cases}\)\(\Leftrightarrow2\le x\le5\)

Bài 1:

a) \(\frac{1}{5}x^4y^3-3x^4y^3\)

= \(\left(\frac{1}{5}-3\right)x^4y^3\)

= \(-\frac{14}{5}x^4y^3.\)

b) \(5x^2y^5-\frac{1}{4}x^2y^5\)

= \(\left(5-\frac{1}{4}\right)x^2y^5\)

= \(\frac{19}{4}x^2y^5.\)

Mình chỉ làm 2 câu thôi nhé, bạn đăng nhiều quá.

Chúc bạn học tốt!

làm hộ mình cái để mai nộp thầy,ai nhanh và đúng thì mình k cho nha

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

\(-\frac{7}{2}+\frac{1}{3}.\left(x-1\right)=-\frac{4}{3}:\frac{7}{3}\)

\(-\frac{7}{2}+\frac{1}{3}.\left(x-1\right)=-\frac{4}{7}\)

\(\frac{1}{3}.\left(x-1\right)=-\frac{4}{7}-\frac{-7}{2}\)

\(\frac{1}{3}.\left(x-1\right)=\frac{41}{14}\)

\(\Rightarrow x-1=\frac{41}{14}:\frac{1}{3}\)

\(\Rightarrow x-1=\frac{123}{14}\)

\(\Rightarrow x=\frac{123}{14}+1\)

\(\Rightarrow x=\frac{137}{14}\)

a, \(5^x+5^{x+1}+5^{x-2}=151\)

\(\Rightarrow5^x.\left(1+5+5^{-2}\right)=151\)

\(\Rightarrow5^x.6,04=151\Rightarrow5^x=25=5^2\)

Vì \(5\ne-1;5\ne0;5\ne1\) nên \(x=2\)

b, \(5^{x-1}+5^{x-2}+5^{x-3}=155\)

\(\Rightarrow5^x.\left(5^{-1}+5^{-2}+5^{-3}\right)=155\)

\(\Rightarrow5^x.0,248=155\Rightarrow5^x=625=5^4\)

Vì \(5\ne-1;5\ne0;5\ne1\) nên \(x=4\)

\(•5^x+5^{x+1}+5^{x-2}=151\\ 5^x\left(1+5+\dfrac{1}{25}\right)=151\\ 5^x=25\\ \Rightarrow x=2\)

\(•5^{x-1}+5^{x-2}+5^{x-3}=155\\ 5^x.\left(\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}\right)=155\\ 5^x=625\\ \Rightarrow x=4\)

\(•5^{2+x}+5^{3+x}=750\\ 5^x\left(25+125\right)=750\\ 5^x=5\\ \Rightarrow x=1\)