3.x^2-5=10x+5 

<=> 3.x^2-10x-10=0

<=> \(x = {5 \pm \sqrt{55} \over 3}\)

\(\frac{x}{2000}+\frac{x+1}{2001}+\frac{x+2}{2002}+\frac{x+3}{2003}=4\)

\(\Leftrightarrow\left(\frac{x}{2000}-1\right)+\left(\frac{x+1}{2001}-1\right)+\left(\frac{x+2}{2002}-1\right)+\left(\frac{x+3}{2003}-1\right)=4-4=0\)

\(\Leftrightarrow\frac{x-2000}{2000}+\frac{x-2000}{2001}+\frac{x-2000}{2002}+\frac{x-2000}{2003}=0\)

\(\Leftrightarrow\left(x-2000\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)=0\)

\(\Leftrightarrow x-2000=0\)  ( do \(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\ne0\) )

\(\Leftrightarrow x=2000\)

Vậy x = 2000

Đây là cách của lớp 7 nha

@@ Học tốt

\(\frac{x}{2000}\)- 1+\(\frac{x+1}{2001}\)-1+\(\frac{x+2}{2002}\)-1+\(\frac{x+3}{2003}\)-1=0

<=>\(\frac{x-2000}{2000}\)+ \(\frac{x-2000}{2001}\)+ \(\frac{x-2000}{2002}\)+ \(\frac{x-2000}{2003}\)=0

<=>\(\left(x-2000\right)\)\(\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)\)=0

Do \(\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)\)khác 0

=> \(x-2000=0\)<=> \(x=2000\)

vì |x| >0 nên nó vô no

nghiệm là gì ?

Ta có: \(x+xy-x^2+y=1\)

<=> \(\left(x+1\right)+\left(1-x^2\right)+\left(xy+y\right)=3\)

<=> (x + 1) + ( 1 + x) ( 1 - x ) + y ( x + 1 ) = 3

<=> ( x + 1 ) ( 1 + 1 - x + y ) = 3

<=> ( x + 1 ) ( 2 - x  + y ) = 3

Chia trường hợp lập bảng rồi làm tiếp nhé!

em có cách khác:

\(x+xy-x^2+y=1\)

\(\Leftrightarrow xy+y=x^2+1-x\)

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

Do y nguyên nên \(\frac{3x}{x+1}\) nguyên 

\(\Rightarrow3x⋮x+1\)

\(\Rightarrow3\left(x+1\right)-3⋮x+1\)

\(\Rightarrow x+1\in\left\{1;3;-1;-3\right\}\)

Tìm được x xong thử vào tìm y nhé !

\(abc=1\ge a^2b^2c^2=1\)

\(\Rightarrow\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}=\frac{b^2c^2}{a\left(b+c\right)}+\frac{a^2c^2}{b\left(c+a\right)}+\frac{a^2b^2}{c\left(a+b\right)}\)

Theo Cauchy-Schwarz ta được: 

\(VP\ge\frac{\left(bc+ab+ac\right)^2}{2\left(ab+ac+bc\right)}=\frac{bc+ab+ac}{2}\ge\frac{3\sqrt[3]{a^2b^2c^2}}{2}=\frac{3}{2}\)

Dấu "=" xảy ra khi a = b = c = 1

\(\)

Bài làm:

Ta có: \(\frac{1}{a^3\left(b+c\right)}=\frac{abc}{a^3\left(b+c\right)}=\frac{bc}{a^2b+a^2c}\)

\(=\frac{b^2c^2}{a^2b^2c+a^2bc^2}=\frac{b^2c^2}{ab+ac}\)

Tương tự: \(\frac{1}{b^3\left(c+a\right)}=\frac{c^2a^2}{ba+bc}\) ; \(\frac{1}{c^3\left(a+b\right)}=\frac{a^2b^2}{ca+cb}\)

=> \(Vt=\frac{a^2b^2}{ca+bc}+\frac{b^2c^2}{ab+ca}+\frac{c^2a^2}{ab+bc}\ge\frac{\left(ab+bc+ca\right)^2}{2\left(ab+bc+ca\right)}\ge\frac{3\sqrt[3]{a^2b^2c^2}}{2}=\frac{3}{2}\)

Dấu "=" xảy ra khi: \(a=b=c=1\)

\(A.\left(2,3x-6,5\right)\left(0,1x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2,3x-6,5=0\\0,1x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2,3x=6,5\\0,1x=-2\end{cases}}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{6,5}{2,3}\\x=-20\end{cases}}\)

\(4\left(2x+7\right)^2=9\left(x+3\right)^2\)

\(\Leftrightarrow\left(4x+14\right)^2=\left(3x+9\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}4x+14=3x+9\\4x+14=-3x-9\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\7x+23=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-5\\x=-\frac{23}{7}\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{-5;-\frac{23}{7}\right\}\)