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Tớ cần gấp lắm ấy ạ 
Tớ camon các cậu nhiều lắm 
\(\left|x-\dfrac{1}{2}\right|+\left|y+\dfrac{2}{3}\right|+\left|x^2+xz\right|=0\)
\(\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|\ge0\forall x\\\left|y+\dfrac{2}{3}\right|\ge0\forall y\\\left|x^2+xz\right|\ge0\forall x;z\end{matrix}\right.\) \(\Rightarrow\left|x-\dfrac{1}{2}\right|+\left|y+\dfrac{2}{3}\right|+\left|x^2+xz\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|=0\\\left|y+\dfrac{2}{3}\right|=0\\\left|x^2+xz\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{2}{3}\\z=-\dfrac{1}{2}\end{matrix}\right.\)
>> Mình không chép lại đề bài nhé ! <<
Cách 1 :
\(A=\left(\dfrac{36-4+3}{6}\right)-\left(\dfrac{30+10-9}{6}\right)-\left(\dfrac{18-14+15}{6}\right)=\dfrac{35}{6}-\dfrac{31}{6}-\dfrac{19}{6}=-\dfrac{15}{6}=-\dfrac{5}{2}\)
Cách 2 :
\(A=6-\dfrac{2}{3}+\dfrac{1}{2}-5+\dfrac{5}{3}-\dfrac{3}{2}-3-\dfrac{7}{3}+\dfrac{5}{2}\)
\(A=\left(6-5-3\right)-\left(\dfrac{2}{3}+\dfrac{5}{3}-\dfrac{7}{3}\right)+\left(\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\right)\)
\(A=-2-0-\dfrac{1}{2}=-\dfrac{5}{2}\)
Cách 1 :
\(\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)
\(=\left(\dfrac{36}{6}-\dfrac{4}{6}+\dfrac{3}{6}\right)-\left(\dfrac{30}{6}+\dfrac{10}{6}-\dfrac{9}{6}\right)-\left(\dfrac{18}{6}-\dfrac{14}{6}+\dfrac{15}{6}\right)\)
\(=\dfrac{35}{6}-\dfrac{31}{6}-\dfrac{19}{6}\)
\(=-\dfrac{5}{2}\)
Cách 2 :
\(\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)
\(=6-\dfrac{2}{3}+\dfrac{1}{2}-5-\dfrac{5}{3}+\dfrac{3}{2}-3+\dfrac{7}{3}-\dfrac{5}{2}\)
\(=\left(6-5-3\right)+\left(\dfrac{-2}{3}+\dfrac{-5}{3}+\dfrac{7}{3}\right)+\left(\dfrac{1}{2}+\dfrac{3}{2}+\dfrac{-5}{2}\right)\)
\(=\left(-2\right)+0+\dfrac{-1}{2}\)
\(=\dfrac{-5}{2}\)
.....
trước tui 1 hôm đó nhớ chia sẻ đề tui được giải cao tui tick cho nha
Tối nay tớ đã cần lắm rồi 
a+1/2=c+2/4=c+1/2=>a=c=>3a=3c
b+2/3=c+2/4=c+1/2=>b=c+1/2-2/3=c-1/6=>2b=2c-1/3
3a-2b+c=3c-2c+1/3+c=2c+1/3=105
=>2c=314/3=>c=157/3
b=c-1/6=157/3-1/6=313/6
a=c=157/3
Đề cậu viết khó nhìn qá :)
Bài 1 :
Ta có :
\(a+b+c=2014\)
\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{9}\)
\(\Leftrightarrow2014\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=2014.\dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{2014}{a+b}+\dfrac{2014}{b+c}+\dfrac{2014}{c+a}=\dfrac{2014}{9}\)
Mà \(a+b+c=2014\) nên :
\(\Leftrightarrow\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}=\dfrac{2014}{9}\)
\(\Leftrightarrow\left(\dfrac{a+b}{a+b}+\dfrac{c}{a+b}\right)+\left(\dfrac{b+c}{b+c}+\dfrac{a}{b+c}\right)+\left(\dfrac{c+a}{c+a}+\dfrac{b}{c+a}\right)=\dfrac{2014}{9}\)
\(\Leftrightarrow3+\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{c+a}=\dfrac{2014}{9}\)
\(\Leftrightarrow\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{c+a}=\dfrac{1987}{9}\)
\(\Leftrightarrow S=\dfrac{1987}{9}\)
Dễ tek mà ko làm đc!
Với mọi x ta có:
|x - 2001| = |2001 - x|
=> A = |x - 2002| + |2001 - x|
Với mọi x ta cũng có:
|x - 2002| + | 2001 - x| \(\ge\)|(x - 2002) + (2001 - x)|
A \(\ge\) |1|
A \(\ge\) 1
Dấu bằng xảy ra <=> (x - 2002).(2001 - x) \(\ge\) 0
=> x - 2002 \(\ge\) 0; 2001 - x \(\ge\) 0 (1)
hoặc x - 2002 \(\le\) 0; 2001 - x \(\le\) 0 (2)
Từ (1) => x > hoặc = 2002; x < hoặc = 2001 => x không có giá trị thoả mãn
Từ (2) => x < hoặc = 2002 ; x > hoặc = 2001 => 2001 \(\le\) x \(\le\) 2002
Vậy 2001 \(\le\) x \(\le\) 2002 thì A có giá trị nhỏ nhất = 1
Đặt P(x)=0
\(\Leftrightarrow x^2-3x-2=0\)
\(\text{Δ}=\left(-3\right)^2-4\cdot1\cdot\left(-2\right)=17>0\)
Do đó; Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{3-\sqrt{17}}{2}\\x_2=\dfrac{3+\sqrt{17}}{2}\end{matrix}\right.\)