Question

\(a.\left|x+2\right|+\left|x-1\right|=3-\left(y+2\right)^2\)

\(b.\left|x-5\right|+\left|1-x\right|=\dfrac{12}{\left|y+1\right|+3}\)

\(c.\left|y+3\right|+5=\dfrac{10}{\left(2x-6\right)^2+2}\)

\(d.\left|x-1\right|+\left|3-x\right|=\dfrac{6}{\left|y-3\right|+3}\)

Answer 1

de bai

Answer 2

tìm x,y

Answer 3

a) Áp dụng bất đẳng thức \(\left|A\right|+\left|B\right|\ge\left|A+B\right|\) ta có :

\(\left|x+2\right|+\left|x-1\right|=\left|x+2\right|+\left|1-x\right|\)

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

Dấu "=" xảy ra \(\Leftrightarrow\left(x+2\right)\left(1-x\right)\ge0\)

\(\Leftrightarrow-2\le x\le1\)

+ \(\left(y+2\right)^2\ge0\forall y\)

\(\Rightarrow3-\left(y+2\right)^2\le3\) (2)

Dấu "=" xảy ra \(\Leftrightarrow\left(y+2\right)^2=0\Leftrightarrow y=-2\)

Từ (1) và (2) suy ra \(\left|x+2\right|+\left|x+1\right|=3-\left(y+2\right)^2=3\)

\(\Leftrightarrow\left\{{}\begin{matrix}-2\le x\le1\\y=-2\end{matrix}\right.\)

b) \(\left|x-5\right|+\left|1-x\right|\ge\left|x-5+1-x\right|=4\) (3)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-5\right)\left(1-x\right)\ge0\)

\(\Leftrightarrow1\le x\le5\)

+ \(\left|y+1\right|\ge0\forall y\) \(\Rightarrow\left|y+1\right|+3\ge3\)

\(\Rightarrow\frac{12}{\left|y+1\right|+3}\le\frac{12}{3}=4\) (4)

Dấu "=" xảy ra \(\Leftrightarrow\left|y+1\right|=0\Leftrightarrow y=-1\)

Từ (3) và (4) suy ra \(\left|x-5\right|+\left|1-x\right|=\frac{12}{\left|y+1\right|+3}=4\)

\(\Leftrightarrow\left\{{}\begin{matrix}1\le x\le5\\y=-1\end{matrix}\right.\)

Câu c,d lm tương tự