de bai

tìm x,y

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

a) (x-1):2/3=-2/5

=>x-1=-4/15

=>x=11/15

b) |x-1/2|-1/3=0

=>|x-1/2|=1/3

=>\(\left\{{}\begin{matrix}x=\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{5}{6}\\x=-\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{1}{6}\end{matrix}\right.\) 

c) Tương Tự câu B

a: ĐKXĐ: \(x^2+y^2\ne0\)

=>\(\left[{}\begin{matrix}x^2\ne0\\y^2\ne0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x\ne0\\y\ne0\end{matrix}\right.\)

b: ĐKXĐ: \(x^2-2x+1\ne0\)

=>\(\left(x-1\right)^2\ne0\)

=>\(x-1\ne0\)

=>\(x\ne1\)

c: ĐKXĐ: \(x^2+6x+10\ne0\)

=>\(x^2+6x+9+1\ne0\)

=>\(\left(x+3\right)^2+1\ne0\)(luôn đúng)

d:ĐKXĐ: \(\left(x+3\right)^2+\left(y-2\right)^2\ne0\)

=>\(\left[{}\begin{matrix}x+3\ne0\\y-2\ne0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x\ne-3\\y\ne2\end{matrix}\right.\)

9) \(\left\{{}\begin{matrix}\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\\\dfrac{3}{2x+y}+\dfrac{2}{2x-y}=32\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{21}{2x+y}+\dfrac{12}{2x-y}=222\\\dfrac{21}{2x+y}+\dfrac{14}{2x-y}=224\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{2x-y}=2\\\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=\dfrac{1}{10}\\2x-y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-2y=\dfrac{9}{10}\\2x+y=\dfrac{1}{10}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{9}{20}\\x=\dfrac{11}{40}\end{matrix}\right.\)

10) \(\left\{{}\begin{matrix}x=2y-1\\2x-y=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x-4y=-2\\2x-y=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2y-1\\3y=7\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{3}\\y=\dfrac{7}{3}\end{matrix}\right.\)

11) \(\left\{{}\begin{matrix}3x-6=0\\2y-x=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3x=6\\y=\dfrac{x+4}{2}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)

12) \(\left\{{}\begin{matrix}2x+y=5\\x+7y=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=5\\2x+14y=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=5\\13y=13\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

13) \(\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{4}{x}-\dfrac{5}{y}=3\end{matrix}\right.\)(ĐKXĐ: \(x,y\ne0\))

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x}-\dfrac{16}{y}=8\\\dfrac{12}{x}-\dfrac{15}{y}=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{1}{y}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\y=1\left(tm\right)\end{matrix}\right.\)

14) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)(ĐKXĐ: \(x,y\ne0\))

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{x}+\dfrac{8}{y}=\dfrac{2}{3}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{7}{y}=\dfrac{1}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=28\left(tm\right)\\y=21\left(tm\right)\end{matrix}\right.\)

15) \(\left\{{}\begin{matrix}2\sqrt{x-1}-\sqrt{y-1}=1\\\sqrt{x-1}+\sqrt{y-1}=2\end{matrix}\right.\)(ĐKXĐ: \(x\ge1,y\ge1\))

\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-1}=3\\\sqrt{x-1}+\sqrt{y-1}=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{y-1}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-1=1\end{matrix}\right.\)\(\Leftrightarrow x=y=2\left(tm\right)\)

a: \(=\dfrac{x-z}{2}\)

b: \(=\dfrac{3x}{4y^3}\)

a) \(\left|3x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|=0\)

Do \(\left|3x-\dfrac{1}{2}\right|,\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|\ge0\forall x,y\)

\(\Rightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{4}y+\dfrac{3}{5}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{12}{5}\end{matrix}\right.\)

b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\le0\)

Do \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|,\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\ge0\forall x,y\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{5}{7}y-\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{7}{10}\end{matrix}\right.\)