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a: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+1+1}{x+1}+\dfrac{2}{y-2}=6\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)
=>x+1=1 và y-2=1/2
=>x=0 và y=5/2
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x-2y}=\dfrac{1}{2}-\dfrac{1}{18}=\dfrac{9}{18}-\dfrac{1}{18}=\dfrac{8}{18}=\dfrac{4}{9}\\\dfrac{2}{2x-y}=\dfrac{1}{18}+\dfrac{1}{x-2y}\end{matrix}\right.\)
=>x-2y=9 và 2/2x-y=1/18+1/9=1/18+2/18=3/18=1/6
=>x-2y=9 và 2x-y=12
=>x=5; y=-2
c: \(\Leftrightarrow\left\{{}\begin{matrix}10\left|x-6\right|+15\left|y+1\right|=25\\10\left|x-6\right|-8\left|y+1\right|=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}23\left|y+1\right|=23\\\left|x-6\right|=1\end{matrix}\right.\)
=>|x-6|=1 và |y+1|=1
=>\(\left\{{}\begin{matrix}x\in\left\{7;5\right\}\\y\in\left\{0;-2\right\}\end{matrix}\right.\)
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Mình giải đơn giản thế này thôi nhé :)
Xét vế trái : \(y^2-2y+3=\left(y^2-2y+1\right)+2=\left(y-1\right)^2+2\ge2\)
Xét vế phải : \(\frac{6}{x^2+2x+4}=\frac{6}{\left(x^2+2x+1\right)+3}=\frac{6}{\left(x+1\right)^2+3}\le2\)
Vậy , phương trình tương đương với : \(\hept{\begin{cases}y^2-2y+3=2\\\frac{6}{x^2+2x+4}=2\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=-1\\y=1\end{cases}}\)
Kết luận tập nghiệm ...............................
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Ta có \(y^2-2y+3=y^2-2y+1+2=\left(y-1\right)^2+2\ge2\)
\(\dfrac{6}{x^2+2x+4}=\dfrac{6}{x^2+2x+1+3}=\dfrac{6}{\left(x+1\right)^2+3}\le2\)
Vậy \(y^2-2y+3=\dfrac{6}{x^2+2x+4}=2\Leftrightarrow\)\(\left\{{}\begin{matrix}y^2-2y+3=2\\\dfrac{6}{x^2+2x+4}=2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left(y-1\right)^2+2=2\\\dfrac{6}{\left(x+1\right)^2+3}=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=-1\end{matrix}\right.\)
\(y^2-2y+3=\left(y-1\right)^2+2\ge2\)
\(\dfrac{6}{x^2+2x+4}=\dfrac{6}{\left(x+1\right)^2+3}\le2\)
So ez
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ta có \(y^2-2y+3=\left(y-1\right)^2+2\ge2\)(1)
\(x^2+2x+4=\left(x+1\right)^2+3\ge3\)
==> \(\frac{6}{\left(x+1\right)^2+3}\le\frac{6}{3}=\)(2)
từ (1) và (2) thì để dấu = xảy ra khi 2 vế cùng =2
khi đó y-1=0 <=> y=1
x+1=0 <=> x=-1
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Câu 1. Ta có phương trình tương đương với \(\sqrt{\left(x-3\right)^2}+\sqrt{\left(x+5\right)^2}=8\leftrightarrow\left|3-x\right|+\left|x+5\right|=8\). Nhớ lại rằng ta luôn có \(\left|A\right|+\left|B\right|\ge\left|A+B\right|,\) với dấu bằng xảy ra khi và chỉ khi \(A\cdot B\ge0\),
Mà \(8=\left(3-x\right)+\left(x+5\right)\to\left(3-x\right)\left(x+5\right)\ge0\leftrightarrow\left(x-3\right)\left(x+5\right)\le0\leftrightarrow-5\le x\le3.\)
Vậy đáp số là \(-5\le x\le3.\)
Câu 2. Ta có
\(VT=y^2-2y+3=\left(y-1\right)^2+2\ge2,VP=\frac{6}{x^2+2x+4}=\frac{6}{\left(x+1\right)^2+3}\le\frac{6}{3}=2\to VP\le VT\)
Do đó để \(VT=VP\) thì các dấu bằng phải xảy ra, ta suy ra ngay \(y=1,x=-1.\) (Ở đây ta kí hiệu VT là vế trái, VP là vế phải). ĐPCM
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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: \(x\ge1,y\ge1\))
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-1}=3\\\sqrt{x-1}+\sqrt{y-1}=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{y-1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-1=1\end{matrix}\right.\)\(\Leftrightarrow x=y=2\left(tm\right)\)
Bài này rất đơn giản
\(y^2-2y+3=\frac{6}{x^2+2x+4}\Leftrightarrow\left(y^2-2y+1\right)+2-\frac{6}{x^2+2x+4}=0\)
\(\Leftrightarrow\left(y-1\right)^2+\frac{2\left(x^2+2x+4\right)-6}{x^2+2x+4}=0\Leftrightarrow\left(y-1\right)^2+\frac{2\left(x^2+2x+1\right)}{x^2+2x+4}=0\)
\(\Leftrightarrow\left(y-1\right)^2+\frac{2\left(x+1\right)^2}{x^2+2x+4}=0\)
Ta có: \(\left(y-1\right)^2\ge0;\frac{2\left(x+1\right)^2}{x^2+2x+4}\ge0\) với mọi x và y
dấu "=" xảy ra khi y=1; x=-1
Vậy (x,y)=(1,-1)
Tick mình nha