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\(\frac{x+1}{x-1}=\frac{x+2}{x-2}\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)=\left(x-1\right)\left(x+2\right)\)
\(\Leftrightarrow x\left(x-2\right)+\left(x-2\right)=x\left(x-1\right)+2\left(x-1\right)\)
\(\Leftrightarrow x^2-2x+x-2=x^2-x+2x-2\)
\(\Leftrightarrow x^2-x-2=x^2+x-2\)
\(\Leftrightarrow-x=x\)
\(\Leftrightarrow2x=0\Leftrightarrow x=0\)
1/a) Ta có: \(A=x^4+\left(y-2\right)^2-8\ge-8\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x=0\\y-2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\y=2\end{cases}}\)
Vậy GTNN của A = -8 khi x=0, y=2.
b) Ta có: \(B=|x-3|+|x-7|\)
\(=|x-3|+|7-x|\ge|x-3+7-x|=4\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x\ge3\\x\le7\end{cases}}\Rightarrow3\le x\le7\)
Vậy GTNN của B = 4 khi \(3\le x\le7\)
2/ a) Ta có: \(xy+3x-7y=21\Rightarrow xy+3x-7y-21=0\)
\(\Rightarrow x\left(y+3\right)-7\left(y+3\right)=0\Rightarrow\left(x-7\right)\left(y+3\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=7\\y=-3\end{cases}}\)
b) Ta có: \(\frac{x+3}{y+5}=\frac{3}{5}\)và \(x+y=16\)
Áp dụng tính chất bằng nhau của dãy tỉ số, ta có:
\(\frac{x+3}{y+5}=\frac{3}{5}\Rightarrow\frac{x+3}{3}=\frac{y+5}{5}=\frac{x+y+8}{8}=\frac{16+8}{8}=\frac{24}{8}=3\)
\(\Rightarrow\hept{\begin{cases}\frac{x+3}{3}=3\Rightarrow x+3=9\Rightarrow x=6\\\frac{y+5}{5}=3\Rightarrow y+5=15\Rightarrow y=10\end{cases}}\)
Bài 3: đề không rõ.
Bài 1:\(a,A=x^4+\left(y-2\right)^2-8\)
Có \(x^4\ge0;\left(y-2\right)^2\ge0\)
\(\Rightarrow A\ge0+0-8=-8\)
Dấu "=" xảy ra khi \(MinA=-8\Leftrightarrow x=0;y=2\)
\(b,B=\left|x-3\right|+\left|x-7\right|\)
\(\Rightarrow B=\left|x-3\right|+\left|7-x\right|\)
\(\Rightarrow B\ge\left|x-3+7-x\right|\)
\(\Rightarrow B\ge\left|-10\right|=10\)
Dấu "=" xảy ra khi \(MinB=10\Leftrightarrow3\le x\le7\Rightarrow x\in\left(3;4;5;6;7\right)\)
\(x^2=\frac{5}{7}x\Leftrightarrow x^2-\frac{5}{7}x=0\)
\(\Leftrightarrow x\left(x-\frac{5}{7}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{5}{7}\end{cases}}\)
\(x^2=\frac{5}{7}x\)
\(\Rightarrow x^2-\frac{5}{7}x=0\)
\(\Rightarrow x\left(x-\frac{5}{7}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-\frac{5}{7}=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{5}{7}\end{cases}}\)
\(\left(\frac{2}{3}\right)^{x+2}=\left(\frac{4}{9}\right)^4\)
\(\left(\frac{2}{3}\right)^{x+2}=\left[\left(\frac{2}{3}\right)^2\right]^4\)
\(\left(\frac{2}{3}\right)^{x+2}=\left(\frac{2}{3}\right)^8\)
\(\Rightarrow x+2=8\)
Vậy \(x=6\)
Bài 1
\(\left(\frac{1}{2}-x\right)^2=\frac{4}{9}\)
\(\Leftrightarrow\left(\frac{1}{2}-x\right)^2=\left(\frac{2}{3}\right)^2\)
\(\Leftrightarrow\frac{1}{2}-x=\frac{2}{3}\)
\(\Leftrightarrow\frac{3}{6}-\frac{4}{6}=x\)
\(\Leftrightarrow x=\frac{-1}{6}\)
Bài 2
Để \(\frac{2x+1}{x-1}\in Z\)
\(\Leftrightarrow\frac{2X-2+3}{X-1}\in Z\)
\(\Leftrightarrow2+\frac{3}{X-1}\in Z\)
\(\Rightarrow3⋮X-1\)
\(\Rightarrow X-1\inƯ\left(3\right)\)
\(\Rightarrow X-1=\left\{-3,-1,1,3\right\}\)
\(\Rightarrow X=\left\{-2,0,2,4\right\}\)
\(3\left(\frac{3}{20\cdot23}+\frac{3}{23\cdot26}+....+\frac{3}{77\cdot80}\right)\)
\(=3\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+.....+\frac{1}{77}-\frac{1}{80}\right)\)
\(=3\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(=\frac{3}{20}-\frac{3}{80}\)
\(< 1\)
Ta có: |x + 1| \(\ge\)0 \(\forall\)x => 5|x + 1| \(\ge\)0 \(\forall\)x
=> 3 + 5|x + 1| \(\ge\) 3 \(\forall\)x
Dấu "=" xảy ra <=> 5|x + 1| = 0 <=> |x + 1| = 0 <=> x = -1
Vậy mmin = 3 khi x = -1
Có: \(|x+1|\ge0\)
\(\Leftrightarrow5|x+1|\ge0\)
\(\Leftrightarrow M\ge3\)
GTNN của M=3 khi x=-1