Tìm các nguyên x,y biết : xy+3xy=6
@Hoàng Lê Bảo Ngọc,@Silver bullet, @Lê Nguyên Hạo,...Giúp với ạ
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\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
\(\Rightarrow\left(x-7\right)^{x+1}.\left[1-\left(x-7\right)^{10}\right]=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}\left(x-7\right)^{x+1}=0\\1-\left(x-7\right)^{10}=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x-7=0\\\left(x-7\right)^{10}=1\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x-7=0\\x-7=1\\x-7=-1\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=7\\x=8\\x=6\end{array}\right.\)
Vậy \(\left[\begin{array}{nghiempt}x=7\\x=8\\x=6\end{array}\right.\) thỏa mãn đề bài
\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
\(\Rightarrow\left(x-7\right)^{x+1}.\left[1-\left(x-7\right)^{x+10}\right]=0\)
\(\Rightarrow x-7=0\) hoặc \(1-\left(x-7\right)^{10}=0\)
+) \(x-7=0\Rightarrow x=7\)
+) \(1-\left(x-7\right)^{10}=0\)
\(\Rightarrow x-7=\pm1\)
+ \(x-7=1\Rightarrow x=8\)
+ \(x-7=-1\Rightarrow x=6\)
Vậy \(x\in\left\{7;8;6\right\}\)
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Đề bài vẫn chưa đúng nhé, đúng ra phải là \(M=\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{abc+ac+c}\)
Ta có : \(M=\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{abc+ac+c}\)
\(=\frac{1}{ab+a+1}+\frac{a}{abc+ab+a}+\frac{ab}{a^2b^2c+a^2bc+abc}\)
\(=\frac{1}{ab+a+1}+\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}=\frac{ab+a+1}{ab+a+1}=1\)
Vì a.b.c = 1
Ta có :
\(\frac{1}{ab+a+1}=\frac{c}{abc+ac+c}=\frac{c}{1+ac+c}\)
\(\frac{1}{bc+b+1}=\frac{ca}{bc.ca+abc+ca}=\frac{ca}{c+ca+1}\)
\(\frac{1}{abc+bc+b}=\frac{ac}{abc.ac+bc.ac+b.ac}=\frac{ac}{ac+c+1}\)
\(\Rightarrow M=\frac{c}{1+ac+c}+\frac{ca}{c+ca+1}+\frac{ac}{ac+c+1}\)
\(\Rightarrow M=\frac{c+2ac}{1+ac+c}\)
\(\Rightarrow M=\frac{bc+2}{b+1+bc}\)
\(\Rightarrow M=\frac{bc++1+abc}{b+1+bc}\)
-_-
Năm ngoái a lm ko ra thế này đâu
\(xy-3x-y=4\Leftrightarrow x\left(y-3\right)-\left(y-3\right)=7\Leftrightarrow\left(x-1\right)\left(y-3\right)=7\)
Phân tích 7 = 1.7 = (-1).(-7) = ...........
Từ đó ghép cặp và tính.
|2x - 3| + x = 2
=> |2x - 3| = 2 - x
+ Với \(x< \frac{3}{2}\) thì |2x - 3| = 3 - 2x
Ta có: 3 - 2x = 2 - x
=> 3 - 2 = -x + 2x
=> x = 1, thỏa mãn \(x< \frac{3}{2}\)
+ Với \(x\ge\frac{3}{2}\) thì |2x - 3| = 2x - 3
Ta có: 2x - 3 = 2 - x
=> 2x + x = 2 + 3
=> 3x = 5
=> \(x=\frac{5}{3}\), thỏa mãn \(x\ge\frac{3}{2}\)
Vậy \(\left[\begin{array}{nghiempt}x=1\\x=\frac{5}{3}\end{array}\right.\)
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