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a) \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{a+b}{2ab}\)
\(\Rightarrow\frac{1}{c}=\frac{a+b}{2ab}\Rightarrow ac+bc=2ab=ac-ab=ab-bc=a\left(c-b\right)=b\left(a-c\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\left(đpcm\right)\)
b) \(\text{Để n nguyên thì P phải nguyên} \)
\(\Rightarrow\frac{2n-1}{n-1}=\frac{2n-2+1}{n-1}=\frac{2\left(n-1\right)+1}{n-1}=\frac{2\left(n-1\right)}{n-1}+\frac{1}{n-1}=2+\frac{1}{n-1}\Rightarrow\frac{1}{n-1}\in Z\)
=> n-1 là ước của 1
=> n-1={-1;1)
=> n={0;2)
c) \(\frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4y-3z}{2}=\frac{12x-8y}{16}=\frac{6z-12x}{9}=\frac{8y-6z}{4}=\)\(\frac{12x-8y+6z-12x+8y-6z}{16+9+4}=0\)
\(\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)
b)\(P=\frac{2n-1}{n-1}=\frac{2n-2+1}{n-1}=\frac{2\left(n-1\right)+1}{n-1}=2+\frac{1}{n-1}\)
P là số nguyên \(\Leftrightarrow2+\frac{1}{n-1}\in Z\Leftrightarrow\frac{1}{n-1}\in Z\Leftrightarrow1⋮n-1\Leftrightarrow n-1\inƯ\left(1\right)\)
\(\Leftrightarrow n-1\in\left\{-1;1\right\}\Leftrightarrow n\in\left\{0;2\right\}\)
c)\(\frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4y-3z}{2}\)
\(\Rightarrow\frac{12x-8y}{16}=\frac{6z-12x}{9}=\frac{8y-6z}{4}=\frac{12x-8y+6z-12x+8y-6z}{16+9+4}=\frac{0}{29}=0\)
\(\Rightarrow12x-8y=0,6z-12x=0,8y-6z=0\)
\(\Rightarrow12x=8y,6z=12x,8y=6z\)
\(\Rightarrow12x=8y=6z\)
\(\Rightarrow\frac{12x}{24}=\frac{8y}{24}=\frac{6z}{24}\)
\(\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)
1
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Ez lắm =)
Bài 1:
Với mọi gt \(x,y\in Q\) ta luôn có:
\(x\le\left|x\right|\) và \(-x\le\left|x\right|\)
\(y\le\left|y\right|\) và \(-y\le\left|y\right|\Rightarrow x+y\le\left|x\right|+\left|y\right|\) và \(-x-y\le\left|x\right|+\left|y\right|\)
Hay: \(x+y\ge-\left(\left|x\right|+\left|y\right|\right)\)
Do đó: \(-\left(\left|x\right|+\left|y\right|\right)\le x+y\le\left|x\right|+\left|y\right|\)
Vậy: \(\left|x+y\right|\le\left|x\right|+\left|y\right|\)
Dấu "=" xảy ra khi: \(xy\ge0\)
Ta có :
\(B=\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{1}{2016}\)
\(B=\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{1}{2016}+1\right)+1\)
\(B=\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2016}+\frac{2017}{2017}\)
\(B=2017.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}\right)\)
\(\Rightarrow\frac{B}{A}=\frac{2017.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}=2017\)
Vậy \(\frac{B}{A}\)là số nguyên
a) Ta có: \(\left(x-1\right)^2\ge\)0 \(\forall\)x
\(\left|y+2\right|\ge0\)\(\forall\) y
=> \(\left(x-1\right)^2+\left|y+2\right|\ge0\)\(\forall\)x,y
=> \(\hept{\begin{cases}\left(x-1\right)^2=0\\y+2=0\end{cases}}\)
=> \(\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
Vậy ...
b) Ta có: \(\frac{1}{2}-\frac{y}{3}=\frac{2}{x}\)
=> \(\frac{3-2y}{6}=\frac{2}{x}\)
=> \(x\left(3-2y\right)=12\)
=> x; 3 - 2y \(\in\)Ư(12) = {1; -1; 2; -2; 3; -3; 4; -4; 6; -6; 12; -12}
Do 3 - 2y là số lẽ , mà x,y \(\in\)Z
=> 3 - 2y \(\in\) {1; -1; 3; -3}
Lập bảng :
| 3 - 2y | 1 | -1 | 3 | -3 |
| x | 12 | -12 | 4 | -4 |
| y | 1 | 2 | 0 | 3 |
Vậy ...
1) \(x^2y+x\left(2y-1\right)=7\)
\(\Leftrightarrow x^2y+2xy-x=7\)
\(\Leftrightarrow xy\left(x+2\right)-x-2=7-2\)
\(\Leftrightarrow xy\left(x+2\right)-\left(x+2\right)=5\)
\(\Leftrightarrow\left(xy-1\right)\left(x+2\right)=5\)
\(\Rightarrow\)xy - 1 và x + 2 là ước của 5 là \(\pm1;\pm5\)
đến đây tự lm đc
2 ) \(B=\frac{255}{1}+\frac{254}{2}+\frac{253}{3}+....+\frac{3}{253}+\frac{2}{254}+\frac{1}{255}\)
\(=\left(\frac{254}{2}+1\right)+\left(\frac{253}{3}+1\right)+....+\left(\frac{2}{254}+1\right)+\left(\frac{1}{255}+1\right)+1\)
\(=\frac{256}{2}+\frac{256}{3}+....+\frac{256}{255}+\frac{256}{256}\)
\(=256\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{255}+\frac{1}{256}\right)=256A\)
\(\Rightarrow\frac{B}{A}=256=16^2\) Là số CP (đpcm)
dễ thì đăng đáp an đi
bọn này tham khảo vs