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Câu 2:
a: \(\Leftrightarrow n-2+7⋮n-2\)
\(\Leftrightarrow n-2\in\left\{1;-1;7;-7\right\}\)
hay \(n\in\left\{3;1;9;-5\right\}\)
b: \(\Leftrightarrow2n-10+11⋮n-5\)
\(\Leftrightarrow n-5\in\left\{1;-1;11;-11\right\}\)
hay \(n\in\left\{6;4;16;-6\right\}\)
d: \(\Leftrightarrow n^2-1+4⋮n-1\)
\(\Leftrightarrow n-1\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(n\in\left\{2;0;3;-1;5;-3\right\}\)
a)Ta có:
\(\left(n+5\right)⋮\left(n-1\right)\)
\(\Rightarrow\left(n-1+6\right)⋮\left(n-1\right)\)
\(\Rightarrow6⋮\left(n-1\right)\)
Ta có bảng sau:
| \(n-1\) | -6 | -3 | -2 | -1 | 1 | 2 | 3 | 6 |
| n | -5 | -2 | -1 | 0 | 2 | 3 | 4 | 7 |
| TM | TM | TM | TM | TM | TM | TM | TM |
b)\(\left(2n-4\right)⋮\left(n+2\right)\)
\(\Rightarrow\left(2n+4-8\right)⋮\left(n+2\right)\)
\(\Rightarrow8⋮\left(n+2\right)\)
Ta có bảng sau:
| n+2 | -8 | -4 | -2 | -1 | 1 | 2 | 4 | 8 |
| n | -10 | -6 | -4 | -3 | -1 | 0 | 2 | 6 |
| TM | TM | TM | TM | TM | TM | TM | TM |
c)Ta có:
\(\left(6n+4\right)⋮\left(2n+1\right)\)
\(\Rightarrow\left(6n+3+1\right)⋮\left(2n+1\right)\)
\(\Rightarrow1⋮\left(2n+1\right)\)
Ta có bảng sau:
| 2n+1 | -1 | 1 |
| 2n | -2 | 0 |
| n | -1 | 0 |
d)Ta có:
\(\left(3-2n\right)⋮\left(n+1\right)\)
\(\Rightarrow\left(-2n-2+5\right)⋮\left(n+1\right)\)
\(\Rightarrow5⋮\left(n+1\right)\)
Ta có bảng sau:
| n+1 | -5 | -1 | 1 | 5 |
| n | -6 | -2 | 0 | 4 |
B2:
a, \(25\times(-\dfrac{1}{5})^2+8^3:\left(\dfrac{4}{3}\right)^3\)
= \(25\times\dfrac{1}{25}+512:\dfrac{64}{3}\)
= \(1+24\)
= 25
b, \(27:\left(\dfrac{3}{2}\right)^3-4^2\times\left(-\dfrac{1}{2}\right)^2\)
= \(27:\dfrac{27}{8}-16\times\dfrac{1}{4}\)
= \(8-4\)
= 4
Câu 1: Lời giải:
a, Đặt \(A=\dfrac{3x+7}{x-1}\).
Ta có: \(A=\dfrac{3x+7}{x-1}=\dfrac{3x-3+10}{x-1}=\dfrac{3x-3}{x-1}+\dfrac{10}{x-1}=3+\dfrac{10}{x-1}\)
Để \(A\in Z\) thì \(\dfrac{10}{x-1}\in Z\Rightarrow10⋮x-1\Leftrightarrow x-1\in U\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
Ta có bảng sau:
| \(x-1\) | \(1\) | \(-1\) | \(2\) | \(-2\) | \(5\) | \(-5\) | \(10\) | \(-10\) |
| \(x\) | \(2\) | \(0\) | \(3\) | \(-1\) | \(6\) | \(-4\) | \(11\) | \(-9\) |
Vậy, với \(x\in\left\{-9;-4;-1;0;2;3;6;11\right\}\)thì \(A=\dfrac{3x+7}{x-1}\in Z\).
Câu 3:
a, Ta có: \(-\left(x+1\right)^{2008}\le0\)
\(\Rightarrow P=2010-\left(x+1\right)^{2008}\le2010\)
Dấu " = " khi \(\left(x+1\right)^{2008}=0\Rightarrow x+1=0\Rightarrow x=-1\)
Vậy \(MAX_P=2010\) khi x = -1
b, Ta có: \(-\left|3-x\right|\le0\)
\(\Rightarrow Q=1010-\left|3-x\right|\le1010\)
Dấu " = " khi \(\left|3-x\right|=0\Rightarrow x=3\)
Vậy \(MAX_Q=1010\) khi x = 3
c, Vì \(\left(x-3\right)^2+1\ge0\) nên để C lớn nhất thì \(\left(x-3\right)^2+1\) nhỏ nhất
Ta có: \(\left(x-3\right)^2\ge0\Rightarrow\left(x-3\right)^2+1\ge1\)
\(\Rightarrow C=\dfrac{5}{\left(x-3\right)^2+1}\le\dfrac{5}{1}=5\)
Dấu " = " khi \(\left(x-3\right)^2=0\Rightarrow x=3\)
Vậy \(MAX_C=5\) khi x = 3
d, Do \(\left|x-2\right|+2\ge0\) nên để D lớn nhất thì \(\left|x-2\right|+2\) nhỏ nhất
Ta có: \(\left|x-2\right|\ge0\Rightarrow\left|x-2\right|+2\ge2\)
\(\Rightarrow D=\dfrac{4}{\left|x-2\right|+2}\le\dfrac{4}{2}=2\)
Dấu " = " khi \(\left|x-2\right|=0\Rightarrow x=2\)
Vậy \(MAX_D=2\) khi x = 2
Bài 1: A
Bài 2: B
Bài 3: D
Bài 4: A
Bài 5:
a) Ta có: \(\left(-25\right)\cdot125\cdot7\cdot\left(-8\right)\cdot\left(-4\right)\)
\(=\left[\left(-25\right)\cdot\left(-4\right)\right]\cdot\left[125\cdot\left(-8\right)\right]\cdot7\)
\(=100\cdot\left(-1000\right)\cdot7\)
\(=-700000\)
b) Ta có: \(49+\left(-16\right)+\left(-49\right)+\left(-84\right)\)
\(=\left(49-49\right)+\left[\left(-16\right)+\left(-84\right)\right]\)
\(=-100\)
c) Ta có: \(31\cdot\left(-109\right)+31\cdot9\)
\(=31\cdot\left(-109+9\right)\)
\(=31\cdot\left(-100\right)=-3100\)
d) Ta có: \(\left(192-37+85\right)-\left(85+192\right)\)
\(=192-37+85-85-192\)
\(=-37\)
Bài 6:
a) Ta có: \(\left|4+x\right|=28\)
\(\Leftrightarrow\left[{}\begin{matrix}4+x=28\\4+x=-28\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=24\\x=-32\end{matrix}\right.\)
Vậy: x\(\in\){24;-32}
b) Ta có: \(3x+7-9x=-11\)
\(\Leftrightarrow-6x=-18\)
hay x=3
Vậy: x=3
c) Ta có: \(12⋮x\)
\(\Leftrightarrow x\inƯ\left(12\right)\)
\(\Leftrightarrow x\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)
mà \(x⋮3\)
nên \(x\in\left\{3;-3;6;-6;12;-12\right\}\)
mà -1<x<5
nên \(x=3\)
Vậy: x=3