Tìm a,b thuộc N:
a/3+b/5=a+b/3+5
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\(\frac{a}{2}-\frac{3}{b}=\frac{5}{6}\)
\(\Rightarrow\frac{3}{b}=\frac{a}{2}-\frac{5}{6}\)
\(\Rightarrow\frac{3}{b}=\frac{3a-5}{6}\)
\(\Rightarrow b\times\left(3a-5\right)=18\)
Ta có bảng
b 1 18 2 9 3 6
3a-5 18 1 9 2 6 3
a \ 2 \ \ \ \
Vậy (a,b) \(\in\){(2;18)}
Chúc bạn hok tốt, k cho mik nha
\(\left(x-1\right)^4=16\)
\(\left(x-1\right)^4=2^4\)
\(\Rightarrow x-1=2\)
\(\Rightarrow x=3\)
vậy \(x=3\)
a, => 1 = 4^x-2^x = 2^x.(2^x-1)
=> 2^x=1 ; 2^x-1=0 ( vì 2^x >= 0 )
=> x=0
b, => (x-1)^4 = 16 = (-2)^4 = 2^4
=> x=-2 hoặc x=2
c, Xét : 1+3+5+....+99 = (1+99).50 : 2 = 2500
=> 2500^2 = (x-2)^2
=> x-2=2500 hoặc x-2=-2500
=> x=2502 hoặc x=2498
Tk mk nha
a) \(\dfrac{1}{a}-\dfrac{1}{b}=\dfrac{1}{a-b}\left(đk:a,b\ne0,a\ne b\right)\Leftrightarrow\dfrac{b-a}{ab}=\dfrac{1}{a-b}\)
\(\Leftrightarrow-\left(a-b\right)^2=ab\Leftrightarrow a^2-ab+b^2=0\)
\(\Leftrightarrow\left(a^2-ab+\dfrac{1}{4}b^2\right)+\dfrac{3}{4}b^2=0\Leftrightarrow\left(a-\dfrac{1}{2}b\right)^2+\dfrac{3}{4}b^2=0\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a-\dfrac{1}{2}b=0\\\dfrac{3}{4}b^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}b\\b=0\end{matrix}\right.\) \(\Leftrightarrow a=b=0\left(ktm\right)\)
Vậy k có a,b thõa mãn
b) \(\dfrac{5}{2a}=\dfrac{1}{6}+\dfrac{b}{3}\left(a\ne0\right)\Leftrightarrow\dfrac{2b+1}{6}-\dfrac{5}{2a}=0\Leftrightarrow\dfrac{a\left(2b+1\right)-15}{6a}=0\)
\(\Leftrightarrow a\left(2b+1\right)-15=0\Leftrightarrow a\left(2b+1\right)=15\)
Do \(a,b\in Z,a\ne0\) nên ta có bảng sau:
a | 1 | -1 | 15 | -15 | 3 | -3 | 5 | -5 |
2b+1 | 15 | -15 | 1 | -1 | 5 | -5 | 3 | -3 |
b | 7(tm) | -8(tm) | 0(tm | -1(tm) | 2(tm) | -3(tm) | 1(tm) | -2(tm) |
Vậy...
a: \(\dfrac{4}{5}-\dfrac{5}{6}< =\dfrac{x}{30}< =\dfrac{1}{3}-\dfrac{3}{10}\)
=>\(\dfrac{24-25}{30}< =\dfrac{x}{30}< =\dfrac{10-9}{30}\)
=>\(\dfrac{-1}{30}< =\dfrac{x}{30}< =\dfrac{1}{30}\)
=>-1<=x<=1
mà x nguyên
nên \(x\in\left\{-1;0;1\right\}\)
b: \(\dfrac{a}{7}+\dfrac{1}{14}=\dfrac{-1}{b}\)
=>\(\dfrac{2a+1}{14}=\dfrac{-1}{b}\)
=>\(\left(2a+1\right)\cdot b=-14\)
mà 2a+1 lẻ (do a là số nguyên)
nên \(\left(2a+1\right)\cdot b=1\cdot\left(-14\right)=\left(-1\right)\cdot14=7\cdot\left(-2\right)=\left(-7\right)\cdot2\)
=>\(\left(2a+1;b\right)\in\left\{\left(1;-14\right);\left(-1;14\right);\left(7;-2\right);\left(-7;2\right)\right\}\)
=>\(\left(a;b\right)\in\left\{\left(0;-14\right);\left(-1;14\right);\left(3;-2\right);\left(-4;2\right)\right\}\)
a. \(7\left(x+3\right)=5\left(x+7\right)\)
\(7x+21=5x+35\)
\(2x=14\)
\(x=7\)