Bài 1 : Tìm số nguyên a biet:
A,|a+5|=3
B,|a-5|=(-2)+9
C,|a|+8=10-|a|
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tinh gia tri bieu thuc
a,A=3a-2b/a-3b voi a/b=10/3
b,B=a-8/a-5-4a-b/3a+3 voi a-b=3,b khac 5,a khac -1
a) Theo đề ta có :
\(A=\frac{3a-2b}{a-3b}\) với \(\frac{a}{b}=\frac{10}{3}\)
* \(\frac{a}{b}=\frac{10}{3}\) \(\Rightarrow a=\frac{10}{3}.b\)
Thay a = \(\frac{10b}{3}\) vào \(\frac{3a-2b}{a-3b}\)
\(\Rightarrow\frac{3a-2b}{a-3b}=\frac{3.\frac{10b}{3}-2b}{\frac{10b}{3}-3b}\) \(=\frac{10b-2b}{\frac{10b}{3}-\frac{9b}{3}}=\frac{8b}{\frac{b}{3}}=8b:\frac{b}{3}=8b.\frac{3}{b}=8.3=24\)
b) Theo đề ta có :
a - b = 3 => a = b + 3
Thay a = b+3 vào \(B=\frac{a-8}{a-5}-\frac{4a-b}{3a+3}\)
\(\Rightarrow B=\frac{b+3-8}{b+3-5}-\frac{4.\left(b+3\right)-b}{3.\left(b+3\right)+3}\) \(=\frac{b-5}{b-2}-\frac{4b+12-b}{3b+9+3}=\frac{b-2-3}{b-2}-\frac{3b+12}{3b+12}\)
\(=\frac{b-2}{b-2}-\frac{3}{b-2}-1\) \(=1-\frac{3}{b-2}-1=0-\frac{3}{b-2}=-\frac{3}{b-2}\)
k đi!!!
Tìm số nguyên a
a)4/5<5/a<10/7 b)2/5<a-1/10<8/15(a-1 là tử, 10 là mẫu)
c)12/7<4/a<8/3. d)5<a^2-15<16
a) \(\dfrac{4}{5}< \dfrac{5}{a}< \dfrac{10}{7}\) \(\left(a\inℤ\right)\)
\(\Leftrightarrow\dfrac{7}{10}< \dfrac{a}{5}< \dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{7.5}{10}< a< \dfrac{5}{4}.5\)
\(\Leftrightarrow\dfrac{7}{2}< a< \dfrac{25}{4}\)
\(\Leftrightarrow a\in\left\{4;5;6\right\}\)
b) \(\dfrac{2}{5}< \dfrac{a-1}{10}< \dfrac{8}{15}\)
\(\Leftrightarrow\dfrac{2.10}{5}< a-1< \dfrac{8.10}{15}\)
\(\Leftrightarrow4< a-1< \dfrac{16}{3}\)
\(\Leftrightarrow5< a< \dfrac{19}{3}\)
\(\Leftrightarrow a\in\left\{6\right\}\)
c) \(\dfrac{12}{7}< \dfrac{4}{a}< \dfrac{8}{3}\)
\(\Leftrightarrow\dfrac{3}{8}< \dfrac{a}{4}< \dfrac{7}{12}\)
\(\Leftrightarrow\dfrac{3.4}{8}< a< \dfrac{7.4}{12}\)
\(\Leftrightarrow\dfrac{3}{2}< a< \dfrac{7}{3}\)
\(\Leftrightarrow a\in\left\{2\right\}\)
d) \(5< a^2-15< 16\)
\(\Leftrightarrow10< a^2< 31\)
\(\Leftrightarrow\sqrt[]{10}< a< \sqrt[]{31}\)
\(\Leftrightarrow a\in\left\{4;5\right\}\)
a: \(\dfrac{x}{6}=\dfrac{8}{3}\)
=>\(x=6\cdot\dfrac{8}{3}=\dfrac{6}{3}\cdot8=8\cdot2=16\)
b: \(\dfrac{5}{x}=\dfrac{4}{9}\)
=>\(x=\dfrac{5\cdot9}{4}=\dfrac{45}{4}\)
c: \(\dfrac{x+3}{-4}=\dfrac{5}{20}\)
=>\(x+3=\dfrac{-4\cdot5}{20}=-1\)
=>x=-1-3=-4
d: \(\dfrac{7}{3+4x}=\dfrac{-2}{9}\)
=>\(4x+3=\dfrac{9\cdot7}{-2}=-\dfrac{63}{2}\)
=>\(4x=-\dfrac{63}{2}-3=-\dfrac{69}{2}\)
=>\(x=-\dfrac{69}{8}\)
f: ĐKXĐ: x<>1
\(\dfrac{3}{x-1}=\dfrac{x-1}{27}\)
=>\(\left(x-1\right)^2=3\cdot27=81\)
=>\(\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=10\left(nhận\right)\\x=-8\left(nhận\right)\end{matrix}\right.\)
Bài 2:
a: \(=7^4\left(7^2+7-1\right)=7^4\cdot55⋮55\)
b: \(5A=5+5^2+...+5^{51}\)
\(\Leftrightarrow4A=5^{51}-1\)
hay \(A=\dfrac{5^{51}-1}{4}\)
Bài 3:
\(S=\left(1^2+2^3+3^3+...+10^2\right)\cdot2=385\cdot2=770\)
\(\left|a+5\right|=3\)
\(\Rightarrow\orbr{\begin{cases}a+5=3\\a+5=-3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}a=3-5\\a=-3-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}a=-2\\a=-8\end{cases}}\)
Vậy a = -2 hoặc -8
a, Ia+5I = 3
=> a+5 = 3 => a= -2
=> a +5=-3 => a= -8
Vậy a= -2 hoặc a= -8