A=(x/x+3 - 2/x-3 + x^2-1/9-x^2):(2- x+5/3+x)
a;rút gọn biểu thức A
b;tìm A biết |x|=1
c;tìm x biết a=1/2
d; tìm các giá trị thuộc z để a thuộc giá trị nguyên
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Câu 1 :
a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)
\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)
\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)
Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)
tương tự
\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)
\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)
\(< =>95-24x+40=6-4x-15x+5\)
\(< =>-24x+135=-19x+11\)
\(< =>5x=135-11=124\)
\(< =>x=\frac{124}{5}\)
a) x - 3/2 = 4/3
x = 4/3 + 3/2
x = 8/6 + 9/6 = 17/6
b) 2/5 * x = 1/3
x = 1/3 : 2/5
x = 1/3 x 5/2 = 5/6
c) x - 4/9 = 3/7 : 9/14
x - 4/9 = 2/3
x = 2/3 + 4/9
x = 6/9 + 4/9 = 10/9
d) 3/5 * x - 1/2 = 1/5
3/5 * x = 1/5 + 1/2 = 7/10
x = 7/10 : 3/5
x = 7/10 * 5/3 = 7/6
a/\(x-\dfrac{3}{2}=\dfrac{4}{3}\)
\(x=\dfrac{4}{3}+\dfrac{3}{2}\)
\(x=\dfrac{17}{6}\)
b/\(\dfrac{2}{5}\times x=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}:\dfrac{2}{5}\)
\(x=\dfrac{5}{6}\)
c/\(x-\dfrac{4}{9}=\dfrac{3}{7}:\dfrac{9}{14}\)
\(x-\dfrac{4}{9}=\dfrac{2}{3}\)
\(x=\dfrac{2}{3}+\dfrac{4}{9}\)
\(x=\dfrac{10}{9}\)
d/\(\dfrac{3}{5}\times x-\dfrac{1}{2}=\dfrac{1}{5}\)
\(\dfrac{3}{5}\times x=\dfrac{1}{5}+\dfrac{1}{2}\)
\(\dfrac{3}{5}\times x=\dfrac{7}{10}\)
\(x=\dfrac{7}{10}:\dfrac{3}{5}\)
\(x=\dfrac{7}{6}\)
a) 2 x 2 = 4 3 x 3 = 9
2 x 4 = 8 3 x 5 = 15
2 x 6 = 12 3 x 7 = 21
2 x 8 = 16 3 x 9 = 27
4 x 4 = 16 5 x 5 = 25
4 x 2 = 8 5 x 7 = 35
4 x 6 = 24 5 x 9 = 45
4 x 8 = 32 5 x 3 = 15
b) 200 x 4 = 800 300 x 2 = 600
200 x 2 = 400 300 x 3 = 900
400 x 2 = 800 500 x 1 = 500
100 x 4 = 400 100 x 3 = 300
`A=2/(1xx3)+2/(3xx5)+2/(5xx7)+2/(7xx9)+2/(9xx11)+2/(11xx13)`
`=1/1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13`
`=1/1-1/13`
`=13/13-1/13`
`=12/13`
a) \(A=\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}+\dfrac{x^2-1}{9-x^2}\right):\left(2-\dfrac{x+5}{x+3}\right)\) (ĐK: \(x\ne\pm3\))
\(A=\left[\dfrac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x^2-1}{\left(x+3\right)\left(x-3\right)}\right]:\left(2+\dfrac{x+5}{x+3}\right)\)
\(A=\dfrac{x^2-3x-2x-6-x^2+1}{\left(x+3\right)\left(x-3\right)}:\dfrac{2\left(x+3\right)-\left(x+5\right)}{x+3}\)
\(A=\dfrac{-5x-5}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{x+3}{x+1}\)
\(A=\dfrac{-5\left(x+1\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)\left(x+1\right)}\)
\(A=\dfrac{-5}{x-3}\)
b) Ta có: \(\left|x\right|=1\)
TH1: \(\left|x\right|=-x\) với \(x< 0\)
Pt trở thành:
\(-x=1\) (ĐK: \(x< 0\))
\(\Leftrightarrow x=-1\left(tm\right)\)
Thay \(x=-1\) vào A ta có:
\(A=\dfrac{-5}{x-3}=\dfrac{-5}{-1-3}=\dfrac{5}{4}\)
TH2: \(\left|x\right|=x\) với \(x\ge0\)
Pt trở thành:
\(x=1\left(tm\right)\) (ĐK: \(x\ge0\))
Thay \(x=1\) vào A ta có:
\(A=\dfrac{-5}{x-3}=\dfrac{-5}{1-2}=\dfrac{5}{2}\)
c) \(A=\dfrac{1}{2}\) khi:
\(\dfrac{-5}{x-3}=\dfrac{1}{2}\)
\(\Leftrightarrow-10=x-3\)
\(\Leftrightarrow x=-10+3\)
\(\Leftrightarrow x=-7\left(tm\right)\)
d) \(A\) nguyên khi:
\(\dfrac{-5}{x-3}\) nguyên
\(\Rightarrow x-3\inƯ\left(-5\right)\)
\(\Rightarrow x\in\left\{8;-2;2;4\right\}\)
a: \(A=\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}+\dfrac{x^2-1}{9-x^2}\right):\left(2-\dfrac{x+5}{x+3}\right)\)
\(=\dfrac{x\left(x-3\right)-2\left(x+3\right)-x^2+1}{\left(x-3\right)\left(x+3\right)}:\dfrac{2x+6-x-5}{x+3}\)
\(=\dfrac{x^2-3x-2x-6-x^2+1}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x+1}\)
\(=\dfrac{-5x-5}{\left(x-3\right)}\cdot\dfrac{1}{x+1}=\dfrac{-5}{x-3}\)
b: |x|=1
=>x=-1(loại) hoặc x=1(nhận)
Khi x=1 thì \(A=\dfrac{-5}{1-3}=-\dfrac{5}{-2}=\dfrac{5}{2}\)
c: A=1/2
=>x-3=-10
=>x=-7
d: A nguyên
=>-5 chia hết cho x-3
=>x-3 thuộc {1;-1;5;-5}
=>x thuộc {4;2;8;-2}