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\(A=\frac{1}{x+5}+\frac{2}{x-5}-\frac{2x+10}{\left(x+5\right)\left(x-5\right)}\) ĐK đề bài
\(=\frac{x-5+2\left(x+5\right)-2x-10}{\left(x+5\right)\left(x-5\right)}=\frac{-\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}=-\frac{1}{x-5}\)
b/ có A=-3 => \(-\frac{1}{x-5}=-3 \Rightarrow x-5=\frac{1}{3}\Rightarrow x=\frac{16}{3}\)
có \(9x^2-42x+49=\left(3x-7\right)^2=\left(\frac{3.16}{3}-7\right)^2=81\)
\(\dfrac{1}{x+5}+\dfrac{1}{x-5}-\dfrac{2x+10}{\left(x+5\right)\cdot\left(x-5\right)}=\dfrac{x-5+x+5-2\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}\)
=\(\dfrac{2x-2}{x+5}=\dfrac{2\left(x-1\right)}{x+5}\)
b, Khi A=-3
thì ta có
\(\dfrac{2\left(x-1\right)}{x+5}=-3\)
\(\Leftrightarrow\) \(\dfrac{2\left(x-1\right)}{x+5}=\dfrac{-3\left(x+5\right)}{x+5}\Leftrightarrow2x-2=-3x-15\Leftrightarrow2x+3x=-15+2\Leftrightarrow5x=-13\Rightarrow x=-\dfrac{13}{5}\)
Bài 2 :
a, Ta có : \(A=\frac{1}{x+5}+\frac{2}{x-5}-\frac{2x+10}{\left(x+5\right)\left(x-5\right)}\)
=> \(A=\frac{x-5}{\left(x+5\right)\left(x-5\right)}+\frac{2\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{2x+10}{\left(x+5\right)\left(x-5\right)}\)
=> \(A=\frac{x-5+2\left(x+5\right)-2x-10}{\left(x-5\right)\left(x+5\right)}\)
=> \(A=\frac{x-5}{\left(x-5\right)\left(x+5\right)}=\frac{1}{x+5}\)
b, - Thay A = -3 ta được phương trình \(\frac{1}{x+5}=-3\)
=> \(-3\left(x+5\right)=1\)
=> \(-3x-15=1\)
=> \(-3x=16\)
=> \(x=-\frac{16}{3}\)
- Thay x = \(-\frac{16}{3}\)vào phương trình trên ta được :
\(9.\left(-\frac{16}{3}\right)^2-42.\left(-\frac{16}{3}\right)+49=529\)
\(a,\frac{\left(2x^2+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}\)
\(=\frac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}\)
\(=\frac{2\left(x-2\right)}{x+2}\)
Với \(x=\frac{1}{2}\)
\(\Rightarrow\frac{2\left(x-2\right)}{x+2}=\frac{2\left(\frac{1}{2}-2\right)}{\frac{1}{2}+2}=\frac{2.-\frac{3}{2}}{\frac{5}{2}}=-3.\frac{2}{5}=\frac{-6}{5}\)
b,Do x = -5; y = 10=> y = -2x
Thay y = -2x vào biểu thức ta được
\(\frac{x^3-x^2\left(-2x\right)+x\left(-2x\right)^2}{x^3+\left(-2x\right)^3}\)
\(=\frac{x^3+2x^3+2x^2}{x^3-8x^3}\)
\(=\frac{3x^3+2x^2}{-7x^3}=\frac{3}{-7}+\frac{2}{-7x}\)
Thay x = -5 là đc
Lời giải của bạn Nhật Linh đúng rồi, tuy nhiên cần thêm điều kiện để A có nghĩa: \(x\ne\pm2\)
Bài 1:
a) \(\dfrac{3x^2-5}{x^2-5x}+\dfrac{5-15x}{5x-25}\)
\(=\dfrac{3x^2-5}{x\left(x-5\right)}+\dfrac{5\left(1-3x\right)}{5\left(x-5\right)}\)
\(=\dfrac{3x^2-5}{x\left(x-5\right)}+\dfrac{1-3x}{x-5}\)
\(=\dfrac{3x^2-5}{x\left(x-5\right)}+\dfrac{x\left(1-3x\right)}{x\left(x-5\right)}\)
\(=\dfrac{3x^2-5+x\left(1-3x\right)}{x\left(x-5\right)}\)
\(=\dfrac{3x^2-5+x-3x^2}{x\left(x-5\right)}\)
\(=\dfrac{-5+x}{x\left(x-5\right)}\)
\(=\dfrac{x-5}{x\left(x-5\right)}\)
\(=\dfrac{1}{x}\)
b) \(\dfrac{4+x^3}{x-3}-\dfrac{2x+2x^2}{x-3}+\dfrac{2x-13}{x-3}\)
\(=\dfrac{\left(4+x^3\right)-\left(2x+2x^2\right)+\left(2x-13\right)}{x-3}\)
\(=\dfrac{4+x^3-2x-2x^2+2x-13}{x-3}\)
\(=\dfrac{x^3-2x^2-9}{x-3}\)
\(=\dfrac{x^3-3x^2+x^2-9}{x-3}\)
\(=\dfrac{x^2\left(x-3\right)+\left(x-3\right)\left(x+3\right)}{x-3}\)
\(=\dfrac{\left(x-3\right)\left(x^2+x+3\right)}{x-3}\)
\(=x^2+x+3\)
c) \(\dfrac{2}{x-5}+\dfrac{x-25}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{2\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}+\dfrac{x-25}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{2\left(x+5\right)+x-25}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{2x+10+x-25}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{3x-15}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{3\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{3}{x+5}\)
d) Đề sai?
Bài 2:
\(A=2\left(x+1\right)+\left(3x+2\right)\left(3x-2\right)-9x^2\)
\(A=2x+2+9x^2-4-9x^2\)
\(A=2x-2\)
\(A=2\left(x-1\right)\)
Thay x = 15 vào A ta được:
\(A=2\left(15-1\right)\)
\(A=2.14=28\)
a ) Rút gọn : \(A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2x+10}{\left(x+5\right)\left(x-5\right)}\)
\(\Leftrightarrow A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}\)
\(\Leftrightarrow A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2}{x-5}\)
\(\Leftrightarrow A=\dfrac{1}{x+5}.\)
Khi \(A=-3\),thì :
\(\dfrac{1}{x+5}=-3\Leftrightarrow x=-\dfrac{16}{3}\)
Ta có : \(9x^2-42x+49\)
\(=\left(3x\right)^2-2.3x.7+49\)
\(=\left(3x-7\right)^2\)
Thay \(x=-\dfrac{16}{3},\) ta có :
\(\left(3.\dfrac{-16}{3}-7\right)^2=\left(-16-7\right)^2=\left(-23\right)^2=529\)