Tìm x \(\in\)N :
\(\frac{4x+5}{x+1}\)
K
Khách
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16 tháng 12 2020
Bài 1.
a)\(\frac{4x-4}{x^2-4x+4}\div\frac{x^2-1}{\left(2-x\right)^2}=\frac{4\left(x-1\right)}{\left(x-2\right)^2}\div\frac{\left(x-1\right)\left(x+1\right)}{\left(x-2\right)^2}=\frac{4\left(x-1\right)}{\left(x-2\right)^2}\times\frac{\left(x-2\right)^2}{\left(x-1\right)\left(x+1\right)}=\frac{4}{x+1}\)
b) \(\frac{2x+1}{2x^2-x}+\frac{32x^2}{1-4x^2}+\frac{1-2x}{2x^2+x}=\frac{2x+1}{x\left(2x-1\right)}+\frac{-32x^2}{4x^2-1}+\frac{1-2x}{x\left(2x+1\right)}\)
\(=\frac{\left(2x+1\right)\left(2x+1\right)}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-32x^3}{x\left(2x-1\right)\left(2x+1\right)}+\frac{\left(1-2x\right)\left(2x-1\right)}{x\left(2x-1\right)\left(2x+1\right)}\)
\(=\frac{4x^2+4x+1}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-32x^3}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-4x^2+4x-1}{x\left(2x-1\right)\left(2x+1\right)}\)
\(=\frac{4x^2+4x+1-32x^3-4x^2+4x-1}{x\left(2x-1\right)\left(2x+1\right)}=\frac{-32x^3+8x}{x\left(2x-1\right)\left(2x+1\right)}\)
\(=\frac{-8x\left(4x^2-1\right)}{x\left(2x-1\right)\left(2x+1\right)}=\frac{-8x\left(2x-1\right)\left(2x+1\right)}{x\left(2x-1\right)\left(2x+1\right)}=-8\)
c) \(\left(\frac{1}{x+1}+\frac{1}{x-1}-\frac{2x}{1-x^2}\right)\times\frac{x-1}{4x}\)
\(=\left(\frac{1}{x+1}+\frac{1}{x-1}+\frac{2x}{x^2-1}\right)\times\frac{x-1}{4x}\)
\(=\left(\frac{x-1}{\left(x-1\right)\left(x+1\right)}+\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{2x}{\left(x-1\right)\left(x+1\right)}\right)\times\frac{x-1}{4x}\)
\(=\left(\frac{x-1+x+1+2x}{\left(x-1\right)\left(x+1\right)}\right)\times\frac{x-1}{4x}\)
\(=\frac{4x}{\left(x-1\right)\left(x+1\right)}\times\frac{x-1}{4x}=\frac{1}{x+1}\)
Bài 3.
N = ( 4x + 3 )2 - 2x( x + 6 ) - 5( x - 2 )( x + 2 )
= 16x2 + 24x + 9 - 2x2 - 12x - 5( x2 - 4 )
= 14x2 + 12x + 9 - 5x2 + 20
= 9x2 + 12x + 29
= 9( x2 + 4/3x + 4/9 ) + 25
= 9( x + 2/3 )2 + 25 ≥ 25 > 0 ∀ x
=> đpcm
B
22 tháng 8 2017
Ta có: x2 – x – 12 = x2 – x – 16 + 4
= (x2 – 16) – (x – 4)
= (x – 4).(x + 4) – (x – 4)
= (x – 4).(x + 4 – 1)
= (x – 4).(x + 3)
KN
18 tháng 8 2021
a. ĐKXĐ : \(x\ne\frac{1}{2};\frac{5}{2};4;-\frac{3}{2};\frac{1\pm\sqrt{43}}{2}\)
\(A=\left(\frac{2x-3}{4x^2-12x+5}+\frac{3x-8}{13x-2x^2-20}-\frac{3}{2x-1}\right):\frac{21+2x-2x^2}{4x^2+4x-3}+\)
\(=\left(\frac{2x-3}{\left(2x-1\right)\left(2x-5\right)}-\frac{3x-8}{\left(2x-5\right)\left(x-4\right)}-\frac{3}{2x-1}\right).\frac{\left(2x-1\right)\left(2x+3\right)}{21+2x-2x^2}+1\)
\(=\frac{\left(2x-3\right)\left(x-4\right)-\left(3x-8\right)\left(2x-1\right)-3\left(2x-5\right)\left(x-4\right)}{\left(2x-1\right)\left(2x-5\right)\left(x-4\right)}.\frac{\left(2x-1\right)\left(2x+3\right)}{21+2x-2x^2}+1\)
\(=\frac{-10x^2+47x-56}{\left(2x-5\right)\left(x-4\right)}.\frac{2x+3}{-2x^2+2x+21}+1\) số to wa
VH
2
IM
13 tháng 9 2016
Vì \(\left|3x+2\right|+\left|x+\frac{3}{5}\right|+\left|\frac{1}{2}-x\right|>0\)
=> 4x > 0
=> x > 0
\(\Rightarrow\left(3x+2\right)+\left(x+\frac{3}{5}\right)+\left(\frac{1}{2}-x\right)=4x\)
\(\Rightarrow\left(3x-x+x\right)+\left(2+\frac{3}{5}-\frac{1}{2}\right)=4x\)
\(\Rightarrow3x+\frac{21}{10}=4x\)
=> x = - 21 / 10
Vậy x = - 21 / 10
7 tháng 5 2019
a)\(10\left(x-7\right)-8\left(x+5\right)=6\cdot\left(-5\right)+24\)
\(10x-10\cdot7-8x-8\cdot5=\left(-30\right)+24\)
\(10x-70-8x-40=-6\)
\(10x-8x=\left(-6\right)+70+40\)
\(2x=104\)
\(x=104\div2\)
\(x=52\)
b)\(2\left(4x-8\right)-7\left(3+x\right)=6\)
\(2\cdot4x-2\cdot8-7\cdot3-7x=6\)
\(8x-16-21-7x=6\)
\(8x-7x=6+16+21\)
\(x=43\)
25 tháng 6 2020
\(\frac{x+2}{4x-1}=\frac{x-5}{4x+1}\) ( đkxđ : \(x\ne\pm\frac{1}{4}\))
<=> \(\frac{\left(x+2\right)\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}=\frac{\left(x-5\right)\left(4x-1\right)}{\left(4x-1\right)\left(4x+1\right)}\)
<=> \(4x^2+9x+2=4x^2-21x+5\)
<=> \(4x^2+9x+2-4x^2+21x-5=0\)
<=> \(30x-3=0\)
<=> \(30x=3\)
<=> \(x=\frac{3}{30}=\frac{1}{10}\)( tmđk )
ko có điều kiện ak
vậy thì tìm kiểu j
\(\frac{4x+5}{x+1}=0\Leftrightarrow4x+5=0\Leftrightarrow x=-\frac{5}{4}\)
( vì phân số bằng 0 khi tử số bằng 0, nếu mẫu số bằng 0 thì phân số không xác định)