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24 tháng 12 2017
Bài 1.
a) ( x - 2)2 - ( x + 3)( x - 3)= 17
=> x2 - 4x + 4 - x2 + 9 - 17 = 0
=> -4x - 4 = 0
=> -4( x + 1 ) = 0
=> x = -1
Vậy,...
b)4( x - 3)2 - ( 2x - 1)( 2x + 1) = 10
=> 4( x2 - 6x + 9) - 4x2 + 1 - 10 = 0
=> - 24x + 36 - 9 = 0
=> -24x + 27 = 0
=> -3( 8x - 9) = 0
=> x = \(\dfrac{9}{8}\)
Vậy,...
c) ( x - 4)2 - ( x - 2)( x + 2)= 36
=> x2 - 8x + 16 - x2 + 4 - 36 = 0
=> -8x - 16 = 0
=> -8( x + 2) = 0
=> x = -2
d) ( 2x + 3)2 - ( 2x + 1)( 2x - 1) = 10
=> 4x2 + 12x + 9 - 4x2 + 1 - 10 = 0
=> 12x = 0
=> x = 0
Vậy,...
Bài 2.
\(\dfrac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}\)
a) ĐKXĐ : ( x + 1)( 2x - 6) # 0
=> 2( x + 1)( x - 3) # 0
=> x # -1 ; x # 3
Vậy,...
b) Để P = 1
=> \(\dfrac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}=1\)
=> \(\dfrac{3x\left(x+1\right)}{2\left(x+1\right)\left(x-3\right)}=\dfrac{3x}{2\left(x-3\right)}=1\)
=> 3x = 2x - 6
=> x = -6 ( thỏa mãn ĐKXĐ)
Vậy,...
Bài 3.
P = \(\dfrac{x}{x-1}+\dfrac{x^2+1}{1-x^2}\)
a) Để P có nghĩa tức P xác định .
ĐKXĐ : x - 1 # 0 => x # 1
* 1 - x2 # 0 => x # 1 ; x # -1
Vậy,...
b) P = \(\dfrac{x}{x-1}+\dfrac{x^2+1}{1-x^2}\)
P = \(\dfrac{x^2+x-x^2-1}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x+1}\)( x# 1; x# -1)
c) Để P = -1 thì :
\(\dfrac{1}{x+1}=-1\)
=> -x - 1 = 1
=> x = -2 ( thỏa mãn ĐKXĐ )
Vậy,...
7 tháng 2 2021
a) Ta có: \(P=\dfrac{x-2}{x^2-1}-\dfrac{x+2}{x^2+2x+1}\cdot\dfrac{1-x^2}{2}\)
\(=\dfrac{x-2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+2}{\left(x+1\right)^2}\cdot\dfrac{-\left(x-1\right)\left(x+1\right)}{2}\)
\(=\dfrac{x-2}{\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x+2\right)\left(x-1\right)}{2\left(x+1\right)}\)
\(=\dfrac{2\left(x-2\right)}{2\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x-1\right)^2\cdot\left(x+2\right)}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{2x-4-\left(x^2-2x+1\right)\left(x+2\right)}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{2x-4-\left(x^3+2x^2-2x^2-4x+x+2\right)}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{2x-4-\left(x^3-3x+2\right)}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{2x-4-x^3+3x-2}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x^3+5x-6}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-\left(x^3-5x+6\right)}{2\left(x-1\right)\left(x+1\right)}\)
AH
Akai Haruma
Giáo viên
28 tháng 6 2019
Bài 2:
a) ĐK: $x\geq \pm \frac{1}{2}; x\neq 0$
\(\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}=\frac{(2x+1)^2-(2x-1)^2}{(2x-1)(2x+1)}.\frac{10x-5}{4x}\)
\(\frac{4x^2+4x+1-(4x^2-4x+1)}{(2x-1)(2x+1)}.\frac{5(2x-1)}{4x}=\frac{8x}{(2x-1)(2x+1)}.\frac{5(2x-1)}{4x}\)
\(=\frac{10}{2x+1}\)
b) ĐK : $x\neq 0;-1$
\(\left(\frac{1}{x^2+x}-\frac{2-x}{x+1}\right):\left(\frac{1}{x}+x-2\right)=\left(\frac{1}{x(x+1)}-\frac{x(2-x)}{x(x+1)}\right):\frac{1+x^2-2x}{x}\)
\(=\frac{1-2x+x^2}{x(x+1)}.\frac{x}{1+x^2-2x}=\frac{x}{x(x+1)}=\frac{1}{x+1}\)
AH
Akai Haruma
Giáo viên
28 tháng 6 2019
Bài 3:
a) ĐKXĐ: \(x\neq \pm 1\)
b)
\(A=\left(\frac{x+1}{2x-2}-\frac{3}{1-x^2}-\frac{x+3}{2x+2}\right).\frac{4x^2-4}{5}\)
\(=\left[\frac{(x+1)^2}{2(x-1)(x+1)}+\frac{6}{2(x-1)(x+1)}-\frac{(x+3)(x-1)}{2(x+1)(x-1)}\right].\frac{4(x^2-1)}{5}\)
\(=\frac{(x+1)^2+6-(x^2+2x-3)}{2(x-1)(x+1)}.\frac{4(x-1)(x+1)}{5}\)
\(=\frac{10}{2(x-1)(x+1)}.\frac{4(x-1)(x+1)}{5}=4\)
24 tháng 5 2022
a)Vì |4x - 2| = 6 <=> 4x - 2 ϵ {6,-6} <=> x ϵ {2,-1}
Thay x = 2, ta có B không tồn tại
Thay x = -1, ta có B = \(\dfrac{1}{3}\)
b)ĐKXĐ:x ≠ 2,-2
Ta có \(A=\dfrac{5}{x+2}+\dfrac{3}{2-x}-\dfrac{15-x}{4-x^2}=\dfrac{10-5x+3x+6}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{16-2x}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{\left(x+2\right)\left(x-2\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{x^2-4}+\dfrac{15-x}{x^2-4}=\dfrac{x-1}{x^2-4}\)c)Từ câu b, ta có \(A=\dfrac{x-1}{x^2-4}\)\(\Rightarrow\dfrac{2A}{B}=\dfrac{\dfrac{\dfrac{2x-2}{x^2-4}}{2x+1}}{x^2-4}=\dfrac{2x-2}{2x+1}< 1\) với mọi x
Do đó không tồn tại x thỏa mãn đề bài
MN
20 tháng 2 2020
\(ĐKXĐ:\hept{\begin{cases}x\ne\pm1\\x\ne-\frac{1}{2}\end{cases}}\)
a) \(A=\left(\frac{1}{x-1}+\frac{x}{x^3-1}\cdot\frac{x^2+x+1}{x+1}\right):\frac{2x+1}{x^2+2x+1}\)
\(\Leftrightarrow A=\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\frac{2x+1}{\left(x+1\right)^2}\)
\(\Leftrightarrow A=\frac{x+1+x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(x+1\right)^2}{2x+1}\)
\(\Leftrightarrow A=\frac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(2x+1\right)}\)
\(\Leftrightarrow A=\frac{x+1}{x-1}\)
b) Thay \(x=\frac{1}{2}\)vào A, ta được :
\(A=\frac{\frac{1}{2}+1}{\frac{1}{2}-1}=\frac{\frac{3}{2}}{-\frac{1}{2}}=-3\)
\(B=\left(\dfrac{1}{x-2}-\dfrac{2x}{4-x^2}+\dfrac{1}{2+x}\right).\dfrac{2}{x-1}\)
\(\Rightarrow B=\dfrac{x+2+2x+x-2}{\left(x-2\right)\left(x+2\right)}.\dfrac{2}{x-1}\)
\(\Rightarrow B=\dfrac{4x}{\left(x-2\right)\left(x+2\right)}.\dfrac{2}{x-1}\)
\(\Rightarrow B=\dfrac{8x}{\left(x-2\right)\left(x+2\right)\left(x-1\right)}\)
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