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5 tháng 3 2022
c: \(=\dfrac{1}{3x-2}-\dfrac{4}{3x+2}+\dfrac{3x-6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+2-12x+8+3x-6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{-6x+4}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{-2}{3x+2}\)
d: \(=\dfrac{x^2-4-x^2+10}{x+2}=\dfrac{6}{x+2}\)
e: \(=\dfrac{1}{2\left(x-y\right)}-\dfrac{1}{2\left(x+y\right)}-\dfrac{y}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{x+y-x+y-2y}{2\left(x-y\right)\left(x+y\right)}=\dfrac{0}{2\left(x-y\right)\left(x+y\right)}=0\)
10 tháng 3 2020
\(a)\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}=\frac{-3}{4}\left(x\ne-3;x\ne2\right)\)
\(\Leftrightarrow\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{1}{x-2}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{x^2-4}{\left(x-2\right)\left(x+3\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{x+3}{\left(x-2\right)\left(x+3\right)}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{x-4}{x-2}=\frac{-3}{4}\)
<=> 4x-16=-3x+6
<=> 4x-16+3x-6=0
<=> 7x-22=0
<=> 7x=22
<=> \(x=\frac{22}{7}\)(TMĐK)
5 tháng 3 2022
Bài 1:
a: \(\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}+\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2-2x+1-x^2-2x-1+4}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-4x+4}{\left(x-1\right)\left(x+1\right)}=\dfrac{-4}{x+1}\)
b: \(=\dfrac{xy\left(x^2+y^2\right)}{x^4y}\cdot\dfrac{1}{x^2+y^2}=\dfrac{x}{x^4}=\dfrac{1}{x^3}\)
c: Đề thiếu rồi bạn
10 tháng 3 2020
\(\frac{x+2}{x+3}-\frac{x+1}{x-1}=\frac{4}{\left(x-1\right)\left(x+3\right)}\left(x\ne-3;x\ne1\right)\)
\(\Leftrightarrow\frac{x+2}{x+3}-\frac{x+1}{x-1}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}-\frac{\left(x+1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{x^2+x-2}{\left(x+3\right)\left(x-1\right)}-\frac{x^2+4x+3}{\left(x-1\right)\left(x+3\right)}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{x^2+x-2-x^2-4x-3-4}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{-3x-9}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{-3\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{-3}{x-1}=0\)
=> PT vô nghiệm
22 tháng 8 2016
(3-12x)(x-1)+(12x-8)(x+2)+x2=52
3(x-1)-12x(x-1)+12x(x+2)-8(x+2)+x2=52
3x-3-12x2+12+12x2+24x-8x-16+x2=52
(3x+24x-8x)+(12-3-16)+(12x2-12x2+x2)=52
19x-7+x2=52
x(19-x)=52+7=59
mà 59 là số ng tố nên x rỗng
Vậy x E \(\theta\)
29 tháng 4 2020
1) \(\frac{x-3}{2}+\frac{4x+1}{3}=\frac{2x-7}{6}\)
<=> 3(x - 3) + 2(4x + 1) = 2x - 7
<=> 3x - 9 + 8x + 2 = 2x - 7
<=> 11x - 7 = 2x - 7
<=> 11x - 7 - 2x = -7
<=> 9x - 7 = -7
<=> 9x = -7 + 7
<=> 9x = 0
<=> x = 0
21 tháng 10 2021
\(A=\left(x-4\right)^2-\left(x+4\right)^2-16\left(x-2\right)\)
\(=x^2-8x+16-x^2-8x-16-16x+32\)
\(=-32x+32\)
Biểu thức phụ thuộc vào giá trị của biến
27 tháng 3 2020
a) \(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{8}+\frac{2x-1}{12}\)
<=> \(\frac{x}{4}+\frac{5}{4}-\frac{2x}{3}+1=\frac{6x}{8}-\frac{1}{8}+\frac{2x}{12}-\frac{1}{12}\)
<=> \(-\frac{4}{3}x=-\frac{59}{24}\)
<=> \(x=\frac{59}{32}\)
Vậy S = { 59/32}
b) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
<=> \(\frac{x^2+14x+40}{12}-\frac{-x^2-2x+8}{4}=\frac{x^2+8x-20}{3}\)
<=> \(\left(\frac{x^2}{12}+\frac{x^2}{4}-\frac{x^2}{3}\right)+\left(\frac{14}{12}x+\frac{2}{4}x-\frac{8}{3}x\right)=-\frac{20}{8}+\frac{8}{4}-\frac{40}{12}\)
<=> \(-x=-8\)
<=> x = 8
Vậy S = { 8 }
(x+1)(x^2+2x+4)-x(x+3)
=x^3+2x^2+4x+x^2+2x+4-x^2-3x
=x^3+2x^2+3x+4
Tại x=-10/3 thì (-10/3)^3+2x(-10/3)^2+3x(-10/3)+4=-562/27
Đặt \(A=\left(x+1\right)\left(x^2+2x+4\right)-x\left(x+3\right)\)
\(=x^3+2x^2+4x+x^2+2x+4-x^2-3x\)
\(=x^3+2x^2+3x+4\)
Khi x=-10/3 thì \(A=\left(-\frac{10}{3}\right)^3+2\cdot\left(-\frac{10}{3}\right)^2+3\cdot\frac{-10}{3}+4\)
\(=-\frac{1000}{27}+2\cdot\frac{100}{9}-10+4=\frac{-1000}{27}+\frac{200}{9}-6\)
\(=-\frac{1000}{27}+\frac{600}{27}-\frac{162}{27}=-\frac{562}{27}\)