biết x:(x^2 +x+1) =1:4.
tính A= (x^5 -4x^3-3x+9):(x^4 +3x^2 +11)
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LL
27 tháng 6 2019
Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
27 tháng 6 2019
a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
LT
3
26 tháng 7 2021
1) Ta có: \(5\left(x-3\right)\left(x-7\right)-\left(5x+1\right)\left(x-2\right)=-8\)
\(\Leftrightarrow5\left(x^2-10x+21\right)-\left(5x^2-10x+x-2\right)=-8\)
\(\Leftrightarrow5x^2-50x+105-5x^2+9x+2+8=0\)
\(\Leftrightarrow-41x=-115\)
hay \(x=\dfrac{115}{41}\)
2) Ta có: \(x\left(x+1\right)\left(x+2\right)-\left(x+4\right)\left(3x-5\right)=84-5x\)
\(\Leftrightarrow x\left(x^2+3x+2\right)-\left(3x^2+7x-20\right)=84-5x\)
\(\Leftrightarrow x^3+3x^2+2x-3x^2-7x+20-84+5x=0\)
\(\Leftrightarrow x^3=64\)
hay x=4
3) Ta có: \(\left(9x^2-5\right)\left(x+3\right)-3x^2\left(3x+9\right)=\left(x-5\right)\left(x+4\right)-x\left(x-11\right)\)
\(\Leftrightarrow9x^3+27x^2-5x-15-9x^3-27x^2=x^2-x-20-x^2+11x\)
\(\Leftrightarrow-5x-15=10x-20\)
\(\Leftrightarrow-5x-10x=-20+15\)
\(\Leftrightarrow x=\dfrac{-5}{-15}=\dfrac{1}{3}\)
NH
11 tháng 11 2020
a)(x+2).(x+3)-(x-2).(x+5)=10
( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10
x^2 +3x+2x+6-x^2 -5x+2x+10-10=0
2x+6=0
2x=-6
x=-3
9 tháng 7 2018
A. \(4\left(x+2\right)-7\left(2x-1\right)+9\left(3x-4\right)=30\)
\(\Leftrightarrow4x+8-14x+7+27x-36=30\)
\(\Leftrightarrow4x-14x+27x=30-8-7+36\)
\(\Leftrightarrow17x=51\)
\(\Leftrightarrow x=3\) . Vậy \(S=\left\{3\right\}\)
B. \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow10x-12x-12x=16-15-16+11\)
\(\Leftrightarrow10x=-4\)
\(\Leftrightarrow x=-\dfrac{2}{5}\) . Vậy \(S=\left\{-\dfrac{2}{5}\right\}\)
Câu C) bạn xem lại đề nha mik tính ko đc
D. \(\left(5x-3\right)4x-2x\left(10x-3\right)=15\)
\(\Leftrightarrow20x^2-12x-20x^2+6x=15\)
\(\Leftrightarrow-6x=15\)
\(\Leftrightarrow x=-\dfrac{5}{2}\) . Vậy \(S=\left\{-\dfrac{5}{2}\right\}\)
28 tháng 4 2018
a) 5 - 4x = 3x - 9
\(\Leftrightarrow5-4x-3x+9=0\)
\(\Leftrightarrow14-7x=0\)
\(\Leftrightarrow7x=14\Leftrightarrow x=2\)
Vậy \(S=\left\{2\right\}\)
b) \(\left(x-4\right)\left(3x+9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
Vậy \(S=\left\{-3;4\right\}\)
c) \(\dfrac{x}{x+4}+\dfrac{12}{x-4}=\dfrac{4x+48}{x\cdot x-16}\)(1)
ĐKXĐ: \(x\ne\pm4\)
\(\left(1\right)\Leftrightarrow\dfrac{x\left(x-4\right)+12\left(x+4\right)-4x-48}{\left(x+4\right)\left(x-4\right)}=0\)
\(\Leftrightarrow x^2-4x+12x+48-4x-48=0\)
\(\Leftrightarrow x^2+4x=0\)
\(\Leftrightarrow x\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-4\left(KTM\right)\end{matrix}\right.\)
Vậy \(S=\left\{0\right\}\)
d) \(4-2x=7-x\)
\(\Leftrightarrow4-2x-7+x=0\)
\(\Leftrightarrow-x-3=0\)
\(\Leftrightarrow-x=3\Leftrightarrow x=-3\)
Vậy \(S=\left\{-3\right\}\)
e) \(\left(x+4\right) \left(8-4x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\8-4x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)
Vậy \(S=\left\{-4;2\right\}\)
f) \(\dfrac{x}{x+5}+\dfrac{11}{x-5}=\dfrac{x+55}{x\cdot x-25}\left(2\right)\)
ĐKXĐ: \(x\ne\pm5\)
\(\left(2\right)\Leftrightarrow\dfrac{x\left(x-5\right)+11\left(x+5\right)-x-55}{\left(x+5\right)\left(x-5\right)}=0\)
\(\Leftrightarrow x^2-5x+11x+55-x-55=0\)
\(\Leftrightarrow x^2+5x=0\)
\(\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-5\left(KTM\right)\end{matrix}\right.\)
Vậy \(S=\left\{0\right\}\)
g) \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=\dfrac{5}{3}+2x\)
\(\Leftrightarrow\dfrac{3\left(3x+2\right)-3x-1-10-12x}{6}=0\)
\(\Leftrightarrow9x+6-3x-1-10-12x=0\)
\(\Leftrightarrow-6x-5=0\)
\(\Leftrightarrow-6x=5\)
\(\Leftrightarrow x=-\dfrac{5}{6}\)
Vậy \(S=\left\{-\dfrac{5}{6}\right\}\)
h) \(2x-\left(3-5x\right)=4\left(x+3\right)\)
\(\Leftrightarrow2x-3+5x-4x-12=0\)
\(\Leftrightarrow3x-15=0\)
\(\Leftrightarrow x=5\)
Vậy \(S=\left\{5\right\}\)
i) \(3x-6+x=9-x\)
\(\Leftrightarrow3x-6+x-9+x=0\)
\(\Leftrightarrow5x-15=0\)
\(\Leftrightarrow x=3\)
Vậy \(S=\left\{3\right\}\)
k)\(2t-3+5t=4t+12\)
\(\Leftrightarrow2t-3+5t-4t-12=0\)
\(\Leftrightarrow3t-15=0\)
\(\Leftrightarrow t=5\)
Vậy \(S=\left\{5\right\}\)
11 tháng 7 2021
9) Ta có: \(\dfrac{2x+5}{x+3}+1=\dfrac{4}{x^2+2x-3}-\dfrac{3x-1}{1-x}\)
\(\Leftrightarrow\left(2x+5\right)\left(x-1\right)+x^2+2x-3=4+\left(3x-1\right)\left(x+3\right)\)
\(\Leftrightarrow2x^2-2x+5x-5+x^2+2x-3-4-3x^2-10x+x+3=0\)
\(\Leftrightarrow-4x=9\)
hay \(x=-\dfrac{9}{4}\)
11 tháng 7 2021
10) Ta có: \(\dfrac{x-1}{x+3}-\dfrac{x}{x-3}=\dfrac{7x-3}{9-x^2}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3-7x}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(x^2-4x+3-x^2-3x-3+7x=0\)
\(\Leftrightarrow0x=0\)(luôn đúng)
Vậy: S={x|\(x\notin\left\{3;-3\right\}\)}
11) Ta có: \(\dfrac{5+9x}{x^2-16}=\dfrac{2x-1}{x+4}+\dfrac{3x-1}{x-4}\)
\(\Leftrightarrow\dfrac{\left(2x-1\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}+\dfrac{\left(3x-1\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{9x+5}{\left(x-4\right)\left(x+5\right)}\)
Suy ra: \(2x^2-9x+4+3x^2+12x-x-4-9x-5=0\)
\(\Leftrightarrow5x^2-7x=0\)
\(\Leftrightarrow x\left(5x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{7}{5}\end{matrix}\right.\)
12) Ta có: \(\dfrac{2x}{2x-1}+\dfrac{x}{2x+1}=1+\dfrac{4}{\left(2x-1\right)\left(2x+1\right)}\)
\(\Leftrightarrow\dfrac{2x\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}+\dfrac{x\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}=\dfrac{4x^2-1+4}{\left(2x-1\right)\left(2x+1\right)}\)
Suy ra: \(4x^2+2x+2x^2-x-4x^2-3=0\)
\(\Leftrightarrow2x^2+x-3=0\)
\(\Leftrightarrow2x^2+3x-2x-3=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=1\end{matrix}\right.\)