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\(P=\left(\frac{x-1}{x+3}+\frac{2}{x-3}+\frac{x^2+3}{9-x^2}\right):\left(\frac{2x-1}{2x+1}-1\right)\)\(\left(đkcđ:x\ne\pm3;x\ne-\frac{1}{2}\right)\)
\(=\left(\frac{\left(x-1\right).\left(x-3\right)+2.\left(x+3\right)-\left(x^2+3\right)}{x^2-9}\right):\left(\frac{2x-1-\left(2x+1\right)}{2x+1}\right)\)
\(=\frac{x^2-4x+3+2x+6-x^2-3}{x^2-9}:\frac{-2}{2x+1}\)
\(=\frac{-2x-6}{x^2-9}.\frac{2x+1}{-2}\)
\(=\frac{-2\left(x+3\right)}{\left(x-3\right).\left(x+3\right)}.\frac{2x+1}{-2}\)
\(=\frac{2x+1}{x-3}\)
b)\(\left|x+1\right|=\frac{1}{2}\Leftrightarrow\orbr{\begin{cases}x+1=\frac{1}{2}\\x+1=-\frac{1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\left(koTMđkxđ\right)\\x=-\frac{3}{2}\left(TMđkxđ\right)\end{cases}}}\)
thay \(x=-\frac{3}{2}\) vào P tâ đc: \(P=\frac{2x+1}{x-3}=\frac{2.\left(-\frac{3}{2}\right)+1}{-\frac{3}{2}-3}=\frac{4}{9}\)
c)ta có:\(P=\frac{x}{2}\Leftrightarrow\frac{2x+1}{x-3}=\frac{x}{2}\)
\(\Rightarrow2.\left(2x+1\right)=x.\left(x-3\right)\)
\(\Leftrightarrow4x+2=x^2-3x\)
\(\Leftrightarrow x^2-7x-2=0\)
\(\Leftrightarrow x^2-2.\frac{7}{2}+\frac{49}{4}-\frac{57}{4}=0\)
\(\Leftrightarrow\left(x-\frac{7}{2}\right)^2-\frac{57}{4}=0\)
\(\Leftrightarrow\left(x-\frac{7}{2}-\frac{\sqrt{57}}{2}\right).\left(x-\frac{7}{2}+\frac{\sqrt{57}}{2}\right)\)
bạn tự giải nốt nhé!!
d)\(x\in Z;P\in Z\Leftrightarrow\frac{2x+1}{x-3}\in Z\Leftrightarrow\frac{2x-6+7}{x-3}=2+\frac{7}{x-3}\in Z\)
\(2\in Z\Rightarrow\frac{7}{x-3}\in Z\Leftrightarrow x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
bạn tự làm nốt nhé
a, \(\left(\dfrac{x^2-4x+3+2x+6-x^2-3}{\left(x+3\right)\left(x-3\right)}\right):\left(\dfrac{2x-1-2x-1}{2x+1}\right)\)
\(=\dfrac{-2x+6}{\left(x+3\right)\left(x-3\right)}:\dfrac{-2}{2x+1}=\dfrac{-2\left(x-3\right)\left(2x+1\right)}{-2\left(x+3\right)\left(x-3\right)}=\dfrac{2x+1}{x+3}\)
b, \(\left|x+1\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}-1\\x=-\dfrac{1}{2}-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\left(ktmđk\right)\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Thay x = -3/2 ta được \(\dfrac{2\left(-\dfrac{3}{2}\right)+1}{-\dfrac{3}{2}+3}=\dfrac{-2}{\dfrac{3}{2}}=-\dfrac{4}{3}\)
a. \(2.\left(5x-8\right)-3.\left(4x-5\right)=4.\left(3x-4\right)+11\Leftrightarrow10x-16-12x+15=12x-16+11\\ \)
\(\Leftrightarrow-2x-1=12x-5\Leftrightarrow14x-4=0\Leftrightarrow x=\frac{2}{7}\)
\(a,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow10x-12x-12x=-16+11+16-15\)
\(\Leftrightarrow-14x=-4\)
\(\Leftrightarrow x=\frac{-4}{-14}=\frac{2}{7}\)
c) \(\left(3x+5\right)^2-2\left(2x+3\right)\left(3x+5\right)+\left(2x+3\right)^2=\left(x+2\right)^3\)
\(\Leftrightarrow\left[\left(3x+5\right)-\left(2x+3\right)\right]^2=\left(x+2\right)^3\)
\(\Leftrightarrow\left(3x+5-2x-3\right)^2=\left(x+2\right)^3\)
\(\Leftrightarrow\left(x+2\right)^2=\left(x+2\right)^3\)
\(\Leftrightarrow\left(x+2\right)^3-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)^2.\left(x+2-1\right)=0\)
\(\Leftrightarrow\left(x+2\right)^2.\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-1\end{cases}}\)
Vậy tập nghiệm của phương trình là: \(S=\left\{-2;-1\right\}\)
a: \(\Leftrightarrow2x^2-8x+3x-12+x^2-7x+10=3x^2-12x-5x+20\)
\(\Leftrightarrow3x^2-12x-2=3x^2-17x+20\)
=>5x=22
hay x=22/5
b: \(\Leftrightarrow24x^2+16x-9x-6-4x^2-16x-7x-28=10x^2-2x+5x-1\)
\(\Leftrightarrow20x^2-16x-34=10x^2+3x-1\)
\(\Leftrightarrow10x^2-19x-33=0\)
hay \(x\in\left\{3;-\dfrac{11}{10}\right\}\)
c: \(\Leftrightarrow x^3+2x^2-5x-10+5x=2x^2+17\)
\(\Leftrightarrow x^3+2x^2-10-2x^2-17=0\)
=>x3=27
=>x=3
d: \(\Leftrightarrow x^3+1-x^3+3x=15\)
=>3x=14
hay x=14/3
làm nốt
d) (2x-1)(3x+2)(3-x)
=(6x2+x-2)(3-x)
=-6x3+17x2+5x-6
e) (x+3)(x2+3x-5)
=x3+6x2+4x-15
f) (xy-2)(x3-2x-6)
=x4y-2x3-2x2y-6xy+4x+12
g) (5x3-x2+2x-3)(4x2-x+2)
=20x5-9x4+19x3-16x2+7x-6
Bài 1:
a) (x-2)(x2+3x+4)
=x(5x+4)-2(5x+4)
= 5x2+4x-10x-8
=5x2-6x-8
tìm x
a.(x−2)\(^3\)−(x−3)(x\(^2\)+3x+9)+6(x+1)\(^2\)
b.(x+2)(x\(^2\)−2x+4x\(^2\)−2x+4)-x(x\(^2\)+2)=15
Câu a thiếu kết quả để tìm
Câu b)
(x + 2)(x2 - 2x + 4x2 - 2x + 4) - x(x2 + 2) = 15
=> (x + 2)(x2 - 4x + 4x2 + 4) - x3 + 2x = 15
=> (x + 2)(5x2 - 4x + 4) - x3 + 2x = 15
=> x(5x2 - 4x + 4) + 2(5x2 - 4x + 4) - x3 + 2x = 15
=> 5x3 - 4x2 + 4x + 10x2 - 8x + 8 - x3 + 2x = 15
=> (5x3 - x3) + (-4x2 + 10x2) + (4x - 8x + 2x) + 8 = 15
=> 4x3 + 6x2 - 2x + 8 = 15
=> 2(2x3 + 3x2 - x + 4) = 15
=> (2x3 + 3x2 - x + 4) = 15/2 => vô nghiệm
a) Rút gọn được VT = 9x + 7. Từ đó tìm được x = 1.
b) Rút gọn được VT = 2x + 8. Từ đó tìm được x = 7 2 .