Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) 1x−1−3x2x3−1=2xx2+x+11x−1−3x2x3−1=2xx2+x+1
Ta có: x3−1=(x−1)(x2+x+1)x3−1=(x−1)(x2+x+1)
=(x−1)[(x+12)2+
a: \(\Leftrightarrow1-x+3x+3=2x+3\)
=>2x+4=2x+3(vô lý)
b: \(\Leftrightarrow\left(x+2\right)^2-2x+3=x^2+10\)
\(\Leftrightarrow x^2+4x+4-2x+3=x^2+10\)
=>4x+7=10
hay x=3/4
d: \(\Leftrightarrow\left(-2x+5\right)\left(3x-1\right)+3\left(x-1\right)\left(x+1\right)=\left(x+2\right)\left(1-3x\right)\)
\(\Leftrightarrow-6x^2+2x+15x-5+3\left(x^2-1\right)=\left(x+2\right)\left(1-3x\right)\)
\(\Leftrightarrow-6x^2+17x-5+3x^2-3=x-3x^2+2-6x\)
\(\Leftrightarrow-3x^2+17x-8=-3x^2-5x+2\)
=>22x=10
hay x=5/11
a: \(=6x^4-9x^3+3x^2-4x^3+6x^2-2x+10x^2-15x+5\)
\(=6x^4-13x^3+19x^2-17x+5\)
b: \(=6x^4-\dfrac{9}{4}x^3-\dfrac{9}{2}x^2-\dfrac{8}{3}x^3+x^2+2x-\dfrac{20}{3}x^2+\dfrac{5}{2}x+5\)
\(=6x^4-\dfrac{59}{12}x^3-\dfrac{67}{6}x^2+\dfrac{9}{2}x+5\)
c: \(=3x^4-\dfrac{9}{8}x^3-\dfrac{3}{4}x^2+8x^3-3x^2-6x-\dfrac{4}{3}x^2+\dfrac{1}{2}x+1\)
\(=3x^4-\dfrac{55}{8}x^3-\dfrac{25}{12}x^2-\dfrac{11}{2}x+1\)
a) 4x -8 ≥ 3(3x-1)-2x +1
⇒4x -8 ≥7x -2
⇒4x -7x ≥ -2 +8
⇒-3x ≥ 6
⇒x≤-2
Vậy bpt có nghiệm là:{x|x≤-2}
b) (x-3)(x+2)+(x+4)2≤ 2x (x+5)+4
⇔ x2+2x - 3x - 6 +x2 + 8x +16≤ 2x2 + 10x +4
⇔ x2 +2x - 3x + x2 + 8x - 2x2- 10x ≤ 4+6-16
⇔ -3x ≤ -6
⇔ x≥ 2
Vậy bpt có tập nghiệm là: {x|x≥2}
a: \(=\dfrac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{2\left(x-1\right)\left(x+1\right)}{5}\cdot2\)
\(=\dfrac{10}{5}\cdot2=4\)
b: \(=\dfrac{x}{x-3}-\dfrac{x\left(x+3\right)}{2x+3}\cdot\dfrac{x^2+6x+9-x^2}{x\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x}{x-3}-\dfrac{3}{x-3}=1\)
1:
Sửa đề: \(\frac{2x}{x+1}-\frac{3}{x-3}=\frac{x^2+3}{\left(x+1\right)\left(x-3\right)}\)
ĐKXĐ: x∉{-1;3}
Ta có: \(\frac{2x}{x+1}-\frac{3}{x-3}=\frac{x^2+3}{\left(x+1\right)\left(x-3\right)}\)
=>\(\frac{2x\left(x-3\right)-3\left(x+1\right)}{\left(x+1\right)\left(x-3\right)}=\frac{x^2+3}{\left(x+1\right)\left(x-3\right)}\)
=>\(2x\left(x-3\right)-3\left(x+1\right)=x^2+3\)
=>\(2x^2-6x-3x-3-x^2-3=0\)
=>\(x^2-9x-6=0\)
=>\(x^2-9x+\frac{81}{4}-\frac{81}{4}-6=0\)
=>\(\left(x-\frac92\right)^2=\frac{81}{4}+6=\frac{81}{4}+\frac{24}{4}=\frac{105}{4}\)
=>\(\left[\begin{array}{l}x-\frac92=\frac{\sqrt{105}}{2}\\ x-\frac92=-\frac{\sqrt{105}}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\sqrt{105}+9}{2}\left(nhận\right)\\ x=\frac{-\sqrt{105}+9}{2}\left(nhận\right)\end{array}\right.\)
2: ĐKXĐ: x<>3; x<>-2
\(\frac{x}{x-3}-\frac{1}{x+2}=\frac{4x+3}{\left(x-3\right)\left(x+2\right)}\)
=>\(\frac{x\left(x+2\right)-\left(x-3\right)}{\left(x-3\right)\left(x+2\right)}=\frac{4x+3}{\left(x-3\right)\left(x+2\right)}\)
=>x(x+2)-(x-3)=4x+3
=>\(x^2+2x-x+3=4x+3\)
=>\(x^2-3x=0\)
=>x(x-3)=0
=>x=0(nhận) hoặc x=3(loại)