(8x-5-3x³-3x²+x⁴):(x-1
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TG
30 tháng 7 2020
a) \(\left(x+1\right)^2-\left(x-1\right)^2-2\left(x+1\right)\left(x-1\right)\)
\(=\left(x^2+2x+1^2\right)-\left(x^2-2x+1^2\right)-2.\left(x^2-1^2\right)\)
\(=x^2+2x+1-x^2+2x-1-2x^2+2\)
\(=4x-2x^2+2\)
b) Check lại đề nha! Thiếu rồi
c) \(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)
\(=\left(9x^2+2.3x.1+1^2\right)-2.\left[3x\left(3x+5\right)+1\left(3x+5\right)\right]+\left(9x^2+2.3x.5+5^2\right)\)
\(=9x^2+6x+1-2\left(9x^2+15x+3x+5\right)+9x^2+30x+25\)
\(=9x^2+6x+1-18x^2-30x-6x-10+9x^2+30x+25\)
\(=16\)
P/s: Ko chắc!
24 tháng 2 2023
a: \(=\dfrac{2x^4+x^3-5x^2-3x-3}{x^2-3}\)
\(=\dfrac{2x^4-6x^2+x^3-3x+x^2-3}{x^2-3}\)
\(=2x^2+x+1\)
b: \(=\dfrac{x^5+x^2+x^3+1}{x^3+1}=x^2+1\)
c: \(=\dfrac{2x^3-x^2-x+6x^2-3x-3+2x+6}{2x^2-x-1}\)
\(=x+3+\dfrac{2x+6}{2x^2-x-1}\)
d: \(=\dfrac{3x^4-8x^3-10x^2+8x-5}{3x^2-2x+1}\)
\(=\dfrac{3x^4-2x^3+x^2-6x^3+4x^2-2x-15x^2+10x-5}{3x^2-2x+1}\)
\(=x^2-2x-5\)
12 tháng 5 2022
*vn:vô nghiệm.
a. \(\left(x^2-2\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2=0\\x^2+x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\pm\sqrt{2}\)
-Vậy \(S=\left\{\pm\sqrt{2}\right\}\).
b. \(16x^2-8x+5=0\)
\(\Leftrightarrow16x^2-8x+1+4=0\)
\(\Leftrightarrow\left(4x-1\right)^2+4=0\) (vô lí)
-Vậy S=∅.
c. \(2x^3-x^2-8x+4=0\)
\(\Leftrightarrow x^2\left(2x-1\right)-4\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\pm2\end{matrix}\right.\)
-Vậy \(S=\left\{\dfrac{1}{2};\pm2\right\}\).
d. \(3x^3+6x^2-75x-150=0\)
\(\Leftrightarrow3x^2\left(x+2\right)-75\left(x+2\right)=0\)
\(\Leftrightarrow3\left(x+2\right)\left(x^2-25\right)=0\)
\(\Leftrightarrow3\left(x+2\right)\left(x+5\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\pm5\end{matrix}\right.\)
-Vậy \(S=\left\{-2;\pm5\right\}\)
NN
1
20 tháng 8 2023
1: =>|x-4|+|x+2|=0
=>x-4=0 và x+2=0
=>\(x\in\varnothing\)
2: =>x^2-x-6=3x+5
=>x^2-4x-11=0
=>x^2-4x+4-15=0
=>(x-2)^2-15=0
=>x=căn 15+2 hoặc x=-căn 15+2
3: =>x^2-x=3x+5
=>x^2-4x-5=0
=>(x-5)(x+1)=0
=>x=-1 hoặc x=5
14 tháng 2 2020
\(\left(3x+1\right)^2-x^2+8x-16=0\)
\(\Leftrightarrow\left(3x+1\right)^2-\left(x-4\right)^2=0\)
\(\Leftrightarrow\left(3x+1+x-4\right)\left(3x+1-x+4\right)=0\)
\(\Leftrightarrow\left(4x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4x-3=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=\frac{-5}{2}\end{cases}}\)
14 tháng 2 2020
\(\left(3x+1\right)^2-x^2+8x-16=0\)
\(\Leftrightarrow\left(3x+1\right)^2-\left(x^2-8x+16\right)=0\)
\(\Leftrightarrow\left(3x+1\right)^2-\left(x-4\right)^2=0\)
\(\Leftrightarrow\left(3x+1+x-4\right)\left(3x+1-x+4\right)=0\)
\(\Leftrightarrow\left(4x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4x-3=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=\frac{-5}{2}\end{cases}}\)
BQ
1
6 tháng 10 2021
\(a,\Rightarrow8x^2-16x-4x+5-8x^2+10x-5x=0\\ \Rightarrow-15x=-5\Rightarrow x=\dfrac{1}{3}\\ b,\Rightarrow9x^2-16-9x^2+12x-4=5\\ \Rightarrow12x=25\\ \Rightarrow x=\dfrac{25}{12}\)
AH
Akai Haruma
Giáo viên
10 tháng 12 2021
1/ ĐKXĐ: $4x^2-4x-11\geq 0$
PT $\Leftrightarrow \sqrt{4x^2-4x-11}=2(4x^2-4x-11)-6$
$\Leftrightarrow a=2a^2-6$ (đặt $\sqrt{4x^2-4x-11}=a, a\geq 0$)
$\Leftrightarrow 2a^2-a-6=0$
$\Leftrightarrow (a-2)(2a+3)=0$
Vì $a\geq 0$ nên $a=2$
$\Leftrightarrow \sqrt{4x^2-4x-11}=2$
$\Leftrightarrow 4x^2-4x-11=4$
$\Leftrightarrow 4x^2-4x-15=0$
$\Leftrightarrow (2x-5)(2x+3)=0$
$\Rightarrow x=\frac{5}{2}$ hoặc $x=\frac{-3}{2}$ (tm)
AH
Akai Haruma
Giáo viên
10 tháng 12 2021
2/ ĐKXĐ: $x\in\mathbb{R}$
PT $\Leftrightarrow \sqrt{3x^2+9x+8}=\frac{1}{3}(3x^2+9x+8)-\frac{14}{3}$
$\Leftrightarrow a=\frac{1}{3}a^2-\frac{14}{3}$ (đặt $\sqrt{3x^2+9x+8}=a, a\geq 0$)
$\Leftrightarrow a^2-3a-14=0$
$\Rightarrow a=\frac{3+\sqrt{65}}{2}$ (do $a\geq 0$)
$\Leftrightarrow 3x^2+9x+8=\frac{37+3\sqrt{65}}{2}$
$\Rightarrow x=\frac{1}{2}(-3\pm \sqrt{23+2\sqrt{65}})$
16 tháng 6 2023
1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35
\(\dfrac{x^4-3x^3-3x^2+8x-5}{x-1}=\dfrac{x^3\left(x-1\right)-3x^2+3x+5x-5}{x-1}=\dfrac{x^3\left(x-1\right)-3x\left(x-1\right)+5\left(x-1\right)}{x-1}=x^3-3x+5\)