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5 tháng 2 2022
Bài 3:
b: \(\Leftrightarrow x^2\left(x+1\right)^2=0\)
hay \(x\in\left\{0;-1\right\}\)
c: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=0\)
=>x-1=0
hay x=1
d: \(\Leftrightarrow6x^2-3x-4x+2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(3x-2\right)=0\)
hay \(x\in\left\{\dfrac{1}{2};\dfrac{2}{3}\right\}\)
HP
21 tháng 9 2021
a. 9x2 - 6x - 3 = 0
<=> 3(3x2 - 2x - 1) = 0
<=> 3(3x2 - 3x + x - 1) = 0
<=> \(3\left[3x\left(x-1\right)+\left(x-1\right)\right]=0\)
<=> 3(3x + 1)(x - 1) = 0
<=> \(\left[{}\begin{matrix}3x+1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=1\end{matrix}\right.\)
b. (2x + 1)2 - 4(x + 2)2 = 9
<=> (2x + 1)2 - \(\left[2\left(x+2\right)\right]^2=9\)
<=> (2x + 1 - 2x - 4)(2x + 1 + 2x + 4) = 9
<=> -3(4x + 5) = 9
<=> 4x + 5 = -3
<=> 5 + 3 = -4x
<=> -4x = 8
<=> -x = 2
<=> x = -2
21 tháng 9 2021
a) \(\Leftrightarrow\left(9x^2-6x+1\right)-4=0\)
\(\Leftrightarrow\left(3x-1\right)^2-4=0\)
\(\Leftrightarrow3\left(x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
b) \(\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\)
\(\Leftrightarrow12x=-24\Leftrightarrow x=-2\)
c) \(\Leftrightarrow3x^2-6x+3-3x^2+15x=21\)
\(\Leftrightarrow9x=18\Leftrightarrow x=2\)
d) \(\Leftrightarrow x^2+6x+9-x^2-4x+32=1\)
\(\Leftrightarrow2x=-40\Leftrightarrow x=-20\)
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 \(S=\left\{-2;\pm5\right\}\)
ND
Nguyễn Đức Trí
VIP
8 tháng 9 2023
d) \(2x^2+5x-7=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\) \(\left(a+b+c=1\right)\)
13 tháng 6 2023
b: =>1/4x+4/5-x-5=1/3x+1-1/2x+1
=>-3/4x+1/6x=2+5-4/5=24/5
=>x=-288/35
c: =>6x^2+3x-30x-15=6x^2+10x-21x-35
=>-27x-15=-11x-35
=>-16x=-20
=>x=5/4
TK
1
KV
16 tháng 10 2023
a) \(6x^2-72x=0\)
\(6x\left(x-12\right)=0\)
\(6x=0\) hoặc \(x-72=0\)
*) \(6x=0\)
\(x=0\)
*) \(x-12=0\)
\(x=12\)
Vậy \(x=0;x=12\)
b) \(-2x^4+16x=0\)
\(-2x\left(x^3-8\right)=0\)
\(-2x=0\) hoặc \(x^3-8=0\)
*) \(-2x=0\)
\(x=0\)
*) \(x^3-8=0\)
\(x^3=8\)
\(x=2\)
Vậy \(x=0;x=2\)
c) \(x\left(x-5\right)-\left(x-3\right)^2=0\)
\(x^2-5x-x^2+6x-9=0\)
\(x-9=0\)
\(x=9\)
d) \(\left(x-2\right)^3-\left(x-2\right)\left(x^2+2x+4\right)=0\)
\(x^3-6x^2+12x-8-x^3+8=0\)
\(-6x^2+12x=0\)
\(-6x\left(x-2\right)=0\)
\(-6x=0\) hoặc \(x-2=0\)
*) \(-6x=0\)
\(x=0\)
*) \(x-2=0\)
\(x=2\)
Vậy \(x=0;x=2\)
19 tháng 4 2023
2:
a: =>x-1=0 hoặc 3x+1=0
=>x=1 hoặc x=-1/3
b: =>x-5=0 hoặc 7-x=0
=>x=5 hoặc x=7
c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)
d: =>x=0 hoặc x^2-1=0
=>\(x\in\left\{0;1;-1\right\}\)
26 tháng 9 2021
e: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow2x=-7\)
hay \(x=-\dfrac{7}{2}\)
f: Ta có: \(x^3-6x^2+12x-19=0\)
\(\Leftrightarrow x^3-6x^2+12x-8-11=0\)
\(\Leftrightarrow\left(x-2\right)^3=11\)
hay \(x=\sqrt[3]{11}+2\)
a, \(x^3+x^2+4\)
\(=x^3+2x^2-x^2-2x+2x+4=0\)
\(=\left(x^3+2x^2\right)-\left(x^2+2x\right)+\left(2x+4\right)\)
\(=x^2\left(x+2\right)-x\left(x+2\right)+2\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-x+2\right)\)