4(3/4x-1)+(12x^2-3x):(-3x)-(2x+1)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
16 tháng 1
a: \(x^3+8x=5x^2+4\)
=>\(x^3-5x^2+8x-4=0\)
=>\(x^3-x^2-4x^2+4x+4x-4=0\)
=>\(x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(x^2-4x+4\right)=0\)
=>\(\left(x-1\right)\left(x-2\right)^2=0\)
=>\(\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
2: \(x^3+3x^2=x+6\)
=>\(x^3+3x^2-x-6=0\)
=>\(x^3+2x^2+x^2+2x-3x-6=0\)
=>\(x^2\cdot\left(x+2\right)+x\left(x+2\right)-3\left(x+2\right)=0\)
=>\(\left(x+2\right)\left(x^2+x-3\right)=0\)
=>\(\left[{}\begin{matrix}x+2=0\\x^2+x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-1+\sqrt{13}}{2}\\x=\dfrac{-1-\sqrt{13}}{2}\end{matrix}\right.\)
3: ĐKXĐ: x>=0
\(2x+3\sqrt{x}=1\)
=>\(2x+3\sqrt{x}-1=0\)
=>\(x+\dfrac{3}{2}\sqrt{x}-\dfrac{1}{2}=0\)
=>\(\left(\sqrt{x}\right)^2+2\cdot\sqrt{x}\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{17}{16}=0\)
=>\(\left(\sqrt{x}+\dfrac{3}{4}\right)^2=\dfrac{17}{16}\)
=>\(\left[{}\begin{matrix}\sqrt{x}+\dfrac{3}{4}=-\dfrac{\sqrt{17}}{4}\\\sqrt{x}+\dfrac{3}{4}=\dfrac{\sqrt{17}}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{\sqrt{17}-3}{4}\left(nhận\right)\\\sqrt{x}=\dfrac{-\sqrt{17}-3}{4}\left(loại\right)\end{matrix}\right.\)
=>\(x=\dfrac{13-3\sqrt{17}}{8}\left(nhận\right)\)
4: \(x^4+4x^2+1=3x^3+3x\)
=>\(x^4-3x^3+4x^2-3x+1=0\)
=>\(x^4-x^3-2x^3+2x^2+2x^2-2x-x+1=0\)
=>\(x^3\left(x-1\right)-2x^2\left(x-1\right)+2x\left(x-1\right)-\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(x^3-2x^2+2x-1\right)=0\)
=>\(\left(x-1\right)\left(x^3-x^2-x^2+x+x-1\right)=0\)
=>\(\left(x-1\right)^2\cdot\left(x^2-x+1\right)=0\)
=>(x-1)^2=0
=>x-1=0
=>x=1
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 1
a.
\(x^3+8x=5x^2+4\)
\(\Leftrightarrow x^3-5x^2+8x-4=0\)
\(\Leftrightarrow\left(x^3-4x^2+4x\right)-\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)^2-\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
b.
\(x^3+3x^2-x-6=0\)
\(\Leftrightarrow\left(x^3+x^2-3x\right)+\left(2x^2+2x-6\right)=0\)
\(\Leftrightarrow x\left(x^2+x-3\right)+2\left(x^2+x-3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-1\pm\sqrt{13}}{2}\end{matrix}\right.\)
2 tháng 1 2023
2: 12-10x=25-30x
=>20x=13
=>x=13/20
3: \(3\left(2x+3\right)-2\left(4x-5\right)=10x+21\)
=>6x+9-8x+10=10x+21
=>10x+21=-2x+19
=>12x=-2
=>x=-1/6
4: \(\Leftrightarrow25x-15-6x+12=11-5x\)
=>19x-3=11-5x
=>24x=14
=>x=7/12
5: \(\Leftrightarrow8-12x-5+10x=4-6x\)
=>4-6x=-2x+3
=>-4x=-1
=>x=1/4
6: \(\Leftrightarrow32x-24-6+9x=13-40x\)
=>41x-30=13-40x
=>81x=43
=>x=43/81
7: \(\Leftrightarrow10x-5+20x=5x-11\)
=>30x-5=5x-11
=>25x=-6
=>x=-6/25
TP
1
30 tháng 10 2023
a: ĐKXD: x<>0
\(\dfrac{14x^3+12x^2-14x}{2x}=\left(x+2\right)\left(3x-4\right)\)
=>\(\dfrac{2x\left(7x^2+6x-7\right)}{2x}=\left(x+2\right)\left(3x-4\right)\)
=>\(7x^2+6x-7=3x^2-4x+6x-8\)
=>\(7x^2+6x-7=3x^2+2x-8\)
=>\(4x^2+4x+1=0\)
=>\(\left(2x+1\right)^2=0\)
=>2x+1=0
=>x=-1/2(nhận)
b: \(\left(4x-5\right)\left(6x+1\right)-\left(8x+3\right)\left(3x-4\right)=15\)
=>\(24x^2+4x-30x-5-\left(24x^2-32x+9x-12\right)=15\)
=>\(24x^2-26x-5-24x^2+23x+12=15\)
=>-3x+7=15
=>-3x=8
=>\(x=-\dfrac{8}{3}\)
TN
6 tháng 7 2016
a/ 3x(12x-4)-9x(4x-3)
=36x2-12x-36x2+27x
=(36x2-36x2)-12x+27x
=15x
b/ x(5-2x)+2x(x-1)
=5x-2x2+2x2-2x
=(5x-2x)-(-2x2+2x2)
=3x
c/ 5x(12x+7)-3x(20x-5)
=60x2+35x-60x2+15x
=(60x2-60x2)+(35x+15x)\
=50x
d/ 3x(2x-7)+2x(5-3x)
=6x2-21x+10x-6x2
=(6x2-6x2)+(10x-21x)
=-11x
e/ đề sai hay sao ý ra số to lắm @@
20 tháng 6 2021
$ a/ 12x(x – 5) – 3x(4x - 10) = 120$
`<=>12x^2-60x-12x^2+30x=120`
`<=>-30x=120`
`<=>x=-4`
Vậy `x=-4`
$b/ 9x(x + 4) – 5x(3x + 2) = 112 - 2x(3x + 1)$
`<=>9x^2+36x-15x^2-10x=112-6x^2-2x`
`<=>-6x^2+26x=112-6x^2-2x`
`<=>28x=112`
`<=>x=4`
Vậy `x=4`
$c/ 3x(1 – x) - 5x(3x + 7) = 154 + 9x(5 – 2x)$
`<=>3x-3x^2-15x^2-35x=154+45x-18x^2`
`<=>-32x-18x^2=154+45x-18x^2`
`<=>77x=-154`
`<=>x=-2`
Vậy `x=-2`
3 tháng 12 2021
d: \(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)}=\dfrac{\left(x+1\right)\left(x+2\right)}{A}\)
hay A=x-2