a,5(2x+3)+/2(2x+3)/+/2x+3/=16
b,/x^2+/6x-2/=x^2+4
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N
0
19 tháng 12 2021
\(a,=\dfrac{x^2+4x+3-2x^2+2x+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ b,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{x^2}\\ c,=\dfrac{75y^2+18xy+10x^2}{30x^2y^3}\\ d,=\dfrac{5x+8-x}{4x\left(x+2\right)}=\dfrac{4\left(x+2\right)}{4x\left(x+2\right)}=\dfrac{1}{x}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)
5 tháng 3 2022
a: (2x+1)(3-x)(4-2x)=0
=>(2x+1)(x-3)(x-2)=0
hay \(x\in\left\{-\dfrac{1}{2};3;2\right\}\)
b: 2x(x-3)+5(x-3)=0
=>(x-3)(2x+5)=0
=>x=3 hoặc x=-5/2
c: =>(x-2)(x+2)+(x-2)(2x-3)=0
=>(x-2)(x+2+2x-3)=0
=>(x-2)(3x-1)=0
=>x=2 hoặc x=1/3
d: =>(x-2)(x-3)=0
=>x=2 hoặc x=3
e: =>(2x+5+x+2)(2x+5-x-2)=0
=>(3x+7)(x+3)=0
=>x=-7/3 hoặc x=-3
f: \(\Leftrightarrow2x^3+5x^2-3x=0\)
\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
hay \(x\in\left\{0;-3;\dfrac{1}{2}\right\}\)
TT
12 tháng 12 2020
Bạn chú ý đăng lẻ câu hỏi! 1/
a/ \(=x^3-2x^5\)
b/\(=5x^2+5-x^3-x\)
c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)
d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)
e/ \(=x^4-x^2+2x^3-2x\)
f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)
23 tháng 8 2021
a, \(x^4-8x^2+16=\left(x^2-4\right)^2\)
b, \(\left(4x+5\right)^2-\left(5x+4\right)^2=\left(4x+5-5x-4\right)\left(4x+5+5x+4\right)=\left(1-x\right)\left(9x+9\right)=9\left(1-x\right)\left(1+x\right)=9\left(1-x^2\right)\)
c, \(\left(2x-3\right)^2-2\left(2x-3\right)\left(x+2\right)+\left(-x-2\right)^2=\left(2x-3-x-2\right)^2=\left(x-5\right)^2\)
23 tháng 8 2021
a) \(x^4-8x^2+16=\left(x^2-4\right)^2\)
b) \(\left(4x+5\right)^2-\left(5x+4\right)^2=\left(4x+5-5x-4\right)\left(4x+5+5x+4\right)=9\left(1-x\right)\left(x+1\right)\)c) \(\left(2x-3\right)^2-2.\left(2x-3\right)\left(x+2\right)+\left(-x-2\right)^2=\left(2x-3-x-2\right)^2=\left(x-5\right)^2\)
14 tháng 8 2021
Bài 1
A= (x-2)(2x-1)-2x(x+3)=2x2-x-4x+2-2x2-6x=-11x+2
HP
14 tháng 8 2021
Bài 1:
a) \(A=\left(x-2\right)\left(2x-1\right)-2x\left(x+3\right)\)
\(A=2x^2-x-4x+2-2x^2-6x\)
\(A=-11x+2\)
b) \(B=\left(3x-2\right)\left(2x+1\right)-\left(6x-1\right)\left(x+2\right)\)
\(B=6x^2+3x-4x-2-6x^2-12x+x+2\)
\(B=-12x\)
c) \(C=6x\left(2x+3\right)-\left(4x-1\right)\left(3x-2\right)\)
\(C=12x^2+18x-12x^2+8x+3x-2\)
\(C=29x-2\)
d) \(D=\left(2x+3\right)\left(5x-2\right)+\left(x+4\right)\left(2x-1\right)-6x\left(2x-3\right)\)
\(D=10x^2-4x+15x-6+2x^2-x+8x-4-12x^2+18x\)
\(D=36x-10\)
NN
6 tháng 12 2017
a) \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)
\(\Leftrightarrow\left(6x^2+21x-2x-7\right)-\left(6x^2-5x+6x-5\right)-16=0\)
\(\Leftrightarrow6x^2+21x-2x-7-6x^2+5x-6x+5-16=0\)
\(\Leftrightarrow18x-18=0\)
\(\Leftrightarrow18x=18\)
\(\Leftrightarrow x=18:18\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
b) \(\left(2x+3\right)^2-2\left(2x+3\right)\left(2x-5\right)+\left(2x-5\right)^2=x^2+6x+64\)
\(\Leftrightarrow\left[\left(2x+3\right)-\left(2x-5\right)\right]^2-\left(x^2+6x+64\right)=0\)
\(\Leftrightarrow\left(2x+3-2x+5\right)^2-x^2-6x-64=0\)
\(\Leftrightarrow8^2-x^2-6x-64=0\)
\(\Leftrightarrow64-x^2-6x-64=0\)
\(\Leftrightarrow-x^2-6x=0\)
\(\Leftrightarrow x\left(-x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=-6\)
LH
6 tháng 12 2017
a) \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)
\(\Leftrightarrow\left(6x^2+21x-2x-7\right)-\left(6x^2-5x+6x-5\right)-16=0\)
\(\Leftrightarrow6x^2+21x-2x-7-6x^2+5x-6x+5-16=0\)
\(\Leftrightarrow18x-18=0\)
\(\Leftrightarrow18x=18\)
\(\Leftrightarrow x=18:18\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
b, \(\left(2x+3\right)^2-2\left(2x+3\right)\left(2x-5\right)+\left(2x- 5\right)^2=x^2+6x+64\)
\(\Leftrightarrow\left[\left(2x+3\right)-\left(2x-5\right)\right]^2- \left(x^2+6x+64\right)=0\)
\(\Leftrightarrow\left(2x+3-2x+5\right)^2-x^2-6x-64=0\)
\(\Leftrightarrow8^2-x^2-6x-64=0\)
\(\Leftrightarrow64-x^2-6x-64=0\)
\(\Leftrightarrow-x^2-6x=0\)
\(\Leftrightarrow x\left(-x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=6\)
a)
5(2x + 3) + \(\left|2\left(2x+3\right)\right|\) + \(\left|2x+3\right|\) = 16
=> 5(2x + 3) + 2 . 2x + 3 + 2x + 3 =16
=> (2x + 3).(5+2+1) = 16
=> (2x +3) . 8 = 16
=> 2x + 3 = 16 : 8 = 2
=> 2x = 2 - 3 = -1
=> x = \(\frac{-1}{2}\)