Tìm x
(2x-1)3-2x(2x+1)2=5x-20x2
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DT
Đỗ Thanh Hải
CTVVIP
16 tháng 4 2021
1) \(\left\{{}\begin{matrix}2x+y=10\\5x-3y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}10x+5y=50\\10x-6y=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}11y=44\\2x+y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=4\\x=3\end{matrix}\right.\)
Vậy hpt có nghiệm (x;y) = (3;4)
2)
a) 3x2 - 2x - 1 = 0
\(\Leftrightarrow3x^2-3x+x-1=0\)
\(\Leftrightarrow3x\left(x-1\right)+\left(x-1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=1\end{matrix}\right.\)
Vậy pt có nghiệm x = 1 hoặc x = 3
b) Đặt x2 = t (t \(\ge\) 0)
Pt trở thành: t2 - 20t + 4 = 0
\(\Delta\) = (-20)2 - 4.1.4 = 400 - 16 = 384
=> pt có 2 nghiệm phân biệt t1 = \(\dfrac{20+8\sqrt{6}}{2}=10+4\sqrt{6}\)
t2 = \(\dfrac{20-8\sqrt{6}}{2}=10-4\sqrt{6}\)
=> x1 = \(\sqrt{10+4\sqrt{6}}=\sqrt{\left(2+\sqrt{6}\right)^2}=2+\sqrt{6}\)
x2 = \(2-\sqrt{6}\)
15 tháng 2 2020
a. (3x - 1).(2x + 7) - (x + 1).(6x - 5) = 16
<=> 6x^2 + 19x - 7 - (6x^2 + x - 5) = 16
<=> 18x - 2 = 16
<=> 18x = 18
<=> x = 1
b. (10x + 9).x - (5x - 1).(2x + 3) = 8
<=> 10x^2 + 9x - (10x^2 + 13x - 3) = 8
<=> -4x + 3 = 8
<=> -4x = 5
<=> x = -5/4
c. (3x - 5).(7 - 5x) + (5x + 2).(3x - 2) - 2 = 0
<=> -15x^2 + 46x - 35 + 15x^2 - 4x - 4 - 2 = 0
<=> 42x - 41 = 0
<=> x = 41/42
8 tháng 1 2022
a: \(P=\left(\dfrac{-\left(x+3\right)}{x-3}+\dfrac{x-3}{x+3}+\dfrac{4x^2}{x^2-9}\right):\dfrac{2x+1-x-3}{x+3}\)
\(=\dfrac{-x^2-6x-9+x^2-6x+9+4x^2}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x-2}\)
\(=\dfrac{4x^2-12x}{x-3}\cdot\dfrac{1}{x-2}=\dfrac{4x}{x-2}\)
b: \(2x^2-5x+2=0\)
=>(x-2)(2x-1)=0
=>x=1/2
Thay x=1/2 vào P, ta được:
\(P=\left(4\cdot\dfrac{1}{2}\right):\left(\dfrac{1}{2}-2\right)=2:\dfrac{-3}{2}=\dfrac{-4}{3}\)
21 tháng 3 2023
a: =>(x^2-2x+1-1)^2+2(x-1)^2=1
=>(x-1)^4-2(x-1)^2+1+2(x-1)^2=1
=>(x-1)^4=0
=>x-1=0
=>x=1
b: =>(x^2+2)^2+3x(x^2+2)+2x^2-20x^2=0
=>(x^2+2)^2+3x(x^2+2)-18x^2=0
=>(x^2+2+6x)(x^2-3x+2)=0
=>\(x\in\left\{-3\pm\sqrt{7};1;2\right\}\)
A
0
MN
2
12 tháng 8 2016
a) x2 - 5x - 6 = 0
=> x2 - 2x - 3x - 6 = 0
=> (x2 - 2x) + (-3x - 6) = 0
=> x(x - 2) - 3 (x - 2) = 0
=> (x - 2) (x - 3) = 0
=> x - 2 = 0 => x = 2
x - 3 = 0 => x = 3
còn lại tương tự nhé!! 46566578768698945635655675656788787868789789879789098089364556546
KB
1
26 tháng 7 2020
a) Ta có: \(3x\left(x+1\right)-2x\left(x+20\right)=-1-x\)
\(\Leftrightarrow3x^2+3x-2x^2-40x+1+x=0\)
\(\Leftrightarrow x^2-36x+1=0\)
\(\Leftrightarrow x^2-36x+324-323=0\)
\(\Leftrightarrow\left(x-18\right)^2=323\)
\(\Leftrightarrow\left[{}\begin{matrix}x-18=\sqrt{323}\\x-18=-\sqrt{323}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=18+\sqrt{323}\\x=18-\sqrt{323}\end{matrix}\right.\)
Vậy: \(x\in\left\{18+\sqrt{323};18-\sqrt{323}\right\}\)
b) Ta có: \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)
\(\Leftrightarrow6x^2+21x-2x-7-\left(6x^2-5x+6x-5\right)-16=0\)
\(\Leftrightarrow6x^2+19x-7-\left(6x^2+x-5\right)-16=0\)
\(\Leftrightarrow6x^2+19x-7-6x^2-x+5-16=0\)
\(\Leftrightarrow18x-18=0\)
\(\Leftrightarrow18x=18\)
hay x=1
Vậy: x=1
c) Ta có: \(\left(10x+9\right)\cdot x-\left(5x-1\right)\left(2x+3\right)=8\)
\(\Leftrightarrow10x^2+9x-\left(10x^2+15x-2x-3\right)-8=0\)
\(\Leftrightarrow10x^2+9x-10x^2-13x+3-8=0\)
\(\Leftrightarrow-4x-5=0\)
\(\Leftrightarrow-4x=5\)
hay \(x=\frac{-5}{4}\)
Vậy: \(x=\frac{-5}{4}\)
\(\left(2x-1\right)^3-2x\left(2x+1\right)^2=5x-20x^2\)
\(\Rightarrow8x^3-12x^2+6x-1-2x\left(4x^2+4x+1\right)-5x+20x^2=0\)
\(\Rightarrow8x^3-12x^2+6x-1-8x^3-8x^2-2x-5x+20x^2=0\)
\(\Rightarrow-x-1=0\)
\(\Rightarrow x=-1\)