Tìm x biết : ( 4x - 1 )2 = 4
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TN
2
T
29 tháng 10 2023
\((4x-1)^2-4(2x-3)^2-x-4=0\\\Leftrightarrow (4x)^2-2\cdot4x\cdot1+1^2-4[(2x)^2-2\cdot2x\cdot3+3^2]-x-4=0\\\Leftrightarrow 16x^2-8x+1-4(4x^2-12x+9)-x-4=0\\\Leftrightarrow 16x^2-8x+1-16x^2+48x-36-x-4=0\\\Leftrightarrow (16x^2-16x^2)+(-8x+48x-x)+(1-36-4)=0\\\Leftrightarrow 39x-39=0\\\Leftrightarrow 39x=39\\\Leftrightarrow x=1\\Vậy:x=1\)
KV
29 tháng 10 2023
(4x - 1)² -4(2x - 3)² - x - 4 = 0
16x² - 8x + 1 - 4(4x² - 12x + 9) - x - 4 = 0
16x² - 8x + 1 - 16x² + 48x - 36 - x - 4 = 0
39x - 39 = 0
39x = 39
x = 39 : 39
x = 1
N
1
4 tháng 8 2018
\(4x^2+4x-3=0\)
\(\left[\left(2x\right)^2+2.2x.1+1\right]-4=0\)
\(\left(2x+1\right)^2-2^2=0\)
\(\left(2x+1-2\right).\left(2x+1+2\right)=0\)
\(\left(2x-1\right).\left(2x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-1=0\\2x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}}\)
Vậy \(\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}\)
\(x^4-3x^3-x+3=0\)
\(x^3.\left(x-3\right)-\left(x-3\right)=0\)
\(\left(x-3\right).\left(x^3-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x^3-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)
Vậy \(\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
\(x^2.\left(x-1\right)-4x^2+8x-4=0\)
\(x^2.\left(x-1\right)-\left[\left(2x\right)^2-2.2x.2+2^2\right]=0\)
\(x^2.\left(x-1\right)-\left(2x-2\right)^2=0\)
\(x^2.\left(x-1\right)-4.\left(x-1\right)^2=0\)
\(\left(x-1\right).\left[x^2-4.\left(x-1\right)\right]=0\)
\(\left(x-1\right).\left[x^2-2.x.2+2^2\right]=0\)
\(\left(x-1\right).\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)
Vậy \(\begin{cases}x=1\\x=2\end{cases}\)
Tham khảo nhé~
N
0
29 tháng 11 2023
a: \(x^3-4x^2-x+4=0\)
=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)
=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(x^2-1\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)
b: Sửa đề: \(x^3+3x^2+3x+1=0\)
=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)
=>\(\left(x+1\right)^3=0\)
=>x+1=0
=>x=-1
c: \(x^3+3x^2-4x-12=0\)
=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)
=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)
=>\(\left(x+3\right)\left(x^2-4\right)=0\)
=>\(\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)
d: \(\left(x-2\right)^2-4x+8=0\)
=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)
=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)
=>\(\left(x-2\right)\left(x-2-4\right)=0\)
=>(x-2)(x-6)=0
=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
12 tháng 3 2022
a, \(-4x+5+2x-1=3\Leftrightarrow-2x=-1\Leftrightarrow x=\dfrac{1}{2}\)
b, \(-2x+2=2\Leftrightarrow x=0\)
c, \(-2x-6=-8\Leftrightarrow x=1\)
24 tháng 11 2021
\(a,\Leftrightarrow x-1=4\Leftrightarrow x=5\\ b,\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\3x+1=4x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\left(tm\right)\end{matrix}\right.\Leftrightarrow x=4\\ c,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=3\\ \Leftrightarrow x+5=9\\ \Leftrightarrow x=4\left(tm\right)\)
\(d,\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{5}+1\right)^2}\\ \Leftrightarrow\left|x-2\right|=\sqrt{5}+1\\ \Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{5}+1\\2-x=\sqrt{5}+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}+3\\x=1-\sqrt{5}\end{matrix}\right.\)
PT
1
TT
Tìm x biết
1. 2(5x-8)-3(4x-5)=4(3x-4)+11
2. (2x+1)2-(4x-1).(x-3)-15=0
3. (3x-1).(2x-7)-(1-3x).(6x-5)=0
2
14 tháng 10 2021
1) \(\Rightarrow10x-16-12x+15=12x-16+11\)
\(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)
2) \(\Rightarrow4x^2+4x+1-4x^2+13x-3-15=0\)
\(\Rightarrow17x=17\Rightarrow x=1\)
3) \(\Rightarrow\left(3x-1\right)\left(2x-7+6x-5\right)=0\)
\(\Rightarrow\left(2x-3\right)\left(3x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
14 tháng 10 2021
2: Ta có: \(\left(2x+1\right)^2-\left(4x-1\right)\left(x-3\right)-15=0\)
\(\Leftrightarrow4x^2+4x+1-4x^2+12x+x-3-15=0\)
\(\Leftrightarrow17x=17\)
hay x=1
\(\left(4x-1\right)^2=2^2\)
\(\Rightarrow\orbr{\begin{cases}4x-1=2\\4x-1=-2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=-\frac{1}{4}\end{cases}}\)
Bài làm
( 4x - 1 )2 = 4
=> ( 4x - 1 )2 = 22
=> 4x - 1 = 2
=> 4x = 3
x = 3/4
Vậy x = 3/4
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