Tìm x biết : x^4= 4x - 1
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\(4x^2+4x-3=0\)
\(\left[\left(2x\right)^2+2.2x.1+1\right]-4=0\)
\(\left(2x+1\right)^2-2^2=0\)
\(\left(2x+1-2\right).\left(2x+1+2\right)=0\)
\(\left(2x-1\right).\left(2x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-1=0\\2x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}}\)
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~
Lời giải:
Áp dụng BĐT dạng $|a|+|b|\geq |a+b|$ ta có:
$|3x-1|+|x+4|\geq |3x-1+x+4|=|4x+3|$
Dấu "=" xảy ra khi $(3x-1)(x+4)\geq 0$
$\Leftrightarrow x\geq \frac{1}{3}$ hoặc $x\leq -4$
Áp dụng bất đẳng thức |a| + |b| >= |a+b| ta có
|3x-1| + |x+4| >= |3x-1+x+4|=|4x+3|
Dấu = xảy ra khi và chỉ khi
3x-1 và x+4 cùng dấu
\(\Leftrightarrow\left[{}\begin{matrix}3x-1,x+4\ge0\\3x-1,x+4\le0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{1}{3}\\x\le-4\end{matrix}\right.\)
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
=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
\((4x-1)^2-4(2x-3)^2-x-4=0\\\Leftrightarrow (4x)^2-2\cdot4x\cdot1+1^2-4[(2x)^2-2\cdot2x\cdot3+3^2]-x-4=0\\\Leftrightarrow 16x^2-8x+1-4(4x^2-12x+9)-x-4=0\\\Leftrightarrow 16x^2-8x+1-16x^2+48x-36-x-4=0\\\Leftrightarrow (16x^2-16x^2)+(-8x+48x-x)+(1-36-4)=0\\\Leftrightarrow 39x-39=0\\\Leftrightarrow 39x=39\\\Leftrightarrow x=1\\Vậy:x=1\)
(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
3) \(x\left(x-4\right)+\left(x-4\right)^2=0\Leftrightarrow\left(x-4\right)\left(x+x-4\right)=0\Leftrightarrow2\left(x-4\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)