\(A=\left(\dfrac{1}{x^2-4x}+\dfrac{2}{16-x^2}+\dfrac{4}{4x+16}\right):\dfrac{1}{4x}\left(x\ne4;x\ne-4;x\ne0\right).\)

\(A=\left(\dfrac{1}{x\left(x-4\right)}+\dfrac{-2}{\left(x+4\right)\left(x-4\right)}+\dfrac{1}{x+4}\right).4x\).

\(A=\dfrac{x+4-2x+x^2-4x}{x\left(x-4\right)\left(x+4\right)}.4x.\)

\(A=\dfrac{x^2-5x+4}{\left(x-4\right)\left(x+4\right)}.4.\)

\(A=\dfrac{\left(x-4\right)\left(x-1\right)}{\left(x-4\right)\left(x+4\right)}.4.\)

\(A=\dfrac{4\left(x-1\right)}{x+4}.\)

chịch ko em

Không ai trả lời luôn

a/ \(A=\sqrt{\left(a-4\right)^2}-3a=\left|a-4\right|-3a\)

+) với a<4: A = 4-a-3a=4-4a

+)với a≥4: A = a-4-3a=-2a - 4

Với a = -3 <4 => A = 4 - 4 . (-3) = 16

b/ \(B=\sqrt{\left(1-2x\right)^2}-2x=\left|1-2x\right|-2x\)

+) nếu x \(\le\frac{1}{2}\) :

\(B=1-2x-2x=-4x+1\)

+) nếu \(x>\frac{1}{2}:B=2x-1-2x=-1\)

với \(x=-\frac{3}{2}< \frac{1}{2}\Rightarrow B=-4\cdot\left(-\frac{3}{2}\right)+1=7\)

c/đk: \(x\ne\pm4\)

\(C=\frac{\sqrt{\left(2x-1\right)^2}}{\left(x-4\right)\left(x+4\right)}\cdot\left(x-4\right)^2=\frac{\left|2x-1\right|\cdot\left(x-4\right)}{x+4}\)

+) nếu \(x\ge\frac{1}{2}:B=\frac{\left(2x-1\right)\left(x-4\right)}{x+4}\)

+) nếu \(x< \frac{1}{2}:B=\frac{-\left(2x-1\right)\left(x-4\right)}{x+4}\)

Với \(x=7\left(>\frac{1}{2}\right):B=\frac{\left(2\cdot7-1\right)\cdot\left(7-4\right)}{7+4}=\frac{39}{11}\)

a) 3-4x = 4.(x-3) hoặc 3-4x = -4.(x-3)

3-4x=4x-12 hoặc 3-4x = -4x +12

8x=15 hoặc -4x+4x=12-3

x=15/8

b) x^2+x+1=4x-1 hoặc  x^2+x+1= -(4x-1)

x^2-3x+2=0 hoặc x^2+5x=0

TH1: x^2-3x+2=0 

x^2-x-2x+2=0

(x^2-x)-(2x-2)=0

x(x-1)-2(x-1)=0

(x-1).(x-2)=0

x=1 hoặc x=2

TH2: x^2+5x=0

x.(x+5)=0

x=0 hoặc x=-5

Các bạn tự đáp số nhé

Lời giải:

a)

$(3-4x)^2=16(x-3)^2=4^2(x-3)^2=(4x-12)^2$

$\Leftrightarrow [(3-4x)-(4x-12)][(3-4x)+(4x-12)]=0$

$\Leftrightarrow (15-8x)(-9)=0$

$\Rightarrow 15-8x=0\Rightarrow x=\frac{15}{8}$

b)

$(x^2+x+1)^2=(4x-1)^2$

$\Leftrightarrow (x^2+x+1)^2-(4x-1)^2=0$

$\Leftrightarrow (x^2+x+1-4x+1)(x^2+x+1+4x-1)=0$

$\Leftrightarrow (x^2-3x+2)(x^2+5x)=0$

$\Leftrightarrow (x-1)(x-2)x(x+5)=0$

\(\Rightarrow \left[\begin{matrix} x-1=0\\ x-2=0\\ x=0\\ x+5=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1\\ x=2\\ x=0\\ x=-5\end{matrix}\right.\)

a) 9 - 24x + 16x² = 16(x² - 6x + 9)

=> 16x² - 24x + 9 = 16x² - 96x + 144

=> -24x + 96x = 144 - 9

=> 72x = 135

=> x = \(\frac{15}{8}\)

b) (x² + x + 1)² = (4x - 1)²

=> x⁴ + x² + 1 + 2x³ + 2x + 2x² = 16x² - 8x + 1

=> x⁴ + 2x³ + 3x² + 2x + 1 = 16x² - 8x + 1

=> x⁴ + 2x³ - 13x² + 10x = 0

=> x⁴ - x³ + 3x³ - 3x² - 10x² + 10x = 0

=> (x³ + 3x² - 10x)(x - 10) = 0

=> x(x² + 3x - 10)(x - 10) = 0

=> x(x - 2)(x+5)(x-10) = 0

=> \(\left[{}\begin{matrix}x=0\\x=2\\x=-5\\x=10\end{matrix}\right.\)

click cho mình nha

a) \(\sqrt[]{x^2-2x+4}=2x-2\)

\(\Leftrightarrow\sqrt[]{x^2-2x+4}=2\left(x-1\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x-1\right)\ge0\\x^2-2x+4=4\left(x-1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x^2-2x+4=4x^2-8x+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\3x^2-6x=0\end{matrix}\right.\) \(\left(1\right)\)

Giải pt \(3x^2-6x=0\)

\(\Leftrightarrow3x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=2\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x=2\)

c) \(\sqrt{x^2-3x+2}=\sqrt[]{x-1}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x^2-3x+2=x-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x^2-4x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x=1\cup x=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)