\(A.x^2-16x=0\)

\(x^2-\left(4x\right)^2=0\)

\(\left(x-4x\right)\left(x+4x\right)=0\)

\(\left(-3x\right)\left(5x\right)=0\)

\(\Rightarrow\) \(-3x=0\) hoặc \(5x=0\)

\(x=\dfrac{0}{-3}\) hoặc \(x=\dfrac{0}{5}\)

Vậy \(x=0\) hoặc \(x=0.\)

B. 4x² - 4x + 1 = 0

(2x)² - (2x)² + 1² = 0

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

1 (4x + 1) =0

=> 1 (4x + 1) =0

4x + 1 = 0

4x = 0-1

Vậy x = \(\dfrac{-1}{4}.\)

C. (3x-1)² - (2x+3)² = 0

(3x -1 -2x +3) (3x -1 +2x +3) = 0

(x + 2)(5x + 2) = 0

=> x + 2 =0 hoặc 5x + 2 =0

x = 0 - 2 hoặc 5x = 0 - 2

Vậy x = -2 hoặc x = \(\dfrac{-2}{5}.\)

Còn về câu d thì mình hơi phân vân, tại mình dốt toán lắm

a/ \(x^2-16x=0\)

\(\Leftrightarrow x\left(x-16\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-16=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=16\end{matrix}\right.\)

Vậy...

b/ \(4x^2-4x+1=0\)

\(\Leftrightarrow\left(2x-1\right)^2=0\)

\(\Leftrightarrow2x-1=0\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy..

c/ \(\left(3x-1\right)^2-\left(2x+3\right)^2=0\)

\(\Leftrightarrow\left(3x-1-2x-3\right)\left(3x-1+2x+3\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(5x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{5}\end{matrix}\right.\)

Vậy...

d/ \(2013x^2-2014x+1=0\)

\(\Leftrightarrow2013x^2-x-2013x+1=0\)

\(\Leftrightarrow x\left(2013x-1\right)-\left(2013x-1\right)=0\)

\(\Leftrightarrow\left(2013x-1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2013x-1=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2013}\\x=1\end{matrix}\right.\)

Vậy..

2x+4x+6x+...+2014x=2015.2016

=>(2+4+6+...+2014).x=2015.2016

tổng trong ngoặc có:(2014-2):2+1=1007(số hạng)

=>tổng= (2014+2).1007:2=1015056

=>1015056.x=4062240

=>x=4030/1007

\(\sqrt{x-2}=3\left(x\ge2\right)\\ \Leftrightarrow x-2=9\Leftrightarrow x=11\left(tm\right)\\ \sqrt{4x^2}+4x+1=3\Leftrightarrow\left|2x\right|=2-4x\\ \Leftrightarrow\left[{}\begin{matrix}2x=2-4x\left(x\ge0\right)\\2x=4x-2\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{3}\)

1. \(\sqrt{x^2-4x+3}=x-2\)

<=> x² - 4x + 3 = (x - 2)²

<=> x² - 4x + 3 = x² - 4x + 4

<=> x² - x² - 4x + 4x = 1

<=> 0 = 1 (Vô lí)

vậy PT có nghiệm là S = \(\varnothing\)

2. \(\sqrt{4x^2-4x+1}=x-1\)

<=> \(\sqrt{\left(2x-1\right)^2}=x-1\)

<=> 2x - 1 = x - 1

<=> 2x - x = -1 + 1

<=> x = 0

1: ta có: \(\sqrt{x^2-4x+3}=x-2\)

\(\Leftrightarrow x^2-4x+3=x^2-4x+4\)(vô lý)

2: Ta có: \(\sqrt{4x^2-4x+1}=x-1\)

\(\Leftrightarrow\left(2x-1-x+1\right)\left(2x-1+x-1\right)=0\)

\(\Leftrightarrow x\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=\dfrac{2}{3}\left(loại\right)\end{matrix}\right.\)

ĐKXĐ: \(3-2x\ge0\Leftrightarrow x\le\dfrac{3}{2}\)

b) ĐKXĐ: \(-1\le x\le3\)

c) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x\ne1\\x\ne3\end{matrix}\right.\).

d) ĐKXĐ: \(x< \dfrac{3}{5}\).

Lời giải:

a. ĐKXĐ: $x\geq 5$

PT $\Leftrightarrow \sqrt{4}.\sqrt{x-5}+\sqrt{x-5}=4+3.\sqrt{\frac{1}{9}}.\sqrt{x-5}$

$\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}=4+\sqrt{x-5}$

$\Leftrightarrow 2\sqrt{x-5}=4$

$\Leftrightarrow \sqrt{x-5}=2$

$\Leftrightarrow x-5=4$

$\Leftrightarrow x=9$ (tm)

b. Sửa đoạn 4x-45 thành 4x-20.

ĐKXĐ: $x\geq 5$

PT $\Leftrightarrow \sqrt{4}.\sqrt{x-5}+\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}\sqrt{4}.\sqrt{x-5}=4$

$\Leftrightarrow 2\sqrt{x-5}+\frac{1}{3}\sqrt{x-5}-\frac{2}{3}\sqrt{x-5}=4$

$\Leftrightarrow \frac{5}{3}\sqrt{x-5}=4$

$\Leftrightarrow \sqrt{x-5}=\frac{12}{5}$

$\Leftrightarrow x-5=\frac{144}{25}=5,76$

$\Leftrightarrow x=10,76$ (tm)

a: Ta có: \(\sqrt{4x^2+4x+3}=8\)

\(\Leftrightarrow4x^2+4x+1+2-64=0\)

\(\Leftrightarrow4x^2+4x-61=0\)

\(\Delta=4^2-4\cdot4\cdot\left(-61\right)=992\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-4-4\sqrt{62}}{8}=\dfrac{-1-\sqrt{62}}{2}\\x_2=\dfrac{-4+4\sqrt{62}}{8}=\dfrac{-1+\sqrt{62}}{2}\end{matrix}\right.\)

1: =>x^2-x=3-x

=>x^2=3

=>x=căn 3 hoặc x=-căn 3

2: =>x^2-4x+3=x^2-4x+4 và x>=2

=>3=4(vô lý)

3: =>2|x-1|=6

=>|x-1|=3

=>x-1=3 hoặc x-1=-3

=>x=-2 hoặc x=4

4: =>|2x-3|=|x-2|

=>2x-3=x-2 hoặc 2x-3=-x+2

=>x=1 hoặc x=5/3

5: =>\(\sqrt{x+2}\left(\sqrt{x-2}+\sqrt{x+2}\right)=0\)

=>x+2=0

=>x=-2

\(a,\Leftrightarrow x-1=4\Leftrightarrow x=5\\ b,\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\3x+1=4x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\left(tm\right)\end{matrix}\right.\Leftrightarrow x=4\\ c,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=3\\ \Leftrightarrow x+5=9\\ \Leftrightarrow x=4\left(tm\right)\)

\(d,\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{5}+1\right)^2}\\ \Leftrightarrow\left|x-2\right|=\sqrt{5}+1\\ \Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{5}+1\\2-x=\sqrt{5}+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}+3\\x=1-\sqrt{5}\end{matrix}\right.\)