ta có 4 x 3 y 2   –   8 x 2 y 3   =   4 x 2 y 2 . x   –   4 x 2 y 2 . 2 y   =   4 x 2 y 2 ( x   –   2 y )    

Vậy 4x³y² – 8x²y³ = 4x²y²(x – 2y)      

Đáp án cần chọn là: C

bấm đúng cho mik đi 

thứ nhất bn đăng sai môn 

thứ hai bn giải r đăng lmj :???

Thứ nhất đang sai môn 

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Lời giải:

a. $\sqrt{x^2}=1$

$\Leftrightarrow |x|=1$

$\Leftrightarrow x=\pm 1$

b. $\sqrt{4x^2-4x+1}=3$

$\Leftrightarrow \sqrt{(2x-1)^2}=3$
$\Leftrightarrow |2x-1|=3$

$\Leftrightarrow 2x-1=\pm 3$

$\Leftrightarrow x=-1$ hoặc $x=2$

3. ĐKXĐ: $x^2\geq 4$

$\sqrt{x^2-4}+\sqrt{x^2+4x+4}=0$

Do $\sqrt{x^2-4}\geq 0; \sqrt{x^2+4x+4}\geq 0$ với mọi $x\in$ ĐKXĐ nên để tổng của chúng bằng $0$ thì:

$\sqrt{x^2-4}=\sqrt{x^2+4x+4}=0$

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

$\Leftrightarrow x=-2$

4. 

PT \(\Leftrightarrow \left\{\begin{matrix} x-3\geq 0\\ x^2-4x+3=(x-3)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 3\\ x^2-4x+3=x^2-6x+9\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq 3\\ 2x=6\end{matrix}\right.\Leftrightarrow x=3\)

Ý 1:

\(\sqrt{x^2}=1\\ \Leftrightarrow\left|x\right|=1\\ Vậy:x=1.hoặc.x=-1\\ S=\left\{\pm1\right\}\)

Ý 2:

\(\sqrt{4x^2-4x+1}=3\\ \Leftrightarrow\sqrt{\left(2x-1\right)^2}=3\\ \Leftrightarrow\left|2x-1\right|=3\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ Vậy:S=\left\{-1;2\right\}\)

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

=>căn x+2=0

=>x+2=0

=>x=-2

4: =>\(\left\{{}\begin{matrix}x>=3\\x^2-4x+3=x^2-6x+9\end{matrix}\right.\Leftrightarrow x=3\)

1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

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

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

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

\(\Leftrightarrow x=1\pm\sqrt{5}\)

3.

ĐK: \(x\ge-9\)

\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)

\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

2.

ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)

\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)

Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)

\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)

Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:

\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)

\(\Leftrightarrow10b+40=3\left(b+8\right)b\)

\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)

TH1: \(b=2\Leftrightarrow...\)

TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)

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: \(\sqrt{4x^2-4x+1}=x-1\)

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

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

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

ối giồi giải mấy bài trẻ con này thì chắc ko bao giờ vươn ra thế giới, há há

\(P=\dfrac{4xy^2-4x^2y+x^3}{4x^3-8x^2y}=\dfrac{x\left(x^2-4xy+4y^2\right)}{4x^2\left(x-2y\right)}=\dfrac{x-2y}{4x}\)

\(Q=\dfrac{2xy-x^2+x-2y}{4x-4x^2}=\dfrac{x\left(2y-x\right)-\left(2y-x\right)}{-4x\left(x-1\right)}=\dfrac{\left(2y-x\right)\left(x-1\right)}{-4x\left(x-1\right)}=\dfrac{x-2y}{4x}\)

Do đó: P=Q

a) x^2+4x+3=x^2+x+3x+3=x(x+1)+3(x+1)=(x+1)(x+3)

b) 4x^2+4x-3=4x^2+4x+1-4=(2x+1)^2-4=(2x+1-2)(2x+1+2)=(2x-1)(2x+3)

c) x^2-x-12=x^2-4x+3x-12=x(x-4)+3(x-4)=(x-4)(x+3)

d) 4x^4+4x^2y^2-8y^4=4(x^4+x^2y^2-2y^4)=4(x^4-x^2y^2+2x^2y^2-2y^4)=4(x^2-y^2)(x^2+2y^2)=4(x-y)(x+y)(x^2+2y^2)

a) \(x^2+4x+3\)

\(=x^2+x+3x+3\)

\(=\left(x^2+x\right)+\left(3x+3\right)\)

\(=x\left(x+1\right)+3\left(x+1\right)\)

\(=\left(x+1\right)\left(x+3\right)\)

c) \(x^2-x-12\)

\(=x^2-4x+3x-12\)

\(=\left(x^2-4x\right)+\left(3x-12\right)\)

\(=x\left(x-4\right)+3\left(x-4\right)\)

\(=\left(x-4\right)\left(x+3\right)\)

\(x^2+4x+3\)

\(=x^2+x+3x+3\)

\(=x\left(x+1\right)+3\left(x+1\right)\)

\(=\left(x+1\right)\left(x+3\right)\)

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

\(3x-4x^2+6-8x=x^2+4x+6\Leftrightarrow5x^2+9x=0\Leftrightarrow x=0;x=-\dfrac{9}{5}\)

đk : x khác 0 ; -1 

\(\Rightarrow x^2+3x+x^2-x-2=2x\left(x+1\right)\Leftrightarrow2x-2=2x\left(voli\right)\)

Vậy pt vô nghiệm