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

Đặt \(x^2-4x=t\)

Phương trình \(\Leftrightarrow\frac{t+12}{t+6}=t+8\Leftrightarrow t+12=\left(t+6\right)\left(t+8\right)\)

\(\Leftrightarrow t+12=t^2+14t+48\Leftrightarrow t^2+13t+36=0\Leftrightarrow\left(t+4\right)\left(t+9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t+4=0\\t+9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}t=-4\\t=-9\end{cases}}}\)

Với \(t=-4\Rightarrow x^2-4x+4=0\Rightarrow\left(x-2\right)^2=0\Rightarrow x=2\)

Với \(t=-9\Rightarrow x^2-4x+9=0\)vô nghiệm vì \(\Delta=16-36=-20< 0\)

Vậy phương trình có nghiệm x=2

Mơn bạn nhìu nha

Em cần làm gì với bài này?

(x² + x)² + 4x² + 4x - 12

= (x² + x)² + 4(x² + x) + 4 - 16

= (x² + x + 2)² - 4²

= (x² + x + 2 - 4)(x² + x + 2 + 4)

= (x² + x - 2)(x² + x + 6)

a: \(\dfrac{x+10}{4x-8}\cdot\dfrac{4-2x}{x+2}\)

\(=\dfrac{x+10}{4\left(x-2\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{-\left(x+10\right)}{2\left(x+2\right)}\)

b: \(\dfrac{1-4x^2}{x^2+4x}:\dfrac{2-4x}{3x}\)

\(=\dfrac{\left(2x-1\right)\left(2x+1\right)}{x\left(x+4\right)}\cdot\dfrac{3x}{2\left(x-2\right)}\)

\(=\dfrac{3\left(2x-1\right)\left(2x+1\right)}{2\left(x-2\right)\left(x+4\right)}\)

c: \(=\dfrac{4y^2}{7x^4}\cdot\dfrac{35x^2}{-8y}=\dfrac{5}{x^2}\cdot\dfrac{-1}{2}\cdot y=\dfrac{-5y}{2x^2}\)

d: \(=\dfrac{\left(x-2\right)\left(x+2\right)}{3\left(x+4\right)}\cdot\dfrac{x+4}{2\left(x-2\right)}=\dfrac{x+2}{6}\)

a.(x+10) /(4*x)-8* 4 -(2*x)/x+2

-(127*x-10)/(4*x)

(5/2-127*x/4)/x

Câu a

a: =(x^2+x-6)(x^2+x-8)

=(x+3)(x-2)(x^2+x-8)

b: =(x^2+x)^2+4(x^2+x)-12

=(x^2+x+6)(x^2+x-2)

=(x^2+x+6)(x+2)(x-1)

c: =x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12

=(x-1)(x^3+3x^2+8x+12)

=(x-1)(x^3+2x^2+x^2+2x+6x+12)

=(x-1)(x+2)(x^2+x+6)

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) \(x^2-x-12\)

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

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

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

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

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

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

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

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

cái đề yêu cầu j z :)

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