Câu C) \(AB^{2=}AD.AC\)nha các bạn

đề là \(\frac{5}{x+2}-3=0\)  à

\(ĐKXĐ:x\ne-2\)

\(\frac{5-3x-6}{x+2}=0\)

\(\frac{-1-3x}{x+2}=0\)

\(-1-3x=0\)

\(x=-\frac{1}{3}\left(TM\right)\)

Giải thích các bước giải:

 a,x+2x−2−x−2x+2=2(x+4)x2−4a,x+2x-2-x-2x+2=2(x+4)x2-4

⇒xx≠≠ −2,2-2,2

⇒x+2x−2−x−2x+2−2(x+4)x2−4=0x+2x-2-x-2x+2-2(x+4)x2-4=0

⇒(x+2)2−(x−2)2−(2x+8)(x−2)(x+2)=0(x+2)2-(x-2)2-(2x+8)(x-2)(x+2)=0

⇒6x−8(x−2)(x+2)=06x-8(x-2)(x+2)=0

⇒6x−8=06x-8=0

⇒x=43x=43

b,x−1x+6−xx−6=−7x2−36b,x-1x+6-xx-6=-7x2-36

⇒xx≠≠ −6,6-6,6

⇒x−1x+6−xx−6+−7x2−36=0x-1x+6-xx-6+-7x2-36=0

⇒(x−6)(x−1)−x(x+6)+7(x−6)(x+6)=0(x-6)(x-1)-x(x+6)+7(x-6)(x+6)=0

⇒−13x+13=0-13x+13=0

⇒x=1x=1

c,5x+2−5x−3+x2+16x2−x−6=0c,5x+2-5x-3+x2+16x2-x-6=0

⇒xx≠≠ −2,3-2,3

⇒5(x−3)−5(x+2)+x2+16(x+2)(x−3)=05(x-3)-5(x+2)+x2+16(x+2)(x-3)=0

⇒x2−9(x+2)(x−3)=0x2-9(x+2)(x-3)=0

⇒x+3x+2=0x+3x+2=0

⇒x+3=0x+3=0

⇒x=−3x=-3

d,x+3x+2−x−2x−3=x−3x2−x−6d,x+3x+2-x-2x-3=x-3x2-x-6

⇒xx≠≠ −2,3-2,3

⇒x+3x+2−x−2x−3=x−3(x+2)(x−3)x+3x+2-x-2x-3=x-3(x+2)(x-3)

⇒x+3x+2−x−2x−3−1x+2=0x+3x+2-x-2x-3-1x+2=0

⇒(x−3)(x+3)−(x+2)(x−2)−(x−3)(x+2)(x−3)=0(x-3)(x+3)-(x+2)(x-2)-(x-3)(x+2)(x-3)=0

⇒−2−x(x+2)(x−3)=0-2-x(x+2)(x-3)=0

⇒1x−3=01x-3=0

⇒xx ∈ ∅

a) \(A=\left(\frac{1}{a}+\frac{1}{a+1}\right):\frac{a}{a^2+a}\)

ĐKXĐ : a \(\ne0;a\ne-1\)

Khi đó A = \(\left(\frac{a+1+a^2}{a\left(a+1\right)}\right):\frac{a}{a\left(a+1\right)}=\frac{a^2+a+1}{a\left(a+1\right)}.\frac{a\left(a+1\right)}{a}=\frac{a^2+a+1}{a}\)

b) Khi (a - 5)(a + 1) = 0

<=> \(\orbr{\begin{cases}a-5=0\\a+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}a=5\left(\text{tmđk}\right)\\a=-1\left(\text{loại}\right)\end{cases}}\)

Thay a = 5 vào M ta được M = \(\frac{5^2+5+1}{5}=\frac{31}{5}\)

Vậy a = 5 thì M = 31/5

c) Ta có M = \(\frac{a^2+a+1}{a}=\frac{a^2-2a+1+3a}{a}=\frac{\left(a-1\right)^2}{a}+3\ge3\)

=> Min M = 3

Dấu "=" xảy ra <=> a - 1= 0 <=> a = 1 (tm)

Vậy Min M = 3 <=> a = 1

| 2-4x | = 4x-2

<=> \(\orbr{\begin{cases}\left|2-4x\right|=-2+4x=4x-2\\\left|2-4x\right|=2-4x=4x-2\end{cases}}\)

<=>\(\orbr{\begin{cases}-2+4x=4x-2\\2-4x=4x-2\end{cases}}\)

<=>\(\orbr{\begin{cases}-2+4x-4x+2=0\\2-4x-4x+2=0\end{cases}}\)

<=>\(\orbr{\begin{cases}0=0\\-8x+4=0\end{cases}}\)

<=> x=\(\frac{-4}{-8}=\frac{1}{2}\)

=> \(S=\left\{\frac{1}{2};\infty\right\}\)

2x-7> 3(x-1)

<=>2x-7>3x-3

<=>2x-3x>-3+7

<=>-x>4

<=>x<4

=>S={x/x<4}

1-2x<4(3x-2)

<=>1-2x<12x-8

<=>-2x-12x<-8-1

<=>-14x<-9

<=>x>\(\frac{9}{14}\)

=>S={\(\frac{9}{14}\)}

-3x+2|-4 -x|> 0

<=>\(\orbr{\begin{cases}-3x+2+4+x>0\\-3x+2-4x-x>0\end{cases}}\)

<=>\(\orbr{\begin{cases}-2x+6>0\\-8x+2>0\end{cases}}\)

<=>\(\orbr{\begin{cases}-2x>-6\\-8x>-2\end{cases}}\)

<=>\(\orbr{\begin{cases}x< 3\\x< \frac{1}{4}\end{cases}}\)

=>S={x/x<3;x/x<\(\frac{1}{4}\)}

4x-1|x-2|< 0

<=>\(\orbr{\begin{cases}4x-1-x+2< 0\\4x-1+x-2< 0\end{cases}}\)

<=>\(\orbr{\begin{cases}3x+1< 0\\3x-3< 0\end{cases}}\)

<=>\(\orbr{\begin{cases}3x< -1\\3x< 3\end{cases}}\)

<=>\(\orbr{\begin{cases}x< \frac{-1}{3}\\x< 1\end{cases}}\)

=>S={x/x<\(\frac{-1}{3}\);x/x<1}

x³ - x² - 8x + 12 = x³ - 2x² + x² - 2x - 6x + 12 = x²(x - 2) + x(x - 2) - 6(x - 2) = (x² + x - 6)(x - 2)

= (x² + 3x - 2x - 6)(x - 2) = (x + 3)(x - 2)(x - 2) = (x + 3)(x - 2)²

a) = ( x⁴ - x² ) + ( x³ - 1 ) = x²( x - 1 )( x + 1 ) + ( x - 1 )( x² + x + 1 ) = ( x - 1 )( x³ + 2x² + x + 1 )

b) = x³( x - 3 ) - ( x - 3 ) = ( x - 3 )( x³ - 1 ) = ( x - 3 )( x - 1 )( x² + x + 1 )

c) = 3( x + y ) - ( x + y )² = ( x + y )( 3 - x - y )

d) = 2( x - y ) - ( x - y )² = ( x - y )( 2 + x + y )

e) = y( x² - y² - 2y - 1 ) = y[ x² - ( y + 1 )² ] = y( x - y - 1 )( x + y + 1 )

f) = xz( x² + 2xy + y² - 9 ) = xz[ ( x + y )² - 3² ] = xz( x + y - 3 )( x + y + 3 )