. là nhân hay chia

. là nhân

lớp mấy vậy mà ko bt

th1: nếu x=-1 thì 4x(x+1)=8(x+1) suy ra 0=0 (thỏa)

th2: nếu x khác -1 thì x+1 khác 0

4x(x+1)=8(x+1)

4x=8 do x+1 khác 0

x=2

vậy x=-1 và 2

4x( x + 1 ) = 8( x + 1 ) 

<=> 4x( x + 1 ) - 8( x + 1 ) = 0

<=> 4( x + 1 )( x - 2 ) = 0

<=> x + 1 = 0 hoặc x - 2 = 0

<=> x = -1 hoặc x = 2

Ta có: \(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}\)

\(>\frac{a}{a+b+c+d}+\frac{b}{a+b+c+d}+\frac{c}{a+b+c+d}+\frac{d}{a+b+c+d}\)

\(=\frac{a+b+c+d}{a+b+c+d}=1\)

Tương tự ta cũng chứng minh được \(\frac{b}{a+b}+\frac{c}{b+c}+\frac{d}{c+d}+\frac{a}{d+a}>1\)

mà \(\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}\right)+\left(\frac{b}{a+b}+\frac{c}{b+c}+\frac{d}{c+d}+\frac{a}{d+a}\right)\)

\(=\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+d}{c+d}+\frac{d+a}{d+a}=4\)

\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}\)là số nguyên 

do đó \(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}=2\)

\(\Leftrightarrow1-\frac{a}{a+b}-\frac{b}{b+c}+1-\frac{c}{c+d}-\frac{d}{d+a}=0\)

\(\Leftrightarrow\frac{b}{a+b}-\frac{b}{b+c}+\frac{d}{c+d}-\frac{d}{d+a}=0\)

\(\Leftrightarrow\frac{b\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}+\frac{d\left(a-c\right)}{\left(c+d\right)\left(d+a\right)}=0\)

\(\Leftrightarrow b\left(c+d\right)\left(d+a\right)-d\left(a+b\right)\left(b+c\right)=0\)(vì \(a\ne c\))

\(\Leftrightarrow\left(b-d\right)\left(ac-bd\right)=0\)

\(\Leftrightarrow ac=bd\)(vì \(b\ne d\))

Khi đó \(abcd=ac.ac=\left(ac\right)^2\)là số chính phương. 

1.Xét \(\Delta ABD\)và \(\Delta ACE\),có:

\(\hept{\begin{cases}\widehat{ADB}=\widehat{AEC}\left(=90^0\right)\\\widehat{BAC}:chung\end{cases}\Rightarrow}\Delta ABD~\Delta ACE\left(g.g\right)\)

\(\Rightarrow\frac{AB}{AC}=\frac{AD}{AE}\)

\(\Rightarrow AB.AE=AC.AD\)

2.Xét \(\Delta ABC\)và \(\Delta ADE\),có :

\(\hept{\begin{cases}\frac{AB}{AC}=\frac{AD}{AE}\\\widehat{BAC}:chung\end{cases}}\Rightarrow\Delta ABC~\Delta ADE\left(c.gc\right)\)

\(\Rightarrow\widehat{ADE}=\widehat{ABC}\)

3.Xét \(\Delta AMC\)và \(\Delta ADM\),có:

\(\hept{\begin{cases}\widehat{AMC}=\widehat{ADM}\left(=90^0\right)\\\widehat{MAC}:chung\end{cases}}\Rightarrow\Delta AMC~\Delta ADM\left(g.g\right)\)

\(\Rightarrow\frac{AM}{AD}=\frac{AC}{AM}\Rightarrow AM^2=AC.AD\)

Tương tự ,ta có: \(AN^2=AB.AE\)

mà AC.AD=AB.AE(theo chứng minh phần a)

\(\Rightarrow AM^2=AN^2\Rightarrow AM=AN\)

a) Ta có  (1 + 2y)² + (1 - 2y)² + 2(1 + 2y)(1 - 2y) 

= (1 + 2y + 1 - 2y)² = 2² = 4

b) Ta có (7x + 2y)² + (7x - 2y)² - 2(49x² - 4y²) 

= (7x + 2y)² + (7x - 2y)² - 2[(7x)² - (2y)²]

= (7x + 2y)² + (7x - 2y)² - 2(7x - 2y)(7x + 2y)

= (7x + 2y - 7x + 2y)² 

= (4y)² = 16y²

\(P\left(x\right)=x^5+x^4-4x^3+x^2-x-2\)

\(=x^5-x^4-x^3+2x^4-2x^3-2x^2-x^3+x^2+x+2x^2-2x-2\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x^2-x-1=0\left(1\right)\\x^3+2x^2-x+2=0\end{cases}}\)

Giải \(\left(1\right)\): \(x^2-x-1=0\Leftrightarrow x^2-x+\frac{1}{4}=\frac{5}{4}\Leftrightarrow\left(x-\frac{1}{2}\right)^2=\frac{5}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}x_1=\frac{1+\sqrt{5}}{2}\\x_2=\frac{1-\sqrt{5}}{2}\end{cases}}\)

Ta thấy \(x_1+x_2=1\)do đó đây là hai nghiệm \(a,b\)thỏa mãn. 

\(ab=x_1x_2=\frac{\left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right)}{2.2}=-1\).

\(\left(3x-2\right)\left(9x^2-6x+4\right)\)

\(\left(3x-2\right)\left(9x^2+6x+4-12x\right)\)

\(\left(3x\right)^3-2^3-12x\left(3x-2\right)\)

\(27x^3-8-36x+24x\)

\(27\left(\frac{1}{3}\right)^3-8-36.\frac{1}{3}+24.\frac{1}{3}\)

\(1-8-12+8=-11\)

Đặt \(a-b=x,b-c=y,c-a=z,x>0,y>0,z< 0\)suy ra \(x+y+z=0\).

Ta có: \(x^3+y^3+z^3-3xyz=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)

\(=\left(x+y+z\right)^3-3\left(x+y\right)z\left(x+y+z\right)-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)

\(\Leftrightarrow x^3+y^3+z^3=3xyz< 0\)

Do đó ta có đpcm.