(x-2)(x² + 2x+4)

= x³ - 2³

= x³ -8 

Ta có :

\(D=\frac{4^4.9^4+2.6^9}{2^{10}.3^8+6^8.20}\)

\(\Rightarrow D=\frac{\left(2^2\right)^4.\left(3^2\right)^4+2.6^9}{2^{10}.3^8+\left(2.3\right)^8.20}\)

\(\Rightarrow D=\frac{2^8.3^8+2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}\)

\(\Rightarrow D=\frac{\left(2^8.3^8\right).\left(1+2.2.3\right)}{\left(2^8.3^8\right).\left(2^2+2^2.5\right)}\)

\(\Rightarrow D=\frac{1+12}{4+4.5}=\frac{13}{24}\)

Ta có :

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

=> ( x - 1 )^x+4 - ( x - 1 )^x+3 = 0

=> ( x- 1 )^x+3 . ( x - 1 - 1 ) = 0

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

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)

Với \(a,b,c\ge0.\)Áp dụng BĐT Cô-si cho các cặp số (a,b);(b,c),(a,c).

Ta được: \(a+b\ge2\sqrt{ab},2\left(b+c\right)\ge2.2\sqrt{bc},3\left(a+c\right)\ge3.2\sqrt{ac}\)

Cộng vế với vế ta được đpcm 

Dấu "=" xảy ra <=> a=b=c=0

Ta có:\(4a+3b+5c\ge2\left(\sqrt{ab}+2\sqrt{bc}+3\sqrt{ca}\right)\)

\(\Leftrightarrow4a+3b+5c-2\sqrt{ab}-4\sqrt{bc}-6\sqrt{ac}\ge0\)

\(\Leftrightarrow\left(a-2\sqrt{ab}+b\right)+\left(3a-6\sqrt{ac}+3c\right)+\left(2b-4\sqrt{bc}+2c\right)\ge0\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2+3\left(\sqrt{a}-\sqrt{c}\right)^2+2\left(\sqrt{b}-\sqrt{c}\right)^2\ge0\)

Đẳng thức xảy ra khi \(a=b=c\)

100001

100001

Theo đề ra, ta có: \(\widehat{xAB}=\widehat{yBA}=120^o\) mà hai góc này ở vị trí so le trong => Ax//By

a) ( x - 1 )³ = 27 

=> ( x - 1 )³ = 3³

=> x - 1 = 3

=> x = 4

b) x² + x = 0

=> x . ( x + 1 ) = 0

\(\Rightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

c) ( 2x + 1 )² = 25 

=> ( 2x + 1 )² = 5² = ( -5 )²

\(\Rightarrow\orbr{\begin{cases}2x+1=5\\2x+1=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=4\\2x=-6\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

d) ( 2x - 3 )² = 36

=> ( 2x - 3 )² = 6² = ( -6 )²

\(\Rightarrow\orbr{\begin{cases}2x-3=6\\2x-3=-6\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=9\\2x=-3\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{9}{2}\\x=\frac{-3}{2}\end{cases}}\)

e) 5^x+2 = 625

=> 5^x+2 = 5⁴

=> x + 2 = 4

=> x = 2

f) ( x - 1 )^x+2 = ( x - 1 )^x+4

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

=> ( x - 1 )^x+2 . [ ( x - 1 )² - 1 ] = 0

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

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

\(\Rightarrow\orbr{\begin{cases}x=1\\x-1\in\left\{-1;1\right\}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x\in\left\{0;2\right\}\end{cases}}\)

g) ( 2x - 1 )³ = 8

=> ( 2x - 1 )³ = 2³

=> 2x - 1 = 2

=> 2x = 3

=> x = \(\frac{3}{2}\)

h) \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}.....\frac{31}{64}=2^x\)

\(\Rightarrow\frac{1}{2.2}.\frac{2}{2.3}.\frac{3}{2.4}.\frac{4}{2.5}.....\frac{31}{2.32}=2^x\)

\(\Rightarrow\frac{1.2.3.4.....31}{\left(2.2\right).\left(2.3\right).\left(2.4\right).....\left(2.32\right)}=2^x\)

\(\Rightarrow\frac{2.3.4.....31}{2^{31}.32.\left(2.3.4.....31\right)}=2^x\)

\(\Rightarrow\frac{1}{2^{31}.2^5}=2^x\)\(\Rightarrow\frac{1}{2^{36}}=2^x\)

\(\Rightarrow2^x=2^{-36}\Rightarrow x=-36\)

a)x=4

b)x=0;x=1

c)x=2

d)x=9/2

g)x=-1/2

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