\(M=2^{2010}-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)\)

\(\Rightarrow M=2^{2010}-\left(2^0+2^1+2^2+...+2^{2008}+2^{2009}\right)\)

Đặt \(N=2^0+2^1+2^2+...+2^{2008}+2^{2009}\)

\(\Rightarrow2N=2^1+2^2+2^3+...+2^{2009}+2^{1010}\)

\(\Rightarrow2N-N=2^{2010}-1\)

\(\Leftrightarrow M=2^{2010}-\left(N\right)=2^{2010}-\left(2^{2010}-1\right)=2^{2010}-2^{2010}+1=1\)

Vậy M = 1

M = 2²⁰¹⁰ - ( 2²⁰⁰⁹ + 2²⁰⁰⁸ + ... + 2¹ + 2⁰ )

Đặt A = 2²⁰⁰⁹ + 2²⁰⁰⁸ + ... + 2¹ + 2⁰

=> 2A = 2( 2²⁰⁰⁹ + 2²⁰⁰⁸ + ... + 2¹ + 2⁰ )

           = 2²⁰¹⁰ + 2²⁰⁰⁹ + ... + 2² + 2¹

2A - A = A

= ( 2²⁰¹⁰ + 2²⁰⁰⁹ + ... + 2² + 2¹ ) - ( 2²⁰⁰⁹ + 2²⁰⁰⁸ + ... + 2¹ + 2⁰ )

=  2²⁰¹⁰ + 2²⁰⁰⁹ + ... + 2² + 2¹ - 2²⁰⁰⁹ - 2²⁰⁰⁸ - ... - 2¹ - 2⁰

= 2²⁰¹⁰ - 2⁰

=> M = 2²⁰¹⁰ - ( 2²⁰¹⁰ - 2⁰ )

         = 2²⁰¹⁰ - 2²⁰¹⁰ + 2⁰

         = 2⁰ = 1

Ta có \(\left|x-4\right|\ge0\forall x\Rightarrow\left|x-4\right|+17\ge17\forall x\)

Dấu " = " xảy ra <=> x - 4 = 0 => x = 4

Vậy GTNN của biểu thức = 17 khi x = 4 

\(\left(x-4\right)+17\)

\(\Leftrightarrow\left(x-4\right)\le0\)

\(\Leftrightarrow\left(x-4\right)+17\le0\)

dấu ''='' xảy ra khi

\(\Leftrightarrow x-4=0\)

\(\Leftrightarrow x=4\)

3.(10.x)=111

=> 10x = 111 : 3

=> 10 x = 37

=> x = 37 : 10

=> x = \(\frac{37}{10}\)

3.(10.x)=111

    (10.x)=111:3

    (10.x)=37

          x=37:10

          x=3,7

vậy x=3,7

5x²-19x²-4

=x²(5-19)-4

= -14x²-4

= -2(7x²+2)

5x² - 19x² - 4 = -14x² - 4 = -7x².2 - 2.2 = 2( -7x² - 2 )

Sử dụng tính chất của dãy tỉ số bằng nhau thì :

\(\frac{x+1}{2}=\frac{y-1}{3}=\frac{z+2}{4}=\frac{x+1+y-1+z+2}{2+3+4}=\frac{x+y+z+2}{9}\)

Do \(\frac{x+1}{2}=\frac{y-1}{3}=\frac{z+2}{4}=\frac{x+y+z+2}{2x+5}\)

Suy ra \(\frac{x+y+z+2}{9}=\frac{x+y+z+2}{2x+5}< =>2x+5=9\)

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

Thế vào thì ta được : \(\hept{\begin{cases}\frac{x+1}{2}=\frac{y-1}{3}< =>\frac{3}{2}=\frac{y-1}{3}\\\frac{x+1}{2}=\frac{z+2}{4}< =>\frac{3}{2}=\frac{z+2}{4}\end{cases}}\)

\(< =>\hept{\begin{cases}2\left(y-1\right)=9\\2\left(z+2\right)=12\end{cases}< =>\hept{\begin{cases}2y-2=9\\2z+4=12\end{cases}}}\)

\(< =>\hept{\begin{cases}2y=11< =>y=\frac{11}{2}\\2z=8< =>z=\frac{8}{2}=4\end{cases}}\)

Vậy ta có bộ số x,y,z thỏa mãn đẳng thức sau : \(\left\{2;\frac{11}{2};4\right\}\)

Áp dụng tính chất dãy tỉ số bằng nhau : 

\(\frac{x+1}{2}=\frac{y-1}{3}=\frac{z+2}{4}=\frac{x+y+z}{2x+5}\frac{x+1+y-1+z+2}{2+3+4}=\frac{x+y+z+2}{9}=\frac{x+y+z}{9}\)(1)

Từ (1) => \(\frac{x+y+z}{2x+5}=\frac{x+y+z}{9}\)

=> 2x + 5 = 9

=> 2x = 4

=> x = 2

Thay x vào (1)

=> \(\frac{2+1}{2}=\frac{y-1}{3}=\frac{z+2}{4}\)

=> \(\frac{y-1}{3}=\frac{z+2}{4}=\frac{3}{2}\)

=> \(\hept{\begin{cases}\frac{y-1}{3}=\frac{3}{2}\\\frac{z+2}{4}=\frac{3}{2}\end{cases}}\Rightarrow\hept{\begin{cases}y=\frac{3}{2}.3+1\\z=\frac{3}{2}.4-2\end{cases}}\Rightarrow\hept{\begin{cases}y=\frac{11}{2}\\z=4\end{cases}}\)

Vậy x = 2 ; y = 11/2 ; z = 4

a) Ta có 3x = 2y = z 

=> \(\frac{3x}{6}=\frac{2y}{6}=\frac{z}{6}\)

\(\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{6}=\frac{x+y+z}{2+3+6}=\frac{99}{11}=9\)

=> \(\hept{\begin{cases}x=18\\y=27\\z=54\end{cases}}\)

b) 6x = 10y = 15z 

=> \(\frac{6x}{30}=\frac{10y}{30}=\frac{15z}{30}\)

=> \(\frac{x}{5}=\frac{y}{3}=\frac{z}{2}=\frac{x+y+z}{5+3+2}=\frac{90}{10}=9\)

=> \(\hept{\begin{cases}x=45\\y=27\\z=18\end{cases}}\)

c) 6x = 4y = 2z

=> \(\frac{6x}{12}=\frac{4y}{12}=\frac{2z}{12}\)

=> \(\frac{x}{2}=\frac{y}{3}=\frac{z}{6}=\frac{x+y+z}{2+3+6}=\frac{27}{11}\)

=> \(\hept{\begin{cases}x=\frac{54}{11}\\y=\frac{81}{11}\\z=\frac{162}{11}\end{cases}}\)

d) x = 3y = 2z

=> \(\frac{x}{6}=\frac{3y}{6}=\frac{2z}{6}\)

=> \(\frac{x}{6}=\frac{y}{2}=\frac{z}{3}\)

=> \(\frac{2x}{12}=\frac{3y}{6}=\frac{4z}{12}=\frac{2x-3y+4z}{12-6+12}=\frac{48}{18}=\frac{8}{3}\)

=> \(\hept{\begin{cases}x=16\\y=\frac{16}{3}\\z=8\end{cases}}\)

ta có:\(\widehat{aOb}\) = 180

\(\Rightarrow\)3 x \(\widehat{aOc}\)=180

\(\Rightarrow\)\(\widehat{aOc}\)=180 : 3 = 60

\(\Rightarrow\)\(\widehat{aOc}\)=\(\widehat{bOd}\)= 60 (2 góc đối đỉnh)

ta có: \(\widehat{aOc}\)+\(\widehat{cOb}\)= 180 (2 góc kề bù)

\(\Rightarrow\)60 + \(\widehat{cOb}\)= 180

\(\Rightarrow\)\(\widehat{cOb}\)= 180 - 60 = 120

\(\Rightarrow\)\(\widehat{aOd}\)=\(cOb\)= 120 (2 goc đối đỉnh)

Vậy \(\widehat{aOc}\)= 60;\(\widehat{cOb}\)= 120;\(\widehat{bOd}\)= 60;\(\widehat{aOd}\)=120

cảm ơn bạn

\(\left|x+5\right|-4=3\)

\(\Rightarrow\left|x+5\right|=7\)

\(\Rightarrow\orbr{\begin{cases}x+5=7\\x+5=-7\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=-12\end{cases}}\)

\(\left|x+5\right|-4=3\Leftrightarrow\left|x+5\right|=7\)

\(\Leftrightarrow\orbr{\begin{cases}x+5=7\\x+5=-7\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-12\end{cases}}\)

vậy x=2 hoặc x=-12

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

\(\Leftrightarrow ac-bc=ac-ad\)

\(\Leftrightarrow ac-ac-bc+ad=0\)

\(\Leftrightarrow-bc+ad=0\)hay như nào =))