19.6849466

39.3698933

đề là tính hả bn?

a = 3x - 1/2 = 0

=>  3x = 1/2

=> x = 1/6

b(x) = |2x - 3| + 11 = 0

=> |2x - 3| = -11

=> b(x) vô nghiệm

c(x) = 2x^3 - 8x = 0

=> x(2x^2 - 8) = 0

=> x = 0 hoặc 2x^2 - 8 = 0

2x^2 - 8 = 0

=> 2x^2 = 8

=> x^2 = 4

=> x = 2 hoặc x = -2

\(a,\frac{-9}{x}=\frac{-9}{\frac{4}{49}}\)

\(\Rightarrow x=\frac{4}{49}\)

\(b,\left|x-2\right|+\left|x+3\right|=0\)

\(\left|x-2\right|\ge0;\left|x+3\right|\ge0\)

\(\Rightarrow\hept{\begin{cases}\left|x-2\right|=0\\\left|x+3\right|=0\end{cases}\Rightarrow\hept{\begin{cases}x-2=0\\x+3=0\end{cases}\Rightarrow}\hept{\begin{cases}x=2\\x=-3\end{cases}vl}}\)

\(c,3x^2+9x+6=0\)

\(\Rightarrow3x^2+3x+6x+6=0\)

\(\Rightarrow3x\left(x+1\right)+6\left(x+1\right)=0\)

\(\Rightarrow\left(3x+6\right)\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x+6=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=-1\end{cases}}}\)

\(d,x^2-7x-8=0\)

\(\Rightarrow x^2+x-8x-8=0\)

\(\Rightarrow x\left(x+1\right)-8\left(x+1\right)=0\)

\(\Rightarrow\left(x-8\right)\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-8=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=8\\x=-1\end{cases}}\)

\(S=\left(\frac{17}{71}+\frac{17.101}{71.101}+\frac{17.10101}{71.10101}\right):\frac{17.1010101}{71.1010101}\)

\(S=\left(\frac{17}{71}+\frac{17}{71}+\frac{17}{71}\right):\frac{17}{71}\)

\(S=3.\frac{17}{71}:\frac{17}{71}=3\)

\(S=\left(\frac{17}{71}+\frac{17.101}{71.101}+\frac{17.10101}{71.10101}\right):\frac{17.1010101}{71.1010101}\)

\(S=\left(\frac{17}{71}+\frac{17}{71}+\frac{17}{71}\right):\frac{17}{71}\)

\(S=3.\frac{17}{71}:\frac{17}{71}\)

\(\Rightarrow S=3\)

Rất vui vì giúp đc bạn <3

a) \(\left(-2\right)^3+2^2+\left(-1\right)^{20}+\left(-2\right)^0\)

\(=-8+4+1+1=-2\)

b) \(\left(3^2\right)^2-\left(-5^2\right)^2+\left[\left(-2\right)^3\right]^2\)

\(=9^2-\left(-25\right)^2+\left(-8\right)^2\)

\(=81-625+64=-480\)

c) Bạn sửa lại đề!

\(A=1+2+2^2+...+2^{51}\)

\(2A=2+2^2+2^3+...+2^{52}\)

\(2A-A=\left(2+2^2+2^3+...+2^{52}\right)-\left(1+2+2^2+...+2^{51}\right)\)

\(A=2^{52}-1\)

\(B=5+5^2+5^3+...+5^{100}\)

\(5B=5^2+5^3+5^4+...+5^{101}\)

\(5B-B=\left(5^2+5^3+5^4+...+5^{101}\right)-\left(5+5^2+5^3+...+5^{100}\right)\)

\(4B=5^{101}-5\)

\(B=\frac{5^{101}-5}{4}\)

đang cần gấp xin mn giúp mik

#) Sửa đề bỏ cái 2^51 nhé.

\(A=1+2^2+2^4+.....+2^{50}\)

\(2^2A=2^2\left(1+2^2+2^4+...+2^{50}\right)\)

\(4A=2^2+2^4+2^6+...+2^{52}\)

\(4A-A=2^{52}-1\)

\(3A=2^{52}-1\)

\(A=\dfrac{2^{52}-1}{3}\)

~ Hok tốt ~
 

#)Giải :

Vì \(\widehat{AOC}\) và \(\widehat{AOD}\)là hai góc kề bù

\(\Rightarrow\widehat{AOC}+\widehat{AOD}=180^o\)

\(\Rightarrow\widehat{AOC}=\frac{\left(180^o+20^o\right)}{2}=100^o\)

\(\Rightarrow\widehat{AOD}=100^o-20^o=80^o\)

\(\Rightarrow\widehat{COB}=\widehat{AOD}=80^o\)(hai góc đối đỉnh)

\(\Rightarrow\widehat{BOD}=\widehat{AOC}=100^o\)(nt)

a) Do Oz là tia p/giác của \(\widehat{xOy}\)nên :

\(\widehat{xOz}=\widehat{zOy}=\frac{\widehat{xOy}}{2}=\frac{100^0}{2}=50^0\)

Ta có: \(\widehat{xOz}=\widehat{tOh}\) (đối đỉnh)

Mà \(\widehat{xOz}=50^0\) => \(\widehat{tOh}=50^0\)

b) Ta có: \(\widehat{tOh}=\widehat{xOz}\) (đối đỉnh)

mà \(\widehat{tỌh}=60^0\) => \(\widehat{xOz}=60^0\)

Do Oz là tia p/giác của \(\widehat{xOy}\)nên :

  \(\widehat{xOz}=\widehat{yOz}=\frac{\widehat{xOy}}{2}=60^0\)

      => \(\widehat{xOy}=2.60^0=120^0\)

c) Do Oz là tia p/giác của \(\widehat{xOy}\)nên :

  \(\widehat{xOz}=\widehat{zOy}=\frac{\widehat{xOy}}{2}\) => \(\widehat{xOy}=2.\widehat{xOz}\)

Mà  \(\widehat{xOz}=\widehat{tOh}\) (đối đỉnh)

=> \(\widehat{xOy}=2.\widehat{tOh}\)

Ta có: \(\widehat{xOy}+\widehat{tOh}=210^0\)

=> \(2.\widehat{tOh}+\widehat{tOh}=210^0\)

=> \(3.\widehat{tOh}=210^0\)

=> \(\widehat{tOh}=210^0:3=70^0\)

  = > \(\widehat{xOy}=2.\widehat{tOh}=2.70^0=140^0\)

\(\frac{\left(5^4-5^3\right)^3}{125^4}=\frac{\left[5^3\left(5-1\right)\right]^3}{125^4}=\frac{\left(4.5^3\right)^3}{\left(5^3\right)^4}=\frac{4^3.5^{3.3}}{5^{3.3}.5^3}=\frac{4^3}{5^3}=\frac{4.4.4}{5.5.5}=\frac{64}{125}\)

\(\frac{\left(5^4-5^3\right)^3}{125^4}\)= \(\frac{\left[5^3.\left(5-1\right)\right]^3}{\left(5^3\right)^4}\)

                             = \(\frac{\left[5^3.4\right]^3}{5^{3.4}}\)

                             = \(\frac{\left(5^3\right)^3.4^3}{5^{3.4}}\)

                             = \(\frac{5^9.4^3}{5^{12}}\)

                              = \(\frac{5^9.4^3}{5^9.5^3}\)

                              = \(\frac{4^3}{5^3}=\left(\frac{4}{5}\right)^3\)