\(x+12=31\cdot2\)
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ZC
0
TK
1
7 tháng 12 2017
\(2^1.2^2.2^3.....2^x=1024\Rightarrow2^{1+2+3+...+x}=2^{10}\)
\(\Rightarrow1+2+3+...+x=1024\Rightarrow x=4\)
H9
HT.Phong (9A5)
CTVHS
13 tháng 7 2023
a) \(\dfrac{25}{37}\times\dfrac{18}{29}+\dfrac{18}{29}\times\dfrac{12}{37}\)
\(=\dfrac{18}{29}\times\left(\dfrac{25}{37}+\dfrac{12}{37}\right)\)
\(=\dfrac{18}{29}\times\dfrac{37}{37}\)
\(=\dfrac{18}{29}\times1\)
\(=\dfrac{18}{29}\)
b) \(\dfrac{31}{85}\times\dfrac{11}{19}+\dfrac{31}{85}\times\dfrac{12}{19}-\dfrac{42}{19}\times\dfrac{31}{85}\)
\(=\dfrac{31}{85}\times\left(\dfrac{11}{19}+\dfrac{12}{19}-\dfrac{42}{19}\right)\)
\(=\dfrac{31}{85}\times\dfrac{-19}{19}\)
\(=\dfrac{31}{85}\times-1\)
\(=-\dfrac{31}{85}\)
c) \(\dfrac{16}{53}:\dfrac{17}{9}-\dfrac{16}{53}:\dfrac{17}{8}\)
\(=\dfrac{16}{53}:\left(\dfrac{9}{17}-\dfrac{8}{17}\right)\)
\(=\dfrac{16}{53}:\dfrac{1}{17}\)
\(=\dfrac{16}{901}\)
c) \(\dfrac{1}{5}\times\dfrac{12}{31}\times\dfrac{4}{3}+\dfrac{19}{31}\times\dfrac{4}{15}\)
\(=\dfrac{4}{15}\times\dfrac{12}{31}+\dfrac{19}{31}\times\dfrac{4}{15}\)
\(=\dfrac{4}{15}\times\left(\dfrac{12}{31}+\dfrac{19}{31}\right)\)
\(=\dfrac{4}{15}\times\dfrac{31}{31}\)
\(=\dfrac{4}{15}\times1\)
\(=\dfrac{4}{15}\)
13 tháng 7 2023
a: =18/29*(25/37+12/37)
=18/29
b: =31/85(11/19+12/19-42/19)
=-31/85
c; =16/53(9/17+8/17)=16/53
d: =4/15(12/31+19/31)=4/15
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 3 2023
\(ac=-12< 0\) nên pt luôn có 2 nghiệm pb trái dấu
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=-12\end{matrix}\right.\)
\(x_1^2-x_2^2-14\left(m+1\right)=0\)
\(\Leftrightarrow\left(x_1-x_2\right)\left(x_1+x_2\right)-14\left(m+1\right)=0\)
\(\Leftrightarrow\left(x_1-x_2\right).2\left(m+1\right)-14\left(m+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}m=-1\\x_1-x_2=7\left(1\right)\end{matrix}\right.\)
Xét (1), kết hợp với Viet ta được: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1-x_2=7\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{2m+9}{2}\\x_2=\dfrac{2m-5}{2}\end{matrix}\right.\)
Thế vào \(x_1x_2=-12\Leftrightarrow\left(\dfrac{2m+9}{2}\right)\left(\dfrac{2m-5}{2}\right)=-12\)
\(\Leftrightarrow4m^2+8m+3=0\Rightarrow\left[{}\begin{matrix}m=-\dfrac{3}{2}\\m=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(m=\left\{-1;-\dfrac{3}{2};-\dfrac{1}{2}\right\}\)
\(x+12=31\cdot2\)
\(x+12=62\)
\(x=62-12\)
\(x=50\)
x + 12 = 31 x 2
x + 12 = 62
x = 62 - 12
x = 40
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