3 mũ x nhân 3 mũ x +5= 3 mũ x + 12
Tính giúp mik ạ
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Gọi tổng số học sinh giỏi lớp 6;7;8 là x
Do x chia 3 dư 2, chia 4 dư 3, chia 53 dư 52
\(\Rightarrow x+1\) chia hết cho 3,4,53
\(\Rightarrow x+1\in BC\left(3;4;53\right)\)
Mà x nhỏ nhất
\(\Rightarrow x+1=BCNN\left(3;4;53\right)\)
\(\Rightarrow x+1=636\)
\(\Rightarrow x=635\)
a: \(\dfrac{7}{2}-\dfrac{3}{2}+5=5+2=7\)
b: \(\dfrac{2}{7}+\dfrac{1}{5}+\dfrac{5}{7}+\dfrac{4}{5}\)
\(=\left(\dfrac{2}{7}+\dfrac{5}{7}\right)+\left(\dfrac{1}{5}+\dfrac{4}{5}\right)\)
=1+1=2
c: \(\dfrac{7}{9}\cdot\dfrac{5}{15}+\dfrac{7}{9}\cdot\dfrac{10}{15}\)
\(=\dfrac{7}{9}\left(\dfrac{5}{15}+\dfrac{10}{15}\right)\)
\(=\dfrac{7}{9}\cdot\dfrac{15}{15}=\dfrac{7}{9}\)
d: \(\dfrac{13}{19}+2025+\dfrac{6}{19}+2024\)
\(=\left(\dfrac{13}{19}+\dfrac{6}{19}\right)+\left(2025+2024\right)\)
=4049+1=4050
e: \(\dfrac{7}{30}+\dfrac{12}{37}+\dfrac{23}{30}+\dfrac{25}{37}\)
\(=\left(\dfrac{7}{30}+\dfrac{23}{30}\right)+\left(\dfrac{12}{37}+\dfrac{25}{37}\right)\)
\(=1+1=2\)
g: \(\dfrac{5}{7}\cdot\dfrac{5}{11}+\dfrac{5}{7}\cdot\dfrac{2}{11}+\dfrac{5}{7}\cdot\dfrac{4}{11}\)
\(=\dfrac{5}{7}\left(\dfrac{5}{11}+\dfrac{2}{11}+\dfrac{4}{11}\right)\)
\(=\dfrac{5}{7}\cdot\dfrac{11}{11}=\dfrac{5}{7}\)
e: \(\dfrac{20}{23}+5+\dfrac{3}{23}+5\)
\(=\left(\dfrac{20}{23}+\dfrac{3}{23}\right)+5+5\)
=1+10=11
f: \(\dfrac{4}{3}+\dfrac{11}{31}+\dfrac{2}{3}+\dfrac{20}{31}\)
\(=\left(\dfrac{4}{3}+\dfrac{2}{3}\right)+\left(\dfrac{11}{31}+\dfrac{20}{31}\right)\)
=2+1=3
h: \(\dfrac{5}{7}\cdot\dfrac{3}{13}+\dfrac{5}{7}\cdot\dfrac{10}{13}\)
\(=\dfrac{5}{7}\left(\dfrac{3}{13}+\dfrac{10}{13}\right)\)
\(=\dfrac{5}{7}\cdot\dfrac{13}{13}=\dfrac{5}{7}\)
a; Giải:
Gọi phân số thứ nhất là \(\dfrac{a}{b}\) thì phân số thứ hai là:
\(\dfrac{2}{5}\) - \(\dfrac{a}{b}\)
Theo bài ra ta có: \(\dfrac{a}{b}\) - (\(\dfrac{2}{5}\) - \(\dfrac{a}{b}\)) = - \(\dfrac{4}{5}\)
\(\dfrac{a}{b}\) - \(\dfrac{2}{5}\) + \(\dfrac{a}{b}\) = - \(\dfrac{4}{5}\)
(\(\dfrac{a}{b}\) + \(\dfrac{a}{b}\)) = - \(\dfrac{4}{5}\) + \(\dfrac{2}{5}\)
2.\(\dfrac{a}{b}\) = - \(\dfrac{2}{5}\)
\(\dfrac{a}{b}\) = - \(\dfrac{2}{5}\) : 2
\(\dfrac{a}{b}\) = - \(\dfrac{1}{5}\)
Phân số thứ hai là: \(\dfrac{2}{5}\) - (- \(\dfrac{1}{5}\)) = \(\dfrac{3}{5}\)
Kết luận:...
b; Giải:
Gọi phân số thứ nhất là: \(\dfrac{a}{b}\)
Phân số thứ hai là: \(\dfrac{12}{5}\) - \(\dfrac{a}{b}\)
Theo bài ra ta có:
\(\dfrac{a}{b}\) : (\(\dfrac{12}{5}\) - \(\dfrac{a}{b}\)) = \(\dfrac{3}{7}\)
\(\dfrac{a}{b}\) = \(\dfrac{3}{7}\) x (\(\dfrac{12}{5}\) - \(\dfrac{a}{b}\))
\(\dfrac{a}{b}\) = \(\dfrac{36}{35}\) - \(\dfrac{3}{7}\) x \(\dfrac{a}{b}\)
\(\dfrac{a}{b}\) + \(\dfrac{3}{7}\) x \(\dfrac{a}{b}\) = \(\dfrac{36}{35}\)
\(\dfrac{a}{b}\) x (1 + \(\dfrac{3}{7}\)) = \(\dfrac{36}{35}\)
\(\dfrac{a}{b}\) x \(\dfrac{10}{7}\) = \(\dfrac{36}{35}\)
\(\dfrac{a}{b}\) = \(\dfrac{36}{35}\) : \(\dfrac{10}{7}\)
\(\dfrac{a}{b}\) = \(\dfrac{18}{25}\)
Phân số thứ hai là:
\(\dfrac{12}{5}\) - \(\dfrac{18}{25}\) = \(\dfrac{42}{25}\)
Kết luận:...
\frac{\sqrt{\left(6.2\right)^{2}-\left(5.9\right)^{2}}}{\sqrt{2.43}}
\(S=\sqrt{4+3\sqrt{4+3\sqrt{4+...}}}\)
\(S=\sqrt{4+3S}\)
\(S^2=4+3S\)
\(S^2-3S-4=0\)
\(\left(S+1\right)\left(S-4\right)=0\)
\(\Rightarrow S=4\) (do \(S>0\))
Bài 2:
a: \(\left|x+\dfrac{1}{5}\right|-\dfrac{1}{2}=\dfrac{9}{10}\)
=>\(\left|x+\dfrac{1}{5}\right|=\dfrac{1}{2}+\dfrac{9}{10}=\dfrac{14}{10}=\dfrac{7}{5}\)
=>\(\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{7}{5}\\x+\dfrac{1}{5}=-\dfrac{7}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=-\dfrac{8}{5}\end{matrix}\right.\)
b: \(\dfrac{5}{4}-3\left|2x+5\right|=\dfrac{3}{4}\)
=>\(3\left|2x+5\right|=\dfrac{5}{4}-\dfrac{3}{4}=\dfrac{2}{4}=\dfrac{1}{2}\)
=>\(\left|2x+5\right|=\dfrac{1}{6}\)
=>\(\left[{}\begin{matrix}2x+5=\dfrac{1}{6}\\2x+5=-\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{6}-5=-\dfrac{29}{6}\\2x=-\dfrac{1}{6}-5=-\dfrac{31}{6}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-\dfrac{29}{12}\\x=-\dfrac{31}{12}\end{matrix}\right.\)
c: \(\left(\dfrac{3}{5}x+\dfrac{1}{2}\right)^2=\dfrac{25}{16}\)
=>\(\left[{}\begin{matrix}\dfrac{3}{5}x+\dfrac{1}{2}=\dfrac{5}{4}\\\dfrac{3}{5}x+\dfrac{1}{2}=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{5}x=\dfrac{5}{4}-\dfrac{1}{2}=\dfrac{3}{4}\\\dfrac{3}{5}x=-\dfrac{5}{4}-\dfrac{1}{2}=-\dfrac{7}{4}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{3}{4}:\dfrac{3}{5}=\dfrac{5}{4}\\x=-\dfrac{7}{4}:\dfrac{3}{5}=-\dfrac{7}{4}\cdot\dfrac{5}{3}=-\dfrac{35}{12}\end{matrix}\right.\)
d: \(3-\left(2x+1\right)^2=2\)
=>\(\left(2x+1\right)^2=3-2=1\)
=>\(\left[{}\begin{matrix}2x+1=1\\2x+1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Bài 1:
a: \(\left(-\dfrac{2}{3}\right)^2\cdot\dfrac{9}{16}-\sqrt{\dfrac{4}{81}}:\dfrac{16}{9}+\left|-0,25\right|\)
\(=\dfrac{4}{9}\cdot\dfrac{9}{16}-\dfrac{2}{9}\cdot\dfrac{9}{16}+\dfrac{1}{4}\)
\(=\dfrac{4}{16}-\dfrac{2}{16}+\dfrac{1}{4}=\dfrac{1}{2}-\dfrac{1}{8}=\dfrac{3}{8}\)
b: \(\left(-2\right)^3+\dfrac{1}{2}:\dfrac{1}{8}-\sqrt{25}+\left|-8\right|\)
\(=-8+\dfrac{1}{2}\cdot8-5+8\)
=4-5=-1
c: \(\left(\dfrac{4}{3}-\dfrac{3}{2}\right)^2-2:\left|-\dfrac{1}{9}\right|+\dfrac{-5}{18}\)
\(=\left(\dfrac{8}{6}-\dfrac{9}{6}\right)^2-2:\dfrac{1}{9}-\dfrac{5}{18}\)
\(=\dfrac{1}{36}-18-\dfrac{5}{18}=\dfrac{1}{36}-\dfrac{10}{36}-18=-\dfrac{9}{36}-18\)
\(=-18-\dfrac{1}{4}=-18,25\)
d: \(\left(-\dfrac{3}{4}\right)^2:\left(-\dfrac{1}{4}\right)^2+9\left(\dfrac{1}{3}\right)^2+\left|-\dfrac{3}{2}\right|\)
\(=\left(-\dfrac{3}{4}:\dfrac{-1}{4}\right)^2+9\cdot\dfrac{1}{9}+\dfrac{3}{2}\)
\(=3^2+1+\dfrac{3}{2}=9+1+\dfrac{3}{2}=10+\dfrac{3}{2}=11,5\)
\(x^2-10x-11=0\)
=>\(x^2-10x+25-36=0\)
=>\(\left(x-5\right)^2-6^2=0\)
=>(x-5-6)(x-5+6)=0
=>(x-11)(x+1)=0
=>\(\left[{}\begin{matrix}x-11=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-1\end{matrix}\right.\)
\(x^2\)\(-2.x.5+5^2\)\(-36\)\(=0\)
\(\Leftrightarrow\)\(\left(x-5\right)^2\)\(-36=0\)
\(\Leftrightarrow\left(x-5^{ }\right)^2\)\(=36\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=6\\x-5=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-1\end{matrix}\right.\)
\(3^x.3^{x+5}=3^{x+12}\)
\(3^{x+x+5}=3^{x+12}\)
\(3^{2x+5}=3^{x+12}\)
\(2x+5=x+12\)
\(2x-x=12-5\)
\(x=7\)