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T
9 tháng 8 2023
a) \(\dfrac{49}{81}=\dfrac{7^x}{9^x}\)(sửa đề)
\(\Leftrightarrow\left(\dfrac{7}{9}\right)^2=\left(\dfrac{7}{9}\right)^x\)\(\Rightarrow x=2\)
b) \(\dfrac{-64}{343}=\left(-\dfrac{4^x}{7^x}\right)\)(sửa đề)
\(\Leftrightarrow\left(-\dfrac{4}{7}\right)^3=\left(-\dfrac{4}{7}\right)^x\) \(\Rightarrow x=3\)
c) \(\dfrac{9}{144}=\dfrac{3^x}{12^x}\)(sửa đề)
\(\Leftrightarrow\left(\dfrac{3}{12}\right)^2=\left(\dfrac{3}{12}\right)^x\Rightarrow x=2\)
d) \(-\dfrac{1}{32}=\left(-\dfrac{1^x}{2^x}\right)\)(sửa đề)
\(\Leftrightarrow\left(-\dfrac{1}{2}\right)^5=\left(-\dfrac{1}{2}\right)^x\Rightarrow x=5\)
Mong bạn xem lại đề bài.
VN
Võ Ngọc Phương
CTVHS
VIP
11 tháng 9 2023
Bài 4:
a) \(\dfrac{4}{3}+\left(1,25-x\right)=2,25\)
\(1,25-x=2,25-\dfrac{4}{3}=\dfrac{9}{4}-\dfrac{4}{3}\)
\(1,25-x=\dfrac{11}{12}\)
\(x=1,25-\dfrac{11}{12}=\dfrac{5}{4}-\dfrac{11}{12}\)
\(x=\dfrac{1}{3}\)
b) \(\dfrac{17}{6}-\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(x-\dfrac{7}{6}=\dfrac{17}{6}-\dfrac{7}{4}=\dfrac{34}{12}-\dfrac{21}{12}\)
\(x-\dfrac{7}{6}=\dfrac{13}{12}\)
\(x=\dfrac{13}{12}+\dfrac{7}{6}=\dfrac{13}{12}+\dfrac{14}{12}\)
\(x=\dfrac{27}{12}=\dfrac{9}{4}\)
c) \(4-\left(2x+1\right)=3-\dfrac{1}{3}=\dfrac{9}{3}-\dfrac{1}{3}\)
\(4-\left(2x+1\right)=\dfrac{8}{3}\)
\(2x+1=\dfrac{8}{3}+4=\dfrac{8}{3}+\dfrac{12}{3}\)
\(2x+1=\dfrac{20}{3}\)
\(2x=\dfrac{20}{3}-1=\dfrac{20}{3}-\dfrac{3}{3}\)
\(2x=\dfrac{17}{3}\)
\(x=\dfrac{17}{3}.\dfrac{1}{2}=\dfrac{17}{6}\)
Bài 15:
a) \(\left(\dfrac{-2}{3}\right)^9:x=\dfrac{-2}{3}\)
\(x=\left(\dfrac{-2}{3}\right)^9:\dfrac{-2}{3}=\left(\dfrac{-2}{3}\right)^{9-1}\)
\(=>x=\left(\dfrac{-2}{3}\right)^8\)
b) \(x:\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^4\)
\(x=\left(\dfrac{4}{9}\right)^4.\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^{4+5}\)
\(=>x=\left(\dfrac{4}{9}\right)^9\)
c) \(\left(x+4\right)^3=-125\)
\(\left(x+4\right)^3=\left(-5\right)^3\)
\(=>x+4=-5\)
\(x=-5-4\)
\(=>x=-9\)
d) \(\left(10-5x\right)^3=64\)
\(\left(10-5x\right)^3=4^3\)
\(=>10-5x=4\)
\(5x=10-4\)
\(5x=6\)
\(=>x=\dfrac{6}{5}\)
e) \(\left(4x+5\right)^2=81\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(4x+5\right)^2=\left(-9\right)^2\\\left(4x+5\right)^2=9^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+5=-9\\4x+5=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=-14\\4x=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-14}{4}\\x=1\end{matrix}\right.\)
Bài 16:
a) \(4-1\dfrac{2}{5}-\dfrac{8}{3}\)
\(=4-\dfrac{7}{5}-\dfrac{8}{3}\)
\(=\dfrac{60-21-40}{15}=\dfrac{-1}{15}\)
b) \(-0,6-\dfrac{-4}{9}-\dfrac{16}{15}\)
\(=\dfrac{-3}{5}+\dfrac{4}{9}-\dfrac{16}{15}\)
\(=\dfrac{\left(-27\right)+20-48}{45}=\dfrac{-55}{45}=\dfrac{-11}{9}\)
c) \(-\dfrac{15}{4}.\left(\dfrac{-7}{15}\right).\left(-2\dfrac{2}{5}\right)\)
\(=\dfrac{7}{4}.\dfrac{-12}{5}\)
\(=\dfrac{-21}{5}\)
\(#Wendy.Dang\)
K
0
3 tháng 9 2021
a) Ta có: \(\dfrac{x}{y}=\dfrac{10}{9}\Rightarrow\dfrac{x}{10}=\dfrac{y}{9}\)
\(\dfrac{y}{z}=\dfrac{3}{4}\Rightarrow\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{y}{9}=\dfrac{z}{12}\)
\(\Rightarrow\dfrac{x}{10}=\dfrac{y}{9}=\dfrac{z}{12}=\dfrac{x-y+z}{10-9+12}=\dfrac{78}{13}=6\)
\(\Rightarrow\left\{{}\begin{matrix}x=6.10=60\\y=6.9=54\\z=6.12=72\end{matrix}\right.\)
b)Ta có: \(\dfrac{x}{y}=\dfrac{9}{7}\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}\)
\(\dfrac{y}{z}=\dfrac{7}{3}\Rightarrow\dfrac{y}{7}=\dfrac{z}{3}\)
\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x-y+z}{9-7+3}=-\dfrac{15}{5}=-3\)
\(\Rightarrow\left\{{}\begin{matrix}x=-3.9=-27\\y=-3.7=-21\\z=-3.3=-9\end{matrix}\right.\)
c) \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{3}\)
\(\Rightarrow\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{z^2}{9}=\dfrac{x^2+y^2+z^2}{9+16+9}=\dfrac{200}{34}=\dfrac{100}{17}\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{900}{17}\\y^2=\dfrac{1600}{17}\\z^2=\dfrac{900}{17}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\pm\dfrac{30\sqrt{17}}{17}\\y=\pm\dfrac{40\sqrt{17}}{17}\\z=\pm\dfrac{30\sqrt{17}}{17}\end{matrix}\right.\)
Vậy\(\left(x;y;z\right)\in\left\{\left(\dfrac{30\sqrt{17}}{17};\dfrac{40\sqrt{17}}{17};\dfrac{30\sqrt{17}}{17}\right),\left(-\dfrac{30\sqrt{17}}{17};-\dfrac{40\sqrt{17}}{17};-\dfrac{30\sqrt{17}}{17}\right)\right\}\)
17 tháng 6 2023
1:
a: =7/5(40+1/4-25-1/4)-1/2021
=21-1/2021=42440/2021
b: =5/9*9-1*16/25=5-16/25=109/25
12 tháng 9 2021
a) \(\dfrac{1}{2}+\dfrac{2}{3}x=\dfrac{1}{4}\\ \Rightarrow\dfrac{2}{3}x=-\dfrac{1}{4}\\ \Rightarrow x=-\dfrac{3}{8}\)
b) \(2\dfrac{2}{3}:x=1\dfrac{7}{9}:0,02\\ \Rightarrow2\dfrac{2}{3}:x=\dfrac{800}{9}\\ \Rightarrow x=\dfrac{3}{100}\)
c) \(x^x-x+1=1\\ \Rightarrow x^x-x=0\\ \Rightarrow x^x=x\\ \Rightarrow x=1\)
d) \(5-\left|3x-1\right|=3\\ \Rightarrow\left|3x-1\right|=2\\ \Rightarrow\left[{}\begin{matrix}3x-1=-2\\3x-1=2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=1\end{matrix}\right.\)
HP
31 tháng 10 2021
1. Ta có: \(\dfrac{x}{-7}=\dfrac{y}{4}\Rightarrow\dfrac{2x}{-14}=\dfrac{3y}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta được:
\(\dfrac{2x-3y}{-14-12}=\dfrac{-78}{-26}=3\)
=> \(\left\{{}\begin{matrix}x=-21\\y=12\end{matrix}\right.\)
2. Ta có:
- \(\dfrac{x}{y}=\dfrac{9}{7}\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}\)
- \(\dfrac{y}{z}=\dfrac{7}{3}\Rightarrow\dfrac{y}{7}=\dfrac{z}{3}\)
=> \(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta được:
\(\dfrac{x-y+z}{9-7+3}=\dfrac{-15}{5}=-3\)
=> \(\left\{{}\begin{matrix}x=-27\\y=-21\\z=-9\end{matrix}\right.\)
MH
8 tháng 10 2021
a) \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)
⇔\(7\left(x-3\right)=5\left(x+5\right)\)
⇔\(7x-21=5x+25\)
⇔\(7x-21-5x-25=0\)
⇔\(2x-46=0\)
⇔\(2x=46\)
⇔\(x=23\)
\(\dfrac{7}{x-1}=\dfrac{x-1}{9}\)
\(\Rightarrow\left(x-1\right)\cdot\left(x-1\right)=7\cdot9\)
\(\Rightarrow\left(x-1\right)^2=63\)
\(\Rightarrow\left[{}\begin{matrix}x-1=\sqrt{63}\\x-1=-\sqrt{63}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\sqrt{7}+1\\x=-3\sqrt{7}+1\end{matrix}\right.\)
7/x-1 = x-1/9 ( ĐK : x khác 1 )
=> (x-1)^2 = 7.9 =63
=> x - 1 = ±\(\sqrt{63}\)
=> x = 1 ±\(\sqrt{63}\) ( tmdk)
Vậy x = 1+√63 hoặc x = 1-√63