tìm x :
a. 11 + (15 – x)=1
b. 2I x+5 I=12
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17 tháng 10 2021
a: Ta có: \(A=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{15\sqrt{x}-11-\left(3x+7\sqrt{x}-6\right)-\left(2x+\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
CN
1
NH
20 tháng 6 2019
a,\(\frac{11}{12}-\left(\frac{5}{42}-x\right)=\frac{15}{28}-\frac{11}{12}\)
\(\Leftrightarrow\frac{11}{12}-\frac{5}{42}+x=\frac{15}{28}-\frac{11}{12}\)
\(\Leftrightarrow x=\frac{15}{28}-\frac{11}{12}-\frac{11}{12}+\frac{5}{42}\)
\(\Leftrightarrow x=\left(\frac{15}{28}+\frac{5}{42}\right)-\left(\frac{11}{12}+\frac{11}{12}\right)\)
\(\Leftrightarrow x=\frac{55}{84}-\frac{11}{6}\)
\(\Leftrightarrow x=\frac{-33}{28}\)
b, \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
7 tháng 3 2023
\(-\dfrac{11}{12}< \dfrac{5}{x}< -\dfrac{11}{15}\)
\(\Rightarrow\dfrac{-55}{60}< \dfrac{5}{x}< \dfrac{-44}{60}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{50}{60}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{5}{-6}\)
Vậy x = -6
PT
1
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
10 tháng 9 2023
\(a,\left(x.\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x.\dfrac{1}{2}=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{1}{3}:\dfrac{1}{2}=\dfrac{2}{3}\\ ---\\ b,\left(x+\dfrac{1}{2}\right)^2=\dfrac{4}{5}=\left(\dfrac{2}{\sqrt{5}}\right)^2=\left(-\dfrac{2}{\sqrt{5}}\right)^2 \\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{2}{\sqrt{5}}\\x+\dfrac{1}{2}=-\dfrac{2}{\sqrt{5}}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{\sqrt{5}}-\dfrac{1}{2}\\x=-\dfrac{2}{\sqrt{5}}-\dfrac{1}{2}\end{matrix}\right.\\ Vậy:x=\pm\dfrac{2}{\sqrt{5}}-\dfrac{1}{2}\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
10 tháng 9 2023
\(c,\left|3x-\dfrac{4}{5}\right|=\dfrac{11}{5}\\ \Rightarrow\left[{}\begin{matrix}3x-\dfrac{4}{5}=\dfrac{11}{5}\\3x-\dfrac{4}{5}=-\dfrac{11}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3x=\dfrac{11}{5}+\dfrac{4}{5}=3\\3x=-\dfrac{11}{5}+\dfrac{4}{5}=-\dfrac{7}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{3}=1\\x=-\dfrac{7}{5}:3=-\dfrac{7}{15}\end{matrix}\right.\\ ---\\ d,\left|2x-2\right|=0\\ \Leftrightarrow2x-2=0\\ \Leftrightarrow2x=2\\ \Leftrightarrow x=1\)
H9
HT.Phong (9A5)
CTVHS
21 tháng 11 2023
a) \(12+x+\left(-5\right)=-18-2x\)
\(\Rightarrow12+x-5=-18-2x\)
\(\Rightarrow x+7+18+2x=0\)
\(\Rightarrow3x=-25\)
\(\Rightarrow x=-\dfrac{25}{3}\)
b) \(\left(-14\right)-x+\left(-15\right)=-10+\left(4-2x\right)\)
\(\Rightarrow-14-x-15=-10+4-2x\)
\(\Rightarrow-x-29=-2x-6\)
\(\Rightarrow-x+2x=-6+29\)
\(\Rightarrow x=23\)
c) \(x-\left(-19\right)-\left(-11\right)=-\left(3x+40\right)\)
\(\Rightarrow x+19+11=-3x-40\)
\(\Rightarrow x+30=-3x-40\)
\(\Rightarrow x+3x=-40-30\)
\(\Rightarrow4x=-70\)
\(\Rightarrow x=-\dfrac{35}{2}\)
OK
3