Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\dfrac{x-2}{15}=\dfrac{9}{5}\)
\(\Leftrightarrow x-2=\dfrac{9\times15}{5}\)
\(\Leftrightarrow x-2=27\)
\(\Leftrightarrow x=29\)
Vậy ...............
b) \(\dfrac{2-x}{16}=\dfrac{-4}{x-2}\)
\(\Leftrightarrow\dfrac{2-x}{16}=\dfrac{4}{2-x}\)
\(\Leftrightarrow\left(2-x\right)^2=64\)
\(\Leftrightarrow\left(2-x\right)^2-64=0\)
\(\Leftrightarrow\left(2-x\right)^2-8^2=0\)
\(\Leftrightarrow\left(2-x-8\right)\left(2-x+8\right)=0\)
\(\Leftrightarrow\left(-x-6\right)\left(-x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x-6=0\\-x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=6\end{matrix}\right.\)
Vậy ..............
c) \(\dfrac{14}{x}=\dfrac{x-1}{4}\)
\(\Leftrightarrow x\left(x-1\right)=56\)
\(\Leftrightarrow x^2-x-56=0\)
\(\Leftrightarrow x^2+7x-8x-56=0\)
\(\Leftrightarrow x\left(x+7\right)-8\left(x+7\right)=0\)
\(\Leftrightarrow\left(x+7\right)\left(x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=8\end{matrix}\right.\)
Vậy ....................
\(a/\frac{7}{9}-\frac{x}{3}=\frac{1}{9}\)
\(\Rightarrow\frac{x}{3}=\frac{7}{9}-\frac{1}{9}\)
\(\Rightarrow\frac{x}{3}=\frac{2}{3}\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
\(b/\frac{1}{x}-\frac{-2}{15}=\frac{7}{15}\)
\(\Rightarrow\frac{1}{x}=\frac{7}{15}+\frac{-2}{15}\)
\(\Rightarrow\frac{1}{x}=\frac{1}{3}\)
\(\Rightarrow x=3\)
Vậy \(x=3\)
\(c/\frac{-11}{14}-\frac{-4}{x}=\frac{-3}{14}\)
\(\Rightarrow\frac{-4}{x}=\frac{-11}{14}-\frac{-3}{14}\)
\(\Rightarrow\frac{-4}{x}=\frac{-4}{7}\)
\(\Rightarrow x=7\)
Vậy \(x=7\)
\(d/\frac{x}{21}-\frac{2}{3}=\frac{5}{21}\)
\(\Rightarrow\frac{x}{21}=\frac{5}{21}+\frac{2}{3}\)
\(\Rightarrow\frac{x}{21}=\frac{19}{21}\)
\(\Rightarrow x=19\)
Vậy \(x=19\)
#Mạt Mạt#
B1
a) \(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):16\frac{2}{3}=0\)
\(1-\left(\frac{43}{8}+x-\frac{173}{24}\right):\frac{50}{3}=0\)
\(1-\left(x-\frac{11}{6}\right).\frac{3}{50}=0\)
\(\left(x-\frac{11}{6}\right).\frac{3}{50}=1-0\)
\(\left(x-\frac{11}{6}\right).\frac{3}{50}=1\)
\(x-\frac{11}{6}=1:\frac{3}{50}\)
\(x-\frac{11}{6}=\frac{50}{3}\)
\(x=\frac{50}{3}+\frac{11}{6}\)
\(x=\frac{37}{2}\)
b) \(\frac{3}{5}+\frac{5}{7}:x=\frac{1}{3}\)
\(\frac{5}{7}:x=\frac{1}{3}-\frac{3}{5}\)
\(\frac{5}{7}:x=-\frac{4}{15}\)
\(x=\frac{5}{7}:\left(-\frac{4}{15}\right)\)
\(x=-\frac{75}{28}\)
c) \(\left(4\frac{1}{2}-\frac{2}{5}.x\right):\frac{7}{4}=\frac{11}{9}\)
\(\left(\frac{9}{2}-\frac{2}{5}.x\right):\frac{7}{4}=\frac{11}{9}\)
\(\frac{9}{2}-\frac{2}{5}.x=\frac{11}{9}.\frac{7}{4}\)
\(\frac{9}{2}-\frac{2}{5}.x=\frac{11}{2}\)
\(\frac{2}{5}.x=\frac{9}{2}-\frac{11}{2}\)
\(\frac{2}{5}.x=-1\)
\(x=-1:\frac{2}{5}\)
\(x=-\frac{5}{2}\)
B2
a) \(\left(\frac{1}{2}+\frac{1}{3}+\frac{2}{6}\right).24:5-\frac{9}{22}:\frac{15}{121}\)
\(=\left(\frac{3}{6}+\frac{2}{6}+\frac{2}{6}\right).24:5-\frac{9}{22}.\frac{121}{15}\)
\(=\frac{7}{6}.24:5-\frac{33}{10}\)
\(=28:5-\frac{33}{10}\)
\(=\frac{28}{5}-\frac{33}{10}\)
\(=\frac{56}{10}-\frac{33}{10}\)
\(=\frac{23}{10}\)
b) \(\frac{5}{14}+\frac{18}{35}+\left(1\frac{1}{4}-\frac{5}{4}\right):\left(\frac{5}{12}\right)^2\)
\(=\frac{25}{70}+\frac{36}{70}+\left(\frac{5}{4}-\frac{5}{4}\right):\frac{25}{144}\)
\(=\frac{61}{70}+0:\frac{25}{144}\)
\(=\frac{61}{70}+0\)
\(=\frac{61}{70}\)
\(a)x+30\%x=-1,31\)
\(\Leftrightarrow x+\frac{3x}{10}=-1,31\)
\(\Leftrightarrow10x+3x=-13,1\)
\(\Leftrightarrow13x=-13,1\Leftrightarrow x=-\frac{131}{130}\)
\(b)\left(x-\frac{1}{2}\right):\frac{1}{3}+\frac{5}{7}=9\frac{5}{7}\)
\(\Leftrightarrow\frac{2x-1}{2}.3+\frac{5}{7}=\frac{68}{7}\)
\(\Leftrightarrow\frac{6x-3}{2}=\frac{63}{7}\)
\(\Leftrightarrow\frac{6x-3}{2}=9\)
\(\Leftrightarrow6x-3=18\)
\(\Leftrightarrow x=\frac{7}{2}\)
a)
Ta có
\(\frac{x}{2}=\frac{y}{5}\Rightarrow\frac{3x}{6}=\frac{y}{5}\)
Áp dụng tc của dãy tỉ só bằng nhau
\(\Rightarrow\frac{3x}{6}=\frac{y}{5}=\frac{3x-y}{6-5}=\frac{10}{1}=10\)
=> x=2.10=20
y=5.10=50
Ta có
\(\frac{x}{2}=\frac{y}{5}\Rightarrow\frac{x^2}{4}=\frac{y^2}{25}=\frac{xy}{10}=\frac{30}{10}=3\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\sqrt{12}\\x=-\sqrt{12}\end{array}\right.\)
\(\left[\begin{array}{nghiempt}y=\sqrt{75}\\y=-\sqrt{75}\end{array}\right.\)
Mà 2;5 cùng dấu
=> x; y cùng dấu
Vậy \(\left(x;y\right)=\left(\sqrt{12};\sqrt{75}\right);\left(-\sqrt{12};-\sqrt{75}\right)\)
a) Ta có: \(\frac{x-2}{15}=\frac{9}{5}\Rightarrow5.\left(x-2\right)=9.15\)
\(\Rightarrow5x-10=135\)
\(\Rightarrow5x=145\)
\(\Rightarrow x=29\)
b) \(\frac{2-x}{16}=\frac{-4}{x-2}\Rightarrow\left(2-x\right).\left(x-2\right)=\left(-4\right).16\)
\(\Rightarrow4x-x^2-4=-64\)
\(\Rightarrow4x-x^2=-60\)
Lập bảng rồi tính ra
c) \(\frac{14}{x}=\frac{x-1}{4}\Rightarrow x.\left(x-1\right)=14.4\)
\(\Rightarrow x.\left(x-1\right)=56\)
Vì 56 = 8 x 7
\(\Rightarrow x=8\)
Câu c) còn có thêm \(x=-7\) nữa nha MMS_Hồ Khánh Châu vì \(x\inℤ\) mà :')