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NT
1
TA
11 tháng 10 2017
\(\left(\frac{1}{7}x-\frac{2}{7}\right)\left(\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
<=> \(\frac{x-2}{7}.\frac{x+3}{5}.\frac{x+4}{3}=0\)
<=> \(\frac{x-2}{7}=0\)hoặc \(\frac{x+3}{5}=0\); \(\frac{x+4}{3}=0\)
Nếu \(\frac{x-2}{7}=0\)<=> \(x-2=0\)<=> \(x=2\)
Nếu \(\frac{x+3}{5}=0\)<=> \(x+3=0\) <=> \(x=3\)
Nếu \(\frac{x+4}{3}=0\)<=> \(x+4=0\)<=> \(x=4\)
Vây x= 2 hoặc 3; 4
TN
4 tháng 10 2017
a) 1/7 - 3/5x = 3/5
3/5x= 1/7 - 3/5
3/5x = -16/35
x= -16/35 : 3/5 = -16/21
b) 3/7 - 1/2x = 5/3
1/2x = 3/7 - 5/3 = -26/21
x= -26/21 : 1/2 = -52/21
NC
7 tháng 8 2018
a) \(\left(x+5\right)^3=64\)
\(\Leftrightarrow\left(x+5\right)^3=4^3\)
\(\Leftrightarrow x+5=4\)
\(\Leftrightarrow x=-1\)
Vậy x = - 1
b) \(x:\left(-\frac{3}{5}\right)^2=-\frac{3}{5}\)
\(\Leftrightarrow x=\left(-\frac{3}{5}\right)^2.\left(-\frac{3}{5}\right)\)
\(\Leftrightarrow x=\left(-\frac{3}{5}\right)^3\)
\(\Leftrightarrow x=-0,216\)
Vậy x = - 0, 216
c) \(\left(\frac{4}{7}\right)^4.x=\left(\frac{4}{7}\right)^6\)
\(\Leftrightarrow x=\left(\frac{4}{7}\right)^6:\left(\frac{4}{7}\right)^4\)
\(\Leftrightarrow x=\left(\frac{4}{7}\right)^2\)
\(\Leftrightarrow\text{x}=\frac{16}{49}\)
Vậy x = 16/49
d) \(\left(-\frac{1}{3}\right)^3x=\frac{1}{81}\)
\(\Leftrightarrow-\frac{1}{27}x=\frac{1}{81}\)
\(\Leftrightarrow x=\frac{1}{81}:\left(-\frac{1}{27}\right)\)
\(\Leftrightarrow x=-\frac{1}{3}\)
Vậy x = - 1/3
12 tháng 7 2021
Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
12 tháng 6 2018
a) \(\left(\frac{1}{7}x-\frac{2}{3}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{1}{7}x-\frac{2}{3}=0\\-\frac{1}{5}x+\frac{3}{5}=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\frac{1}{7}x=\frac{2}{3}\\-\frac{1}{5}x=-\frac{3}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{14}{3}\\x=3\end{cases}}\)
b)\(\frac{1}{10}x-\frac{4}{5}x+1=0\)
\(\Leftrightarrow x.\left(\frac{1}{10}-\frac{4}{5}\right)+1=0\)
\(\Rightarrow-\frac{7}{10}x=-1\)
\(\Rightarrow x=\frac{10}{7}\)
c)\(\left(2x-\frac{1}{3}\right).\left(5x+\frac{2}{7}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{3}=0\\5x+\frac{2}{7}=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=\frac{1}{3}\\5x=-\frac{2}{7}\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{6}\\x=-\frac{2}{35}\end{cases}}\)
A
12 tháng 6 2018
a, (1/7 . x - 2/3) . (-1/5 . x + 3/5) = 0
Suy ra : 1/7 .x -2/3 = 0 hoặc -1/5 .x + 3/5 =0
Vậy : 1/7 .x = 2/3 hoặc -1/5 .x = 3/5
x =2/3 : 1/7 hoặc x = 3/5 : (-1/5)
x = 14/3 hoặc x = -3
b, 1/10 .x - 4/5 .x + 1 =0
x . (1/10 - 4/5) + 1 = 0
x . (-7/10) + 1 = 0
x . -7/10 =0 +1 = 1
x = 1 : (-7/10)
x = -10/7
c, (2x - 1/3 ) . (5x +2/7) = 0
Suy ra : 2x - 1/3 = 0 hoặc 5x + 2/7 = 0
Vậy : 2x = 1/3 hoặc 5x = 2/7
x = 1/3 : 2 hoặc x = 2/7 : 5
x = 1/6 hoặc x = 2/35
2 tháng 8 2020
a ) \(\frac{3}{7}-\left(\frac{2}{5}+x+\frac{3}{2}\right)=\frac{5}{14}-\left|\frac{4}{35}-\frac{\left(-11\right)}{70}\right|\)
=> \(\frac{3}{7}-\left(\frac{2}{5}+x+\frac{3}{2}\right)=\frac{5}{14}-\left|\frac{4}{35}+\frac{11}{70}\right|\)
=> \(\frac{3}{7}-\left(\frac{2}{5}+x+\frac{3}{2}\right)=\frac{5}{14}-\left|\frac{19}{70}\right|\)
=> \(\frac{3}{7}-\left(\frac{2}{5}+x+\frac{3}{2}\right)=\frac{5}{14}-\frac{19}{70}=\frac{3}{35}\)
=> \(\frac{2}{5}+x+\frac{3}{2}=\frac{3}{7}-\frac{3}{35}=\frac{12}{35}\)
=> \(\frac{2}{5}+x=\frac{12}{35}-\frac{3}{2}=-\frac{81}{70}\)
=> \(x=-\frac{81}{70}-\frac{2}{5}=-\frac{109}{70}\)
b) \(\frac{3}{4}\left(x-8\right)=\frac{5}{7}\left(4-\frac{1}{2}\right)\)
=> \(\frac{3}{4}x-6=\frac{5}{2}\)
=> \(\frac{3}{4}x=\frac{17}{2}\)
=> \(x=\frac{17}{2}:\frac{3}{4}=\frac{34}{3}\)
Câu c,d tự làm nhé
KN
2 tháng 8 2020
a. \(\frac{3}{7}-\left(\frac{2}{5}+x+\frac{3}{2}\right)=\frac{5}{14}-\left|\frac{4}{35}-\frac{-11}{70}\right|\)
\(\Rightarrow\frac{3}{7}-\left(\frac{19}{10}+x\right)=\frac{5}{14}-\left|\frac{4}{35}+\frac{11}{70}\right|\)
\(\Rightarrow\frac{3}{7}-\frac{19}{10}-x=\frac{5}{14}-\left|\frac{19}{70}\right|=\frac{5}{14}-\frac{19}{70}\)
\(\Rightarrow-\frac{103}{70}-x=\frac{3}{35}\)
\(\Rightarrow x=-\frac{103}{70}-\frac{3}{35}\)
\(\Rightarrow x=-\frac{109}{70}\)
b. \(\frac{3}{4}\left(x-8\right)=\frac{5}{7}\left(4-\frac{1}{2}\right)\)
\(\Rightarrow\frac{3}{4}\left(x-8\right)=\frac{5}{7}.\frac{7}{2}=\frac{5}{2}\)
\(\Rightarrow x-8=\frac{10}{3}\)
\(\Rightarrow x=\frac{34}{3}\)
c. \(\frac{3}{2}-4\left(\frac{1}{4}-x\right)=\frac{2}{3}-7x\)
\(\Rightarrow\frac{3}{2}-1+4x=\frac{2}{3}-7x\)
\(\Rightarrow\frac{1}{2}=\frac{2}{3}-7x-4x=\frac{2}{3}-11x\)
\(\Rightarrow11x=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}\)
\(\Rightarrow x=\frac{1}{66}\)
d. \(4\left(\frac{1}{2}-x\right)-5\left(x-\frac{3}{10}\right)=\frac{7}{4}\)
\(\Rightarrow2-4x-5x+\frac{3}{2}=\frac{7}{4}\)
\(\Rightarrow2-9x=\frac{1}{4}\)
\(\Rightarrow9x=\frac{7}{4}\)
\(\Rightarrow x=\frac{7}{36}\)
MA
26 tháng 9 2016
a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
\(\Rightarrow\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=-4x+1\end{cases}}\Rightarrow\orbr{\begin{cases}4x-\frac{3}{2}x-1=\frac{1}{2}\\-4x-\frac{3}{2}x+1=\frac{1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{5}{2}x=\frac{3}{2}\\-\frac{11}{2}x=-\frac{1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
26 tháng 9 2016
phần b ở đề bài mình ghi sai, là bằng 0 chứ ko phải bằng 10