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NT
Tìm x biết: \(\frac{x+1}{9}+\frac{x+4}{6}+\frac{x+5}{5}=\frac{x+2}{8}+\frac{x+3}{7}+\frac{x+6}{4}.\)
4
13 tháng 8 2017
\(\frac{x+1}{9}+\frac{x+4}{6}+\frac{x+5}{5}=\frac{x+2}{8}+\frac{x+3}{7}+\frac{x+6}{4}\)
\(\Rightarrow\frac{x+1}{9}+\frac{x+4}{6}+\frac{x+5}{5}+3=\frac{x+2}{8}+\frac{x+3}{7}+\frac{x+6}{4}+3\)
\(\Rightarrow\left(\frac{x+1}{9}+1\right)+\left(\frac{x+4}{6}+1\right)+\left(\frac{x+5}{5}+1\right)=\left(\frac{x+2}{8}+1\right)\)\(+\left(\frac{x+3}{7}+1\right)+\left(\frac{x+6}{4}\right)\)
\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{6}+\frac{x+10}{5}=\frac{x+10}{8}+\frac{x+10}{7}+\frac{x+10}{4}\)
\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{6}+\frac{1}{5}\right)=\left(x+10\right)\left(\frac{1}{8}+\frac{1}{7}+\frac{1}{4}\right)\)
\(\Rightarrow\left(x+10\right)\frac{43}{90}=\left(x+10\right)\frac{29}{56}\)
\(\Rightarrow x+10=0\)
\(\Rightarrow x=-10\)
S
13 tháng 8 2017
cộng 3 vào cả hai vế nên phương trình vẫn bằng nhau
Ta có \(\frac{x+1}{9}+1+\frac{x+4}{6}+1+\frac{x+5}{5}+1=\frac{x+2}{8}+1+\frac{x+3}{7}+1+\frac{x+6}{4}+1\)
\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{6}+\frac{x+10}{5}=\frac{x+10}{8}+\frac{x+10}{7}+\frac{x+10}{4}\)
\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{6}+\frac{x+10}{5}-\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{4}=0\)
\(\Leftrightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{6}+\frac{1}{5}-\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)
mà \(\frac{1}{9}+\frac{1}{6}+\frac{1}{5}-\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)
\(\Rightarrow x+10=0\)
\(\Leftrightarrow x=-10\)
MY
0
NT
7
5 tháng 1 2019
\(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\)
\(\left(x-\frac{2}{9}\right)^3=\frac{64}{729}\)
\(\Rightarrow x-\frac{2}{9}=\frac{4}{9}\Leftrightarrow x=\frac{6}{9}=\frac{2}{3}\)
Chúc em học tốt!!!
5 tháng 1 2019
\(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\)
\(\left(x-\frac{2}{9}\right)^3=\left(\left(\frac{2}{3}\right)^2\right)^3\)
\(\left(x-\frac{2}{9}\right)^3=\left(\frac{4}{9}\right)^3\)
\(\Rightarrow x-\frac{2}{9}=\frac{4}{9}\)
\(\Rightarrow x=\frac{4}{9}+\frac{2}{9}\)
\(\Rightarrow x=\frac{2}{3}\)
Vậy \(x=\frac{2}{3}\)
G
3 tháng 3 2018
Câu 2 đây:
\(|x^2+|x-1||=x^2+2\)
\(\Rightarrow\orbr{\begin{cases}x^2+\left|x-1\right|=x^2+2\\x^2+\left|x-1\right|=-x^2-2\left(l\right)\end{cases}}\)
\(\Rightarrow\left|x-1\right|=2\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
KT
3 tháng 3 2018
a) \(M=\left(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+0,5}{1\frac{1}{6}-0,875+0,7}\right):\frac{2012}{2013}\)
\(=\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{2}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\right):\frac{2012}{2013}\)
\(=\left(\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{2\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}{7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\right):\frac{2012}{2013}\)
\(=\left(\frac{2}{7}-\frac{2}{7}\right):\frac{2012}{2013}\)
\(=0\)
8 tháng 7 2018
/X/ = \(\frac{10}{3}\)
suy ra: X = \(\frac{10}{3}\)hay X = \(\frac{-10}{3}\)
/X - \(\frac{5}{6}\)/ = \(\frac{1}{2}\)
suy ra : X - \(\frac{5}{6}\)= \(\frac{1}{2}\)hay X - \(\frac{5}{6}\)= \(\frac{-1}{2}\)
X = \(\frac{1}{2}+\frac{5}{6}\) hay X = \(\frac{-1}{2}+\frac{5}{6}\)
X = \(\frac{4}{3}\) hay X = \(\frac{1}{3}\)
/ X + \(\frac{4}{9}\)/ - \(\frac{1}{2}\)= \(\frac{3}{2}\)
/ X + \(\frac{4}{9}\)/ = \(\frac{3}{2}+\frac{1}{2}\)
/ X + \(\frac{4}{9}\)/ = \(2\)
suy ra : X + \(\frac{4}{9}\)= \(2\)hay X + \(\frac{4}{9}\)= \(-2\)
X = \(2-\frac{4}{9}\)hay X = \(\left(-2\right)-\frac{4}{9}\)
X = \(\frac{14}{9}\) hay X = \(\frac{-22}{9}\)
8 tháng 7 2018
a. |x| = \(10/3\)
=> x = 10/3 hoặc x = -10/3
b/ |x - 5/6| = 1/2
*TH1: x - 5/6 = 1/2
=> x = 1/2 + 5/6
=> x = 4/3
*TH2: -x + 5/6 = 1/2
=> -x = 1/2 - 5/6
=> -x = -1/3
=> x = 1/3
Vậy...
Phần c mk không chắc nên để lại ạ
C
24 tháng 9 2019
a) Đặt \(x-1=a\)
\(pt\Leftrightarrow\frac{13}{a}+\frac{5}{2a}=\frac{6}{3a}\)
\(\Leftrightarrow\frac{31}{2a}=\frac{6}{3a}\)
\(\Leftrightarrow\frac{31}{2}=2\)(vô lí)
Vậy pt vô nghiệm
24 tháng 9 2019
a) \(\frac{13}{x-1}+\frac{5}{2x-2}=\frac{6}{3x-3}\)
\(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}=\frac{6}{3\left(x-1\right)}\)
\(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}=\frac{2}{x-1}\)
\(\frac{31}{2\left(x-1\right)}=\frac{2}{x-1}\)
\(\frac{31}{2}=2\)
=> không có x thỏa mãn đề bài.
b) \(\frac{1}{x-1}+\frac{-2}{3}\left(\frac{3}{4}-\frac{6}{5}\right)=\frac{5}{2-2x}\)
\(\frac{1}{x-1}+\frac{-2}{3}.\frac{-9}{20}=\frac{5}{2\left(1-x\right)}\)
\(\frac{1}{x-1}-\frac{-18}{60}=\frac{5}{2\left(1-x\right)}\)
\(\frac{1}{x-1}+\frac{3}{10}=\frac{5}{2\left(1-x\right)}\)
\(10\left(1-x\right)+3\left(x-1\right)\left(1-x\right)=25\left(x-1\right)\)
\(7-4x-3x^2=25x-25\)
\(7-4x-3x^2-25x+25=0\)
\(32-29x-3x^2=0\)
\(3x^2+29x-30=0\)
\(3x^2+32x-3x-32=0\)
\(x\left(3x+32\right)-\left(3x+32\right)=0\)
\(\left(3x+32\right)\left(x-1\right)=0\)
\(\orbr{\begin{cases}3x+32=0\\x-1=0\end{cases}}\)
\(\orbr{\begin{cases}x=-\frac{32}{3}\\x=1\end{cases}}\)
Ta có: 3x . 2 = 6 . 9
3x . 2 = 54
3x = 54 : 2
3x = 27
3x = 33
=> x = 3