tìm x
\(\frac{x+10}{8}+\frac{x+11}{7}=-2\)
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\(\frac{2-x}{x+3}=\frac{6}{5}\)
<=>\(5\left(2-x\right)=6\left(x+3\right)\)
<=>\(10-5x=6x+18\)
<=>\(\left(-5x\right)-6x=18-10\)
<=>\(-11x=8\)
<=>\(x=\frac{-8}{11}\)
a) |x - 1,7| = 2,3
Xét 2 trường hợp:
TH1: x - 1,7 = -2,3
x = -2,3 +1,7
x = -0,6
TH2: x - 1,7 = 2,3
x = 2,3 + 1,7
x = 4
Vậy: Tự kl :<
\(\left(\frac{1}{x}-\frac{2}{3}\right)^2-\frac{1}{16}=0\)
\(\Leftrightarrow\left(\frac{1}{x}-\frac{2}{3}\right)^2-\left(\frac{1}{4}\right)^2=0\)
\(\Leftrightarrow\left(\frac{1}{x}-\frac{2}{3}+\frac{1}{4}\right)\left(\frac{1}{x}-\frac{2}{3}-\frac{1}{4}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{x}-\frac{5}{12}\right)\left(\frac{1}{x}-\frac{11}{12}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{x}-\frac{5}{12}=0\\\frac{1}{x}-\frac{11}{12}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{x}=\frac{5}{12}\\\frac{1}{x}=\frac{11}{12}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{12}{11}\\x=\frac{12}{5}\end{cases}}\)
Vậy....
\(\left(\frac{1}{x}-\frac{2}{3}\right)^2-\frac{1}{16}=0\)
\(\Rightarrow\left(\frac{1}{x}-\frac{2}{3}\right)^2=\frac{1}{16}\)
\(\Rightarrow\left(\frac{1}{x}-\frac{2}{3}\right)^2=\left(\frac{1}{4}\right)^2\)
\(\Rightarrow\frac{1}{x}-\frac{2}{3}=\frac{1}{4}\)
\(\Rightarrow\frac{1}{x}=\frac{11}{12}\)
\(\Rightarrow x=\frac{11}{12}\)
X2-\(\frac{7}{9}\)X=0 <=> X(X-\(\frac{7}{9}\))=0
=> x=0 hoặc x-\(\frac{7}{9}\)=0
x-\(\frac{7}{9}\)=0 <=>X=0+\(\frac{7}{9}\)=\(\frac{7}{9}\)
=> X=0 hoặc \(\frac{7}{9}\)
B1:
a)x=-3/5*9/25 =>x=-27/125
b)x=(4/7)6:(4/7)4 =>x=(4/7)2=16/49
c)(x/4)2=4:(x/2)
(x/4)2=8/x
x2/16=8/x2
x3=128
x=5,039
B2
M=23.10+22.10/23.4+22.11
=230+220/212+222
=230+28+222
=28(222+1+214)
=2
Bài 1:
\(A=\left(\frac{-5}{11}+\frac{7}{22}-\frac{4}{33}-\frac{5}{44}\right):\left(38\frac{1}{122}-39\frac{7}{22}\right)\)
\(=\frac{-49}{132}:\left(-\frac{879}{671}\right)=\frac{2989}{105408}\)
Bài 2:
\(\frac{4}{5}-\left(\frac{-1}{8}\right)=\frac{7}{8}-x\)
<=> \(\frac{7}{8}-x=\frac{27}{40}\)
<=> \(x=\frac{7}{8}-\frac{27}{40}=\frac{1}{5}\)
Vậy...
X = 18 nha
ta có
\(\frac{x+1\text{0}}{8}+\frac{x+11}{7}=-2\)
\(\Leftrightarrow\frac{7\left(x+1\text{0}\right)}{7\cdot8}+\frac{8\left(x+11\right)}{8\cdot7}=-\frac{112}{56}\)
\(\Leftrightarrow\frac{7x+7\text{0}+8x+88}{56}=-\frac{112}{56}\)
\(\Leftrightarrow15x+158=-112\)
\(\Leftrightarrow15x=-27\text{0}\)
\(\Leftrightarrow x=-18\)
Vậy ...........
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