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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 :<
a, |x - 1,7| = 2,3
=> x - 1,7 = 2,3 hoặc x - 1,7 = -2,3
=> x = 4 hoặc x = -0,6
câu b tương tự câu a
c, |x - 1| = 2x - 3
=> x - 1 = 2x - 3 hoặc x - 1 = 3 - 2x
=> x - 2x = -3 + 1 hoặc x + 2x = 3 + 1
=> -x = -2 hoặc 3x = 4
=> x = 2 hoặc x = 4/3
\(\left|x-3,2\right|+\left|2x-\frac{1}{5}\right|=x+3.\)
ĐK : \(x+3\ge0\Leftrightarrow x\ge-3\)
Th1 : \(x-3,2+2x-\frac{1}{5}=x+3\)
\(x-3,2+2x=x+\frac{16}{5}\)
\(x+2x=x+\frac{32}{5}\)
\(2x=\frac{32}{5}\)
\(\Leftrightarrow x=3,2\)(tm)
\(x-3,2+2x-\frac{1}{5}=3-x\)
\(x-3,2+2x=3-x+\frac{1}{5}\)
\(x-3,2+2x=\frac{16}{5}-x\)
\(x+2x=\frac{16}{5}-x+3,2\)
\(x+2x=\frac{32}{5}-x\)
\(2x=\frac{32}{5}-x-x\)
\(2x=\frac{32}{5}-2x\)
\(4x=\frac{32}{5}\)
\(x=1,6\)(tm)
Vậy \(x=1,6\)hoặc \(x=3,2\)
a)\(-\frac{2}{5}+\frac{2}{3}x+\frac{1}{6}x=-\frac{4}{5}\Leftrightarrow\frac{5}{6}x=-\frac{2}{5}\Leftrightarrow x=-\frac{12}{25}\)
Vậy nghiệm là x = -12/25
b)\(\frac{3}{2}x-\frac{2}{5}-\frac{2}{3}x=-\frac{4}{15}\Leftrightarrow\frac{5}{6}x=\frac{2}{15}\Leftrightarrow x=\frac{4}{25}\)
Vậy nghiệm là x = 4/25
c)\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\right)\)\(\Leftrightarrow x=-1\)
Vậy nghiệm là x = -1
x-3/-3 = -27/x-3
=> (x-3)(x-3)=(-3)(-27)
=> (x-3)^2 = 81=9^2
=> x-3=9 hoặc x-3=-9
=> x=12 hoặc x=-6
Ta có :\(\frac{x-3}{-3}=\frac{-27}{x-3}\)
\(\Rightarrow\left(x-3\right)\left(x-3\right)=\left(-3\right).\left(-27\right)\) ( Tính chất tỉ lệ thức )
\(\Rightarrow\left(x-3\right)^2=81\)
\(\Rightarrow\left(x-3\right)^2=\left(\pm9\right)^2\)
\(\Rightarrow x-3=\pm9\)
\(\Rightarrow x=-6;12\)
\(\frac{25}{5^x}=\frac{1}{125}\Rightarrow25.125=5^x.1\)
\(3125=5^x\)
\(5^5=5^x\)
\(\Rightarrow x=5\)
x+x+1+x+2+.........................+x+2003=2004
(x+x+x+...................+x)+(1+2+3+...................+2003)=2004
2004x+2007006=2004
2004x=2004:2007006=2/2003
x=2/2003:2004
a)\(\left(0,25^{10}\right).4^{10}.\sqrt{5^2-3^2}=\left(0,25.4\right)^{10}.\sqrt{25-9}=1^{10}.\sqrt{16}=1.4=4\)
b)\(\frac{\left(-3\right)^6.15^5+9^3.\left(-15\right)^6}{\left(-3\right)^{10}.5^5.2^3}=\frac{3^6.15^5+3^6.15^6}{3^{10}.5^5.2^3}=\frac{3^6.15^5.\left(1+15\right)}{3^{10}.5^5.2^3}\)\(=\frac{3^{11}.5^5.16}{3^{10}.5^5.2^3}=3.2=6\)
2)a)\(4-\left|x+\frac{2}{3}\right|=-1\Rightarrow\left|x+\frac{2}{3}\right|=5\Rightarrow\orbr{\begin{cases}x+\frac{2}{3}=5\\x+\frac{2}{3}=-5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{13}{3}\\x=\frac{-17}{3}\end{cases}}\)
b)\(\frac{x-2}{-9}=\frac{16}{2-x}\Rightarrow\left(x-2\right)^2=144\Rightarrow\orbr{\begin{cases}x-2=12\\x-2=-12\end{cases}\Rightarrow\orbr{\begin{cases}x=14\\x=-10\end{cases}}}\)
c)\(\frac{2}{3}x+\frac{1}{7}=\frac{5}{3}\Rightarrow\frac{2}{3}x=\frac{32}{21}\Rightarrow x=\frac{16}{7}\)
\(\frac{x+5}{3}=\frac{x-1}{4}\)
\(\Rightarrow\left(x+5\right).4=\left(x-1\right).3\)
\(\Rightarrow4x+20=3x-3\)
\(\Rightarrow4x-3x=-3-20\Rightarrow x=-23\)
\(\frac{x+5}{3}=\frac{x-1}{4}\)
\(\Rightarrow\left(x+5\right)\cdot4=\left(x-1\right)\cdot3\)
\(4x+20=3x-3\)
\(4x-3x=-3-20\)
\(x=-23\)
Vậy \(x=-23\)