50%/0,7=-2/2x+1
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\(a,=\dfrac{13}{50}\cdot\dfrac{50}{13}\cdot\left(-\dfrac{31}{2}\right)\cdot\dfrac{169}{2}=-\dfrac{5239}{2}\\ b,=\dfrac{-\dfrac{49}{100}\cdot\left(-125\right)}{-\dfrac{343}{27}\cdot\dfrac{81}{16}\cdot\left(-1\right)}=\dfrac{\dfrac{245}{4}}{\dfrac{1029}{16}}=\dfrac{245}{4}\cdot\dfrac{16}{1029}=\dfrac{20}{21}\)
a) \(\dfrac{13}{50}.\left(-15.5\right):\dfrac{13}{50}.84\dfrac{1}{2}=\dfrac{13}{50}.-75:\dfrac{13}{50}.\dfrac{169}{2}=-\dfrac{75.169}{2}=-\dfrac{12675}{2}\)
b) \(\dfrac{\left(-0,7\right)^2.\left(-5\right)^3}{\left(-2\dfrac{1}{3}\right)^3.\left(1\dfrac{1}{2}\right)^4.\left(-1\right)^5}=\dfrac{0,49.\left(-125\right)}{-\dfrac{343}{27}.\dfrac{81}{16}.\left(-1\right)}=-\dfrac{\dfrac{245}{4}}{\dfrac{1029}{16}}=\dfrac{20}{21}\)
a: Ta có: \(8x+11-3=5x+x-3\)
\(\Leftrightarrow8x+8=6x-3\)
\(\Leftrightarrow2x=-11\)
hay \(x=-\dfrac{11}{2}\)
b: Ta có: \(2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\)
\(\Leftrightarrow2x\left(x^3+6x^2+12x+8\right)-8x^2=2\left(x^3-8\right)\)
\(\Leftrightarrow2x^4+12x^3+24x^2+16x-8x^2-2x^3+16=0\)
\(\Leftrightarrow2x^4+10x^3+16x^2+16x+16=0\)
\(\Leftrightarrow2x^4+4x^3+6x^3+12x^2+4x^2+8x+8x+16=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x^3+6x^2+4x+8\right)=0\)
\(\Leftrightarrow x+2=0\)
hay x=-2
c: Ta có: \(\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\)
\(\Leftrightarrow2x^2-3x+2x-3-2x^2-10x+x+5=0\)
\(\Leftrightarrow-10x+2=0\)
\(\Leftrightarrow-10x=-2\)
hay \(x=\dfrac{1}{5}\)
d: Ta có: \(\dfrac{1}{10}-2\cdot\left(\dfrac{1}{2}t-\dfrac{1}{10}\right)=2\left(t-\dfrac{5}{2}\right)-\dfrac{7}{10}\)
\(\Leftrightarrow\dfrac{1}{10}-t+\dfrac{1}{5}=2t-5-\dfrac{7}{10}\)
\(\Leftrightarrow-t-2t=-\dfrac{57}{10}-\dfrac{3}{10}=-6\)
hay t=2
Ta có: \(\left\{{}\begin{matrix}80x+160y=20\\2x+6y=0,7\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}4x+8y=1\left(1\right)\\4x+12y=1,4\left(2\right)\end{matrix}\right.\)
Trừ (1) cho (2) vế theo vế ta được:
\(12y-8y=1,4-1\) <=> \(4y=0,4\)
<=> \(y=0,1\) => \(x=\dfrac{0,7-6.0,1}{2}\) = 0,05
\(2x-4=6\)
\(\Leftrightarrow2x=10\)
<=> x = 5
Vậy S = {5}
\(2,3-2\left(0,7+2x\right)=3,6-1,7\)
\(\Leftrightarrow2,3-1,4-4x-3,6+1,7=0\)
<=> -1 - 4x = 0
<=> -4x = 1
\(\Leftrightarrow x=-\dfrac{1}{4}\)
Vậy:..
chuyên hang tư x sang 1 bên ,sô sang 1 bên va lam nhu bt /cai nay dê ma
ủa bn cũng trl linh tinh mà?!
\(\frac{50%}{0,7}=\frac{-2}{2x+1}\)
\(\Rightarrow\frac{5}{7}=\frac{-2}{2x+1}\)
=> 5 . ( 2x + 1 ) = -2 . 7
=> 10x + 1 = -14
=> 10x = -15
=> x = -1,5
Vậy x = -1,5