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\(\frac{x+2}{x}=\frac{1}{2}\)
\(\Rightarrow2.\left(x+2\right)=x\)
\(\Rightarrow2x+4=x\)
\(\Rightarrow2x-x=-4\)
\(\Rightarrow x=-4\)
\(b,\frac{x+3}{x+4}=\frac{3}{5}\)
\(\Rightarrow5.\left(x+3\right)=3.\left(x+4\right)\)
\(\Rightarrow5x+15=3x+12\)
\(\Rightarrow5x-3x=12-15\)
\(\Rightarrow2x=-3\)
\(\Rightarrow x=-\frac{3}{2}\)
\(\frac{x+5}{6}=\frac{6}{x+5}\)
\(\Rightarrow\left(x+5\right).\left(x+5\right)=6.6\)
\(\Rightarrow\left(x+5\right)^2=6^2\)
\(\Rightarrow\orbr{\begin{cases}x+5=6\\x+5=-6\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=-11\end{cases}}\)
Vậy x = 1 hoặc x= - 11
\(\frac{x+1}{3}=\frac{12}{x+1}\)
\(\Rightarrow\left(x+1\right).\left(x+1\right)=3.12\)
\(\Rightarrow\left(x+1\right)^2=36\)
\(\Rightarrow\left(x+1\right)^2=6^2\)
\(\Rightarrow\orbr{\begin{cases}x+1=6\\x+1=-6\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x=-7\end{cases}}\)

c) \(\dfrac{x+1}{35}+\dfrac{x+2}{34}+\dfrac{x+3}{33}=\dfrac{x+4}{32}+\dfrac{x+5}{31}+\dfrac{x+6}{30}\)
\(\Rightarrow\dfrac{x+1}{35}+1+\dfrac{x+2}{34}+1+\dfrac{x+3}{33}+1=\dfrac{x+4}{32}+1+\dfrac{x+5}{31}+1+\dfrac{x+6}{30}+1\)
\(\Rightarrow\dfrac{x+1+35}{35}+\dfrac{x+2+34}{34}+\dfrac{x+3+33}{33}=\dfrac{x+4+32}{32}+\dfrac{x+5+31}{31}+\dfrac{x+6+30}{30}\)
\(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}=\dfrac{x+36}{32}+\dfrac{x+36}{31}+\dfrac{x+36}{30}\)
\(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}-\dfrac{x+36}{32}-\dfrac{x+36}{31}-\dfrac{x+36}{30}=0\)
\(\Rightarrow\left(x+36\right)\left(\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\right)=0\)
\(\Rightarrow x+36=0\left(\text{vì }\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\ne0\right)\)
\(\Rightarrow x=-36\)
Vậy ...
a/ Ta có: \(-4\dfrac{3}{5}.2\dfrac{4}{3}\le x\le-2\dfrac{3}{5}:1\dfrac{6}{15}\)
\(\Rightarrow\dfrac{-23}{5}.\dfrac{10}{3}\le x\le\dfrac{-13}{5}:\dfrac{21}{15}\)
\(\Rightarrow\dfrac{-46}{3}\le x\le\dfrac{-13}{5}.\dfrac{15}{21}\)
\(\Rightarrow\dfrac{-46}{3}\le x\le\dfrac{-13}{7}\)
\(\Rightarrow-15,\left(3\right)\le x\le-1,\left(857142\right)\)
Vì x \(\in\) Z nên x \(\in\left\{-1;-2;-3;...;-15\right\}\)
Chúc bạn học tốt!!!

c)
\(\left|\dfrac{2}{3}x-\dfrac{1}{2}\right|-1=\dfrac{5}{6}\)
\(\Leftrightarrow\left|\dfrac{4x-3}{6}\right|=1+\dfrac{5}{6}=\dfrac{11}{6}\)
\(\Leftrightarrow\left|4x-3\right|=11\Leftrightarrow\left[{}\begin{matrix}4x-3=11=>x=\dfrac{11+3}{4}=\dfrac{7}{2}\\4x-3=-11=>x=\dfrac{-8}{2}=-4\end{matrix}\right.\)

\(\frac{3}{7}=\frac{y}{21}\Rightarrow y=\frac{3\times21}{7}=9\)
\(\frac{3}{x-5}=\frac{-4}{x+2}\)
\(\Leftrightarrow3.\left(x+2\right)=-4\left(x-5\right)\)
\(\Leftrightarrow3x+6=-4x+20\)
\(\Leftrightarrow3x+4x=20-6\)
\(\Leftrightarrow7x=14\)
\(\Leftrightarrow x=2\)

a, x : [(1800 + 600) : 30] = 560 : (315 - 35)
<=> x : (2400 : 30) = 560 : 280
<=> x : 80 = 2
<=> x = 160
b, x - 6 : 2 - (48 - 24) : 2 : 6 - 3 = 0
<=> x - 3 - 24 : 2 : 6 - 3 = 0
<=> x - 3 - 2 - 3 = 0
<=> x - 8 = 0
<=> x = 8
c, 390 - (x - 7) = 169 : 13
<=> 390 - (x - 7) = 13
<=> x - 7 = 377
<=> x = 384
d, (x - 140) : 7 = 33 - 23.3
<=> (x - 140) : 7 = 33 - 24
<=> (x - 140) : 7 = 3
<=> x - 140 = 21
<=> x = 161
@Đỗ Thị Huyền Trang
Thử \(x = 1\):
\(6^{1} + 2^{1} + 3^{1} = 6 + 2 + 3 = 11\)Thử \(x = 2\):
\(6^{2} + 2^{2} + 3^{2} = 36 + 4 + 9 = 49\)Thử \(x = 3\):
\(6^{3} + 2^{3} + 3^{3} = 216 + 8 + 27 = 251\)Thử \(x = 4\):
\(6^{4} + 2^{4} + 3^{4} = 1296 + 16 + 81 = 1393\)✅ Khớp!
✅ Đáp án là: \(\boxed{4}\)