
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.


Câu 1:
Ta có: \(\dfrac{x-4}{y-3}=\dfrac{4}{3}\)
=> \(3.\left(x-4\right)=4.\left(y-3\right)\)
=>\(3x-12=4y-12\)
=>\(3x=4y\) (1)
Ta có: \(x-y=5\)
=> \(y=y+5\) Thay vào (1) ta có:
\(3.\left(y+5\right)=4.\)y
=>\(3y+15=4y\)
=> \(15=4y-3y\)
=> 15 = y
=> y =15
ta có: x = y +5
=> x = 15 +5
=> x =20
Câu 2:
\(B=\dfrac{10}{56}+\dfrac{10}{140}+\dfrac{10}{260}+...+\dfrac{10}{1400}\)
\(B=\dfrac{5}{28}+\dfrac{6}{70}+\dfrac{5}{130}+...+\dfrac{5}{700}\)
\(B=\dfrac{5}{4.7}+\dfrac{5}{7.10}+\dfrac{5}{10.13}+...+\dfrac{5}{25.28}\)
\(B=5,\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)
\(3B=5.\left(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{25.28}\right)\)
\(3B=5.\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)
\(3B=5.\left(\dfrac{1}{4}-\dfrac{1}{28}\right)\)
\(3B=5.\dfrac{3}{14}\)
\(B=\dfrac{15}{14}:3=\dfrac{5}{14}\)
Câu 3:
38 - (|x+10|+13) = \(\left(-6\right)^{20}:\left(9^9.4^{10}\right)\)
=> \(38-\left(\left|x+10\right|+13\right)=\left(2.3\right)_{ }^{20}:\)\(\left[\left(3^2\right)^9.\left(2^2\right)^4\right]\)
=>\(38-\left(\left|x+10\right|+13\right)=2^{20}.3^{20}:\left(3^{18}.2^{20}\right)\)
=> \(38-\left(\left|x+10\right|+13\right)=\dfrac{3^{20}.2^{20}}{3^{18}.2^{20}}\)
=> \(38-\left(\left|x+10\right|+13\right)=9\)
=> |x +10| + 13 = 38 -9
=> |x+10| +13 = 29
=> |x+10| = 29 -13
=> |x+10| = 16
\(\Rightarrow\left[{}\begin{matrix}x+10=16\\x+10=-16\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-26\end{matrix}\right.\)

a,|x|+3=5
\(\Leftrightarrow\left|x\right|=5-3=2\)
\(\Rightarrow x=\left\{{}\begin{matrix}2\\-2\end{matrix}\right.\)
b,|x+3|=5
\(\Rightarrow\left\{{}\begin{matrix}x+3=5\\x+3=-5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2\\-8\end{matrix}\right.\)
c,|x-7|+13=25
<=>|x-7|=25-13=12
\(\Rightarrow\left\{{}\begin{matrix}x-7=12\\x-7=-12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=19\\x=-5\end{matrix}\right.\)
d,\(26-\left|x+9\right|=-13\)
\(\Leftrightarrow\left|x+9\right|=26-\left(-13\right)=39\)
\(\Rightarrow\left\{{}\begin{matrix}x+9=39\\x+9=-39\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=30\\x=-48\end{matrix}\right.\)
e,8-|x|=15
<=>|x|=8-15=-7
\(\Rightarrow\left\{{}\begin{matrix}x=-7\\x=7\end{matrix}\right.\)
f,6-|-3+x|=-15
\(\Leftrightarrow\left|-3+x\right|=6-\left(-15\right)=21\)
\(\Rightarrow\left\{{}\begin{matrix}-3+x=21\\-3+x=-21\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=24\\x=-18\end{matrix}\right.\)
a, |x| + 3 = 5
|x| = 5 - 3
|x| = 2
|x| = 2 hoặc |x| = -2
Vậy |x| thuộc {2; -2}
b,|x + 3| = 5
|x + 3| = 5 hoặc |x + 3| = -5
x = 5 - 3 x = (-5) - 3
x = 2 x = -8
Vậy x thuộc {2; -8}
c,|x - 7| + 13 = 25
|x - 7| = 25 - 13
|x - 7| = 12
|x - 7| = 12 hoặc |x - 7| = -12
x = 12 + 7 x = (-12) + 7
x = 19 x = -5
Vậy x thuộc {19 ; -5}

a) Ta có:+) \(\frac{12}{16}=\frac{-x}{4}\) <=> 12.4 = 16.(-x)
<=> 48 = -16x
<=> x = 48 : (-16) = -3
+) \(\frac{12}{16}=\frac{21}{y}\) <=> 12y = 21.16
<=> 12y = 336
<=> y = 336 : 12 = 28
+) \(\frac{12}{16}=\frac{z}{-80}\) <=> 12. (-80) = 16z
<=> -960 = 16z
<=> z = -960 : 16 = -60
b) Ta có: \(\frac{x+3}{7+y}=\frac{3}{7}\) <=> (x + 3).7 = 3(7 + y)
<=> 7x + 21 = 21 + 3y
<=> 7x = 3y
<=> \(\frac{x}{3}=\frac{y}{7}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{3}=\frac{y}{7}=\frac{x+y}{3+7}=\frac{20}{10}=2\)
=> \(\hept{\begin{cases}\frac{x}{3}=2\\\frac{y}{7}=2\end{cases}}\) => \(\hept{\begin{cases}x=2.3=6\\y=2.7=14\end{cases}}\)
Vậy ...

ko chép lại đầu bài nha
=(53x+800):7020=(49x+147):6000
=(53x+800).6000=70
x=145,0369515
lưu ý:mik làm theo cách suy luận nhé @@@

a) x - 13+x = -24-7 + x
2x-13 = -31 +x
2x-x = -31+13
x = -18
b) 12.20 = 15x
240 = 15x
x = 16
c) 42(-5) = 14x
-210 = 14x
x = -15

Bài 1:
\(g.\frac{5}{11}+\frac{6}{11}=\frac{5+6}{11}=\frac{11}{11}=1\)\(\)
\(e.\frac{-17}{25}.\frac{20}{33}+\frac{-17}{25}.\frac{13}{33}+\frac{-3}{25}=\frac{-17}{25}.\left(\frac{20}{33}+\frac{13}{33}\right)+\frac{-3}{25}\)
\(=\frac{-17}{25}.1+\frac{-3}{25}=\frac{-17}{25}+\frac{-3}{25}=\frac{-17-3}{25}=\frac{-20}{25}=\frac{-4}{5}\)
\(d.\frac{5}{7}.\frac{19}{23}+\frac{5}{7}.\frac{5}{23}-\frac{5}{7}.\frac{1}{23}=\frac{5}{7}\left(\frac{19}{23}+\frac{5}{23}-\frac{1}{23}\right)\)
\(=\frac{5}{7}\left(\frac{19+5-1}{23}\right)=\frac{5}{7}.1=\frac{5}{7}\)
\(c.\left(-11\right).\frac{9}{22}=\frac{\left(-11\right).9}{22}=\frac{-99}{22}=\frac{-9}{2}\)
\(b.\frac{5}{6}-\frac{1}{8}=\frac{5.4}{6.4}-\frac{1.3}{8.3}=\frac{20}{24}-\frac{3}{24}=\frac{17}{24}\)
\(a.\frac{2}{3}+\frac{1}{5}-\frac{1}{6}=\frac{2.10}{3.10}+\frac{1.6}{5.6}-\frac{1.5}{6.5}=\frac{20}{30}+\frac{6}{30}-\frac{5}{30}\)
\(=\frac{20+6-5}{30}=\frac{21}{30}=\frac{7}{10}\)
Bài 2:
\(a.\frac{3}{4}+x=\frac{11}{12}\)
\(\Leftrightarrow x=\frac{11}{12}-\frac{3}{4}\)
\(\Leftrightarrow x=\frac{11}{12}-\frac{9}{12}\)
\(\Leftrightarrow x=\frac{2}{12}=\frac{1}{6}\)
\(b.x-\frac{11}{12}=0,5\)
\(\Leftrightarrow x=\frac{1}{2}-\frac{11}{12}\)
\(\Leftrightarrow x=\frac{6}{12}+\frac{11}{12}\)
\(\Leftrightarrow x=\frac{17}{12}\)
Đáp án là B