tìm a\b biết
a. a\bx 5\6=5\9
b.a\b-2\7=3\8
c.a\b:12\19=19\36
d.a\b+1\4=4\5
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1b)\(\frac{7}{19}x\frac{8}{11}+\frac{3}{11}:\frac{19}{7}-\frac{2}{-19}=\frac{7}{19}x\frac{8}{11}+\frac{3}{11}x\frac{7}{19}+\frac{2}{19}=\left(\frac{8}{11}+\frac{3}{11}\right)\frac{7}{19}+\frac{2}{19}=\frac{7}{19}+\frac{2}{19}=\frac{9}{19}\)
c)\(4\left(\frac{4}{9}+\frac{7}{11}-\frac{4}{9}\right)=4\frac{7}{11}\)
từ rồi làm tiếp
a, Ta thấy : \(\left\{{}\begin{matrix}\left(2a+1\right)^2\ge0\\\left(b+3\right)^2\ge0\\\left(5c-6\right)^2\ge0\end{matrix}\right.\)\(\forall a,b,c\in R\)
\(\Rightarrow\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2\ge0\forall a,b,c\in R\)
Mà \(\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2\le0\)
Nên trường hợp chỉ xảy ra là : \(\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2=0\)
- Dấu " = " xảy ra \(\left\{{}\begin{matrix}2a+1=0\\b+3=0\\5c-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{1}{2}\\b=-3\\c=\dfrac{6}{5}\end{matrix}\right.\)
Vậy ...
b,c,d tương tự câu a nha chỉ cần thay số vào là ra ;-;
a: \(A=\dfrac{19}{9}+\dfrac{4}{11}+\dfrac{2}{3}=\dfrac{209}{99}+\dfrac{44}{99}+\dfrac{66}{99}=\dfrac{319}{99}\)
b: \(B=\dfrac{-50}{60}+\dfrac{-35}{60}+\dfrac{12}{60}=\dfrac{-73}{60}\)
c: \(C=\dfrac{-27}{36}+\dfrac{132}{36}+\dfrac{10}{36}=\dfrac{115}{36}\)
d: \(D=\dfrac{-19}{3}+\dfrac{2}{3}-\dfrac{4}{5}=\dfrac{-17}{3}-\dfrac{4}{5}=\dfrac{-85-12}{15}=-\dfrac{97}{15}\)
câu 1:
a) 500-(300)-190+(-210)
= 500-300-190-210
= 200 - 210 -190
=-10 - 190
=-200
b) (-3)3 .5+12.(-6)
= -27.5 -72
=-135 - 72
=-207
c) 15.(-19-4)-19.(15-4)
= 15.(-23) - 19.11
=-345 - 209
=-554
câu 2: tìm x thuộc Z
a) 3x-2=3
=> 3x=3/2
=> x=1/2
b) x chia hết cho 5 và -7<x<11
=> x thuộc {-5;0;5;10}
Câu 1:
a) Ta có: \(500-\left(300\right)-190+\left(-210\right)\)
\(=500-300-190-210\)
\(=\left(500-300\right)-\left(190+210\right)\)
\(=200-400=-200\)
b) Ta có: \(\left(-3\right)^3\cdot5+12\cdot\left(-6\right)\)
\(=\left(-3\right)^3\cdot5-3\cdot4\cdot3\cdot2\)
\(=-5\cdot3^3-3^2\cdot8\)
\(=3^2\cdot\left(-5\cdot3-8\right)\)
\(=9\cdot\left(-15-8\right)=9\cdot\left(-23\right)=-207\)
c) Ta có: \(15\cdot\left(-19-4\right)-19\cdot\left(15-4\right)\)
\(=-15\cdot19-15\cdot4-15\cdot19+19\cdot4\)
\(=-30\cdot19+4\cdot4\)
\(=-2\cdot\left(15\cdot19+2\cdot4\right)\)
\(=-2\cdot\left(285+8\right)=-586\)
A.a/B = 5/9 / 5/6
a/B = 2/3
B. a/b = 3/8 +2/7
a/b = 37/56
C. a/b = 19/36 x 12/19
a/b = 1/3
D. a/b = 4/5-1/4
a/b = 11/20
Tìm a/b
A.a/b x 5/6 = 5/9
a/b = 5/9 : 5/6
a/b = 30/45 = 2/3
tk mk nha
cảm ơn mọi người rất nhiều ạ !!!!! hihi
b: \(B=\dfrac{5}{2}-\dfrac{7}{2}+\dfrac{3}{8}+\dfrac{6}{8}+\dfrac{-6}{11}-\dfrac{5}{11}=-2-1+\dfrac{9}{8}=\dfrac{9}{8}-3=-\dfrac{15}{8}\)
c: \(C=\left(\dfrac{4}{3}+\dfrac{7}{3}+\dfrac{1}{3}\right)+\left(\dfrac{2}{5}+\dfrac{3}{5}\right)=4+1=5\)
d: \(D=\dfrac{4}{19}\left(\dfrac{-5}{6}-\dfrac{7}{12}\right)-\dfrac{40}{57}\)
\(=\dfrac{4}{19}\cdot\dfrac{-17}{12}-\dfrac{40}{57}=-1\)
e: \(E=\dfrac{1}{3}\left(\dfrac{4}{5}-\dfrac{9}{5}\right)+\dfrac{2}{3}=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)
Lời giải:
a.
$\frac{7}{4}+\frac{5}{-6}+\frac{21}{8}=\frac{42}{24}+\frac{-20}{24}+\frac{63}{24}=\frac{85}{24}$
b.
$\frac{4}{9}+\frac{-7}{12}+\frac{8}{15}$
$=\frac{80}{180}+\frac{-105}{180}+\frac{96}{180}=\frac{71}{180}$
c.
$=\frac{-1}{3}+\frac{5}{6}+\frac{19}{12}+2$
$=\frac{-4}{12}+\frac{10}{12}+\frac{19}{12}+2=\frac{25}{12}+2=\frac{49}{12}$
a. a / b x 5 / 6 = 5 /9
a / b = 5 / 9 : 5 / 6
a / b = 2 / 3
b. a / b - 2 / 7 = 3 /8
a / b = 3 / 8 + 2 / 7
a / b = 37 / 56
c. a / b : 12 / 19 = 19 / 36
a / b = 19 / 36 x 12 / 19
a / b = 1 / 3
d. a / b + 1 / 4 = 4 / 5
a / b = 4 / 5 - 1 / 4
a / b = 11 / 20