giúp tui: 3+4+5+6+⋯+x=52 tầm lớp 5 ạ
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\(\left(2x-6\right)\times\left(5-x\right)=0\)
\(2x-6=0;5-x=0\)
\(x=3;x=5\)
Vậy: \(x=3;x=5\)
(2x−6)×(5−x)=0
2x−6=0;5−x=02x−6=0;5−x=0
x=3;x=5x = 3, x = 5
Vậy x = 3 và 5.

a,\(4:\dfrac{9}{5}:\dfrac{10}{3}\)
\(=\dfrac{4\times5\times3}{9\times10}\)
\(=\dfrac{2\times2\times5\times3}{3\times3\times5\times2}=\dfrac{2}{3}\)
b,\(\dfrac{1}{2}\times\dfrac{3}{4}:\dfrac{6}{5}\)
\(=\dfrac{3\times5}{2\times4\times6}\)
\(=\dfrac{3\times5}{2\times4\times3\times2}=\dfrac{5}{16}\)
a) 4 :9/5:10/3
=4×5/9:10/3
=4/1×5/9:10/3
=20/9×3/10
=60/90
=60:30/90:30
=2/3

52x - 3 - 2.52 = 52.3
52x - 3 = 52.3 + 52.2
52x - 3 = 52.(3 + 2)
52x - 3 = 52.5
52x - 3 = 53
2x - 3 = 3
2x = 3 + 3 = 6
x = 6 : 2 = 3
Vậy x = 3

3: Số học sinh giỏi là 40*1/5=8 bạn
Số học sinh trung bình là 32*3/8=12 bạn
Số học sinh khá là 32-12=20 bạn
1:
a: -1/3+7/6=7/6-2/6=5/6
b: 5/7-3/5=25/35-21/35=4/35
c: 0,75*4/5=4/5*3/4=3/5

Mik tìm đc TH1 là ra -2
TH2 là ra -8 nhưng mà TH3 ra -5 nhg mik k bt lm bạn nào biết chỉ mik với ạ
Mik cần trng sáng nay ạ

a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)
\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{-5}{12}\)
b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)
\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{5}\)
c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)
\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-77}{120}\)
d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)
\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{-7}{20}\)
e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-59}{105}\)
g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-13}{12}\)

Ta có: \(\dfrac{-4}{15}< \dfrac{5x-1}{18}< \dfrac{5}{12}\)
\(\Leftrightarrow\dfrac{-48}{180}< \dfrac{10\left(5x-1\right)}{180}< \dfrac{75}{180}\)
Suy ra: \(-48< 10\left(5x-1\right)< 75\)
\(\Leftrightarrow10\left(5x-1\right)\in\left\{-40;-30;-20;-10;0;10;20;30;40;50;60;70\right\}\)
\(\Leftrightarrow5x-1\in\left\{-4;-3;-2;-1;0;1;2;3;4;5;6;7\right\}\)
\(\Leftrightarrow5x\in\left\{-3;-2;-1;0;1;2;3;4;5;6;7;8\right\}\)
\(\Leftrightarrow x\in\left\{0;1\right\}\)(Vì x nguyên)
3 + 4 + 5 + 6 + ... + x = 52
Suy ra [(3 + x) * (x - 2)] : 2 = 52
(3 + x ) * (x - 2) = 52 * 2 = 104
(3 + x) * (x - 2) = 13 * 8
Suy ra x = 10
Vậy x = 10
3 + 4 + 5 + 6 + ... + 10 = 52